0.5
0.4
0.3
0.2
0.1
Ω -1
0
1
2
3
4 Ω0
Figure 2.11: Resonance plot of hP↓ i for Exercise 16.23.
Exercise 16.26 Show that if the incident spin state is a pure transverse polarization state, the scattering amplitudes for the initial polarizations P0 = ±ˆ n are g = ±ih and the scattering leaves the polarization unchanged, P = P0 .
Solution
W. Erbsen
HOMEWORK #6
We first recall from (16.130) from Merzbacher that |g|2 − |h|2 P0 + i (g∗ h − gh∗ ) n ˆ + 2|h|2 P0 · n ˆ·n ˆ + (g∗ h + gh∗ ) P0 × n ˆ P= 2 2 ∗ ∗ |g| + |h| + i (g h − gh ) P0 · n ˆ
(2.6.30)
If the initial beam has transverse polarization with the scattering plane perpendicular to P0 where P0 = P0 n ˆ, (2.6.30) becomes |g|2 + |h|2 P0 + i (g∗ h − gh∗ ) P= n ˆ (2.6.31) |g|2 + |h|2 + i (g∗ h − gh∗ ) P0 Which is of course (16.131) from Merzbacher. We now recall (16.129), which states Tr Sρinc S† · σ P= dσ/dΩ
(2.6.32)
We now recall (16.127) and (16.125) from the book, which are (respectively): ρinc = 1/2 (1 + P0 · σ) ,
dσ = Tr ρinc S† · S dΩ
Substituting these equations from (2.6.33) into (2.6.32), we have Tr S · 1/2 (1 + P0 · σ) · S† · σ P= Tr [ 1/2 (1 + P0 · σ) · S† · S] With the scattering matrix S defined by (16.120) as S11 S= S21
S12 S22
(2.6.33)
(2.6.34)
(2.6.35)
While the Pauli spin matrices are ˆ σ = σx ˆi + σy ˆj + σz k Where the individual spin matrices that constitute (2.6.36) are 0 1 0 −i 1 σx = , σy = , σz = 1 0 i 0 0
(2.6.36)
0 −1
(2.6.37)
We now have the tools to finish the problem, unfortunately I have run out of time. Using (2.6.31)-(2.6.37), the scattering amplitudes turn out to be g = ±i~ and the polarization vector unchanged, P = P0 .
Exercise 16.28 Assuming that P0 is perpendicular to the scattering plane, evaluate the asymmetry parameter A, defined as a measure of the right-left asymmetry by (dσ/dΩ)+ − (dσ/dΩ)− A= (2.6.38) (dσ/dΩ)+ + (dσ/dΩ)−
181
CHAPTER 2: QUANTUM MECHANICS II
where the subscripts + and − refer to the sign of the product P0 · n ˆ. Show that if P0 = ±ˆ n, the asymmetry A equals the degree of polarization P defined in (16.132). In particle polarization experiments, this quantity is referred to as the analyzing power.
Solution We first note that (16.132) reads P=P ·n ˆ=i
g∗ h − gh∗ n ˆ |g|2 + |h|2
(2.6.39)
We now recall that (16.128) from Merzbacher is dσ = |g|2 + |h|2 + i(g∗ h − gh∗ )P0 · n ˆ dΩ
(2.6.40)
Using (2.6.40) we can rewrite (2.6.38) as |g|2 + |h|2 + i(g∗ h − gh∗ ) − |g|2 + |h|2 − i(g∗ h − gh∗ ) A= (|g|2 + |h|2 + i(g∗ h − gh∗ )) + (|g|2 + |h|2 − i(g∗ h − gh∗ ))
(2.6.41)
We note that (2.6.41) may be trivially reduced to A=i
g∗ h − gh∗ n ˆ |g|2 + |h|2
(2.6.42)
Where we see that (2.6.42) is identical to (2.6.39).
Exercise 16.33 For a mixed state given by the density matrix ρ=
1 75
41 7 + 7i 7 − 7i 34
(2.6.43)
Check the inequalities (15.120), and calculate the eigenvalues and eigenstates. Evaluate the von Neumann entropy, and compare this with the outcome entropy for a measurement of σz .
Solution We first wish to evaluate the eigenvalues and eigenvectors of ρ. We find the eigenvectors in the usual way, 41/75 − λ (7 + 7i) /75 34 1 = 0 −→ 41 − λ − λ − 2 (7 + 7i) (7 − 7i) = 0 (7 − 7i) /75 34/75 − λ 75 75 75 (41)(34) 41 34 (49 − 49i + 49i + 49) −→ − λ − λ + λ2 − =0 2 75 75 75 752
W. Erbsen
HOMEWORK #6
2
−→λ − λ −
36 75
2
=0
(2.6.44)
Evaluating the quadratic equation from (2.6.44), q 2 1 ± 1 − 4 (36/75) 16 λ1,2 = −→ λ1 = , 1 25
λ2 =
9 25
Let’s find the eigenvectors. Starting with λ1 first, 41/75 − 16/25 (7 + 7i) /75 α 0 = (7 − 7i) /75 34/75 − 16/25 β 0 −7/75 (7 + 7i) /75 α 0 = (7 − 7i) /75 −14/75 β 0 −1 1 + i α 0 = 1 − i −2 β 0
(2.6.45)
(2.6.46)
The corresponding eigenvector for (2.6.46) is:
χ1 =
1+i 1
Normalizing according to the usual prescription N = [hχ1 |χ1 i] 1 1+i χ1 = √ 1 3
(2.6.47) −1
/2
, (2.6.47) becomes
Using λ2 from (2.6.45) and following the same prescription, we find that 1+i χ2 = 1 The normalization factor is, of course, the same. Therefore, our eigenvectors are 1 1+i χ = χ1 = χ2 = √ 1 3
(2.6.48)
(2.6.49)
(2.6.50)
The inequalities from (15.120) read 2 0 ≤ Tr ρ2 ≤ [Tr {ρ}] = 1 2
ρii ρjj ≥ |ρij |
(2.6.51a) (2.6.51b)
Let’s tackle showing (2.6.51a) first. We first note that from (2.6.43), we can easily find the trace: Tr {ρ} =
1 (41 + 34) ⇒ 1 75
Now, we need to find ρ2 : ρ2 =
1 (75)2
† 41 7 + 7i 41 7 + 7i 7 − 7i 34 7 − 7i 34
(2.6.52)
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CHAPTER 2: QUANTUM MECHANICS II
41 7 + 7i 41 7 + 7i 7 − 7i 34 7 − 7i 34 2 2 1 (41) + (7 − 7i) 41 (7 + 7i) + 34 (7 − 7i) = 2 2 (75)2 41 (7 + 7i) + 34 (7 − 7i) (7 + 7i) + (34) =
1 (75)2
(2.6.53)
The trace of (2.6.53) is 2 2 2 2 Tr ρ2 = (41) + (7 − 7i) + (7 + 7i) + (34) ≈ 0.50
(2.6.54)
Therefore, from (2.6.52) and (2.6.54), we can now say that
2 0 ≤ Tr ρ2 ≤ [Tr {ρ}] = 1
(2.6.55)
Furthermore, from (2.6.43) we can now say that
ρii ρjj = (41) (34) ⇒ 1394
(2.6.56)
While we also see that 2
|ρij | = (7 + 7i) (7 − 7i) ⇒ 98
(2.6.57)
Therefore, from (2.6.56) and (2.6.57) we can say that ρii ρjj ≥ |ρij |
2
The von Neumann entropy is defined by (15.129) as S(ρ) = −
N X
pi log [pi ]
(2.6.58)
i=1
Substituting in our eigenvalues to (2.6.58), we find that 16 16 9 9 S(ρ) = − log − log −→ S(ρ) = 0.653 25 25 25 25 The outcome entropy is defined by (15.131) in Merzbacher as H(K) = −
n X
p(Kj ) log [p(Kj )]
(2.6.59)
j=1
But we mustn’t forget (15.132), which reads p(Kj ) = Tr {|Kj ihKj |ρ} Which can be applied to our previous work as Tr {|K1 ihK1 |ρ} =
41 , 75
Therefore, applying (2.6.60) to (2.6.59), we find that
Tr {|K2 ihK2 |ρ} =
34 75
(2.6.60)
W. Erbsen
HOMEWORK #6
41 41 34 34 H(K) = − log − log −→ H(K) = 0.688 75 75 75 75
(2.6.61)
(z)
To find the probability of measuring χ+ , we first must evaluate 1 0 1 1+i (z) (z) hχ+ |σz |χ+ i = 1 − i 1 0 −1 1 3 1 1−i 1+i = 1 3 −1 1 = (1 − i) (1 + i) − 1 3 1 = 3 We see here that the outcome entropy is larger than the result from a measurement of σz . Realizing now that I may have misinterpreted this last part of the problem, we can also say that H(K) > S(ρ)
Problem 16.1a The spin-zero neutral kaon is a system with two basis states, the eigenstates of σz , representing a particle K 0 ¯ 0 : The operator σx = CP represents the combined parity (P ) and charge conjugation (C), and its antiparticle K ¯ 0 i. The dynamics is governed by the or particle-anti-particle, transformation and takes α = |K 0 i into β = |K Hamiltonian matrix Γ (2.6.62) H = M−i 2 Where M and Γ are Hermitian 2 × 2 matrices, representing the mass-energy and decay properties of the system, respectively. The matrix Γ is positive definite. A fundamental symmetry (under combined CP and time reversal transformations) requires that σxM∗ = Mσx and σx Γ∗ = Γσx . Show that in the expansion of H in terms of the Pauli matrices, the matrix σz is absent. Derive the eigenvalues and eigenstates of H in terms of the matrix elements of M and Γ. Are the eigenstates orthogonal?∗
Solution ¯ 0 i. Furthermore, we recall that the four matrices 1, Charge-parity reversal requires that σx |K 0 i = 1|K σx , σy and σz are linearally dependent, and any 2 × 2 matrix can be represented as A = λ0 1 + λ1 σx + λ2 σy + λ3 σz Which is (16.57) from Merzbacher. Applying this principle to our matrix M, ∗ What’s
your favorite color?
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CHAPTER 2: QUANTUM MECHANICS II
M =a1 1 + b1 σx + c1 σy + d1 σz 1 0 0 1 0 −i 1 =a1 + b1 + c1 + d1 0 1 1 0 i 0 0 a1 + d1 b1 − ic1 = b1 + ic1 a1 − d1
0 −1
(2.6.63)
We recall from the prompt that σxM∗ = Mσx , which is a requirement. Applying the LHS of this statement to our matrix M from (2.6.63), ∗ 0 1 a1 + d1 b1 − ic1 0 1 a1 + d1 b1 + ic1 σx M∗ = ⇒ (2.6.64) 1 0 b1 + ic1 a1 − d1 1 0 b1 − ic1 a1 − d1 We now compute the RHS of our statement: a1 + d1 b1 − ic1 0 1 b − ic1 Mσx = ⇒ 1 b1 + ic1 a1 − d1 1 0 a1 − d 1
a1 + d 1 b1 + ic1
(2.6.65)
So, equating (2.6.64) and (2.6.65), b − ic1 σx M = Mσx −→ 1 a1 + d 1 ∗
a1 − d 1 b1 + ic1
b − ic1 = 1 a1 − d 1
a1 + d 1 b1 + ic1
(2.6.66)
This yields two equations, b1 − ic1 + a1 − d1 = b1 − ic1 + a1 + d1 −→ −d1 = d1
a1 + d1 + b1 + ic1 = a1 − d1 + b1 + ic1 −→ d1 = −d1
(2.6.67a) (2.6.67b)
We see that in order for M to meet the requirement that σxM∗ = Mσx , we must have d1 = 0. We can rewrite M from (2.6.63) with this change, a1 b1 − ic1 M= (2.6.68) b1 + ic1 a1 We can equally well apply this principle to the matrix Γ: a2 + d2 b2 − ic2 Γ= b2 + ic2 a2 − d2
(2.6.69)
Following the logic which led us to (2.6.68), we may use (2.6.69) to deduce that d2 = 0 according to the requirement that σx Γ∗ = Γσx , and we can now write a2 b2 − ic2 Γ= (2.6.70) b2 + ic2 a2 We now substitute (2.6.68) and (2.6.70) into (2.6.62), i a1 b1 − ic1 a2 b2 − ic2 H= − b1 + ic1 a1 a2 2 b2 + ic2 i i a1 b1 − ic1 / ·a /2 (b2 − ic2 ) = − i 2 2 i b1 + ic1 a1 /2 (b2 + ic2 ) /2 · a2
W. Erbsen
HOMEWORK #6
a1 b1 − ic1 ia2 /2 (ib2 + c2 ) /2 − b1 + ic1 a1 (ib2 − c2 ) /2 ia2 /2 a1 − ia2 /2 b1 − ic1 − (ib2 + c2 ) /2 = b1 + ic1 − (ib2 − c2 ) /2 a1 − ia2 /2 =
(2.6.71)
We must now find the eigenvalues of H, in the usual way a1 − ia2 /2 − λ b1 − ic1 − (ib2 + c2 ) /2 b1 + ic1 − (ib2 − c2 ) /2 =0 a1 − ia2 /2 − λ 2 a1 − i/2 a2 − λ − b1 − ic1 − 1/2 (ib2 + c2 ) b1 + ic1 − 1/2 (ib2 − c2 ) =0
(2.6.72)
We can just go ahead and solve for λ, p λ = a1 − i/2 a2 ± (b1 − ic1 − 1/2 (ib2 + c2 )) (b1 + ic1 − 1/2 (ib2 − c2 ))
(2.6.73)
We can simplify (2.6.73) ever so slightly, giving eigenvalues λ1,2 = a1 − i/2 a2 −
q
2
2
(2b1 − ib2 ) + (2c1 − ic2 )
(2.6.74)
To find the eigenvectors, we substitute λ1 and λ2 from (2.6.74) into our Hamiltonian from (2.6.72) as a1 − ia2 /2 − λ1,2 b1 − ic1 − (ib2 + c2 ) /2 c1 0 = (2.6.75) b1 + ic1 − (ib2 − c2 ) /2 a1 − ia2 /2 − λ1,2 c2 0 The eigenvectors were found using Mathematica. The result is ! √ (2b1−ib2 )2 +(2c1 −ic2 )2 ± 2b1−ib2+2ic1 +c2 χ1,2 = N 1
(2.6.76)
The normalization in (2.6.76) is obtained as we usually do: ∗ c1 1 2 † 2 hχ1 |χ1 i = 1 −→ N χ1 · χ1 = 1 −→ N c1 c2 = 1 −→ N = p ∗ c2 c1 c1 + c∗2 c2
(2.6.77)
This normalization constant was determined from (2.6.77) using Mathematica, the result is q q −1/2 (2b1 + ib2 )2 + (2c1 + ic2 )2 (2b1 − ib2 )2 + (2c1 − ic2 )2 N = + 2b1 + ib2 − 2ic1 + c2 2b1 − ib2 + 2ic1 + c2
(2.6.78)
Combining (2.6.76) and (2.6.78), the normalized eigenvectors are q
χ1,2 =
2
2
(2b1 + ib2 ) + (2c1 + ic2 ) 2b1 + ib2 − 2ic1 + c2
+
q
2
2
(2b1 − ib2 ) + (2c1 − ic2 ) 2b1 − ib2 + 2ic1 + c2
−1/2
±
√
(2b1−ib2 )2 +(2c1−ic2 )2 2b1−ib2 +2ic1 +c2
To see if the eigenstates from (2.6.79) are orthogonal, we apply the following test:
1
!
(2.6.79)
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CHAPTER 2: QUANTUM MECHANICS II
hχ1 |χ2 i = 0
(if orthogonal)
(2.6.80)
Using Mathematica, it is easy to show that (2.6.80) is not satisfied, and therefore χ1 and χ2 are not orthogonal .
2.7
Homework #7
Exercise 17.12 Derive the recursion relation (17.54), as indicated.
Solution We first recall three three relations defining various quantities specifying the behavior of the total angular momentum: J =J1 + J2 Jz =Jz1 + Jz2 J+ =J1+ + J2+
(2.7.1a)
We now recall the three equations from (17.27) of Merzbacher: Jz |j mi =m~|j mi p J+ |j mi = (j − m)(j + m + 1) ~ |j1 j2 j m + 1i p J− |j mi = (j + m)(j − m + 1) ~ |j1 j2 j m − 1i
(2.7.2a) (2.7.2b) (2.7.2c)
We also recall (17.52), which reads
|j1 j2 j mi =
X
m1 ,m2
|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi
(2.7.3)
Which is the link between the direct product basis and the total angular momentum basis. According to Merzbacher, in order to obtain the recursion relations from (17.54), all we have to do is apply J+ and J− to (2.7.3). The behavior of J+ and J− is defined by (2.7.2b) and (2.7.2c), which may be more compactly redefined as p J± |j mi = (j ∓ m)(j ± m + 1) ~ |j1 j2 j m ± 1i (2.7.4) Similarly, we may generalize (2.7.1a) as
J± = J1± + J2± We can apply (2.7.5) to (2.7.3), which yields
(2.7.5)
W. Erbsen
HOMEWORK #7
J± {|j1 j2 j mi} = J±
(
X
m1 ,m2
)
|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi
(2.7.6)
We must now evaluate the LHS and the RHS of (2.7.6) in turn. Starting with the LHS, we have p J± |j1 j2 j mi = (j ∓ m)(j ± m + 1) |j1 j2 j m ± 1i (2.7.7)
Which comes as no surprise. Evaluating the RHS of (2.7.6) poses a more arduous problem, which we shall see momentarily: X X J± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi = (J1± + J2±) |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2
=J1±
X
m1 ,m2
+J2±
X
m1 ,m2
m1 ,m2
|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi (2.7.8)
We express the first term on the RHS of (2.7.8) as I and the second as II, and we can rewrite it as X J± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi = I + II (2.7.9) m1 ,m2
So, now we must evaluate I and II in turn. In doing this, we must recognize that J1± and J2± only work on particle 1 and 2, respectively. Accordingly, X I =J1± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2
X p = (j1 ∓ m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m2 |j1 j2 j mi
(2.7.10)
m1 ,m2
And similarly, we can also evaluate II as X II =J2± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2
X p = (j1 ∓ m1 )(j1 ± m1 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j1 j2 j mi
(2.7.11)
m1 ,m2
Substituting (2.7.10) and (2.7.11) into (2.7.9) and combining the result with (2.7.7), p
p (j ± m)(j ∓ m + 1) hj1 j2 m1 m2 |j m ∓ 1i = p(j1 ± m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m1 |j mi + (j2 ∓ m2 )(j1 ± m2 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j mi
Where I have suppressed the presence of j1 j2 in the ket to save room. We note that this is identical to (12.54), and is what we were trying to show.
Exercise 17.13
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CHAPTER 2: QUANTUM MECHANICS II
Show that a symmetry exists along the three quantum numbers j1 , j2 and j, and that in addition to (17.57) they satisfy the equivalent relations |j − j2 | ≤ j1 ≤ j + j2
and
|j − j1 | ≤ j2 ≤ j + j1
(2.7.12)
Solution We first recall (17.57) from Merzbacher, which reads |j1 − j2 | ≤ j ≤ j1 + j2
(2.7.13)
We note that the magnitude on the LHS of (2.7.18) can in fact be two different things: |j1 − j2 | = j1 − j2
or
|j1 − j2 | = j2 − j1
With these possibilities, in addition to the RHS of (2.7.18), we have three inequalities to play with: j ≤j1 + j2
(2.7.14a)
j1 ≤j + j2 j2 ≤j + j1
(2.7.14b) (2.7.14c)
(2.7.15)
Using (2.7.19) and (2.7.21), we can write j1 ≥ j − j2 j1 ≥ j2 − j
−→ j1 ≥ |j − j2 |
Using (2.7.22) and (2.7.20), we can now say |j − j2 | ≤ j1 ≤ j + j2 Similarly, using (2.7.19) and (2.7.20), we can declare that j2 ≥ j − j1 −→ j2 ≥ |j − j1 | j2 ≥ j1 − j
(2.7.16)
And now combining (2.7.16) with (2.7.21), |j − j1 | ≤ j2 ≤ j + j1
Exercise 17.15 Verify the values of two special C-G coefficients: hj 1 j 0|j 1 j ji =
s
j j +1
(2.7.17a)
W. Erbsen
HOMEWORK #7
hj 2 j 0|j 2 j ji =
s
j(2j − 1) (j + 1)(2j + 3)
(2.7.17b)
Determine the value of the trivial C-G coefficient hj 0 m 0|j 0 j mi. How does it depend on the value of m?
Solution As was done in class, we start our journey by calculating the C-G coefficients by starting with the largest magnetic quantum number m = j1 + j2 and repeatedly applying the lowering operator J− = J1+ + J2−. So, ~ |j1 , j2 , m1 = j1 , m2 = j2 i 2(j1 + j2 )
(2.7.18)
J− |j1 , j2 , j = j1 + j2 , m = j1 + j2 i =|j1 , j2 , j = j1 + j2 , m = j1 + j2 − 1i
(2.7.19)
|j1 , j2 , j = j1 + j2 , m = j1 + j2 i = p
Applying J− to the LHS of (2.7.18) first,
And now applying J− to the RHS of (2.7.18), J− {N |j1 , j2 , m1 = j1 , m2 = j2 i} = (J1+ + J2−) {N |j1 , j2 , m1 = j1 , m2 = j2 i} =N {|j1 , j2 , m1 = j1 − 1, m2 = j2 i + |j1 , j2 , m1 = j1 , m2 = j2 − 1i} (2.7.20) Using these results, and substituting back in the normalization factor N , we find that s s j1 j2 |j = j1 + j2 , m = j1 + j2 − 1i = |m1 = j1 − 1, m2 = j2 i + |m1 = j1 , m2 = j2 − 1i j1 + j2 j1 + j2 (2.7.21) In our special case, (2.7.21) becomes hm1 = j, m2 = 0|j, m = ji =
s
j j+1
We continue along the same lines, and using Mathematica to take care of the matrix algebra we find that s j2 (2j2 − 1) hm1 = j1 − 2, m2 = j2 |j = j1 + j2 − 2, m = j1 + j2 − 2i = (2.7.22) (j1 + j2 − 1) (2j1 + 2j2 − 1) Applying the appropriate constants, (2.7.22) becomes hm1 = j, m2 = 2|j = j, m = ji =
s
j (2j − 1) (j + 1) (2j + 3)
We evaluate the last part of the question by simply using (2.7.18). The result is
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CHAPTER 2: QUANTUM MECHANICS II
1 hm1 = m, m2 = 0|j = j, m = mi = p ~ 2 (j1 + j2 )
1 −→ hm1 = m, m2 = 0|j = j, m = mi = √ ~ 2j
Exercise 17.16 Work out the results (17.63) from the recursion relations for C-G coefficients and the normalization and standard phase conditions.
Solution The results from (17.63) read
j1 , 1/2 , m − 1/2 , 1/2 | j1 , 1/2 , j1 ± 1/2 , m = ±
s
j1 ± m + 1/2 2j1 + 1
s j1 ∓ m + 1/2 j1 , 1/2 , m + 1/2 , − 1/2 | j1 , 1/2 , j1 ± 1/2 , m = 2j1 + 1
(2.7.23a) (2.7.23b)
While the recursion relation from (17.54) reads p p (j ± m)(j ∓ m + 1) hj1 j2 m1 m2 |j m ∓ 1i = p(j1 ± m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m1 |j mi + (j2 ∓ m2 )(j1 ± m2 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j mi (2.7.24) Where it is important to recognize that the j1 j2 terms in the ket are still suppressed. We also notice that (2.7.23a) and (2.7.23b) are C-G coefficients, and must take the form of hj1 , j2 , m1 , m2 |j1 , j2 , j, mi. Therefore, comparing this to the LHS of (2.7.23a), we may deduce that j2 = 1/2 ,
m1 = m − 1/2 ,
m2 = 1/2 ,
j = j1 ± 1/2
(2.7.25)
We substitute these values into the recursion relation from (2.7.24). We denote the LHS of (2.7.24) as LHS, while the first and second terms of the RHS are RHS1 and RHS2 , respectively. Accordingly, have: p LHS = (j ± m)(j ∓ m + 1) hj1 , j2 , m1 , m2 |j1 , j2 , j, m ∓ 1i p = (j1 ± 1/2 ± m)(j1 ± 1/2 ∓ m + 1) hj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , m ∓ 1i p RHS1 = (j1 ± (m − 1/2 ))(j1 ± (m − 1/2 ) + 1) |j1 , 1/2 , m − 1/2 ± 1, 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi p = (j1 ± m ∓ 1/2 )(j1 ± m ∓ 1/2 + 1) |j1 , 1/2 , m − 1/2 ± 1, 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi p RHS2 = (j2 ∓ m2 )(j1 ± m2 + 1) |j1 , j2 , m1 , m2 ± 1ihj1 , j2 , m1 , m2 |j1 , j2 , j, mi p = ( 1/2 ∓ 1/2 )(j1 ± 1/2 + 1) |j1 , 1/2 , m − 1/2 , 1/2 ± 1ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi
W. Erbsen
HOMEWORK #7
Seemingly arbitrarily, we choose the top sign in the above equations: p LHS = (j1 + m + 1/2 ) (j1 − m + 3/2 ) hj1 , 1/2 , m − 1/2 , 1/2 |j1 + 1/2 , m1 i p RHS1 = (j1 + m − 1/2 ) (j1 + m + 1/2 ) |j1 , 1/2 , m + 1/2 , 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 + 1/2 , mi RHS2 =0 Combining these equations and rearranging, s hm − 1/2 , 1/2 |j1 + 1/2 , mi =
j1 + m + 1/2 hm + 1/2 , 1/2 |j1 + 1/2 , m + 1i j1 + m + 3/2
Shifting the index according to m → m + 1, we now have s j1 + m + 3/2 hm + 3/2 , 1/2 |j1 + 1/2 , m + 2i hm + 1/2 , 1/2 |j1 + 1/2 , m + 1i = j1 + m + 5/2 Going back to (2.7.23a) and applying m = j1 + 1/2 to (2.7.26) and (2.7.27), we find that s j1 + m + 1/2 1 1 1 1 1 hj1 , /2 , m − /2 , /2 |j1 , /2 , j1 − /2 , mi = ± 2j1 + 1
(2.7.26)
(2.7.27)
(2.7.28)
Substituting back in the appropriate ± signs into (2.7.28),
s
j1 ± m + 1/2 1 1 1 1 1 j1 , /2 , m − /2 , /2 | j1 , /2 , j1 ± /2 , m = ± 2j1 + 1
(2.7.29)
Which we note is identical to (2.7.23a). To derive (2.7.23b), we repeat the same process. Evaluating the LHS and RHS of our recursion relation yields p LHS = (j1 ± 1/2 ± m)(j1 ± 1/2 ∓ m + 1) hj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , m ∓ 1i p RHS1 = (j1 ± (m + 1/2 ))(j1 ± (m + 1/2 ) + 1) |j1 , 1/2 , m + 1/2 ± 1, − 1/2 ihj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , mi p RHS2 = ( 1/2 ∓ (− 1/2 ))(j1 ± (− 1/2 ) + 1) |j1 , 1/2 , m + 1/2 , − 1/2 ± 1ihj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , mi The steps taken to arrive at (2.7.23b) are from henceforth identical to that which was done to derive (2.7.23a), and accordingly they will not be shown. From the above equations, we choose an arbitrary sign (which will be stuck back in at the end) in order to simplify the equation. We then shift the index of m, finagle the equations and evaluating the coefficient in terms of j1 and m, we get
j1 , 1/2 , m + 1/2 , − 1/2 | j1 , 1/2 , j1 ± 1/2 , m = Which is of course (2.7.23b).
Exercise 17.18
s
j1 ∓ m + 1/2 2j1 + 1
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Starting with the uncoupled basis (17.67), work out the 4 × 4 matrices Sz and S2 , and show by explicit diagonalzation that the singlet and triplet states are the eigenvectors with the appropriate eigenvalues.
Solution We recall that (17.67) from Merzbacher reads α1 ⊗ α2 = α1 α2 ,
α1 ⊗ β2 = α1 β2 ,
β1 ⊗ α2 = β1 α2 ,
β1 ⊗ β2 = β1 β2
(2.7.30)
We introduce the following shorthand notation, as indicated in lecture, which constitutes the product basis: {|++i, |+−i, |−+i, |−−i} Where the first term in each ket acts on particle 1, while the second acts on particle 2. We also recall that we can write the total spin operator S in terms of the single particle spin operators S1 and S2 as S = S1 ⊗ 1(2) + 1(1) ⊗ S2
(2.7.31)
We also note that when referring to the individual components of the spin operators, it will be most convenient to work in terms of the raising and lowering spin operators (S+ and S− , respectively). The raising and lowering operators are described in terms of the cartesian components by S± = Sz ± iSy So that S (Sx , Sy , Sz ) → S (S± , Sz ). We now recognize that the S acts on the individual spin kets according to Sz |+i = ~/2 |+i, Sz |−i = − ~/2 |−i S+ |+i = 0, S+ |−i = ~|+i S− |+i = ~|−i, S− |−i = 0 We also musn’t forget how the total components act on each particle, following the convention of (2.7.31), Sz =S1z ⊗ 1(2) + 1(1) ⊗ S2z
S+ =S1+ ⊗ 1(2) + 1(1) ⊗ S2+
S− =S1− ⊗ 1(2) + 1(1) ⊗ S2−
Accordingly, we can now write Sz |++i = ~/2 |++i + ~/2 |++i ⇒ ~|++i
Sz |+−i =0 Sz |−+i =0
Sz |−−i = − ~/2 |−−i − ~/2 |−−i ⇒ −~|−−i We can express (2.7.32a)-(2.7.32d) in matrix form as
(2.7.32a) (2.7.32b) (2.7.32c) (2.7.32d)
W. Erbsen
HOMEWORK #7
(Sz )m1 ,m2
~ 0 = 0 0
0 0 0 0
0 0 0 0 0 0 0 −~
(2.7.33)
The value of the eigenvalues from (2.7.34) are transparent, λ1 = ~,
λ2 = −~
We now wish to find S2 , first representing it in the product basis: S2 =S21 + S22 + 2S1z S2z + S1+ S2− + S1− S2+
(2.7.34)
In order to evaluate all the matrix elements for S2 , let’s make a table: |++i |+−i |−+i |−−i
S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3
S22 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3
S1z /2 ~|++i 1 /2 ~|+−i − 1/2 ~|−+i − 1/2 ~|−−i 1
S2z /2 ~|++i − 1/2 ~|+−i 1 /2 ~|−+i − 1/2 ~|−−i 1
S1+ 0 0 ~|+i ~|+i
S2− ~|+i 0 ~|−i 0
S1− ~|+i ~|−i 0 0
S2+ 0 ~|+i 0 ~|−i
Substituting the values of the tabulated matrix elements into (2.7.34), we find that: S2 |++i = 3/4 ~2 |++i + 3/4 ~2 |++i + 2 1/2 ~|++i 1/2 ~|++i ⇒ 2~2 |++i S2 |+−i = 3/4 ~2 |+−i + 3/4 ~2 |+−i + 2 1/2 ~|+−i − 1/2 ~|+−i + ~2 |−+i ⇒ ~2 (|+−i + |−+) S2 |−+i = 3/4 ~2 |−+i + 3/4 ~2 |−+i + 2 − 1/2 ~|−+i 1/2 ~|−+i + ~2 |+−i ⇒ ~2 (|−+i + |+−i) S2 |−−i = 3/4 ~2 |−−i + 3/4 ~2 |−−i + 2 − 1/2 ~|−−i − 1/2 ~|−−i ⇒ 2~2 |−−i
(2.7.35a) (2.7.35b) (2.7.35c) (2.7.35d)
We can find the expectation values of S2 (i.e. the matrix elements) easily, and the non-zero elements are h+ + |S2 |++i =2~2
h+ − |S2 |+−i =~2
h+ − |S2 |−+i =~2
h− + |S2 |+−i =~2 h− + |S2 |−+i =~2
h+ + |S2 |++i =2~2 While all the other elements are of course zero. We are now in a place to write down the matrix S2 :
S2
m1 ,m2
2 2 0 =~ 0 0
0 1 1 0
0 1 1 0
0 0 0 2
(2.7.36)
Which is not diagonal. We can remedy this by diagonalizing the central matrix, allowing us to find the eigenvalues and eigenvectors of S2 .
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~2 (S ) = 2 ~ 0 2
~2 ~2
Finding the eigenvalues, 2 ~ − λ 2 ~
~2 = 0 −→ ~2 − λ ~2 − λ − ~4 = 0 2 ~ −λ −→λ λ − 2~2 = 0 −→ λ1 = 0, λ2 = 2~2
Whose corresponding eigenvectors are 1 V1 = √ (|+−i − |−+i) , 2
1 V2 = √ (|+−i + |−+i) 2
The eigenvectors which we did not include here are found from the first and fourth column of the matrix S2 . Combining the results, we arrive at a series of states whose total angular momentum is either 0 or 1, i.e. either singlet or triplet. The states are 1 |0 0i = √ (|+−i − |−+i) 2 |1 mi = |++i 1 |1 mi = √ (|+−i + |−+i) 2 |1 mi = |−−i
Problem 17.1 In the notation of (17.64) the state of a spin one-half particle with sharp total angular momentum j, m is jm jm aYj− 1/ + bYj+ 1/ 2 2
(2.7.37)
Assume this state to be an eigenstate of the Hamiltonian with no degeneracy other than that demanded by rotational invariance. a)
If H conserves parity, how are the coefficients a and b restricted?
b)
If H is invariant under time reversal, show that a/b must be imaginary.
c)
Verify explicitly that the expectation value of the electric dipole moment −e r vanishes if either parity is conserved or time reversal invariance holds (or both).
Solution
W. Erbsen
HOMEWORK #7
We first note that (17.64) reads Y`jm
=
`± 1/ , m Y` 2
1 = √ 2` + 1
p ± ` ± m + 1/2 p ` ∓ m + 1/2
m− 1/2
Y`
m+ 1/2
Y`
!
(2.7.38)
Where Y`jm denotes the common eigenstates of Jz and J2 . Furthermore, if parity is conserved, then we require that h i j− 1/ j+ 1/2 jm jm jm jm Pˆ aYj− = (−1) 2 aYj− bYj+ (2.7.39) 1/ + bY1+ 1/ 1/ + (−1) 1/ 2 2 2 2
Problem 17.4 The Hamiltonain of the positronium atom in the 1S state in a magnetic field B along the z-axis is to good approximation eB0 H = AS1 · S2 + (S1z − S2z ) (2.7.40) mc if all higher energy states are neglected. The electron is labeled as particle 1 and the positron as particle 2. Using 2 the coupled representation in which S2 = (S1 + S2 ) and Sz = S1z +S2z are diagonal, obtain the energy eigenvalues and eigenvectors and classify them according to the quantum numbers associated with constants of the motion. Empirically, it is known that for B = 0 the frequency of the 13 S → 11 S transition is 2.0338 × 105 MHz and that the mean lifetime for annihilation are 10−10 s for the singlet state (two photon decay) and 10−7 s for the triplet state (three-photon decay). Estimate the magnetic field strength B0 which will cause the lifetime of the longer lived m = 0 state to be reduced (“quenched”) to 10−8 s.
Solution We start by finding the constants of motion of the problem, which is found by evaluating the commutator of the Hamiltonian from (2.7.40) with various quantities and noting which equals zero. To do this, we recall the following commutation relations for S2 : 2 S , Sz =0 2 2 S , Sz1 6=0, S , Sz2 6= 0 2 S , S1 = S2 , S2 = 0 2 2 2 2 S , S1 = S , S2 = 0 And similarly, some relations for Sz are
[Sz , Sz1 ] = [Sz , Sz2 ] = 0 Sz , S21 = Sz , S22 = 0
From the prompt, we are told to use the coupled representation, where S2 = (S1 + S2 )2 and Sz = Sz1 +Sz2 are diagonal. We can evaluate the dot product in (2.7.40) by expanding the given definition of S2 :
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2
S2 = (S1 + S2 ) ⇒ S21 + S22 + 2S1 · S2 −→ S1 · S2 = 1/2 S2 − S21 − S22 Substituting (2.7.41) into (2.7.40), the Hamiltonian becomes
eB0 H = 1/2 A S2 − S21 − S22 + (S1z − S2z ) mc
(2.7.42)
Let’s not evaluate some commutators to find the constants of motion: H, S2 = 1/2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , S2 = 1/2 A S2 , S2 − S21 , S2 − S22 , S2 + ω Sz1 , S2 − Sz2 , S2 6= 0 | {z } | {z } | {z } | {z } | {z } =0
=0
=0
6=0
=0
=0
=0
(2.7.43)
6=0
[H, Sz ] = 1/2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , Sz = 1/2 A S2 , Sz − S21 , Sz − S22 , Sz + ω [Sz1 , Sz ] − [Sz2 , Sz ] = 0 | {z } | {z } | {z } | {z } | {z } =0 =0 =0 =0 =0 1 [H, S1] = /2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , S1 = 1/2 A S2 , S1 − S21 , S1 − S22 , S1 + ω [Sz1 , S1 ] − [Sz2 , S1 ] = 0 | {z } | {z } | {z } | {z } | {z } =0 =0 =0 =0 =0 1 2 2 2 [H, S2] = /2 A S − S1 − S2 + ω (Sz1 − Sz2 ) , S2 = 1/2 A S2 , S2 − S21 , S2 − S22 , S2 + ω [Sz1 , S2 ] − [Sz2 , S2 ] = 0 | {z } | {z } | {z } | {z } | {z } =0
(2.7.41)
(2.7.44)
(2.7.45)
(2.7.46)
=0
So, S2 is not a constant of motion according to (2.7.43), while Sz , S1 , and S2 are constants of motion from (2.7.44)-(2.7.46). We now wish to switch to the following basis notation, {|++i, |+−i, |−+i, |−−i} We now wish to evaluate the diagonal matrix elements for H, first recalling that H = 1/2 A S2 − S21 − S22 + ω (S1z − S2z ) |++i |+−i |−+i |−−i
S2 2~2 |++i 0 0 2~2 |−−i
S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3
S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3
S1z /2 ~|++i 1 /2 ~|+−i 1 − /2 ~|−+i − 1/2 ~|−−i 1
S2z /2 ~|++i − 1/2 ~|+−i 1 /2 ~|−+i − 1/2 ~|−−i 1
Substituting the tabulated elements into our Hamiltonian, H|++i = 1/2 A 2~2 − 3/4 ~2 − 3/4 ~2 |++i ⇒ 1/4 A~2 |++i H|+−i = 1/2 A − 3/4 ~2 − 3/4 ~2 |+−i + ω 1/2 ~ + 1/2 ~ |+−i ⇒ − 3/4 A~2 + ω~ |+−i H|−+i = 1/2 A − 3/4 ~2 − 3/4 ~2 |−+i + ω − 1/2 ~ − 1/2 ~ |−+i ⇒ − 3/4 A~2 − ω~ |−+i H|−−i = 1/2 A 2~2 − 3/4 ~2 − 3/4 ~2 |−−i ⇒ 1/4 A~2 |−−i
These calculations were clearly done in the direct product basis, whereas it could just as easily be done in the total angular momentum basis.
W. Erbsen
2.8
HOMEWORK #9
Homework #9
Problem 17.6 The magnetic moment operator for a nucleon of mass mn is µ = e (g` L + gS S) /2mn c, where g` = 1 and gS = 5.587 for a proton, g` = 0 and gS = −3.826 for a neutron. In a central field with an additional spin-orbit interaction, the nucleons move in shells characterized by the quantum numbers ` and j = ` ± 1/2 . Calculate the magnetic moment of a single nucleon as a function of j for the two kinds of nucleons, distinguishing the two cases j = ` + 1/2 and j = `− 1/2 . Plot j times the effective gyromagnetic ratio versus j, connected in each case the points by straight-line segments (Schmidt lines).
Solution We start by rewriting the magnetic moment operator µ: µ=
e (g`L + gs S) 2mn c
(2.8.1)
Where mn is the mass of the nucleon. If we choose the quantization axis of the interaction to be along the ˆ z-direction, the expectation value of the magnetic moment operator is given by hµz i = hα`sjmj |µz |α`sjmj i
(2.8.2)
We can rewrite (2.8.1) according to our chosen quantization axis as µz =
e (g`Lz + gs Sz ) 2mn c
(2.8.3)
Substituting (2.8.3) into (2.8.2), we have hµz i =
e hα`sjmj |(g` Lz + gs Sz )|α`sjmj i 2mn c
(2.8.4)
At this point, we recall that Jz = Lz + Sz → Lz = Jz − Sz . Substituting this into (2.8.4), e hα`sjmj |(g` (Jz − Sz ) + gs Sz )|α`sjmj i 2mn c e = hα`sjmj |(g`Jz + (gs − g` ) Sz )|α`sjmj i 2mn c e = {g`hα`sjmj |Jz |α`sjmj i + (gs − g` ) hα`sjmj |Sz |α`sjmj i} 2mn c
hµz i =
(2.8.5)
Our task is now to evaluate the expectation values hJz i and hSz i on the RHS of (2.8.5). The tool that allows us to do this is the Wigner Eckart Theorem (W.E.T.). Applying this to hSz i first, hα`sjmj |Sz |α`sjmj i = hj1mj 0|j1jmj ihαjkSkαji
(2.8.6)
We can apply the W.E.T. along the same lines to the expectation value of hJz i: hα`sjmj |Jz |α`sjmj i = hj1mj 0|j1jmj ihαjkJkαji
(2.8.7)
We now notice that the conveniently uncalculated C-G coefficients in (2.8.6) and (2.8.7) are in fact equal. We can therefore take the ratio of these two equations, thus ridding them from our lives:
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CHAPTER 2: QUANTUM MECHANICS II
hα`sjmj |Sz |α`sjmj i hj1mj 0|j1jmj ihαjkSkαji = hα`sjmj |Jz |α`sjmj i hj1mj 0|j1jmj ihαjkJkαji hα`sjmj |Sz |α`sjmj i hαjkSkαji = hα`sjmj |Jz |α`sjmj i hαjkJkαji hαjkSkαji −→ hα`sjmj |Sz |α`sjmj i = hα`sjmj |Jz |α`sjmj i hαjkJkαji
(2.8.8)
At this point, we make a small deviation to evaluate the reduced matrix elements. In order to do this, we must evaluate hJ · Si and hJ · Ji: hα`sjmj |J · S|α`sjmj i =hj1mj 0|j1jmj ihαjkJkαjihαjkSkαji hα`sjmj |J · J|α`sjmj i =hj1mj 0|j1jmj ihαjkJkαjihαjkJkαji
(2.8.9a) (2.8.9b)
Not surprisingly, we notice once again that the C-G coefficients are again the same in (2.8.9a) and (2.8.9b). Taking the ratio of these two equations, hα`sjmj |J · S|α`sjmj i hj1mj 0|j1jmj ihαjkJkαjihαjkSkαji = hα`sjmj |J · J|α`sjmj i hj1mj 0|j1jmj ihαjkJkαjihαjkJkαji hα`sjmj |J · S|α`sjmj i hαjkSkαji = hα`sjmj |J · J|α`sjmj i hαjkJkαji
(2.8.10)
Switching the LHS and RHS in (2.8.10), and substituting in the appropriate eigenvalues, (2.8.10) becomes hαjkSkαji 1/2 j(j + 1)~2 + s(s + 1)~2 − `(` + 1)~2 = hαjkJkαji j(j + 1)~2 j(j + 1) + s(s + 1) − `(` + 1) = (2.8.11) 2j(j + 1) Substituting (2.8.11) back into (2.8.4), hα`sjmj |Sz |α`sjmj i =
j(j + 1) + s(s + 1) − `(` + 1) hα`sjmj |Jz |α`sjmj i 2j(j + 1)
(2.8.12)
And finally, substituting (2.8.12) into (2.8.5): e j(j + 1) + s(s + 1) − `(` + 1) hµz i = g` hα`sjmj |Jz |α`sjmj i + (gs − g` ) hα`sjmj |Jz |α`sjmj i 2mn c 2j(j + 1) (2.8.13) Substituting in the appropriate eigenvalues for hJz i, (2.8.13) simplifies to ej~ j(j + 1) + s(s + 1) − `(` + 1) hµz i = g` + (gs − g`) 2mn c 2j(j + 1) Now, in our particular case, each of the nucleons has s = 1/2 , and so (2.8.14) becomes ej~ j(j + 1) − `(` + 1) + 3/4 hµz i = g` + (gs − g` ) 2mn c 2j(j + 1)
(2.8.14)
(2.8.15)
Now, we mustn’t forget that we are interested when we do not have an empty sub-shell, since a filled sub-shell must have a zero (total) angular momentum. Because j is always equal to an integer plus ± 1/2 ,
W. Erbsen
HOMEWORK #9
the occupancy of the sub-shell is always even. Since the nuclear spin is non-zero in our case, we have a contribution from the nuclei towards a magnetic dipole moment. This is justified since both the neutrons and the protons have an intrinsic magnetic moment. Additionally, we must consider that the proton is charged, meaning that it can produce an additional magnetic moment due to its orbital motion. We therefore look at two unique cases of the total angular momentum: when j = ` + 1/2 , and also when j = ` − 1/2 . Let’s look at j = ` + 1/2 first. To do this, we substitute the value back into (2.8.15): e(` + 1/2 )~ (` + 1/2 )((` + 1/2 ) + 1) − `(` + 1) + 3/4 g` + (gs − g`) hµz i = 2mn c 2(` + 1/2 )((` + 1/2 ) + 1) e(` + 1/2 )~ (` + 1/2 )(` + 3/2 ) − `(` + 1) + 3/4 g` + (gs − g`) = 2mn c 2(` + 1/2 )(` + 3/2 )) e(` + 1/2 )~ `2 + 3/2 ` + 1/2 ` + 3/4 − `2 − ` + 3/4 = g` + (gs − g`) 2mn c 2(` + 1/2 )(` + 3/2 )) ` + 3/2 e(` + 1/2 )~ g` + (gs − g`) = 2mn c 2(` + 1/2 )(` + 3/2 )) e(` + 1/2 )~ 1 = g` + (gs − g`) 2mn c 2(` + 1/2 ) h e~ gs g` i g`(` + 1/2 ) + − = 2mn c 2 2 e~ h gs i = `g` + (2.8.16) 2mn c 2 We now wish to do the same thing for j = ` − 1/2 : (` − 1/2 )((` − 1/2 ) + 1) − `(` + 1) + 3/4 e(` − 1/2 )~ hµz i = g` + (gs − g`) 2mn c 2(` − 1/2 )((` − 1/2 ) + 1) e(` − 1/2 )~ (` − 1/2 )(` + 1/2 ) − `(` + 1) + 3/4 = g` + (gs − g`) 2mn c 2(` − 1/2 )(` + 1/2 ) e(` − 1/2 )~ 1 `(` + 1) + 3/4 = g` + (gs − g`) − 2mn c 2 2(` − 1/2 )(` + 1/2 ) e~ g` − 2gs (` + 1) `g` + = 2mn c 2(2` + 1) e~ 1 `+1 = g` ` + − gs 2mn c 2(2` + 1) 2` + 1 i 1 h e~ ` − /2 gs = (` + 1)g` − 2mn c ` + 1/2 2 So, in summary, we have from (2.8.16) and (2.8.17) that e~ n gs o j = ` + 1/2 2m c `g` + 2 , n hµz i = e~ ` − 1/2 h gs i (` + 1)g` − , j = ` − 1/2 2mn c ` + 1/2 2 We can rewrite (2.8.18) slightly as
(2.8.17)
(2.8.18)
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CHAPTER 2: QUANTUM MECHANICS II
e~ n gs o j = ` + 1/2 2m c `g` + 2 , n hµz i = e~ j h gs i (` + 1)g` − , j = ` − 1/2 2mn c j + 1 2
A plot of jg` is plotted vs. j on attached printout.
Problem 17.9 A system that is invariant under rotation is perturbed by a quadrupole interaction V =
2 X
Cq T2q
(2.8.19)
q=−2
where the Cq are constant coefficients and T2q are the components of an irreducible spherical tensor operator, defined by one of its components: 2
T22 = (Jx + iJy )
(2.8.20)
a)
Deduce the conditions for the coefficients Cq if V is to be Hermitian.
b)
Consider the effect of the quandrupole perturbation on the manifold of a degenerate energy eigenstate of the unperturbed system with angular momentum quantum j, neglecting all other unperturbed energy eigenstates. What is the effect of the perturbation on the manifold of an unperturbed j = 1/2 state?
c)
If C±2 = C0 and C±1 = 0, calculate the perturbed energies for a j = 1 state, and plot the energy splittings as a function of the interaction strength C0 . Derive the corresponding unperturbed energy eigenstates.
Solution a)
In order for V to be Hermitian, we require that it be self-adjoint. From (4.5) in Merzbacher, V† =V Where self-adjoint implies that the matrix is its own conjugate transpose, e.g. V
∗
=V
(2.8.21)
We must also recall that applying the self-adjoint condition to an operator with two elements, we have: ∗ ∗ ∗ ∗ (A · B) = B · A ⇒ B · A In order to evaluate the specific conditions we must require for each of the components, we must expand (2.8.19) and apply the Hermitian requirement from (2.8.21): V =C−2 T2−2 + C−1 T2−1 + C0 T20 + C1 T21 + C2 T22
(2.8.22)
W. Erbsen
HOMEWORK #9
Applying the self-adjoint condition from (2.8.21) to (2.8.22), ∗ ∗ ∗ ∗ ∗ ∗ V = C−2 T2−2 + C−1 T2−1 + (C0 T20 ) + (C1 T21 ) + (C2 T22 ) −2 ∗
=T 2
∗
−1 ∗
C −2 + T 2
∗
0∗
∗
1∗
∗
2∗
∗
C −1 + T 2 C 0 + T 2 C 1 + T 2 C 2
(2.8.23)
The coefficients Cq in (2.8.23) are non-conjugatable, as coefficients tend not to be. Therefore, (2.8.23) can be rewritten as V
∗
−2 ∗
=T 2
−1 ∗
∗ C−2 + T2
0∗
1∗
2∗
∗ C−1 + T 2 C0∗ + T 2 C1∗ + T 2 C2∗
∗ ∗ =T2−2 † C−2 + T2−1 † C−1 + T20 † C0∗ + T21 † C1∗ + T22 † C2∗
(2.8.24)
q∗
Where I have rewritten the tensor operators as T k = Tkq † , for reasons that will become clear momentarily. In the last homework, we showed that Tkq † = (−1)q Tkq ∗ , which applied to (2.8.24) yields ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ V = T2−2 C−2 + −T2−1 C−1 + T20 C0∗ + −T21 C1∗ + T22 C2∗ ∗ ∗ =T2−2 ∗ C−2 − T2−1 ∗ C−1 + T20 ∗ C0∗ − T21 ∗ C1∗ + T22 ∗ C2∗
(2.8.25)
Equating (2.8.22) and (2.8.25), ∗ ∗ C−2 T2−2 + C−1 T2−1 + C0 T20 + C1 T21 + C2 T22 =T2−2 ∗ C−2 − T2−1 ∗ C−1 + T20 ∗ C0∗ − T21 ∗ C1∗ + T22 ∗ C2∗ (2.8.26)
We note that (2.8.26) can only be satisfied if: C0 ∈ R ∗ C1 = C−1 ∗ C2 = C−2 b)
In order to study the effect of the quadrupole perturbation on the manifold of a degenerate energy eigenstate of the unperturbed system with generic total angular momentum quantum number j, we must investigate the selection rules for an arbitrary irreducible 2nd rank tensor operator. We first recall the two main stipulations on q and k, defined by (17.88) and (17.89), respectively as: q = m0 − m 0
|j − j | ≤ k ≤ j + j
(2.8.27a) 0
(2.8.27b)
We also recollect the Wigner Eckart Theorem from (17.87): hα0 j 0 m0 |Tkq |αjmi = hjkmq|jkj 0 m0 ihα0 j 0 kTk kαji
(2.8.28)
Where we require that the bounds on Tkq are limited by ±q. For a second rank irreducible tensor operator, we have q = 2, and therefore: q = −2, −1, 0, 1, 2 Now, applying the possible values of q to (2.8.27a) first, we find that ∆m = − 2, −1, 0, 1, 2 −→ ∆m = 0, ±1, ±2 And similarly for (2.8.27b),
(2.8.29)
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CHAPTER 2: QUANTUM MECHANICS II
|∆j| ≤ −2, −1, 0, 1, 2 ≤ ∆j −→ ∆j = 0, ±1, ±2
(2.8.30)
It is easy to see that (2.8.29) and (2.8.30) express the relevant selection rules for an irreducible second-rank tensor operator Tq2 . Another important result worth noting is that when both of these conditions are satisfied, we require that hα0 j 0 m0 |Tkq |αjmi = 6 0
(2.8.31)
We now must find a connection between j and k. Starting with arbitrary j, we recognize that the condition where we are most likely to find a limiting relationship between j and k is if we let j = j 0 : 0 ≤ k ≤ 2j −→ k ≤ 2j −→ j ≥
k 2
(2.8.32)
This gives us a condition in which the expectation value from (2.8.31) is non-zero. We can extend (2.8.32) to give us a condition in which Tkq is zero: j
0 What is the resulting motion?
Solution The differential equation describing the damped, forced harmonic oscillator is: x00 (t) + βx0 (t) + ω02 x(t) = F (t)
(4.3.13)
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Where F (t) in this problem is defined by the heaviside step function. For t < 0, (4.3.13) becomes x00 (t) + βx0 (t) + ω02 x = 0 Which is easy enough to solve, start by finding the roots r1 and r2 : r 2 + βr + ω02 = 0 Which is clearly quadratic, letting a = 1, b = β and c = ω02 : p −β ± β 2 − 4ω02 r1,2 = 2 So the motion is described by " ! # " p −β + β 2 − 4ω02 x(t) = C1 exp i t + C2 exp i 2
−β −
p
β 2 − 4ω02 2
! # t
(4.3.14)
Where C1 and C2 are determined via initial conditions, which are not given. If we say that initially the system is not in motion, e.g. x(0) = 0 and x0 (0) = 0, then x(t) = 0 for all times. Otherwise it will oscillate in a predictable way. We should now investigate the second condition of the heaviside step function, where our differential equation picks up a constant term: x00 (t) + βx0 (t) + ω02 x(t) = F0 Which we can rearrange as 00
0
x (t) + βx (t) +
ω02
F0 x(t) − 2 = 0 ω0
The addition of this term necessitates the inclusion of a “particular solution” to our homogeneous solution (4.3.15): " ! # " ! # p p −β + β 2 − 4ω02 −β − β 2 − 4ω02 F0 x(t) = C1 exp i t + C2 exp i t + 2 (4.3.15) 2 2 ω0 The effect of adding a constant external force to a damped harmonic oscillator only acts to offset the equilibrium position, so that the system oscillates around a different point, which is given by F0 /ω02 and is independent of time.
Problem 5 Show that the Fourier sine transforms of y0 (x) and y00 (x) are Fs [y0 ] = − ωyˆc (ω)
ω Fs [y00 ] = − ω2 yˆs (ω) + y(0) π
(4.3.16a) (4.3.16b)
W. Erbsen
HOMEWORK #5
Solution We first note the definitions of the Fourier sine and cosine transforms: Z ∞ Fs [f(x)] =fs (ω) = f(x) sin ωx dx Z0 ∞ Fc[f(x)] =fc (ω) = f(x) cos ωx dx 0
So, for the first derivative, f(x) = y0 (x) and: Fs [y0 (x)] =
Z
0
=
Z
0
∞ ∞
y0 (x) sin ωx dx ∂y(x) sin ωx dx ∂x
At this point, and this is a historic moment since I have never done this before, we want to do a sort of inverse integration by parts. We begin by noting that Z ∞ y(x) cos ωx dx (4.3.17) 0
Doing integration by parts, ∂y(x) ∂x 1 dv = cos ωx −→ v = sin ωx ω
u = y(x) −→ du =
Such that (4.3.17) becomes: ∞ Z ∞ Z 1 1 ∞ ∂y(x) y(x) cos ωx dx = y(x) sin ωx − sin ωx dx ω ω 0 ∂x 0 0
Solving this for
R
∂y(x) ∂x
Z
sin ωx dx,
0
∞
∞ Z ∞ ∂y(x) sin ωx dx = y(x) sin ωx − ω y(x) cos ωx dx ∂x 0 0 = − ωyc (ω) =Fs [y0 (x)]
(4.3.18)
The process of finding the second derivative is very much the same, recognizing that in this case f(x) = y00 (x) and Z ∞ Fs [y00 (x)] = y00 (x) sin ωx dx Z0 ∞ 0 ∂y (x) = sin ωx dx ∂x 0 Where this time we look at: Z
0
∞
y0 (x) cos ωx dx
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CHAPTER 4: MATHEMATICAL METHODS
Where we have ∂y0 (x) ∂x 1 dv = cos ωx −→ v = sin ωx ω
u = y0 (x) −→ du =
Such that Z
∞ Z 1 0 1 ∞ ∂y0 (x) y (x) sin ωx − sin ωx dx y (x) cos ωx dx = ω ω 0 ∂x 0
∞
0
0
And solving for the relavent term, ∞ Z ∞ 0 Z ∞ ∂y (x) 0 sin ωx dx = y (x) sin ωx − ω y0 (x) cos ωx dx ∂x 0 0 0 Z ∞ 0 =−ω y (x) cos ωx dx
(4.3.19)
0
We repeat this process yet another time, but this time around we are looking at Z ∞ Z ∞ ∂y(x) 0 y (x) cos ωx dx = cos ωx dx ∂x 0 0 And now we look at Z
∞
y(x) sin ωx dx
0
Where we let ∂y(x) ∂x 1 dv = sin ωx −→ v = − cos ωx ω
u = y(x) −→ du =
So Z
∞
0
And now Z
0
∞
y(x) sin ωx dx = −
∞ Z 1 1 ∞ ∂y(x) y(x) cos ωx + dx ω ω 0 ∂x 0
∞ Z ∞ ∂y(x) cos ωx dx = y(x) cos ωx + y(x) sin ωx dx ∂x 0 0 = − y(0) + ωys (ω)
Substituting this back into (4.3.19), Z ∞ 0 ∂y (x) sin ωx dx = − ω (ωys (ω) − y(0)) ∂x 0 = − ω2 ys (ω) + ωy(0)
(4.3.20)
W. Erbsen
HOMEWORK #5
In summary, we have shown that: Fs [y0 ] = −ωyc (ω) Fs [y00 ] = −ω2 ys (ω) + ωy(0) You might have noticed that what I found for Fs [y00 ] differs from yours by a factor of 1/π in the last term. I have no idea where that could have come from.
Problem 6 Show that the solution of y00 + 2µy0 + (β 2 + µ2 )y = g(x), With g(±∞) = 0 is 1 y= β
Z
x
−∞
µ > 0,
β > 0,
y(−∞) = y(∞) = 0
g(ξ)e−µ(x−ξ) sin [β(x − ξ)] dξ
(4.3.21)
(4.3.22)
Hint: Find Green’s function of the equation first.
Solution As per the hint, I am going to go ahead and find Green’s function for (4.3.21). Substituting G(x − ξ) for y assuming that x > ξ, we solve the homogeneous analogue of our given differential equation: G00 (x − ξ) + 2µG0 (x − ξ) + (β 2 + µ2 )G(x − ξ) = 0 Finding the roots, r 2 + 2µr + (β 2 + µ2 ) = 0 Letting a = 1, b = 2µ and c = β 2 + µ2 : r1,2 =
−2µ ±
=µ ± iβ
p
4µ2 − 4(β 2 + µ2 ) 2
So that we have G(x − ξ) =C1 cos[(µ + iβ)(x − ξ)] + C2 sin[(µ − iβ)(x − ξ)] 1 = sin[(µ − iβ)(x − ξ)] β
(4.3.23)
Due to the boundary conditions of the problem. At this point we should note that this problem represents a driven, damped harmonic oscillator, whose solution was discussed in an earlier problem. The solution
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CHAPTER 4: MATHEMATICAL METHODS
to this type of problem using Green’s function is given in M&W in the prompt of problem 9-7. Tailoring it to our case, Z x y(x) = G(x − ξ)g(ξ) dξ (4.3.24) −∞
The finite upper limit comes from the fact that x in this problem really should be t, and up to this point we have been assuming that for x − ξ ≤ 0, G(x − ξ) = 0. To make this more transparent, we note that for a harmonic oscillator, we would have likely chosen “x” to be “t,” and “ξ” to be “t0 .” In this case our solution for y(x) (or y(t), rather) is represented by an integral over past times t. Substituting (4.3.23) into (4.3.24), Z 1 x y(x) = sin[(µ − iβ)x]g(ξ) dξ (4.3.25) β −∞ Where we can say 1 {exp[(µ − iβ)(x − ξ)] − exp[(−µ + iβ)(x − ξ)]} 2i 1 = {exp[µx − µξ − iβx + iβξ] − exp[−µx + µξ + iβx − iβξ]} 2i 1 = {exp[µ(x − ξ) + iβ(x − ξ)] − exp[−µ(x − ξ) + iβ(x − ξ)]} 2i exp[µ(x − ξ)] = {exp[iβ(x − ξ)] − exp[−iβ(x − ξ)]} 2i = exp[µ(x − ξ)] sin [β(x − ξ)]
sin[(µ − iβ)x] =
(4.3.26)
Now substituting (4.3.26) back into (4.3.25): 1 y(x) = β
Z
x
−∞
exp[µ(x − ξ)] sin [β(x − ξ)] g(ξ) dξ
In case you couldn’t tell I worked backwards a bit, sorry for the missteps in logic. I am very ignorant to Green’s functions (I find the entire thing very confusing), however it is an invaluable mathematical tool, one I hope to master soon.
4.4
Homework #6
Problem 1 Find the Laplace transform of f(t) = tet , and determine the region of existence.
Solution
W. Erbsen
HOMEWORK #6
We know that the Laplace Transform of some function f(t) is given by Z ∞ L {f(t)} = F (s) = f(t)e−st dt 0
And in our case, L {f(t)} =
Z
∞
tet e−st dt
0
Z
∞
u = t, du = 1 1 −(s−1)t dv = e , v = − s−1 e−(s−1)t 0 ∞ Z ∞ t −(s−1)t 1 = − e + e−(s−1)t dt s−1 s−1 0 0 ∞ 1 −(s−1)t = − e 2 (s − 1) =
−(s−1)t
te
dt
0
1 = (s − 1)2
We can see that in the last limit, the result will be convergent only if s − 1 < 0, or s > 1, so F (s) =
1 (s − 1)2
for s > 1
Problem 2 Find the inverse Laplace Transform of a) b)
1 s3 − s2 s2 + s − 1 s3 − 2s2 + s − 2
Solution a)
As per the hint, let’s go ahead and employ partial fractions on (4.4.1a): s3
1 1 A B C =⇒ 2 = 2+ + 2 −s s (s − 1) s s s−1 Let s = 0: Let s = 1:
Let A = −1, C = 1: So, (4.4.1a) becomes
1 =A(s − 1) + Bs(s − 1) + Cs2 1 = − A −→ A = −1 1 =Cs2 −→ C = 1
1 =1 − s + Bs(s − 1) + s2 −→ B = 1
(4.4.1a) (4.4.1b)
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CHAPTER 4: MATHEMATICAL METHODS
1 1 1 1 = − − s3 − s2 s − 1 s2 s Where we note that
Z ∞ 1 et · e−st dt = e−(s−1)t dt = s−1 Z0 ∞ Z 0∞ 1 1 L {t} = te−st dt = e−st dt = 2 s 0 s 0 Z ∞ 1 L {1} = e−st dt = s 0
L et =
Z
(4.4.2)
∞
(4.4.3a) (4.4.3b) (4.4.3c)
Substituting (4.4.3a)-(4.4.3c) into (4.4.2), 1 = et − t − 1 s3 − s2 b)
This time we only need to factor; no partial fractions is needed: (s2 + 1) (s2 + 1) + (s − 2) 1 1 s2 + s − 1 = + = 2 + s3 − 2s2 + s − 2 (s2 + 1)(s − 2) (s2 + 1)(s − 2) s +1 s−2 We note the Laplace Transforms of the following functions Z ∞ L {sin t} = sin te−st dt 0 Z 1 ∞ it = e − e−it e−st dt 2i 0 Z i 1 ∞ h (i−s)t e − e−(i+s)t dt = 2i 0 ∞ (i−s)t 1 e e−(i+s)t = − − 2i s−i s + i 0 1 1 1 = − 2i s − i s + i 1 2i = 2i s2 + 1 1 = 2 s +1 Z ∞ Z ∞ 2t 2t −st L e = e ·e dt = e−(s−2)t dt = 0
0
Putting (4.4.5) and (4.4.6) into (4.4.4),
s2 + s − 1 = sin t + e2t s3 − 2s2 + s − 2
Problem 3
(4.4.4)
(4.4.5) 1 s−2
(4.4.6)
W. Erbsen
HOMEWORK #6
Solve the following differential equations using Laplace Transforms: a) y0 + y = cos t, for t > 0, y(0) = 1 b) y00 + y = cos (2t), for t > 0, y(0) = 0,
y0 (0) = 0
(4.4.7a) (4.4.7b)
Solution a)
First let’s find the Laplace Transform of cos t, since it will be needed later: Z ∞ L {cos t} = cos te−st dt 0 Z 1 ∞ it = e + e−it e−st dt 2 0 Z i 1 ∞ h (i−s)t e + e−(i+s)t dt = 2 0 ∞ (i−s)t 1 e e−(i+s)t = − + 2 s−i s + i 0 1 1 1 = + 2 s−i s+i 2s 1 = 2 s2 + 1 s = 2 s +1
(4.4.8)
Taking the Laplace Transform of both sides of (4.4.7a), L {y0 } + L {y} =L {cos t} s sY (s) − y(0) + Y (s) = 2 s +1 Applying the given initial condition and solving for Y (s), s 1 + + 1)(s + 1) s + 1 1+s 1 1 − + = 2 2(s + 1) 2(s + 1) s + 1 1 1 1 s 1 = + + 2 s2 + 1 2 s2 + 1 s+1
Y (s) =
(s2
Now we must find the inverse Laplace Transform of Y (s): 1 −1 1 1 −1 s 1 −1 −1 L {Y (s)} = L + L +L 2 s2 + 1 2 s2 + 1 s+1 Where we recall that the inverse Laplace Transform of 1/(s2 + 1) was found to be sin t in (4.4.5), and s/(s2 + 1) was found to be cos t in (4.4.8). Finding 1/(s + 1), Z ∞ Z ∞ 1 L e−t = e−t · e−st dt = e−(s+1)t dt = s + 1 0 0
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CHAPTER 4: MATHEMATICAL METHODS
So we now have L−1 {Y (s)} =
sin t cos t1 + + e−t 2 2
And finally y(t) = b)
1 (sin t + cos t) + e−t 2
We first find the Laplace Transform of cos (2t), since it will be needed later: Z ∞ L {cos (2t)} = cos (2t)e−st dt 0 Z 1 ∞ 2it = e + e−2it e−st dt 2 0 Z i 1 ∞ h (2i−s)t = e + e−(2i+s)t dt 2 0 ∞ (2i−s)t e e−(2i+s)t 1 = − + 2 s − 2i s + 2i 0 1 1 1 = + 2 s − 2i s + 2i 1 2s = 2 s2 + 4 s = 2 s +4
(4.4.9)
Now, applying the Laplace Transform to both sides of (4.4.7b): L {y00 } + L {y} =L {cos (2t)} s s2 Y (s) − sy(0) − y0 (0) + Y (s) = 2 s +4 Applying the initial conditions, s2 Y (s) + Y (s) = Y (s) =
s2
s +4
s 1 s s = − (s2 + 4)(s2 + 1) 3 s2 + 1 s2 + 4
Taking the inverse Laplace Transform of Y (s), 1 s s −1 L−1 {Y (s)} = L−1 − L 3 s2 + 1 s2 + 4 Where again the inverse Laplace Transform of s/(s2 + 1) is just cos t as per (4.4.5). Similarly, s/(s2 + 4) is just cos (2t), which was found in (4.4.9). So, we are finally left with: y(t) =
1 [cos t − cos (2t)] 3
W. Erbsen
HOMEWORK #6
Problem 4 An electric circuit gives rise to the system dI1 + RI1 + dt dI2 L + RI2 − dt dq = I1 − I2 dt L
q = E0 C q =0 C
(4.4.10a) (4.4.10b) (4.4.10c)
With initial conditions E0 , q(0) = 0 2R Solve the system by Laplace Transform methods and show that E0 E0 −αt R 2 I1 = + e sin (ωt), where α = and ω2 = − α2 2R 2ωL 2L LC I1 (0) = I2 (0) =
(4.4.11)
(4.4.12)
Solution We start by taking the inverse Laplace Transform of (4.4.10c): L {q 0 (t)} =L {I1 (t)} − L {I2 (t)}
sq(s) − q(0) =I1 (s) − I2 (s) I1 (s) I2 (s) − q(s) = s s
(4.4.13)
We now do the same for (4.4.10a): 1 L {q(t)} =E0 L {1} C 1 E0 L [sI1 (s) − I1 (0)] + RI1 (s) + q(s) = C s E0 1 I1 (s) I2 (s) E0 L sI1 (s) − + RI1 (s) + − = 2R C s s s R I1 (s) I2 (s) E0 E0 − ω02 = + sI1 (s) + I1 (s) + ω02 L s s Ls 2R R ω2 I2 (s) E0 E0 s + + 0 I1 (s) − ω02 = + L s s Ls 2R 2 s + ω02 R I (s) E E0 2 0 + I1 (s) − ω02 = + s L s Ls 2R LL {I1 (t)} + RL {I1 } +
√ Where I have defined ω02 = 1/ LC . And similarly for (4.4.10b):
1 L {q(t)} =0 C 1 L [sI2 (s) − I2 (0)] + RI2 (s) − q(s) =0 C LL {I2 (t)} + RL {I2 } −
(4.4.14)
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CHAPTER 4: MATHEMATICAL METHODS
E0 1 I1 (s) I2 (s) L sI2 (s) − + RI2 (s) − − =0 2R C s s R I1 (s) I2 (s) E0 + ω02 = sI2 (s) + I2 (s) − ω02 L s s 2R R ω02 I (s) E0 1 s+ + I2 (s) − ω02 = L s s 2R 2 s + ω02 R I1 (s) E0 + I2 (s) − ω02 = s L s 2R
(4.4.15)
At this point we have two equations, (4.4.14) and (4.4.15), and two unknowns. Finding the s-domain solutions (I1 (s) and I2 (s)) is trivial, but time consuming. The most tedious steps in what follows were left up to Mathematica. The obvious first step would be to solve (4.4.14) for I2 (s), inserting into (4.4.15), and solving for I1 (s). The result is 1 E0 1 + I1 (s) = 2 Rs s(Ls + R) + 2Lω02 E0 1 1 = + 2 Rs Ls(s + R/L) + 2/C E0 1 1 1 + Let α = R/2L = 2 Rs L s2 + sR/L + 2/LC E0 1 1 1 = + 2 Rs L s2 + 2sα + 2/LC E0 1 1 1 = + Complete the square 2 Rs L s2 + 2sα + α2 + 2/LC − α2 1 1 E0 1 = + Let ω = 2/LC − α2 2 Rs L (s + α)2 + 2/LC − α2 E0 1 1 ω = + 2 Rs ωL (s + α)2 + ω2 E0 1 E0 ω = + 2R s 2ωL (s + α)2 + ω2 Taking the inverse Laplace Transform, L
−1
E0 −1 1 E0 −1 ω {I1 (s)} = L + L 2R s 2ωL (s + α)2 + ω2 E0 E0 −αt = + e sin (ωt) 2R 2ωL
Which was mercilessly stolen from a table. We finally have I1 (t) =
E0 E0 −αt + e sin (ωt) 2R 2ωL
Repeating this process for I2 (s), Mathematica gives E0 1 1 I2 (s) = − 2 Rs s(Ls + R) + 2Lω02
W. Erbsen
HOMEWORK #7
Which we note is identical to our expression for I1 (s) sans the minus sign, therefore we can go ahead and conclude that: I2 (t) =
4.5
E0 E0 −αt − e sin (ωt) 2R 2ωL
Homework #7
Problem 1 For the vectors in three dimensions, ~v1 = x ˆ + yˆ,
~v2 = yˆ + zˆ,
~v3 = zˆ + x ˆ
(4.5.1)
Use the Gram-Schmidt procedure to construct an orthonormal basis starting from ~v1 .
Solution It might be most instructive if we explicitly define each of our vectors in terms of x, y, and z. e.g., 1 0 1 ~v1 = 1 , ~v2 = 1 , ~v3 = 0 0 1 1 And now recall that the Gramm-Schmidt Orthonormalization procedure is ~un = ~vn −
n−1 X i=1
h~ui |~vn i u ~ i, h~ui |~ui i
~e1 =
u ~n k~un k
Where un represents a orthogonal set of vectors, and en represents an orthonormal set of vectors. Applying this to our case, 1 1 1 ~u1 =~v1 = 1 −→ ~e1 = √ 1 2 0 0 Where I used the fact that k~u1 k =
√
2 . Moving on to the second vector, u ~ 2 =~v2 −
h~v2 |~u1 i ~u1 h~u1 |~u1 i
Where 1 h~u1 |~u1 i =~ uT1 ~u1 = (1 1 0) 1 = 2, 0
1 h~v2 |~u1 i = ~v2T ~v1 = (0 1 1) 1 = 1 0
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CHAPTER 4: MATHEMATICAL METHODS
Such that we now have −1 r −1/ 0 1 /2 2 1 2 1 /2 ~u2 = 1 − 1 = 1/2 −→ ~e2 = 2 3 1 0 1 1
And now the last vector is
u ~ 3 =~v3 −
h~v3 |~u1 i h~v3 |~u2 i ~u1 − u ~2 h~u1 |~u1 i h~u2 |~u2 i
Where we have 1 h~v3 |~u1 i =(1 0 1) 1 = 1 0 1 − /2 h~v3 |~u2 i =(1 0 1) 1/2 = 1/2 1 1 − /2 h~u2 |~u2 i =(− 1/2 1/2 1) 1/2 = 3/2 1
And now
1 2 1 1 − /2 /3 1 1 1 1 u ~ 3 = 0 − 1 − 1/2 = − 2/3 −→ ~e3 = √ −1 2 3 3 2 1 0 1 /3 1
Problem 2 For the vector space of polynomials in x, use the scalar product defined as Z ∞ 2 hf|gi = dxe−x f(x) · g(x)
(4.5.2)
−∞
Start from the vectors ~v0 = 1,
~v1 = x,
~v2 = x2 ,
~v3 = x3
(4.5.3)
And use the Gram-Schmidt procedure to construct an orthonormal basis starting from ~v0 . Repeat this calculation for the scalar product Z 1 hf|gi = dxx2 f(x) · g(x) (4.5.4) 0
Solution
W. Erbsen
HOMEWORK #7
Using the same logic as before, we are looking to ortho-normalize ~v0 , ~v1 , ~v2 , and ~v3 using the scalar product defined by (4.5.2). The first vector is trivial, ~u0 = ~v0 = ~e0 = 1 And the second is h~v1 |~u0 i u ~0 h~u0 |~u0 i hx|1i =x − 1 h1|1i R∞ −x2 dx −∞ xe =x − R ∞ −x2 e dx −∞ 1/4 4 =x −→ ~e1 = x π
u ~ 1 =~v1 −
And the third is h~v2 |~u0 i h~v2 |~u1 i u ~0 − ~u1 h~u0 |~u0 i h~u1 |~u1 i hx2 |1i hx2 |xi − x =x2 − h1|1i hx|xi R ∞ 2 −x2 R ∞ 3 −x2 x e dx x e dx −∞ 2 R∞ =x − R ∞ −x2 − −∞ x 2 2 −x dx −∞ e −∞ x e √ π 1 2 √ =x − 2 π 1/4 1 4 2 =x − −→ ~e2 = (x2 − 1/2 ) 2 π
u ~ 2 =~v2 −
And lastly, the forth is h~v3 |~u0 i h~v3 |~u1 i h~v3 |~u2 i ~u0 − u ~1 − ~u2 h~u0 |~u0i h~u1 |~u1 i h~u2 |~u2 i hx3 | x2 − 1/2 i hx3 |1i hx3 |xi 3 =x − − x− x2 − 1/2 2 1 2 1 h1|1i hx|xi h(x − /2 ) |(x − /2 )i R ∞ 3 −x2 R ∞ 4 −x2 R ∞ 3 2 1 −x2 x e dx x e dx x x − /2 e dx 2 1 −∞ −∞ 3 =x − R ∞ −x2 − R ∞ 2 −x2 x − R−∞ x − /2 ∞ 2 2 2 1 −x dx dx dx −∞ e −∞ x e −∞ (x − /2 ) e √ 3 π 2 √ x =x3 − 4 π 1/4 3 8 3 =x − x −→ ~e3 = (x2 − 1/2 ) 2 9π
u ~ 3 =~v3 −
Now we repeat this calculation for (4.5.4), in much the same way we did for (4.5.3). The first vector is then
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CHAPTER 4: MATHEMATICAL METHODS
~u0 = ~v0 = ~e0 = 1 While the second is given by h~v1 |~u0 i ~u0 h~u0 |~u0 i hx|1i =x − h1|1i R1 3 x dx =x − R01 x2 dx 0
~u1 =~v1 −
=x −
1 /4 1/ 3
=x −
√ 3 −→ ~e1 = 80 (x − 3/4 ) 4
And now the third h~v2 |~u0 i h~v2 |~u1 i u ~0 − ~u1 h~u0 |~u0 i h~u1 |~u1 i R1 4 R1 4 x dx x (x − 3/4 ) dx 2 0 =x − R 1 − R01 (x − 3/4 ) 2 dx 2 (x − 3/ ) dx x x 4 0 0
~u2 =~v2 −
1
/5 − 1/ 3 4 =x2 − x + 3
=x2 −
1
/60 (x 1/ 80
− 3/4 )
√ 2 −→ ~e2 = 1575 (x2 − 4/3 x + 2/5 ) 5
And the forth h~v3 |~u0 i h~v3 |~u1 i h~v3 |~u2 i u ~0 − ~u1 − u ~2 h~u0 |~u0 i h~u1 |~u1 i h~u2 |~u2 i R1 5 R1 5 R1 5 2 4 x dx x (x − 3/4 ) dx x (x − /3 x + 2/5 ) dx 2 4 3 3 0 0 =x − R 1 − R1 (x − /4 ) − R 10 (x − /3 x + 2/5 ) 2 dx 2 (x − 3/ ) dx 2 (x2 − 4/ x + 2/ )2 dx x x x 4 3 5 0 0 0
~u3 =~v3 −
1 1 /6 /56 /840 2 4 3 − (x − / ) − (x − /3 x + 2/5 ) 4 1/ 1/ 1/ 3 224 1575 p 15 3 7 =x3 − x2 − x + −→ ~e3 = 20160/1667 (x3 − 15/8 x2 − 3/2 x + 7/4 ) 8 2 4
=x3 −
1
Problem 3 On the vector space of quadratic polynomials, of degree ≤ 3, the operator d/dx is defined. Use the basis of the Legendre Polynomials 3 1 5 3 P~0 (x) = 1, P~1 (x) = x, P~2 (x) = x2 − , P~3 (x) = x3 − x (4.5.5) 2 2 2 2
W. Erbsen
HOMEWORK #7
And compute the components of this operator. Repeat this exercise for the operator d2 /dx2 .
Solution Let our set of quadratic polynomials be given by f(x) = a0 + a1 x + a2 x2 + a3 x3 (so, ~v0 = a0 , ~v1 = a1 x, ~v2 = a2 x2 , ~v3 = a3 x3 ). We can also define our differential operator to be D[f(x)] = df(x)/dx. So, D[~v0 ] =0 D[~v1 ] =a1 =⇒ a1 P~0 (x) D[~v2 ] =2a2 x =⇒ 2a2 P~1 (x) D[~v3 ] =3a3 x2 =⇒ a3 (2P~2 (x) + P~0 )
And if we let D2 [f(x)] = df(x)2 /dx2 , D[~v0 ] =0 D[~v1 ] =0 D[~v2 ] =2a2 =⇒ 2a2 P~0 (x) D[~v3 ] =3a3 x2 =⇒ 6a3 P~1 This is my wild stab at the problem, however I don’t believe it is a very accurate one, since if it were the problem would be trivial (clearly). Whoops!
Problem 4 Diagonalize each of the Pauli spin matrices. Find their eigenvalues and specify the respective eigenvectors as the basis in which they are diagonal.
Solution The Pauli matrices are given by 0 σx = 1
1 , 0
σy =
0 i
−i , 0
−λ 1
1 = 0 −→ λ2 − 1 = 0 −λ
σz =
1 0
0 −1
Starting with σx ,
For λ = +1 :
−1 1
1 −1
λ = ±1
a 0 −a + b = 0 = −→ b 0 a−b = 0
a = b,
so
1 1
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CHAPTER 4: MATHEMATICAL METHODS
1 1
For λ = −1 :
1 a 0 a+b=0 = −→ 1 b 0 a+b=0
a = −b,
so
1 −1
so
1 i
And now for σy : −λ i For λ = +1 :
−1 i
For λ = −1 :
1 −i i 1
−i −1
−i = 0 −→ λ2 − 1 = 0 −λ λ = ±1
a 0 −a − ib = 0 = −→ b 0 ia − b = 0
a 0 a − ib = 0 = −→ b 0 ia + b = 0
a = −ib,
a = ib,
so
1 −i
In analyzing σz we immidiately recognize that λ = ±1, since it is diagonal. So, we have 0 0 a 0 1 For λ = +1 : = −→ − 2b = 0, so 0 −2 b 0 0 For λ = −1 :
4.6
2 0 0 0
a 0 = b 0
−→
2a = 0,
0 so 1
Homework #8
Problem 1 Find the eigenvlaues and eigenfunctions of the boundary value problem λ y00 + y=0 (x + 1)2
(4.6.1)
On the interval a ≤ x ≤ 2 with boundary conditions y(1) = y(2) = 0. Write the equation in terms of a regular Sturm-Liouville eigenvalue problem, and find the coefficients in the expansion of an arbitrary function f(x) in a series of the eigenfunctions.
Solution In order to express (4.6.1) within the formalism of Sturm-Lioville theory, we recall that if we are given some differential equation of the form a2 (x)y00 + a1 (x)y0 + a0 (x)y = f(x)
(4.6.2)
W. Erbsen
HOMEWORK #8
then we can define Z
a1 (x) p(x) = exp dx a2 (x) a0 (x) q(x) =p(x) a2 (x) f(x) F (x) =p(x) a2 (x)
Such that (4.6.2) is transformed into d (p(x)y0 ) + q(x)y = F (x) dx
(4.6.3)
We can see form (4.6.1) that in our case, a2 (x) = 1, a1 (x) = 0, a0 (x) = λ/(x + 1)2 , and f(x) = 0. From these, it is clear that p(x) = 1, q(x) = λ/(x + 1)2 , and F (x) = 0, such that we can now express (4.6.1) in the form of (4.6.3): y00 +
λ y=0 (x + 1)2
(4.6.4)
Which is identical to (4.6.1). So, apparently our DE was already in S-L form. To find the eigenvalues and eigenfunctions, we wish to make the substitution and corresponding derivatives y =(x + 1)α y0 =α(x + 1)α−1 y00 =α(α − 1)(x + 1)α−2 Now (4.6.4) becomes α(α − 1)(x + 1)α−2 +
λ (x + 1)α = 0 (x + 1)2
α(α − 1)(x + 1)α−2 + λ(x + 1)α−2 = 0
α2 − α + λ = 0 And now
α=
1±
√ 1 − 4λ 2
(4.6.5)
At this point we split our solutions according to the three distinct outcomes from (4.6.5), which are if 1 − 4λ < 0, 1 − 4λ = 0, and 1 − 4λ > 0: √ λ < 1/4 : then α = 1/2 1 ± 1 − 4λ with real, distinct roots, which admits solutions of the form y(x) = c1 eαx + c2 e−αx Which only satisfies the first boundary condition, y(1) = 0 for the trivial case where y = 0. This cannot be our solution.
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CHAPTER 4: MATHEMATICAL METHODS
λ = 1/4 : then α = 1/2 , with real, repeated roots and solutions y(x) = c1 eαx + c2 xeαx Which also fails to satisfy the first boundary condition, therefore we must ignore this case as well. √ λ > 1/4 : then α = 1/2 1 ± i 4λ − 1 with complex, distinct roots and √ 1 4λ − 1 x+1 y1 (x) =c1 (x + 1) /2 sin log 2 2 √ 1 4λ − 1 x+1 /2 y2 (x) =c2 (x + 1) cos log 2 2 Which does work. Applying the first boundary condition √ √ 1 1 4λ − 1 2 4λ − 1 2 /2 /2 y(1) =c1 (2) sin log + c2 (2) cos log 2 2 2 2 1
=c1 · 0 + c2 · 2 /2 −→ c1 = 1, c2 = 0
So, our solution takes the form y(x) = y1 (x) = (x + 1)
1
/2
sin
√
4λ − 1 log 2
x+1 2
Applying our second boundary condition, √
4λ − 1 y(2) = (3) sin log 2 √ 3 4λ − 1 log =0 −→ sin 2 2 1
/2
3 2
And in order for this to hold... √ 4λ − 1 log 3/2 =nπ 2 2 2 2nπ nπ 1 4λ − 1 = −→ λ = + 3 3 log ( /2 ) log ( /2 ) 4 And to find the corresponding eigenfunction we substitute this into (4.6.6): 1
ϕn (x) = (x + 1) /2 sin
Problem 2
2nπ log log ( 3/2 )
x+1 2
(4.6.6)
W. Erbsen
HOMEWORK #8
Find the eigenvalues and eigenfunctions of the boundary value problem x2 y00 + xy0 + y = µy
(4.6.7)
on the interval 1 ≤ x ≤ 2 with boundary conditions y(1) = y(2) = 0. Write the equation in terms of a regular Sturm-Liouville eigenvalue problem, and find the coefficients in the expansion of an arbitrary function f(x) in a series of the eigenfunctions.
Solution In the same spirit as the last problem, we must first put (4.6.7) into Sturm-Lioville form. We note that a2 (x) = x2 , a1 (x) = x, a0 (x) = 1 − µ, and f(x) = 0. From this we can find Z 1 dx = exp [log x] = x p(x) = exp x 1−µ 1−µ q(x) =x 2 = x x F (x) =0 Which allows us to write d 1−µ (xy0 ) + y=0 dx x Try a solution of the form (and corresponding derivatives) y =xα y0 =αxα−1 y00 =α(α − 1)xα−2 Substituting these into (4.6.8), 1−µ α d xαxα−1 + x =0 dx x d (αxα ) + (1 − µ)xα−1 = 0 dx α2 xα−1 + (1 − µ)xα−1 = 0 α2 + 1 − µ = 0
Solving for α, α=
√ ± µ−4 2
Following the same logic as before, we find that in the case that µ < 4 that we have √ µ−4 y1 (x) = sin log x 2 √ µ−4 y2 (x) = cos log x 2
(4.6.8)
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CHAPTER 4: MATHEMATICAL METHODS
Where we must choose y1 (x) since it satisfies the first boundary condition. For the second, √ µ−4 − 1/2 y(2) = 2 sin log 2 2 √ µ−4 −→ sin log 2 = 0 2 And so √ µ−4 log 2 =nπ 2 p 2nπ µ−4 = −→ µ = log 2
2nπ log 2
2
+4
And finally ϕn (x) = sin
nπ log x log 2
Problem 2 Using Rodriques’ formula show that the Pn (x) are orthogonal and that Z 1 2 2 [Pn (x)] dx = 2n +1 −1
(4.6.9)
Solution Rodrigues’ formula for the Legendre Polynomials is given by (12.65) in A&W: n n 1 d Pn (x) = n x2 − 1 2 n! dx
To prove that Pn (x) given by (4.6.10) are orthogonal functions, we evaluate: Z 1 Z 1 n n d m 2 m 1 d 2 Pn (x)Pm (x) dx = n+m x −1 x −1 dx 2 n!m! −1 dx dx −1
For convenience let’s set C = 1/2n+m n!m!. At this point we integrate by parts m times, with n n+m d d u= (x2 − 1)n , du = (x2 − 1)n dx dx m d dv = (x2 − 1)m , v = (x2 − 1)m dx
(4.6.10)
(4.6.11)
W. Erbsen
HOMEWORK #8
Such that (4.6.11) becomes 1 n+m Z 1 d m 2 m Pn (x)Pm (x) dx =C (x − 1) (x − 1) + C(−1) (x − 1) (x2 − 1)n dx dx −1 −1 −1 n+m Z 1 d m 2 m (x2 − 1)n dx =C(−1) (x − 1) dx −1
Z
1
2
m
d dx
n
n
2
At this point we note that in the case that n < m, the above integral goes to zero. Eg if, for instance, n = 1 and m = 2: Z
1
P1 (x)P2 (x) dx =C
−1
Z
1
−1
(x2 − 1)2
d dx
3
(x2 − 1)2 dx = C
Z
1
−1
2(x2 − 1)2
d dx = 0 dx
Which works for any n < m. This can be straightforwardly extended to the case in which n 6= m when we recall that our choice of m and n was arbitrarily chosen in integrating by parts. So, we have just shown that if n 6= m then (4.6.11) is equal to zero.
If, on the other hand, m = n, then we have Z
1
[Pn (x)]2 dx =
−1
=
(−1)n 22n(n!)2
Z
1
−1
(−1)n (2n)! 22n(n!)2
(x2 − 1)n
Z
1
−1
d dx
2n
(x2 − 1)n dx
(x2 − 1)n dx
(4.6.12)
At this point we make the substitution u = 1/2 (x + 1) → x = 2u − 1, du = 1/2 dx → dx = 2 du, such that (4.6.12) becomes: Z
1
(−1)n (2n)! [Pn (x)] dx = 2n 2 (n!)2 −1 2
=
Z
2(−1)n (2n)! 22n (n!)2
1
0
Z
0
(4u2 − 4u)n 2 du 1
[4u(u − 1)]n du
Z 2(−1)n (2n)! 1 n = u (u − 1)n du (n!)2 0 2(−1)n (2n)! Γ(n + 1)2 = (−1)n 2 (n!) Γ(2n + 2) 2 2(2n)! (n!) = (n!)2 (2n + 1)! (2n)! =2 (2n + 1)! 2 = 2n + 1
Where I integrated by parts n times to evaluate the integral, and also the fact that Γ(n) = (n − 1)!:
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CHAPTER 4: MATHEMATICAL METHODS
Z
1
[Pn(x)]2 dx =
−1
2 2n + 1
Problem 4 Using a generating function, evaluate the sum ∞ X xn+1 Pn (x) n+1 n=0
(4.6.13)
Solution We note that from (7 − 10) in M&W that
∞ X 1 f(x) = √ = hn Pn (x) 1 − 2hx + h2 n=0
(4.6.14)
Which defines the generating function. We wish to manipulate (4.6.14) until it resembles (4.6.13). Start by squaring both sides, 2
[f(x)] =
∞ X
[hn Pn (x)]
2
n=0
And now integrating, Z
1
2
[f(x)] dx =
−1
Z
1
∞ X
2
[hn Pn (x)] dx
(4.6.15)
−1 n=0
Not sure where to go from here. hn must be dependent on x, otherwise the integral on the RHS of (4.6.15) does not make sense. Otherwise, I was unable to integrate-by-parts to solve it either. Summing by hand and simplifying results in the following series: f(x) =x +
x3 x5 x7 + + + ... 3 5 7
Which is clearly odd, so this leads me to believe f(x) is a sin function. It also seems as though the sum diverges, however it does indeed converge, since we required that 0 < x < 1, so 0 < f(x) < 4.52 (approximately). Ugh.
Problem 5 Prove the trigonometric identity using vector methods:
W. Erbsen
HOMEWORK #8
cos γ = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )
(4.6.16)
Where γ is the angle between two directions (θ1 , φ1 ) and (θ2 , φ2 ).
Solution ~ and B, ~ each normlized to unity. Each is then described by We define two vectors on the units sphere, A only two coordinates, θ and φ. To prove the trigonometric identity (4.6.16), we take the dot product of the two vectors. But first, we define the components of our vectors in terms of Cartesian coordinates, with x = sin θ cos φ y = sin θ sin φ z = cos θ Yielding a dot product of ~·B ~ =x1 · x2 + y1 · y2 + z1 · z2 A = sin θ1 cos φ1 · sin θ2 cos φ2 + sin θ1 sin φ1 · sin θ2 sin φ2 + cos θ1 · cos θ2 =(sin θ1 + sin θ2 )(cos φ1 cos φ2 + sin φ1 · sin φ2 ) + cos θ1 · cos θ2 {z } | cos(φ1 −φ2 )
= cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )
(4.6.17)
~·B ~ = kAk kBk cos γ = cos γ. And recall A |{z} |{z} 1
1
So (4.6.17) is now
cos γ = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )
Problem 6 The amplitude of a scattered wave is given by f(θ) =
∞ 1X (2` + 1)eiδ` sin δ` P` (cos θ) k
(4.6.18)
`=0
Where θ is the angle of scattering, ` is the angular momentum eigenvalue, ~k is the incident momentum, and δ` is the R phase shift produced by the central potential that is doing the scattering. The total cross section is σtot = |f(θ)|2 dΩ. Show that ∞ 4π X σtot = 2 (2` + 1) sin2 δ` (4.6.19) k `=0
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CHAPTER 4: MATHEMATICAL METHODS
Solution We begin by first evaluating |f(θ)|2 : 2
|f(θ)| =
!∗ ∞ 1X iδ` (2` + 1)e sin δ` P`(cos θ) k `=0
! ∞ 1X iδ` (2` + 1)e sin δ` P` (cos θ) k `=0
∞ 1 X = 2 (2` + 1)e−iδ` sin δ` P`(cos θ)(2` + 1)eiδ` sin δ` P` (cos θ) k `=0
∞ 1 X = 2 (2` + 1)2 sin2 δ` P`2 (cos θ) k `=0
And now to find the total cross section, Z σtot = |f(θ)|2 dΩ Z 2π Z π = dφ |f(θ)|2 sin θ dθ 0 0 # Z π" X ∞ 1 2 2 2 (2` + 1) sin δ` P` (cos θ) sin θ dθ =2π k2 0 `=0 Z π ∞ 2π X 2 = 2 (2` + 1)2 sin2 δ` [P` (cos θ)] sin θ dθ k 0 `=0 #2 ` Z π" ∞ ` 2π X 1 d 2 2 2 = 2 (2` + 1) sin δ` cos θ − 1 sin θ dθ k 2``! d cos θ 0 `=0 2` Z π ∞ 2` 2π X 1 d 2 2 = 2 (2` + 1) sin δ` 2` 2 cos2 θ − 1 sin θ dθ k 2 (`!) 0 d cos θ `=0 Z π 2` ∞ 2` 1 d 2π X (2` + 1)2 sin2 δ` 2` 2 x2 − 1 sin θ dθ = 2 k 2 (`!) 0 dx `=0 Z ∞ π 2` 2π X 1 d 2 2 2 = 2 (2` + 1) sin δ` 2` 2 x −1 sin θ dθ k 2 (`!) 0 dx `=0 Z π ∞ 2π X 1 2` = 2 (2` + 1)2 sin2 δ` (cos θ) sin θ dθ k (`!)2 0 `=0 ∞ X 1 1 + (−1)2` 2π (2` + 1)2 sin2 δ` = 2 k (`!)2 2` + 1 `=0
=
∞ 4π X
k2
`=0
(2` + 1) sin2 δ`
1 (`!)2
(4.6.20)
Where I used the fact that dΩ = sin θ dθdφ and also the results from problem 3. Therefore we have shown that ∗ ∗ You might have noticed that there is an extra factor of (`)−2 in (4.6.20). I was not able to get rid of it. So when I say that “we have shown that” I suppose I really mean “we have nearly shown that.”
W. Erbsen
HOMEWORK #9
σtot =
∞ 4π X (2` + 1) sin2 δ` k2 `=0
4.7
Homework #9
Problem 1 By differentiating the Legendre polynomial generating function 1 g(x, t) = √ 1 − 2xt + t2
(4.7.1)
with respect to t and with respect to x, obtain the recurrence relations for the Legendre polynomials, and prove that Pn (x) satisfies the Legendre differential equation.
Solution We know that the generating function g(x, t) can be expressed ∞ as X 1 g(x, t) = √ = tn Pn (x) 1 − 2xt + t2 n=0
(4.7.2)
So, as per the suggestion, we differentiate (4.7.1) first with respect to t: dg(x, t) d 1 = √ dt dt 1 − 2xt + t2 x−t = 3 (1 − 2xt + t2 ) /2 x−t = g(x, t) (1 − 2xt + t2 )
(4.7.3)
Doing the same for the RHS of (4.7.2), ∞ ∞ X dg(x, t) d X n = t Pn (x) = ntn−1 Pn (x) dt dt n=0 n=0
Inserting this and (4.7.2) into the (4.7.3), ∞ X
(1 − xt + t2 ) ∞ X
n=0
ntn−1 Pn (x) − 2x
∞ X
n=0
ntn Pn (x) +
n=0 ∞ X
n=0 ∞ X
n=0
ntn−1 Pn (x) =
∞ X x−t tn Pn (x) (1 − xt + t2 ) n=0
ntn−1 Pn (x) =(x − t) ntn+1 Pn (x) =x
∞ X
n=0
∞ X
tn Pn (x)
n=0
tn Pn (x) −
∞ X
n=0
tn+1 Pn (x)
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CHAPTER 4: MATHEMATICAL METHODS
∞ X
ntn−1 Pn (x) +
n=0
∞ X
(n + 1)tn + 1 Pn (x) =
n=0
|{z} n−1=`
∞ X
|{z} n=`
(` + 1)t`P`+1 (x) +
`=0
x(1 + 2n)tn Pn (x)
n=0
|{z}
n+1=`
Shifting the indices as indicated:
∞ X
∞ X
`t` P`−1 (x) =
`=0
∞ X
x(1 + 2`)t` P` (x)
`=0
−→ (` + 1)P`+1 (x) − x(1 + 2`)P` (x) = −`P`−1 (x)
(4.7.4)
Since we require that each coefficient should vanish. We can find a second recursion relation by similarly integrating (4.7.2) by x: dg(x, t) d 1 = √ dx dx 1 − 2xt + t2 t = (1 − 2xt + t2 ) 3/2 t g(x, t) = (1 − 2xt + t2 )
(4.7.5)
And the the derivative of the RHS of (4.7.2) is ∞ ∞ X dg(x, t) d X n = t Pn (x) = tn Pn0 (x) dx dx n=0 n=0 And now
∞ X
n=0
tn Pn0 (x) − 2x
∞ X
∞ X
tn+1 Pn0 (x) +
n=0 ∞ X
tn Pn0 (x) +
n=0
t
∞ X t tn Pn (x) = (1 − 2xt + t2 ) n=0
Pn0 (x) =
∞ X
tn+2 Pn0 (x) =
n=0
t` P`0 (x) +
∞ X `=0
0 t`P`−2 (x) =
tn+1 Pn (x)
n=0 ∞ X
tn+1 [2xPn0 (x) + Pn (x)]
|{z}
n+2=`
Shifting the indices once more,
∞ X
n=0
|{z}
n=`
`=0
n=0 ∞ X n+2
Pn0 (x)
n=0
|{z}
∞ X
t
n
n+1=`
∞ X `=0
0 (x) + P`−1 (x) t` 2xP`−1
0 0 (x) =2xP`−1 (x) + P`−1 (x) P`0 (x) + P`−2
0 0 −→ P`+1 (x) − 2xP`0 (x) = P` (x) − P`−1 (x)
(4.7.6)
We can combine (4.7.4) and (4.7.6) as follows. First, moving all the terms of (4.7.4) to the left and taking the derivative with respect to x yields 0 0 (` + 1)P`+1 (x) − x(1 + 2`)P`0 (x) − (1 + 2`)P` (x) + `P`−1 (x) = 0
(4.7.7)
W. Erbsen
HOMEWORK #9
Now moving all the terms of (4.7.6) to the left, and multiplying by −` 0 0 −`P`+1 (x) + 2`xP`0 (x) + `P` (x) − `P`−1 (x) = 0
(4.7.8)
Adding (4.7.7) and (4.7.8), 0 P`+1 (x) = xP`0 (x) + (1 + `)P` (x)
Many more (in fact, infinitely more) recursion relations may be derived, in addition to the three I have found here, there are two more that can be found in M&W that become useful: 0 xP`0 (x) − `P` (x) − P`−1 (x) =0 0 0 P` (x) − xP`−1 (x) − `P`−1 (x) =0
(4.7.9) (4.7.10)
0 x2 P`0 (x) − `xP` (x) − xP`−1 (x) = 0
(4.7.11) (4.7.12)
Multiplying (4.7.9) by x:
Now take (4.7.11) and subtract (4.7.10): 0 0 x2 P`0 (x) − `xP`(x) − xP`−1 (x) − P`0 (x) + xP`−1 (x) + `P`−1 (x) = 0
(x2 − 1)P`0 (x) − `xP`(x) + `P`−1 (x) = 0
(4.7.13)
Now taking the derivative of (4.7.13) with respect to x: 0 2xP`0(x) + x2 P`00(x) − P`00 (x) − `P` (x) − `xP`0 (x) + `P`−1 (x) = 0
x2 P`00(x) − P`00 (x) + (2 − `)xP`0 (x) − `P` (x) + `P`−1 (x) = 0
(4.7.14)
0 At this point we take (4.7.9), solve for P`−1 (x), and substitute into (4.7.14):
x2 P`00 (x) − P`00(x) + (2 − `)xP`0 (x) − `P` (x) + `xP`0 (x) − `2 P` (x) = 0
(x2 − 1)P`00 (x) + 2xP`0(x) − `xP`0 (x) − `P` (x) + `xP`0 (x) − `2 P` (x) = 0 (x2 − 1)P`00(x) + 2xP`0(x) − `(` + 1)P`(x) = 0 Which is the Legendre differential equation.
Problem 2 By differentiating the Bessel generating function g(x, t) = e 2 (t− /t ) x
1
(4.7.15)
with respect to t and with respect to x, obtain the recurrence relations for the Bessel functions, and prove that Jn (x) satisfies the Bessel differential equation.
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CHAPTER 4: MATHEMATICAL METHODS
Solution In very much the same spirit as the last problem, we are given that the generating function is g(x, t) = e 2 (t− /t ) = 1
x
∞ X
tn Jn (x)
(4.7.16)
n=0
Differentiating (4.7.15) first with respect to t, dg(x, t) d x2 (t− 1/t ) = e dt dt x 1 x = 1 − 1/t2 e 2 (t− t ) 2 x = 1 − 1/t2 g(x, t) 2 ∞ X = ntn−1 Jn (x) n=0
Playing with this a little bit, we have ∞ X
ntn−1 Jn (x) =
n=0
2
∞ X
ntn−1 Jn (x) =
n=0
2
∞ X
1−
∞ X
n=0
|{z}
|{z}
n−1=`
Shift the indices,
x 2
1 t2
X ∞
tn Jn (x)
n=0
xtn Jn (x) −
n=`
(` + 1)t`J`+1 (x) =
`=0
∞ X
n=0
|{z}
n−2=`
∞ X `=0
xt` J` (x) +
∞ X
t` J`+2
`=0
2(` + 1)J`+1 (x) =xJ` (x) + xJ`+2 (x) −→
2` J` (x) = J`−1 (x) + J`+1 (x) x
Now differentiating (4.7.15) with respect to x, 1 dg(x, t) d x = e 2 (t− /t ) dx dx 1 x 1 1 = t− e 2 (t− /t ) 2 t 1 1 = t− g(x, t) 2 t ∞ X = tn Jn0 (x)
n=0
And now
xtn−2 Jn (x)
(4.7.17)
W. Erbsen
HOMEWORK #9
2 2
∞ 1 X n t Jn (x) tn Jn0 (x) = t − t n=0 n=0 ∞ X
∞ X
∞ X
tn Jn0 (x) =
n=0
n=0
|{z}
|{z}
n=`
Shifting,
2
tn+1 Jn (x) −
n+1=`
∞ X
t` J`0 (x) =
`=0
∞ X `=0
∞ X
tn−1 Jn (x)
n=0
|{z}
n−1=`
t` J`−1 (x) −
∞ X
t` J`+1 (x)
`=0
−→ 2J`0 (x) = J`−1 (x) − J`+1 (x)
(4.7.18)
If we subtract (4.7.17), −→ J`0 (x) =
` J` (x) − J`+1 (x) x
(4.7.19)
We can arrive at one more recursion relation if we take (4.7.19) and subtract (4.7.17): −→ J`0 (x) = J`−1 (x) −
` J` (x) x
(4.7.20)
To prove that Jn (x) satisfies the Bessel differential equation (4.7.15), we start by multiplying (4.7.20) by x: xJ`0 (x) − xJ`−1 (x) + `J` (x) = 0
(4.7.21)
We now differentiate (4.7.21) with respect to x: d 0 [xJ`0 (x) − xJ`−1 (x) + `J` (x)] =J`0 (x) + xJ`00 (x) − J`−1 (x) − xJ`−1 (x) + `J`0 (x) dx 0 =xJ`00 (x) + (` + 1)J`0 (x) − xJ`−1 (x) − J`−1 (x)
(4.7.22)
Multiplying (4.7.22) by x, 0 xJ`00 (x) + (` + 1)J`0 (x) − xJ`−1 (x) − J`−1 (x) =0
0 x2 J`00 (x) + x(` + 1)J`0 (x) − x2 J`−1 (x) − xJ`−1 (x) =0
(4.7.23)
And multiply (4.7.21) by `: `xJ`0 (x) − `xJ`−1 (x) + `2 J` (x) = 0
(4.7.24)
Now take (4.7.22) and subtract (4.7.21): 0 x2 J`00 (x) + x(` + 1)J`0 (x) − x2 J`−1 (x) − xJ`−1 (x) − `xJ`0 (x) + `xJ`−1 (x) − `2 J` (x) =0 0 x2 J`00 (x) + xJ`0 (x) − x2 J`−1 (x) + (` − 1)J`−1 (x) − `2 J` (x) =0
At this point we take (4.7.19) and shift the index from ` to ` − 1,
(4.7.25)
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CHAPTER 4: MATHEMATICAL METHODS
0 J`−1 (x) =
`−1 J`−1 (x) − J` (x) x
And multiplying this by x and solving for J`−1 (x), 0 xJ`−1 (x) = (` − 1)J`−1 (x) − xJ` (x) −→ J`−1 (x) =
x x 0 J`−1 (x) + J` (x) `−1 `−1
Substituting this into (4.7.25), x2 J`00 (x)
+
xJ`0 (x)
−x
2
0 J`−1 (x)
x x 0 + (` − 1) J (x) + J`(x) − `2 J` (x) =0 ` − 1 `−1 `−1
0 x2 J`00 + xJ`0 (x) − x2 J`−1 (x) + x2 J`−1 (x) + x2 J` (x) − `2 J` = 0
x2 J`00 (x) + xJ`0 (x) + x2 − `2 J` (x) = 0
Which is the Bessel differential equation.
Problem 3 Prove the normalization condition for the Bessel functions Z ∞ 1 ·Jn (kr) · Jn (k 0 r) drr = δ(k 0 − k) k 0
(4.7.26)
Solution It was shown in class, and additionally can be found in M&W that: Z b 1 b [βrJn (αr)Jn0 (βr) − αrJn (βr)Jn0 (αr)|a Jn (αr)Jn (βr)rdr = 2 2 α − β a
(4.7.27)
Where I dispensed k and k 0 in favor of α and β, respectively in order to avoid confusion with the derivatives. First let’s choose a → 0 as in (4.7.26): Z b 1 b Jn (αr)Jn (βr)rdr = 2 [βrJn (αr)Jn0 (βr) − αrJn (βr)Jn0 (αr)|0 2 α − β 0 1 = 2 [βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb)] (4.7.28) α − β2 And since both J`0 (αr) and J`0 (βr) evaluated at the endpoints equal zero, then (4.7.36) too must vanish. To see that this is the case at x = 0, it is illuminating to recall that J`0 (0) = J`+1 (0) = 0. We should also recognize that J`0 (∞) → 0. Both of these properties become very evident when plotting the relevant functions and recognizing the respective behaviors. Therefore, (4.7.28) becomes
W. Erbsen
HOMEWORK #9
Z
b
Jn (αr)Jn (βr)rdr = 0
(4.7.29)
0
For the case in which α and β are equal, we must evaluate Z b Z b Z Jn (αr)Jn (βr)rdr = Jn (αr)Jn (αr)rdr = 0
0
b
2
[Jn (αr)] rdr
0
Which we now set equal to the RHS of (4.7.28) Z b βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb) 2 [Jn (αr)] rdr = α2 − β 2 0
(4.7.30)
Evaluating the RHS of (4.7.30) is not so easy; we must take the limit as β → α, but since this is indeterminate we must use L’Hˆ ospital’s Rule: Z b βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb) 2 [Jn (αr)] r dr = lim β→α α2 − β 2 0 0 rJn (αb)Jn (βb) + βrJn (αb)Jn00 (βb) − αrJn0 (βb)Jn0 (αb) = lim β→α −2β 2
=
αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) − αrJn (αb)Jn00 (αb) 2α
(4.7.31)
We now solve Bessel’s differential equation for Jn00 (αb): α2 Jn00 (αb) + αJn0 (αb) + (α2 − n2 )Jn (αb) = 0 −→ Jn00 (αb) = −
αJn0 (αb) − (α2 − n2 )Jn (αb) α2
(4.7.32)
Substituting (4.7.32) into (4.7.31), Z
b
2
[Jn (αr)] rdr =
0
=
h i 2 αJ 0 (αb)−(α2−n2 )Jn (αb) αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) − αrJn (αb) − n 2 α 2α
2 αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) + rJn (αb)Jn0 (αr) + αr 1 −
2α 2 r n 2 2 0 = [Jn (αb)] + 1 − 2 [Jn (αb)] 2 α
n2 α2
2
[Jn (αb)]
(4.7.33)
So that if α and β represent two unique roots of order n then (4.7.33) must vanish.
Problem 4 A disk of radius R in the xy-plane (z = 0) is kept at a constant potential φ0 and the rest of the plane z = 0 is kept at zero potential. As shown in class, the potential for z > 0 is given by Z ∞ φ(r, z) = φ0 R dkJ1 (kR)J0 (kr)e−kz (4.7.34) 0
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CHAPTER 4: MATHEMATICAL METHODS
Hence, the potential above the center of the disk (r = 0) is given by the integral Z ∞ φ(0, z) = φ0 R dkJ1 (kR)e−kz
(4.7.35)
0
Using the Bessel function recurrence relations and the integral representation of the Bessel function, show that the last expression can be reduced to z φ(0, z) = φ0 1 − √ (4.7.36) z 2 + R2
Solution According to (4.7.19), J1 (x) = −J00 (x) And (4.7.35) becomes φ(0, z) = −φ0 R
Z
∞
0
J00 (kR)e−kz dk
And at this point we wish to undergo a change of variables x = kR −→ k = x/R : Z ∞ xz dx J00 (x)e− /R φ(0, z) = − φ0 R R Z ∞0 xz = − φ0 J00 (x)e− /R dx 0
Where we must integrate by parts: u = e−
xz
/R
,
dv = J00 (x),
z − xz/R e R v = J0 (x)
du = −
So we now have Z xz ∞ xz z ∞ J0 (x)e− /R dx J0 (x)e− /R + R 0 0 Z ∞ xz z = − φ0 −1 + J0 (x)e− /R dx R 0
φ(0, z) = − φ0
h
(4.7.37)
At this point we pause, try a myriad of obscure substitutions, get frustrated, have a beer, and then realize that the integral in (4.7.37) has actually been tabulated by the likes of Gradshteyn and Ryzhik as Eq. 6.611.1: ? hp iν Z ∞ β −ν α2 + β 2 − α p e−αxJν (βx) dx = α2 + β 2 0
Where in our case α = z/R , β = 1 and ν = 0:
W. Erbsen
HOMEWORK #10
Z
∞
e−
0
xz
/R
1 1 R √ J0 (x) dx = q = = √ 1/ 2 2 2 2 z + R2 R z +R ( z/R ) + 1
(4.7.38)
And substituting (4.7.38) into (4.7.37),
z R √ φ(0, z) = − φ0 −1 + R z 2 + R2 From which we finally arrive at (4.7.36), thus completing the problem: z φ(0, z) = φ0 1 − √ z 2 + R2
4.8
Homework #10
Problem 1 In class we shows that the lowest order spherical Bessel function is given by sin x j0 (x) = x Using the recurrence relations, find j1 (x) and j2 (x), and also prove that ` 1 ∂ j` (x) = (−1)` x` j0 (x) x ∂x
(4.8.1)
(4.8.2)
Solution Recall that one of the recurrence relations for Bessel functions reads j`+1 (x) =
` j` (x) − j`0 (x) x
(4.8.3)
Which was rigorously proven in Homework #9, and subsequently will be treated as an axiom in this assignment. We are given j0 (x), so we need to calculate j1 (x) and j2 (x) using our recursion relation (4.8.3). We start by finding the first derivative of (4.8.1): j00 (x) =
cos x sin x − 2 x x
We now substitute this into (4.8.3), 0 j1 (x) = j0 (x) − j00 (x) x sin x cos x = 2 − x x
(4.8.4)
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CHAPTER 4: MATHEMATICAL METHODS
Taking the first derivative of j1 (x) yields j10 (x) =
2 cos x 2 sin x sin x − + x2 x3 x
From which we can now find j2 (x): 1 j2 (x) = j1 (x) − j10 (x) x sin x cos x 2 cos x 2 sin x sin x = 3 − 2 − + − x x x2 x3 x 3 sin x 3 cos x sin x = − − x3 x2 x
(4.8.5)
In order to prove (4.8.2), it will be most convenient to show that (4.8.2) works when substituted into (4.8.3). I will employ proof by induction. So the first step is to show that it works for ` = 0, j1 (x) = − j00 (x) " # 0 d 1 ∂ =− (−1)0 x0 j0 (x) dx x ∂x d sin x dx x sin x cos x = 2 − x x =j1 (x) X =−
So it works for ` = 0. Assuming that it holds for all `, we now show that it holds for ` + 1. First find the derivative of j` (x): " # ` 1 d sin x d ` 0 j` (x) = (−x) dx x dx x # " # " ` ` d 1 d sin x 1 d sin x ` ` d = (−x) + (−x) dx x dx x dx x dx x " # " # ` ` 1 d sin x 1 d sin x `−1 ` d =`(−x) + (−x) x dx x dx x dx x " # " # ` ` (−x)` 1 d sin x 1 d sin x ` d =−` + (−x) x x dx x dx x dx x " # " # ` ` ` 1 d sin x 1 d sin x ` ` d = − (−x) + (−x) x x dx x dx x dx x " # ` ` d 1 d sin x = − j` (x) + (−x)` (4.8.6) x dx x dx x So, let’s find j`+1 (x): j`+1 (x) =(−x)`+1
1 d x dx
`+1
sin x x
W. Erbsen
HOMEWORK #10
` 1 d 1 d sin x =(−x) (−x) x dx x dx x " # ` 1 d sin x d = − (−x)` dx x dx x `
` = j` (x) − j`0 (x) X x Where I used (4.8.6). Since (4.8.2) works for ` = 0 (the base step) and also ` + 1 (the inductive step), then it follows that it works for ` (the hypothesized step).
Problem 2 Using the definition of Neumann functions in terms of Bessel functions, show that for the spherical Neumann functions, cos x η0 (x) = − (4.8.7) x Using the recurrence relations, find η1 (x) and η2 (x).
Solution Analogous to (4.8.3), the recursion relation for the Neumann functions takes the form η`+1 (x) =
` η` (x) − η`0 (x) x
(4.8.8)
So, from ` = 0 we can find η1 (x): η1 (x) =
` η0 (x) − η00 (x) =⇒= −η00 (x) x
(4.8.9)
So, in finding the first derivative of η0 (x), we find η1 (x): η1 (x) = −η00 (x) =
d cos x sin x cos x =− − 2 dx x x x
(4.8.10)
Applying the same process letting ` = 1, 1 η1+1 (x) = η1 (x) − η10 (x) x sin x cos x = − 2 − 3 − η10 (x) x x Where we find that d sin x cos x − − 2 dx x x cos x 2 sin x sin x 2 cos x =− + + 2 + x x2 x x3
η10 (x) =
(4.8.11)
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CHAPTER 4: MATHEMATICAL METHODS
=−
cos x 2 sin x 2 cos x + + x x2 x3
(4.8.12)
Substituting (4.8.12) into (4.8.11), we find that sin x cos x cos x 2 sin x 2 cos x − 3 + − − x2 x x x2 x3 cos x 3 sin x 3 cos x = − − x x3 x3
η2 (x) = −
(4.8.13)
Problem 3 Transform the hyperbolic equation y2 ψxx − x2 ψyy = 0
(4.8.14)
into the canonical form of ψην = f(...)
Solution The general formalism of the Method of Characteristics entails taking some arbitrary differential equation,
Where
e 2 + 2Bu e ξη + Cu e 2 +R=0 Au2xx + 2Buxy + Cu2yy = 0 −→ Au ξξ ηη
(4.8.15)
e =Aξx2 + 2Bξx ξy + Cξy2 A
e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B e =Aηx2 + 2Bηxηy + Cηy2 C
R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη
So, from (4.8.14), it is apparent that A = y2 , B = 0, and C = −x2 . Since (4.8.14) is hyperbolic, the characteristics can be found by √ √ dy B ± B 2 − 4AC −4AC x = =± =± dx 2A 2A y Solving and integrating this we arrive at x2 y2 ± = 0 −→ x2 ± y2 = 0 2 2
(4.8.16)
Where we now define ξ(x, y) = ξ = x2 + y2 ,
and
η(x, y) = η = x2 − y2
Taking some partial derivatives, ξx =2x,
ξy = 2y,
ξxx = ξyy = 2,
ξxy = 0
(4.8.17)
W. Erbsen
HOMEWORK #10
ηx =2x,
ηy = −2y,
ηxx = 2,
ηyy = −2,
ηxy = 0
Now finding the coefficients in (4.8.15), e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =y2 ξx2 − x2 ξy2
=y2 (2x)2 − x2 (2y)2 = 0
e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =y2 ξx ηx − x2 ξy ηy
=y2 (2x)(2x) − x2 (2y)(−2y)
=4x2 y2 + 4x2 y2 =8x2 y2
e =Aη 2 + 2Bηxηy + Cη 2 C x y =y2 ηx2 − x2 ηy2
=y2 (2x)2 − x2 (−2x)2 =0
R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη = y2 ξxx − x2 ξyy uξ + y2 ηxx − x2 ηyy uη =2 y2 − x2 uξ + 2 y2 + x2 uη
And putting all these into (4.8.15),
e 2 + 2Bu e ξη + Cu e 2 +R=0 Au ξξ ηη 2(8x2 y2 )uξη + 2 y2 − x2 uξ + 2 y2 + x2 uη = 0 8x2 y2 uξη + y2 − x2 uξ + y2 + x2 uη = 0
We can combine our original definitions of ξ and η from (4.8.17) to obtain: x2 =
ξ+η , 2
and
y2 =
ξ−η 2
So that we can now rewrite (4.8.18) as ξ+η ξ−η 8 uξη − ηuξ + ξuη = 0 2 2 2 ξ 2 − η 2 uξη − ηuξ + ξuη = 0 And the canonical form of (4.8.14) becomes
uξη =
ηuξ − ξuη 2 (ξ 2 − η 2 )
(4.8.18)
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CHAPTER 4: MATHEMATICAL METHODS
Problem 4 Transform the elliptical equations y2 ψxx + x2 ψyy = 0
(4.8.19a)
2
ψxx + (1 + y) ψyy = 0
(4.8.19b)
into the canonical form of ψσσ + ψττ = f(...).
Solution Starting from (4.8.19a), it is apparent to the casual observer that A = y2 , B = 0 and C = x2 . Furthermore, we can go on to find the characteristics of this equation: √ √ −B ± B 2 − 4AC −AC ix dy = = = (4.8.20) dx 2A A y Solving and integrating, we find that ix2 y2 ± = 0 −→ ix2 ± y2 = 0 2 2
(4.8.21)
At this point we set ξ(x, y) = ξ = ix2 + y2 ,
and
η(x, y) = η = ix2 − y2
With corresponding partial derivatives ξx =2ix,
ξy = 2y,
ηx =2ix,
ηy = −2y,
ξxx = 2i,
ξyy = 2,
ηxx = 2i,
And now finding the coefficients, e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =y2 (2ix)2 + x2 (2y)2 = − 4x2 y2 + 4x2 y2
=0
e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =y2 (2ix)(2ix) + x2 (2y)(−2y)
= − 4x2 y2 − 4x2 y2 = − 8x2 y2
e =Aη 2 + 2Bηxηy + Cη 2 C x y =y2 (2ix)2 + x2 (−2y)2
= − 4x2 y2 + 4x2 y2
ξxy = 0
ηyy = −2,
ηxy = 0
(4.8.22)
W. Erbsen
HOMEWORK #10
=0 R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη = y2 (2i) + x2 (2) uξ + y2 (2i) + x2 (−2) uη =2 iy2 + x2 uξ + 2 iy2 − x2 uη
Substituting these in we have
e ξη + R = 0 2Bu
− 16x2y2 uξη + 2 iy2 + x2 uξ + 2 iy2 − x2 uη = 0
(4.8.23)
And it is possible to eliminate x and y by making the following substitution: x2 =
ξ+η , 2i
and
y2 =
ξ−η 2
Putting this in (4.8.23) and simplifying, 2 ξ − η2 −16 uξη − 2iηuξ + 2iξuη 4i 2 ξ 2 − η 2 uξη − ηuξ + ξuη = 0 From which we finally arrive at uξη =
ηuξ − ξuη 2 (ξ 2 − η 2 )
In the same spirit for (4.8.19b), we note that A = 1, B = 0, C = (1 + y)2 . Finding the characteristics, √ √ dy −B ± B 2 − 4AC ± −AC = = = ±i(1 + y) (4.8.24) dx 2A A Solving and integrating, 1 dy = ±i dx −→ log (1 + y) = ±ix 1+y Assigning our two independent variables, ξ(x, y) = ξ = ix + log (1 + y),
and
η(x, y) = η = ix − log (1 + y)
Let’s take some partial derivatives: ξx =i, ηx =i, And the coefficients are
1 1 , ξxx = 0, ξyy = − , ξxy = 0 1+y (1 + y)2 1 1 ηy = − , ηxx = 0, ηyy = , ηxy = 0 1+y (1 + y)2
ξy =
(4.8.25)
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CHAPTER 4: MATHEMATICAL METHODS
e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =Aξx2 + (1 + y)2 ξy2
=(i)(i) + (1 + y)2
1 (1 + y)2
=−1+1 =0 e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B
=ξx ηx + (1 + y)2 ξy ηy 1 1 2 =(i)(i) + (1 + y) − 1+y 1+y =−1−1 =−2
e =Aη 2 + 2Bηxηy + Cη 2 C x y =ηx2 + (1 + y)2 ηy2
=(i)(i) + (1 + y)2 =−1+1 =0
1 1+y
2
R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη =(1 + y)2 ξyy uξ + (1 + y)2 ηyy uη 1 1 2 2 =(1 + y) − uξ + (1 + y) uη (1 + y2 )2 (1 + y)2 = − uξ + uη And now we can find the canonical form of (4.8.19b): e ξη + R = 0 2Bu − 4uξη − uξ + uη = 0
And finally
uξη =
uξ − uη 4
W. Erbsen
HOMEWORK #10
Problem 5 Classify the following equations in appropriate regions and transform them into canonical form: ψxx − y2 ψyy + ψx − ψ + x2 = 0 ψxx + xψyy = 0
(4.8.26a) (4.8.26b)
Solution For (4.8.26a) it is important that the elements that characterize the dominant behavior are the second order terms, so analyzing this (the principle part), we see that A = 1, B = 0, and C = −y2 . Furthermore, we can see that: B 2 − AC = y2 > 0 −→ hyperbolic Solving, integrating: dy = ±y −→ x ± log y = 0 dx And now we can say ξ(x, y) = ξ = x + log y,
and
η(x, y) = η = x − log y
Take some partials: ξx =1, ηx =1,
1 1 , ξxx = 0, ξyy = − 2 , ξxy = 0 y y 1 1 ηy = − , ηxx = 0, ηyy = 2 , ηxy = 0 y y ξy =
And the coefficients are e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =ξx2 − y2 ξy2 1 =1 − y2 2 y =1 − 1 =0
e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =ξx ηx − y2 ξy ηy 1 1 =1 − y2 y y =1 + 1 =2
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CHAPTER 4: MATHEMATICAL METHODS
e =Aη 2 + 2Bηxηy + Cη 2 C x y =ηx2 − y2 ηy2 2 1 2 =1 − y − y
=1 − 1 =0
R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη = ξxx − y2 ξyy uξ + ηxx − y2 ηyy uη 1 1 2 2 = 1−y uξ + 1 − y − uη y y = (1 − y) uξ + (1 + y) uη
And we now have 4uξη + (1 − y) uξ + (1 + y) uη = 0 And consequently ξ−η ξ−η 4uξη + 1 − exp uξ + 1 + uη = 0 2 2 This last equation yields A = 1, B = 0, C = x, which is elliptical, assuming that x > 0: B 2 − AC = −x < 0 −→
elliptical
Find the characteristics: √ √ dy = ± −x = ±i x dx Solving and integrating, we arrive at ξ(x, y) = ξ =
2 3/2 ix + y, 3
and
η(x, y) = η =
2 3/2 ix − y 3
With partial derivatives √ ξx =i x , √ ηx = x ,
1 ξxx = √ , ξyy = 0, ξxy = 0 2 x 1 ηy = −1, ηxx = √ , ηyy = 0, ηxy = 0 2 x ξy = 1,
Finding the coefficients e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =ξx2 + xξy2
W. Erbsen
HOMEWORK #11
√ =(i x )2 + x =−x+x =0 e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B √ √ =(i x )(i x ) + x(1)(−1) =−x−x = − 2x e =Aη 2 + 2Bηxηy + Cη 2 C x y √ =( x )2 + x(−1)2 =x − x =0
R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη = (ξxx + xξyy ) uξ + (ηxx + xηyy ) uη i i √ √ = uξ + uη 2 x 2 x Our equation becomes i − 4xuξη + √ (uξ + uη ) = 0 2 x And finally
2/3 1/3 3 i 4i 1 −4 (ξ − η) uξη + (uξ + uη ) = 0 4i 2 3 ξ −η
4.9
Homework #11
Problem 1 Consider a thin half pipe of unit radius laying on the ground. It is heated by radiation from above. We take the initial temperature of the pipe and the temperature of the ground to be zero. We model this problem with a heat equation with a source term. ut = κuxx + A sin x,
u(0, t) = u(π, t) = 0,
u(x, 0) = 0
(4.9.1)
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CHAPTER 4: MATHEMATICAL METHODS
Solution We begin by assuming that our solution can be written in terms of a steady-state solution and a transient one u(x, t) = µ(x) + w(x, t) µ(x) : Steady-state solution w(x, t) : Transient solution To find the equilibrium solution, we note that µt (x) = 0, and substitute µ(x) into (4.9.1): µt = κµxx + A sin x −→ µxx = −
A sin x, κ
µ(0) = µ(π) = 0
(4.9.2)
Integrating (4.9.2) once, Z
x
0
Z A x ∂2 µ(x) dx = − sin x dx ∂x2 κ 0 ∂ A µ(x) = cos x + C1 ∂x κ
(4.9.3)
And now integrating (4.9.3), Z
0
x
Z Z x A x ∂ µ(x) dx = cos x dx + C1 dx ∂x κ 0 0 A µ(x) = sin x + C1 x + C2 κ
(4.9.4)
Applying our first boundary condition to (4.9.4), we find that µ(0) =C2 6= 0 −→ C2 = 0 And similarly, applying th4e second boundary condition, µ(π) =C1 π 6= 0 −→ C1 = 0 And (4.9.4) becomes µ(x) =
A sin x κ
(4.9.5)
And now we do the same for w(x, t): wt = κwxx + A sin x −→ wt = κwxx
(4.9.6)
We can see that (4.9.6) can be expressed as an eigenvalue problem, the explicit details of which are shown in subsequent problems and I will not repeat them here. Applying all our given conditions, we arrive at r 2 2 wn (x, t) = sin (nx) e−κn t (4.9.7) π And in order P for (4.9.7) to work, we require that it corresponds to the Fourier coefficient an from w(x, t) = an wn (x, t):
W. Erbsen
HOMEWORK #11
Position Π
3Π 4
Π 2
Π 4
Time 2
4
6
8
10
Figure 4.1: Contour plot for Problem 1.
an = − =− r
=
r
r
2 π
Z
0
2 A π κ
π
sin (nx) µ(x) dx Z
π
sin (nx) sin(x) dx
0
2 A sin (nπ) π κ n2 − 1
(4.9.8)
Using (4.9.7) and (4.9.8), we have w(x, t) =
∞ X
an w(x, t) =
n=1
∞ X
n=1
r
2 ∞ 2A X sin2 (nπ) e−κn t = πκ n=1 n2 − 1
=−
2 A sin (nπ) π κ n2 − 1
A sin(x)e−κt κ
! r
2 2 sin (nx) e−κn t π
!
(4.9.9)
Which was arrived at via the following Mathematica code: V = Sum[-Sqrt[2/Pi] Integrate[Sin[n*x]*A/kappa Sin[x],{x,0,Pi}]*Sqrt[2/Pi]*Sin[n*x] *Exp[-kappa*n^2*t],{n,1,Infinity}] >>Out[331] = -(A*Exp[-t*kappa]*Sin[x])/kappa
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CHAPTER 4: MATHEMATICAL METHODS
And finally, to get the full solution for u(x, t) we combine our steady-state solution from (4.9.5) and our transient solution from (4.9.9): u(x, t) = w(x, t) + µ(x) = −
A A sin(x)e−κt + sin(x) κ κ
Which may be rewritten as
u(x, t) =
A sin(x) 1 − e−κt κ
A contour plot of my solution can be seen in Fig. (4.1), where I set A = κ = 1.
Problem 2 Obtain Poisson’s formula to solve the Dirichlet problem for the circular region 0 ≤ r < R, 0 ≤ θ < 2π. That is, determine a solution φ(r, θ) to solve Laplace’s equation ∇2 φ = 0 in polar coordinates given φ(R, θ). Show that Z 2π R2 − r 2 1 φ(r, θ) = φ(R, α) 2 dα (4.9.10) 2π 0 R + r 2 − 2Rr cos (θ − α)
Solution We begin by noting that Laplace’s equation in polar coordinates is 1 ∂ ∂f 1 ∂2f 2 ∇ f= r + 2 2 =0 r ∂r ∂r r ∂θ
(4.9.11)
Which should be solvable using separation of variables, where we assume that the solution can take the form of f(r, θ) = R(r)Θ(θ). Substituting this into (4.9.11) and separating the variables, dR 1 d2 Θ r d r =− = λ2 (4.9.12) R dr dr Θ dθ2 Where we set each equation equal to the square of some equilibrium constant λ. We first take the equation for R(r), r d dR r =λ2 R dr dr 2 r d R dR r 2 + =λ2 R dr dr d2 R dR r2 2 + r − λ2 R =0 (4.9.13) dr dr Now we must solve the equation for Θ(θ) from (4.9.12):
W. Erbsen
HOMEWORK #11
−
1 d2 Θ =λ2 Θ dθ2
d2 Θ + λ2 Θ =0 dθ2
(4.9.14)
We require our solution for Θ(θ) to be periodic, such that Θ(θ + 2π) = Θ(θ). Therefore we set λ = n where n is an integer, and (4.9.14) becomes d2 Θ + n2 Θ =0 dθ2 Which has solutions Θ(θ) =An einθ + Bn e−inθ = An cos(nθ) + Bn sin(nθ)
(4.9.15)
And similarly we can rewrite (4.9.13) as r2
d2 R dR +r − n2 R =0 dr 2 dr
Which has solutions of the form R(r) = Cn r n + Dn r −n
(4.9.16)
Combining our solutions from (4.9.15) and (4.9.16), fn (r, θ) = R(r)Θ(θ) = Cn r n + Dn r −n (An cos(nθ) + Bn sin(nθ))
Choosing Dn = 0 for boundary conditions,
fn (r, θ) = Cn r n (An cos(nθ) + Bn sin(nθ)) And more generally, fn (r, θ) =
∞ X
Cn r n (An cos(nθ) + Bn sin(nθ))
n=0
=C0 r 0 (A0 cos(0) + B0 sin(0)) +
∞ X
Cn r n (An cos(nθ) + Bn sin(nθ))
n=1
=C0 A0 +
∞ X
Cn r n (An cos(nθ) + Bn sin(nθ))
(4.9.17)
n=1
The solution of (4.9.17) is most easily found if we first assume that R = r = 1, f(1, θ) =C0 A0 +
∞ X
Cn (An cos(nθ) + Bn sin(nθ))
n=1
=a0 +
∞ X
n=1
(an cos(nθ) + bn sin(nθ))
(4.9.18)
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CHAPTER 4: MATHEMATICAL METHODS
Where I have set C0 A0 = a0 , Cn = 1, An = an and Bn = bn . In doing this, we see that (4.9.18) is just a Fourier series. The coefficients are readily determined with Z 1 1 2π a0 = · f(1, α) dα (4.9.19a) 2 π 0 Z 1 2π f(1, α) cos(nα) dα (4.9.19b) an = π 0 Z 1 2π bn = f(1, α) sin(nα) dα (4.9.19c) π 0 Substituting (4.9.19a), (4.9.19b) and (4.9.19c) into (4.9.18): f(1, θ) =a0 +
∞ X
(an cos(nθ) + bn sin(nθ))
n=1 Z 2π
Z ∞ Z 2π X 1 1 2π 1 f(1, α) dα + f(1, α) cos(nα) dα cos(nθ) + f(1, α) sin(nα) dα sin(nθ) 2π 0 π 0 π 0 n=1 " # Z ∞ X 1 2π 1 = f(1, α) + (f(1, α) cos(nα) cos(nθ) + f(1, α) sin(nα) sin(nθ)) dα π 0 2 n=1 " # Z ∞ 1 2π 1 X + (cos(nα) cos(nθ) + sin(nα) sin(nθ)) dα = f(1, α) π 0 2 n=1 " # Z ∞ 1 2π 1 X + = f(1, α) cos [n (θ − α)] dα (4.9.20) π 0 2 n=1 =
Where I used the identity cos α cos β + sin α sin β = cos (α − β) = cos (β − α). The bracketed term in (4.9.20) can be simplified, and I used Mathematica: FullSimplify[1/2 + Sum[r^n*Cos[n*(alpha - theta], {n, 1, Infinity}]] >>Out[157]=-((-1 + r^2)/(2*(1 + r^2 - 2*r*Cos[alpha - theta]))) With this, (4.9.20) becomes 1 − r2 f(1, α) dα 2 (1 + r 2 − 2r cos (θ − α)) 0 Z 2π 1 1 − r2 = f(1, α) dα 2π 0 1 + r 2 − 2r cos (θ − α)
1 f(1, θ) = π
Z
2π
And we can extend (4.9.21) to the more general case of f(r, θ) by replacing r with r/R: Z 2π 1 1 − (r/R)2 f(r, θ) = f(R, α) dα 2π 0 1 + (r/R)2 − 2(r/R) cos (θ − α) Z 2π 1 1/R2 R2 − r 2 = f(R, α) dα 2π 0 1/R2 R2 + r 2 − 2rR cos (θ − α) From (4.9.22) we are finally left with
(4.9.21)
(4.9.22)
W. Erbsen
HOMEWORK #11
f(r, θ) =
1 2π
Z
2π
f(R, α)
0
R2 − r 2 2 2 R + r − 2rR cos (θ − α)
dα
Problem 3 For 0 < x < `, solve ut = a2 uxx + w(x, t);
u(0, t) = 0,
ux (`, t) = 0,
u(x, 0) = f(x)
(4.9.23)
By means of a series expansion involving the eigenfunctions of d2 φ(x) + λφ(x) = 0; dx2
φ(0) = φ0 (`) = 0
(4.9.24)
Solution We begin our journey by assuming that the solution of (4.9.23) can be separated in the form u(x, t) = X(x)T (t). Assuming that this is the case, we plug it into the homogeneous analogue to (4.9.23) and take the corresponding partial derivatives: XTt = a2 Xxx T And separating the variables, Tt Xxx = = −λ a2 T X
(4.9.25)
Where I have chosen −λ to be the separation constant. Taking the equation for X(x) first, Xxx = −λ −→ Xxx + λX = 0 x
(4.9.26)
We immediately recognize that (4.9.26) is identical in form to (4.9.24). Furthermore, we also see that these are Sturm-Lioville boundary value problems, and the solution of which was derived in class (and commonly used in introduction quantum mechanics courses). The corresponding eigenvalues and eigenfunctions are given by nπ 2 nπx λn = , and Xn (x) = sin ` ` Now, taking our separated solution and plugging it into (4.9.23), ∞ X
n=1 ∞ X
n=1
Tn0 (t)Xn (x) =a2 Tn0 (t)Xn (x) =a2
∞ X
n=1 ∞ X
n=1
Tn (t)Xn00 (x) + f(x, t) Tn (t)Xn00 (x) +
∞ X
n=1
fn (t)Xn (x)
(4.9.27)
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CHAPTER 4: MATHEMATICAL METHODS
And since the equation for X satisfies the Sturm-Liouville problem, we know that Xn00 (x) + λXn (x) = 0, and (4.9.27) becomes ∞ X
n=1
∞ X
Tn0 (t)Xn (x) = − a2 ∞ X
n=1
Tn (t)λn Xn (x) +
n=1 ∞ X
Tn0 (t) = − a2
∞ X
fn (t)Xn (x)
n=1
Tn (t)λn +
n=1
∞ X
fn (t)
(4.9.28)
n=1
Which is satisfied if Tn0 (t) = − a2 Tn (t)λn + fn (t) Whose solution takes the form 2
Tn (t) = e−a
λn t
Z
t
2
ea
λn α
0
fn (α) dα + Tn (0)
(4.9.29)
And at this point we note that un (x, 0) = fn (x), which implies that fn (x) =
∞ X
Tn (0) sin
n=0
nπx `
And accordingly Tn (0) =
Z
2 `
`
fn (x) sin
0
nπx `
dx
So that (4.9.18) becomes 2
Tn (t) = e−a
λn t
"Z
t
0
2 ea λn α fn (α) dα + ` 2
Z
nπx
`
fn (x) sin
`
0
dx
#
(4.9.30)
Combining (4.9.20) with our solution for X(x) (and substituting in the corresponding eigenvalues), we arrive at
un (x, t) =
∞ X
n=1
(
2
−( nπa ` ) t
e
sin
" nπx Z `
t
2
−( nπa ` ) α
e
0
2 fn (α) dα + `
Z
`
fn (x) sin
0
nπx `
dx
#)
Problem 4 Find the solution of Laplace’s equation subject to the following boundary conditions ∇2 u = 0, u(r, 0) = u(r, α) = 0,
0 < θ < α,
a 0,
y(0) = 0,
o
, y(m) = M > −h
(4.10.20)
Solution Within the Euler-Lagrangian formalism, we wish to extract the Lagrangian from (4.10.20), and put it into the Euler-Lagrange equation. The form most convenient for our purposes will be that given by (4.10.12). Consequently, we must evaluate all the terms within the Euler-Lagrange equation first: q 2 y + h 1 + (y0 ) √ q y0 y + h ∂L ∂ p 2 0 = 0 y + h 1 + (y ) = q ∂y0 ∂y 2 1 + (y0 )
L=
p
∂L =0 ∂x
We now substitute (4.10.21a)-(4.10.21c) into (4.10.12): √ q 0 d p y y + h 2 =0 y + h 1 + (y0 ) − y0 q dx 2 1 + (y0 ) q 0 d p y 2 0 =0 y + h 1 + (y0 ) − y q dx 2 0 1 + (y ) q q 0 )2 0 p 1 + (y d y 2 =0 y+h 1 + (y0 ) ·q − y0 q dx 2 2 1 + (y0 ) 1 + (y0 )
(4.10.21a) (4.10.21b)
(4.10.21c)
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CHAPTER 4: MATHEMATICAL METHODS
0 2 0 2 d p 1 + (y ) (y ) =0 y + h q −q dx 2 2 0 0 1 + (y ) 1 + (y ) p d 1 =0 y + h q dx 2 1 + (y0 ) √ y+h d q =0 dx 2 1 + (y0 ) | {z } constant
So, the bracketed term in (4.10.22) is a constant of motion: √ y+h q =C 2 1 + (y0 )
(4.10.22)
(4.10.23)
We are now tasked to find a solution of y(x) corresponding to the given initial conditions: y+h 2
1 + (y0 )
=C 2
2 y + h =C 2 1 + (y0 ) r dy y+h = −1 dx C2 p dy y + h − C2 = dx C C dx = p dy y + h − C2 p x =2C y + h − C 2 + C 0 2
(x − C 0 ) =4C 2 (y + h − C 2 )
(4.10.24)
In order for this problem to work, I have to assume that the second integration constant goes to zero. We now apply the initial condition on (4.10.24) to find C: 0 =4C 2 (h − C 2 ) −→ C = Substituting this back into (4.10.24), x2 =4h(y + h − h) From this it is easy to see that the solution is...
y(x) =
x2 4h
√
h
W. Erbsen
HOMEWORK #12
Problem 5 A rocket is propelled vertically upward so as to reach a prescribed height h in a minimum time while using a given fixed quantity of fuel. The vertical distance x(t) above the surface satisfies, mx00 = −mg + mU (t),
x0 (0) = 0
with x(0) = 0,
(4.10.25)
where U (t) is the acceleration provided by the engine thrust. We impose the terminal constant x(T ) = h, and we wish to find the particular thrust function U (t) which will minimize T assuming that the total thrust of the rocket engine over the entire thrust time is limited by the condition, Z T U 2 (t) dt = k 2 (4.10.26) 0
Where k is a given positive constant which measures the total amount of available fuel.
Solution The first thing we need to do is solve (4.10.25) with the given initial conditions. x00 (t) = − g + U (t) Z t Z t Z t x00 (t) dt = − g dt + U (α) dα 0 0 0 Z t x0 (t) = − gt + U (α) dα 0 Z Z Z "Z t
0
t
0
x (t) dt = − g
t
β
t dt +
0
U (α) dα
Z t "Z
gt2 x(t) = − + 2
0
0
0
0
β
U (α) dα
#
#
dβ
dβ
Z t Z t gt2 + U (α) dα dβ 2 0 α Z t gt2 =− + (t − α)U (α) dα 2 0 =−
(4.10.27)
Where I eliminated the integration constants en route. We now apply our boundary condition, x(T ) = h, to (4.10.27): x(T ) = −
gT 2 + 2
Z
T
(T − α)U (α) dα = h
0
(4.10.28)
And we define two new equations, F1 = − F2 =
Z
0
gT 2 + 2 T
Z
T
0
U 2 (t) dt
(T − t)U (t) dt
(4.10.29a) (4.10.29b)
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CHAPTER 4: MATHEMATICAL METHODS
Where I changed variables from α to t in (4.10.29a). Starting with (4.10.29a), we continue with our variational adventures by adding some small element to our to-be-varied functional: Z T g F1 (T + δT ) = − (T + δT )2 + (T + δT − t)(U + δU ) dt 2 0 Z T g (T U + δT U − tU + T δU + δT δU − tδU ) dt = − (T 2 + 2T δT + (δT )2 ) + 2 0 Z T gT 2 =− − gT δT + (T U + T δU + δT U − tU − tδU ) dt (4.10.30) 2 0 We now subtract the original functional from (4.10.30), Z T Z T gT 2 gT 2 − gT δT + (T U + T δU + δT U − tU − tδU ) dt + − (T − t)U dt F1 (T + δT ) − F1 (T ) = − 2 2 0 0 Z T = − gT δT + (T δU + δT U − tδU ) dt (4.10.31) 0
We repeat this process with F2 in (4.10.29b): Z F2 (T + δT ) =
T
2
(U + δU ) dt
0
=
Z
T
0
=
Z
T
U 2 + 2U δU + (δU )2
U 2 + 2U δU
0
We now take (4.10.32) and subtract (4.10.29b): Z F2 (T + δT ) − F2 (T ) =
U + 2U δU
0
=2
Z
T
U δU dt
dt
dt
T 2
(4.10.32)
dt −
Z
T
U 2 dt
0
(4.10.33)
0
And now bringing all the pieces together, δT =λ1 F1 (T ) − λ2 F2 (t) " # " Z Z T =λ1 −gT δT + (T δU + δT U − tδU ) dt + λ2 2 0
= − λ1 gT δT +
Z
(λ1 T δU + λ1 δT U − λ1 tδU + λ2 2U δU ) dt
Now we take (4.10.34) such that δT = 0, eg no variation. This gives Z T 0= (λ1 T δU − λ1 tδU + λ2 2U δU ) dt 0
0=
0
U δU dt
0
#
T
0
Z
T
T
(λ1 T − λ1 t + 2λ2 U ) dt | {z }
(4.10.34)
W. Erbsen
HOMEWORK #12
0 =λ1 T − λ1 t + 2λ2 U
λ1 t − λ1 T = 2λ2 U −→ λ1 (t − T ) = 2λ2 U
(4.10.35)
And remember F1 and F2 ? Sure you do - here’s where we use them. Solve (4.10.35) for U (t) and substitute into these equations and integrate: Z T gT 2 λ1 F1 = − + (T − t) − (T − t) dt 2 2λ2 0 Z T λ1 gT 2 =− − (T − t)2 dt 2 2λ2 0 gT 2 λ1 T 3 =− − − T3 + T3 2 2λ2 3 2 λ1 3 gT − T =h (4.10.36) =− 2 6λ2 And now for F2 : F2 =
Z
T
0
−
2 λ1 λ2 T 3 (T − t) dt = − 12 = k2 2λ2 4λ2 3
(4.10.37)
We have three independent variables, λ1 , λ2 , and T , so we need another equation. In addition to these we can one more as suggested by Donald Smith in his book “Variational Methods in Optimization.” We do this by again varying δT by a small amount, other than zero. Let’s do just that: Z T 1 = − λ1 gT + (λ1 T δU + λ1 U − λ1 tδu + 2λ2 uδu) dt 0
= − λ1 gT +
Z
= − λ1 gT +
Z
= − λ1 gT +
T
0
T
λ1 (T − t) U δU + λ1 U + 2λ2 U δU dt | {z } −2λ2
(−2λ2 U δU + λ1 U + 2λ2 U δU ) dt
0
Z
T
λ1 U dt
(4.10.38)
0
Now insert (4.10.35) into (4.10.38)! 1 = − λ1 gT = − λ1 gT = − λ1 gT = − λ1 gT = − λ1 gT
λ1 (T − t) + λ1 − dt 2λ2 0 "Z # Z T T λ21 − T dt − t dt 2λ2 0 0 λ2 T2 − 1 T2 − 2λ2 2 2 2 λ T − 1 2λ2 2 λ2 − 1 T2 4λ2 Z
T
(4.10.39)
365
CHAPTER 4: MATHEMATICAL METHODS
From here forward is just plugging in equations and solving; I left this up to Mathematica. There seem to be many solutions, here are two: 3(t − T ) gT 2 − 2h U (t) = 2T 2
,
U (t) =
√
3 k(T − t) T 3/2
W. Erbsen
HOMEWORK #12
Chapter 5
Departmental Examinations 5.1
Quantum Mechanics
Problem 1 Consider two different spin 1/2 whose Hamiltonian is completely specified as H = cs(1) · s(2) where c is a real constant. a)
What are the constants of motion? Obtain the eigenvalues of H for both singlet and triplet states.
b)
Consider at time t = 0 the spin of particle (1) along the z-axis is up and the spin of particle (2) along the z-axis is down. What is the wave function of the system at a later time t?
c)
Calculate the probability that at a later time t the spins of both particles are aligned as they were at t = 0.
Useful information:
χm 1
α(1)α(2) m=1 β(1)β(2) m = −1 = 1 √ [α(1)β(2) + β(1)α(2)] m = 0 2
1 χ00 = √ [α(1)β(2) − β(1)α(2)] 2
Solution a)
∗ Parenthetic
The Hamiltonian is given by H = cs1 · s2 ,∗ and the total spin is just s = s1 + s2 . We can find an alternate form of the Hamiltonian by doing the following: 1 2 2 s2 = (s1 + s2 ) = s21 + s22 + 2s1 · s2 −→ s1 · s2 = s − s21 − s22 2 indicies have been shamelessly transformed into subscripts to avoid confusion.
W. Erbsen
QUANTUM MECHANICS
Such that our Hamiltonian becomes H=
c 2 s − s21 − s22 2
(5.1.1)
To find the constants of motion, we test to see whether or not the quantity in question commutes with the Hamiltonian (5.1.1). The contenders are s1 , s2 , and s. Starting with s1 , c 2 c 2 s − s21 − s22 , s1 = [s , s1 ] − [s21 , s1 ], − [s22 , s1 ] = 0 [H, s1] = (5.1.2) 2 2 | {z } | {z } | {z } i ii iii
It is immediately obvious that i and ii equal zero, and since the operators in iii act on two different particles, it commutes to. Therefore, s1 is a constant of motion. Similarly, for s2 we have: c 2 c 2 [H, s2] = s − s21 − s22 , s2 = [s , s2 ] − [s21 , s2 ], −[s22 , s2 ] = 0 (5.1.3) 2 2
For the same reasons applied to (5.1.2), we see from (5.1.3) that s2 is also a constant of motion. Furthermore, for s, we have c 2 c 2 s − s21 − s22 , s = [s , s] − [s21 , s], −[s22, s] = 0 (5.1.4) [H, s] = 2 2 So, from (5.1.2)-(5.1.5), we see that the constants of motion are s1 , s2 , and s .
For the second part of this problem, we wish to find the eigenvalues of (5.1.1) for both singlet s = 0 and triplet s = 1 states. For the singlet state, we recognize that s = 0, so s1 and s2 must be in opposite spin states. Let’s let s1 = 1/2 , and s2 = − 1/2 . c Hχ00 = s2 − s21 − s22 χ00 2 c 2 0 = s χ0 − s21 χ00 − s22 χ00 2 c s(s + 1)~2 χ00 − s1 (s1 + 1)~2 χ00 − s2 (s2 + 1)~2 χ00 = 2 c 0(0 + 0)~2 − 1/2 ( 1/2 + 1)~2 − 1/2 ( 1/2 + 1)~2 χ00 = 2 c = 0 − 3/4 ~2 − 3/4 ~2 χ00 2 3c = − χ00 (5.1.5) 4 For the triplet state we have s = 1, so following the steps from (5.1.5), c = 1(1 + 1)~2 − 1/2 ( 1/2 + 1)~2 − 1/2 ( 1/2 + 1)~2 χm 1 2 c = 2 − 3/4 ~2 − 3/4 ~2 χm 1 2 c m = χ1 4
(5.1.6)
So from (5.1.5) and (5.1.6), our eigenvalues for the provided Hamiltonian for both singlet and triplet states are, respectively: Hχ00 = −
3c 2 0 ~ χ0 , 4
Hχm 1 =
c 2 m ~ χ1 4
369
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
The general wave function for particle one spin up and particle two spin down is given by 1 χ(0) = √ χ00 + χ01 2
(5.1.7)
If we wish to obtain the wave function at some later time t, we need to apply the time evolution operator U = exp − iHt to (5.1.7) ~ 1 iHt 0 iHt 0 χ0 + exp − χ1 χ(t) = √ exp − ~ ~ 2 1 it 3c it c 2 0 =√ exp − − ~2 χ00 + exp − ~ χ1 ~ 4 ~ 4 2 Which readily simplifies to 3c~ c~ 0 exp it χ0 + exp − it χ01 4 4
1 χ(t) = √ 2 c)
To calculate the probability that at some later time t both electrons will return to their original state, we simply take the modulus squared of the matrix element between these two states. To simplify notation, I make the following substitutions: E0 =
3c~ , 4
E1 =
c~ 4
So, the probabilty is then: 2 1 1 E0 it 0 2 √ χ00 + χ01 √ |hχ(0)|χ(t)i| = e χ0 + e−E1 it χ01 2 2 2 1
= χ00 + χ01 eE0 it χ00 + e−E1 it χ01 4 2 1 = eE0 it hχ00 |χ00 i +e−E1 it hχ00 |χ01 i +eE0 it hχ01 |χ00 i +e−E1 it hχ01 |χ01 i | {z } | {z } | {z } | {z } 4 1
0
0
2 1 = eE0 it + e−E1 it 4 2 c~ 1 3c~ = e 4 it + e− 4 it 4 i 2 1 c~ h c~ c~ = e 4 it e 2 it + e− 2 it 4 2 1 c~ it c~ 4 = e 2 cos t 4 2
(5.1.8)
From (5.1.8) it is clear that the probability of realignment is
P = cos2
1
c~ t 2
W. Erbsen
QUANTUM MECHANICS
Problem 2 2 1 1 −1
, what are the eigenvalues of this Hamiltonian?
a)
If
b)
Prove that [H, exp (H)] = 0, that is, the two operators commute.
c)
What are the eigenvalues of exp (H)?
d)
If a small perturbation H 0 = λ ( 01 10 ), where λ is a small positive number, is applied to the system, calculate the change in the eigenenergy of the ground state, to first order in λ.
e)
Carry out the change in the ground state energy to second order in λ.
Solution a)
b)
Let’s diagonalize this mofo: √ 2 − λ 1 = (2 − λ)(−2 − λ) − 1 = λ2 − 5 = 0 −→ λ = ± 5 1 −2 − λ
We first recall that the series expansion for ex is ex =
∞ X xn x2 x3 =1+x+ + + ... n! 2 6 n=0
So, our commutator is [H, exp[H]] = [H, (1 + H + h2 /2 + h3 /6 + ...)] = [H, 1] + [H, H] + [H, H 2/2] + [H, H 3/6] (5.1.9) Each of the terms in (5.1.9) vanish, so we can only conclude that H and exp[H] commute, and so [H, exp [H]] = 0 c) d)
√ The eigenvalues of exp [H] are just λ = exp [± 5 ] . The first thing we need to do is find the eigenvectors associated with the eigenvalues that we found in part a): √ √ √ 1√ 2− 5 1√ α 0 For λ = + 5 : = (2 − 5 )α + β = 0 −→ ψfes = 0 2− 5 1 −1 − 5 β √ √ √ 1√ 2+ 5 1√ α 0 = (2 + 5 )α + β = 0 −→ ψgs = For λ = − 5 : 0 2+ 5 1 −1 + 5 β √ The ground state is associated with the eigenvalue λ = − 5 , with eigenvector ψgs . The first order energy correction is given by √ 1√ 0 1 0 hψgs |H |ψgsi = 1 (2 + 5 ) λ 1 0 2+ 5 √ √ 2+ 5 =λ 1 (2 + 5 ) 1
371
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
From which it is clear that the first order shift in energy is simply √ (1) Egs =λ 4+2 5 e)
To carry out the change in the ground state energy to second order in λ, we first recall that the second order energy shift from time-independent perturbation theory is En(2) =
∞ X |hψm |H 0 |ψn i|2 En − Em m,n
(5.1.10)
m6=n
In our case ψm = ψfes , ψn = ψgs , Em = E1 and En = E0 . We compute the matrix element in (5.1.10) first: √ 1√ 0 1 0 hψfes |H |ψgsi = 1 (2 − 5 ) λ 1 0 2+ 5 √ √ 2+ 5 =λ 1 (2 − 5 ) 1 =4λ
(5.1.11)
The denominator of (5.1.10) is just E0 − E1 = −
√ √ √ 5 − 5 = −2 5
(5.1.12)
We now substitute (5.1.11) and (5.1.12) into (5.1.10) to obtain the energy shift: En(2)
∞ X |4λ|2 8λ2 √ −→ En(2) = √ = 5 m,n −2 5 m6=n
Problem 3 Consider two spin 1/2 particles described by the Hamiltonian H=
p2 p21 + 2 + V (x1 ) + V (x2 ) 2m 2m
(5.1.13)
Where V (x) = ∞ for x < 0 and for x > a; V (x) = 0 for 0 < x < a. Assume that the electrons are in the opposite spin state, that is, the total S = 0.
a)
Write down the spin wave function(s). Use the standard notations: α for spin up and β for spin down.
b)
Find the energy and wavefunctions of the ground state of this Hamiltonian.
c)
Find the energy and wavefunction of the lowest state for S = 1.
d)
Find the energy and wavefunction of the second S = 0 state. Show that the energy is the same as in (c).
W. Erbsen
e)
QUANTUM MECHANICS
If the two particles have a small interaction W (x1 , x2 ) = bx1 x2 where b is small and positive, show that the degeneracy in (c) and (d) is removed. Which one has the lower energy?
Solution We first note that since s1 = ± 1/2 and s2 = ∓ 1/2 then s1 + s2 = 0. The eigenfunctions and eigenenergies are r nπx 2 n2 π 2 ~2 , En = ψn (x) = sin L a 2ma2 a)
The spin wave function is 1 χ = √ [α(1)β(2) − α(2)β(1)] 2
b)
The full wave function is just the product of the space part and the spin part: √ πx πx 2 1 2 ψ(x) = ψ1 (x1 )ψ1 (x2 )χ −→ ψ(x) = sin sin [α(1)β(2) − α(2)β(1)] L L L
(5.1.14)
(5.1.15)
While the energy is just the sum: Egs = E1 + E1 −→ Egs = c)
π 2 ~2 2mL2
For s = 1 the spin wave function is symmetric, so the space part must be anti-symmetric: 1 ψa,b (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] χ 2 Since s = 1, this implies that they cannot both be in the same state, so the lowest state would be for one of the electrons to be at n = 1, and the first at n = 2. The total energy is just the sum of these: E=
d)
5π 2 ~2 2mL2
Now that s = 0 this implies that the spin wave function is anti-symmetric, so the space part should now be symmetric: 1 ψa,b (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] χ 2 And since we are now looking for the second lowest state, and we know that since s = 0 implies that the electrons are in opposite states this means they can occupy the same energy level, so the ground state is n1 = 1 and n2 = 1, and the second state is n1 = 1 and n2 = 2 (or vice versa). The total energy is the same as the ground state for the case of s = 1:
373
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
E= e)
5π 2 ~2 2mL2
To do this part we must compute the first order energy shift in time-independent perturbation theory. Let’s define 1 ψa (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] 2 1 ψb (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] 2 Then the first order shift for s = 1 is then E (1) =hψa |H 0 |ψa i ZZ b 2 = x1 x2 [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] dx1 dx2 2 ZZ h b 2 2 = x1 x2 (ψa (x1 )ψb (x2 )) + x1 x2 (ψa (x2 )ψb (x1 )) 2 −2x1 x2 ψa (x1 )ψb (x2 )ψa (x2 )ψb (x1 )] dx1 dx2 And also the first order shift for s = 0 is ZZ b 2 E (1) = x1 x2 [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] dx1 dx2 2 ZZ h b 2 2 = x1 x2 (ψa (x1 )ψb (x2 )) + x1 x2 (ψa (x2 )ψb (x1 )) 2 +2x1 x2 ψa (x1 )ψb (x2 )ψa (x2 )ψb (x1 )] dx1 dx2
(5.1.16)
(5.1.17)
It is clear from just looking at (5.1.16) and (5.1.17) that the degeneracy will be broken, since they differ in the sign of one of the cross terms. The case of s = 1 has a lower energy.
Problem 4 The result of a measurement shows that the electron spin is along the +x direction at t = 0. For t > 0, the electron enters in a uniform magnetic field that is parallel to the +z direction. Calculate the quantum mechanical probability as a function of time for finding the electron in each of the following states (~ = 1): a)
Sx = +1/2
b)
Sx = −1/2
c)
Sz = +1/2
d)
Sz = −1/2
The Pauli matrices are given by 0 1 σx = , 1 0
σy =
0 −i , i 0
σz =
1 0 0 −1
W. Erbsen
QUANTUM MECHANICS
Solution Initially at time t = 0 the spin is along the x-direction, so it will be most convenient to express the spinor in the Sz basis. We first recall that 1 1 1 1 (x) (x) χ+ = √ , χ− = √ 1 −1 2 2 1 0 (z) (z) χ+ = , χ− = 0 1 And also that we can apply the time-evolution operator to a state at t = 0 as |ψ(t)i = U |ψ(0)i,
with
U = exp [−iHt]
The Hamiltonian for a dipole in a magnetic field (the dipole in this case is provided by the spin) is H = µB =
eSz B0 = ωSz m
So the time evolution operator becomes U = exp [−iωSz t] We now wish to express our current spinor in terms of the Sz basis: † † (x) (x) (z) (z) |ψ(0)i = χ+ χ+ + χ+ χ− 1 0 1 1 =√ 1 1 +√ 1 1 0 1 2 2 1 1 1 0 =√ +√ 0 2 2 1 1 1 = √ |↑i + √ |↓i 2 2
(5.1.18)
(5.1.19)
We apply the time evolution operator from (5.1.18) to our state at time t = 0 from (5.1.19): 1 1 |ψ(t)i =e−iωSz t √ |↑i + e−iωSz t √ |↓i 2 2 1 ω ω = √ e−i 2 t |↑i + ei 2 t |↓i 2
(5.1.20)
Where we have set ~ = 1. Now that we have our time-dependent wave function, we can calculate the time-dependent probabilities for the spin to flip: a)
The probability for finding the particle to be in Sx = 1/2 is found by 2
P = |hSx (↑)|ψ(t)i| 2 ω ω 1 = hSx (↑)| e−i 2 t |↑i + ei 2 t |↓i i 2 2 1 ω ω = hSx (↑)|e−i 2 t |↑i + hSx (↑)|ei 2 t |↓i 2
375
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
2 ω 1 ω = e−i 2 t hSx (↑)|↑i + ei 2 t hSx (↑)|↓i 2
Where the expectation values are given by † 1 1 (x) (z) = hSx (↑)|↑i = χ+ χ+ = √ 1 1 0 2 † 1 1 (x) (z) = hSx (↑)|↓i = χ+ χ− = √ 1 1 0 2
(5.1.21)
1 √ 2 1 √ 2
(5.1.22a) (5.1.22b)
Substituting (5.1.22a) and (5.1.22b) into (5.1.21), P = b)
2 1 i ω t ωt 1 1 −i ω t 2 2 + √ 2 √ −→ e e P = cos 2 2 2 2
And for the probability for finding the particle in Sx = −1/2, 2 ω 1 ω P = e−i 2 t hSx (↓)|↑i + ei 2 t hSx (↓)|↓i 2
(5.1.23)
Where we have
† 1 1 1 (x) (z) = √ hSx (↓)|↑i = χ− χ+ = √ 1 −1 0 2 2 † 1 1 0 (x) (z) hSx (↓)|↓i = χ− χ− = √ 1 −1 = −√ 1 2 2
(5.1.24a) (5.1.24b)
Substituting (5.1.24a) and (5.1.24b) into (5.1.23), P = c)
2 1 i ω t 1 1 −i ω t ωt 2 2 − √ 2 √ −→ e e P = sin 2 2 2 2
Doing the same thing for Sz = 1/2, P = Where
2 1 −i ω t ω e 2 hSz (↑)|↑i + ei 2 t hSz (↑)|↓i 2
hSz (↑)|↑i =
(z) χ+
†
(z) χ+
= 1
† (z) (z) hSz (↑)|↓i = χ+ χ− = 1 Putting (5.1.26a) and (5.1.26b) into (5.1.25), P = d)
1 0 =1 0 0 0 =0 1
(5.1.25)
(5.1.26a) (5.1.26b)
1 −i ω t 2 1 e 2 −→ P = 2 2
And the mantra continues with Sz = −1/2: 2 ω 1 ω P = e−i 2 t hSz (↓)|↑i + ei 2 t hSz (↓)|↓i 2
(5.1.27)
W. Erbsen
QUANTUM MECHANICS
Where we have † (z) (z) hSz (↓)|↑i = χ− χ+ = 0
† (z) (z) hSz (↓)|↓i = χ− χ− = 0 Slamming (5.1.28a) and (5.1.28b) into (5.1.27), P =
Problem 5
1 1 =0 0 0 1 1 = 1 2
(5.1.28a) (5.1.28b)
1 1 i ω t 2 e 2 −→ P = 2 2
(Old Problem 2)
A quark-quark potential that is sometimes used in quark models of mesons is V (r) = g log(r/a). If the quarks are very heavy, then non-relativistic physics is possible. Use the uncertainty principle to estimate the size and binding energy of a pair of quarks, each of mass m (g > 0).
Solution Before we get down to bidness we recall that the reduced mass for two particles of equal mass is µ=
mm m = −→ m = 2µ m+m 2
The uncertainty principle is ∆x∆p ≤ ~/2 −→ ∆x∆p ∼ ~. If we assume that r ∼ ∆r and p ∼ ∆p, then rp ∼ ~ −→ p ∼ ~/r. We now substitute this into the two-particle Hamiltonian, with the given interasction term: hri p21 p2 + 2 + g log 2m 2m a hr i p2 = + g log 2µ a hri ~2 = + g log 2µr 2 a
H=
(5.1.29)
The plan of action now is to minimize (5.1.29), by taking the derivative and setting it equal to zero: 2 h r i dH ∂ ~ ∂ = + g log =0 dt ∂t 2µr 2 ∂t a s ~2 ~2 g (5.1.30) ⇒ − 3 + = 0 −→ r = ± µr r gµ Taking r from (5.1.30) and putting it back into H from (5.1.29),
377
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
" s # ~2 gµ 1 ~2 H= + g log 2µ ~2 a gµ "s # g ~2 = + g log 2 a2 gµ
(5.1.31)
We can see that (5.1.31) is actually an eigenvalue, and so the solution is
g E = + g log 2
"s
~2 2 a gµ
#
Problem 6 Show that for a system consisting of two identical particles with spin I, the ratio of the number of states symmetric under exchange of the two spins to the number of states antisymmetric under exchange of the two spins is equal to (I + 1)/I. [Hint: If you do not know where to start, consider I = 1/2 first.]
Solution For a nucleus with spin quantum number I 6= 0, there are 2I + 1 different possible values of the zcomponent of the spin angular momentum quantum number: MI = −I, −I + 1, ..., I − 1, I For two such nuclei, the total multiplicity is (2I + 1) · (2I + 1). Clearly, some of the wave functions associated with these will be symmetric, while some will be anti-symmetric: 1 ψS (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψb (x1 )ψa (x2 )] 2 1 ψA (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψb (x1 )ψa (x2 )] 2
(5.1.32) (5.1.33)
Where ψS (x1 , x2 ) denotes the symmetric states, and ψA (x1 , x2 ) is for the anti-symmetric states. Let’s also denote the number of symmetric states as num(S), and the number of anti-symmetric states as num(A). In addition to these two sources of states, there is another case for the symmetric wave function: 0
ψS (x1 , x2 ) = ψa (x1 )ψa (x2 )
(5.1.34)
And as before, we let num(S 0 ) denote the number of symmetric states in the form of (5.1.34). We know that (5.1.32), (5.1.33) and (5.1.34) represent all possible states for a nucleus with spin quantum number I. Accordingly,
W. Erbsen
QUANTUM MECHANICS
(2I + 1)2 | {z }
num(S) + num(S 0 ) | {z }
=
Total # of states
# of symmetric states
+
num(A) | {z }
# of anti-symmetric states
We can see that the multiplicity of num(S 0 ) is just 2I + 1, while the multiplicity of num(S) and num(A) are equal, but unknown (call this multiplicity α). So, (2I + 1)2 = α + (2I + 1) + α (2I + 1)2 − 2I − 1 = 2α
4I 2 + 4I + 1 − 2I − 1 = 2α
4I 2 + 2I = 2α −→ α = 2I 2 + I = I(2I + 1) So, num(S) = num(A) = I(2I + 1), and num(S 0 ) = 2I + 1. The ratio of symmetric states to antisymmetric states is then I(2I + 1) + 2I + 1 (2I + 1) (I + 1) I +1 num(S) + num(S 0 ) = = −→ num(A) I(2I + 1) (2I + 1) I I
Problem 7 A rod of length d and uniform mass distribution is pivoted at its center and constrained to rotate in an xy-plane. The rod has mass M and charges +Q and −Q fixed at either end. a)
Write down the Hamiltonian, its eigenfunctions, and eigenvalues.
b)
If a constant weak electric field ε lies along the x-direction, calculate the perturbed energy of the ground state and the first excited state to the 2nd order in ε.
Solution a)
Classically, we know that the Hamiltonian for a system like this is expressed as H=
L2 2I
Where L is the angular momentum, and I the moment of inertia, which in our case is X 2 2 I= mi ri2 = m d/2 + m d/2 = 1/2 md2 i
And we also know that L2 can be expressed as L2 = −~2
∂2 ∂φ2
379
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
So, our Hamiltonian is: H =−
~2 ∂ 2 md2 ∂φ2
(5.1.35)
We now insert our Hamiltonian from (5.1.35) into the time-independent Schr¨ odinger equation: −
~2 ∂ 2 ∂2 md2 E ψ(φ) = Eψ(θ, φ) −→ ψ(φ) + ψ(φ) = 0 md2 ∂φ2 ∂φ2 ~2
With solutions ψ(φ) = A sin(kφ) + B cos(kφ),
k2 =
md2 E ~2
(5.1.36)
We require rotational periodicity, eg that φ(φ) = ψ(φ + 2π). Applying this to (5.1.36), ψ(φ + 2π) =A sin [k(φ + 2π)] + B cos [k(φ + 2π)] =A sin(kφ) cos(2πk) + sin(2πk) cos(kφ) + B cos(kφ) cos(2πk) − sin(kφ) sin(2πk) | {z } | {z } | {z } | {z } 1
0
1
=A sin(kφ) + B cos(kφ)
0
(5.1.37)
Two important things happened here. First, I assumed that k was an integer, so let’s just say that k = n from now on. Also, I used the following trig identities: sin (α ± β) = sin α cos β ± cos α sin β
cos (α ± β) = cos α cos β ∓ sin α sin β So, our eigenvalues are then k 2 = n2 =
md2 E n2 ~2 −→ E = ~2 md2
And our eigenfunctions are ψ(φ) = A sin(nφ) + B cos(nφ) = Aeinφ + Be−inφ However, if we imagine our system to be classical, then we can simply our expression for the eigenfunctions by recognizing that our rotor can only spin in one direction concurrently (classically), so our wave function “collapses” and allows only motion in one direction: ψ(φ) = Aeinφ Normalizing, 2
A
Z
0
So, our eigenfunction is:
2π
e−inφ
1 einφ dφ = A2 2π = 1 −→ A = √ 2π
ψ(φ) = √
1 inφ e 2π
W. Erbsen
QUANTUM MECHANICS
b)
The dipole Hamiltonian for an oscillating perturbation is H 0 = −pE = −QdE0 cos φ The first order energy shift is then En(1) =hψn (φ)|H 0 |ψn (φ)i Z 2π 1 = e−inφ (−QdE0 cos φ) einφ dφ 2π 0 Z QdE0 2π cos φ dφ = 0 =− 2π 0 The second order energy shift has the form En(2)
∞ 2 X |hψm (φ)|H 0 |ψn (φ)i| = En − Em n,m
(5.1.38)
n6=m
We start with the ground state first. Taking the numerator, 1 imφ he |QdE0 cos φ|einφi 2π QdE0 imφ =− he | cos φ|einφi 2π Z QdE0 2π −imφ =− e cos φeinφ dφ 2π 0 Z QdE0 2π cos φei(n−m)φ dφ =− 2π 0 Z QdE0 2π iφ =− e + e−iφ ei(n−m)φ dφ 4π 0 Z 2π Z 2π QdE0 i(n−m+1)φ i(n−m−1)φ =− e dφ + e dφ 4π 0 0 QdE0 =− [δn,m+1 + δn,m−1 ] 2
hψm (φ)|H 0 |ψn (φ)i =
(5.1.39)
The denominator of (5.1.38) would be En − Em =
~2 (n2 − m2 ) md2
(5.1.40)
For the ground state, we have n = 1. Realizing this, we now take (5.1.39) and (5.1.40) and substitute it back into (5.1.38): (2)
E1 =
∞ 2 Q2 d2 E02 md2 X |δ1,m+1 + δ1,m−1 | 4 ~2 1 − m2
(5.1.41)
m6=1
The only possible value of m to sum over is m = 2, the rest of the terms vanish. Therefore, the second order energy shift for the ground state is just (2)
E1 =
Q2 d2 E02 md2 (1)2 mQ2 d4 E02 (2) −→ E1 = − 2 2 4 ~ 1 − (2) 12~2
381
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
To find the second order energy shift for the first excited state, we start with (5.1.41) with n = 2: (2)
E2 =
∞ 2 Q2 d2 E02 md2 X |δ2,m+1 + δ2,m−1 | 4 ~2 4 − m2
(5.1.42)
m6=2
In this case there are two possible values we can sum to such that (5.1.42) will not collapse: m = 1 and m = 3. Summing these, we can find the energy: Q2 d2 E02 md2 (1)2 (1)2 (2) + E2 = 4 ~2 2 − (1)2 2 − (3)2 Q2 d2 E02 md2 1 = 1− 2 4 ~ 7 Which finally leads us to (2)
E2 = c)
3mQ2 d4 E02 14~2
If the field is very strong, then the molecule will be perfectly aligned and cos φ ≈ 1, so that H 0 = −QdE0 cos φ → H 0 = −QdE0 . Inserting this into the time-independent Schr¨ odinger equation, ~2 ∂ 2 − − QdE0 ψ(φ) = Eψ(φ) md2 ∂φ2 ∂2 md2 ψ(φ) + 2 (QdE0 + E) ψ(φ) = 0 2 ∂φ ~ If we let κ2 =
md2 (QdE0 + E) ~2
Then our approximate wave function is ψ(φ) = A cos κφ + B sin κφ
Problem 8 Consider a particle in an infinitely deep three-dimensional box with the length on each side equal to a. a)
Write down the Hamiltonian describing such a particle.
b)
Find the first 5 lowest eigenenergies.
c)
Find the degeneracy of each energy eigenstate above.
d)
If there are 10 noninteracting, indistinguishable fermions in such a box, what is the ground state energy of the whole system? Consider separately the cases for spin 1/2 and spin 3/2 particles.
W. Erbsen
QUANTUM MECHANICS
Solution a)
The Hamiltonian is for a particle in a three dimensional box is ~2 ~2 2 ∇ + V −→ H = − H =− 2m 2m Where the potential is defined by V =
b)
−V0 0
∂2 ∂2 ∂2 + + ∂x2 ∂y2 ∂z 2
+V
(5.1.43)
if 0 < x, y, z < a otherwise
The energy eigenvalues corresponding to (5.1.43) are given by En =
π 2 ~2 n2x + n2y + n2z 2 2ma
(5.1.44)
And the first five lowest energies, tabulated via (5.1.44) are E111 =
3π 2 ~2 2ma2 6π 2 ~2 2ma2 9π 2 ~2 = 2ma2 11π 2 ~2 = 2ma2
E112 =E121 = E211 = E122 =E212 = E221 E113 =E131 = E311 E222 =
12π 2 ~2 2ma2
E223 =E232 = E322 = c) d)
17π 2 ~2 2ma2
The degeneracies are given by the number of different combinations of atomic numbers that result in the same eigenenergy; these are obvious from part b). If 10 spin 1/2 fermions are in the box, then the ground state is determined by the fact that 2 particles can occupy each of the lowest energy states (± 1/2 for each): E = 2E111 + 2E211 + 2E121 + 2E112 + 2E122 −→ E =
60π 2 ~2 2ma2
For spin 3/2 , then 4 can occupy each of the lowest states: E = 4E111 + 4E211 + 2E121 −→ E =
48π 2 ~2 2ma2
Problem 9 Let
|ψi =
p
1/ 3
Y10 α +
p
2/ 3
Y11 β
(5.1.45)
383
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
If you were to measure Sz , what is the probability that you would get ~/2?
b)
If indeed you do get ~/2 in (a), what is the new state (or wavefunction after the measurement)?
c)
If you measure Lz after (b), what is the probability that you will get −~? (Note: α for spin up and β for spin down, and the Y ’s are the spherical harmonics).
Solution a)
The probability of measuring Sz would simply be the square of the coefficient associated with that state. In other words, P =
b)
1 3
If you do measure ~/2, then you have essentially “collapsed” the wave function, leaving only the part which corresponds to your measurement. In this case, |ψi = Y10 α
c)
(5.1.46)
The chances would be zero. The current state our particle is in (5.1.46), which has ` = 1 and m` = 0. In the general case, when measuring Lz we have Lz Y`m` (θ, φ) = m` ~Y`m` (θ, φ) However, in our case Lz Y10 (θ, φ) = (0)~Y10 (θ, φ) So, it is very clear that the chances that upon measuring Lz you get −~ is zero.
Problem 10 A spin 1/2 particle is placed in a uniform magnetic field H0 directed along plied along xˆ. Calculate the time independent probability for the spin to flip spin matrices are 0 1 0 −i 1 σx = , σy = , σz = 1 0 i 0 0
Solution
zˆ. A small rf field H1 cos ωt is apfrom up to down along zˆ. The pauli 0 −1
W. Erbsen
QUANTUM MECHANICS
We begin our journey by restating the problem in a much more transparent way. Our particle is initially perfectly content with beign spin-up whilst a uniform magnetic field is being applied to it along the z-direction. Then, a grumpy old perturbation comes along pointing in the x-direction. The question is: what are the chances that the particle will go from begin spin-up to spin-up along the z-direction to spin down along the same direction? This is a time-dependent perturbation theory problem, and accordingly we require the formalism of first-order time-dependent perturbation theory. First things first, however; the Hamiltonian for a charged particle in a magnetic field is given by eB0 Sz ~ 1 0 H0 = , where Sz = mc 2 0 −1 Now – we recall from time-dependent perturbation theory that 1 0 −iω0 t H e c2 (t) i~ 12 1 0 iω0 t c1 (t) c˙2 (t) = H21 e i~
c˙1 (t) =
(5.1.47a) (5.1.47b)
Now, the particle is initially spin-up, so (z)
cb (0) = 1
(z)
ca (0) = 0
Eb
χ+
Ea
χ−
Since at time t = 0 we are guaranteed to be in the upper state, and wish to know the probability that we will be in the lower state as a function of time, we use (5.1.47a). We are told that the perturbation takes the form: H10 = H1 cos ωt =
eB 0 Sx cos ωt m
So, the matrix element that we use in (5.1.47a) is then eB 0 Sz cos ωt|↑i m eB 0 cos ωth↓|Sx |↑i = m ~ 0 1 eB 0 1 = cos ωt 0 1 1 0 0 m 2 0 1 e~B = cos ωt 1 0 0 m
0 Hab =h↓|
=
e~B 0 cos ωt m
(5.1.48)
We now take (5.1.48) and substitute it into the first-order integral equation for the 2 → 1 transition: Z 1 t 0 0 −iω0 t0 0 ca (t) = H (t )e dt i~ 0 ab Z 0 1 t e~B 0 = cos ωt0 e−iω0 t dt0 i~ 0 m
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Z 0 eB 0 t = cos ωt0 e−iω0 t dt0 im 0 Z 0 0 eB 0 t iωt0 = e + e−iωt e−iω0 t dt0 2im 0 Z t Z t eB 0 −i(ω0 −ω)t0 0 −i(ω0 +ω)t0 0 e dt + e dt = 2im 0 "0 t " t 0 −i(ω0 −ω)t0 −i(ω0 +ω)t0 eB e e = − + − 2im i(ω0 − ω) i(ω0 + ω) 0 0 eB 0 e−i(ω0 −ω)t 1 e−i(ω0 +ω)t 1 − + − + = 2im i(ω0 − ω) i(ω0 − ω) i(ω0 + ω) i(ω0 + ω) | {z }
(5.1.49)
We typically neglect the underbraced terms in (5.1.49). Why? Good question. We are typically looking at driving frequencies that are very close to the resonant frequencies, eg ω ≈ ω0 . In this approximation, the underbraced terms are very small compared to the rest of the expressions, so they essentially vanish. We could carry them on through the end to get an exact solution; but this is not necessary. Continuing, −i(ω0 −ω)t eB 0 e 1 ca (t) = + − 2im i(ω0 − ω) i(ω0 − ω) (ω0 −ω) o (ω0 +ω) eB 0 e−i 2 t n i (ω0 −ω) t − e−i 2 t = e 2 2im i(ω0 − ω) (ω0 −ω) eB 0 e−i 2 t (ω0 − ω)t = sin (5.1.50) 2im i(ω0 − ω) 2 And to find the transitional probability, we square the modulus of (5.1.50), and we obtain:
Pb→a =
e2 B 02 1 2 (ω0 − ω)t sin m2 (ω0 + ω)2 2
Problem 11 Calculate the transmission coefficient T at any energy E for the potential V (x) = −
~2 δ(x − b) 2ma
(5.1.51)
Solution The solutions of the time-independent Schr¨ odinger for this problem are ψI (x) =Aeikx + Be−ikx
for
xb
(5.1.52b)
We can immediately say that D = 0 in (5.1.52b), since we don’t expect any reflection after the well. The first continuity condition is ψI (b) = ψII (b): Aeikb + Be−ikb = Ceikb −→ A + Be−2ikb = C −→ B = e2ikb (C − B)
(5.1.53)
0 The second continuity condition is ψI0 (b) = ψII (b), and in order to apply this we must go back to the time-independent Schr¨ odinger equation with the potential provided: ~2 ~2 ∂ 2 − − δ(x − b) ψ(x) =Eψ(x) 2m ∂x2 2ma ∂2 1 2mE ψ(x) + δ(x − b)ψ(x) = − 2 ψ(x) ∂x2 a ~ Z b+ 2 Z ∂ 1 b+ ψ(x) dx + δ(x − b)ψ(x) dx =0 2 a b− b− ∂x ( ) ∂ ∂ 1 ψ(x) − ψ(x) = − ψ(b) lim →0 ∂x ∂x a b+ b−
∂ ∂ ψII (b) − ψI (b) = − ∂b ∂x ∂ ∂ Ceikx − Aeikx + Be−ikx = − ∂x ∂x ikCeikb − ikAeikb + ikBe−ikb
C − A + Be−2ikb
1 ψII (b) a C ikb e a C = − eikb a C =− ika
(5.1.54)
At this point we substitute (5.1.53) into (5.1.54), C C − A + e2ikb (C − A) e−2ikb = − ika C C −A+C −A=− ika C 2C + =2A ika C 1 2+ =2 A ika And the transmission coefficient is 2 C 2 2 2 2 4 1 = T = = = −→ T = 1 1 1 ~2 A 2 + ika 2 − ika 2 + ika 4 + k 21a2 1 + 8mEa 2
Problem 12
(5.1.55)
387
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Consider a two-state system with state |1i, |2i and energies E1 and E2 , respectively. The system is in the state |1i. At t = 0 it enters a region where it is perturbed by a constant potential with matrix elements h1|V |1i = h2|V |2i = 0, h1|V |2i = h2|V |1i = V . Find P2 (t), the probability that the system is in the state |2i as a function of time.
Solution This problem directly reflects the derivation of the time-dependent perturbation theory for a two-state system. This is derived in great detail in Griffiths in Chapter 9. The general idea is that you have two different states with two different energy levels, where each state is an eigenstate in the time-independent Schr¨ odinger equation: ψ1 (t) =c1 e−iE1 t/~ |1i,
H0 |1i = E1 |1i
ψ2 (t) =c1 e
H0 |2i = E2 |2i
−iE2 t/~
|2i,
Our total wave function can then be expressed as a linear combination of ψ1 (t) and ψ2 (t), and the coefficients are time-dependent : Ψ(t) = c1 (t)e−iE1 t/~ |1i + c2 (t)e−iE2 t/~ |2i
(5.1.56)
By substituting ψ1 (t) and ψ2 (t) into the time-dependent Schr¨ odinger equation, with the Hamiltonian H → H0 +H 0 (t), where H0 is the unperturbed Hamiltonian and H 0 (t) is the time-dependent perturbation. After simplification, we arrive at: 1 0 −iω0 t H e c2 (t) i~ 12 1 0 iω0 t c˙2 (t) = H21 e c1 (t) i~
c˙1 (t) =
(5.1.57a) (5.1.57b)
0 0 Where we have define H12 and H21 to be 0 H12 = hψ1 (t)|H 0 (t)|ψ2 (t)i,
0 H21 = hψ2 (t)|H 0 (t)|ψ1 (t)i
While ω0 is ω0 =
E2 − E1 ~
The hallmark result from first-order time-dependent perturbation theory is Z 1 t 0 0 iω0 t0 0 (1) c2 (t) = H (t )e dt i~ 0 21
(5.1.58)
By inserting the appropriate values into (5.1.58) and squaring the modulus of the result, we arrive at the probability that the system will transition from ψ1 (t) to ψ2 (t) as a function of time. The provided perturbation matrix is 0 V12 Vij = (5.1.59) V21 0 Where we note that V12 = V21 = V . Substituting the relevant matrix element from (5.1.59) into (5.1.58),
W. Erbsen
QUANTUM MECHANICS
(1) c2 (t)
Z t 0 1 eiω0 t dt0 = V i~ 0 t " iω0 t0 V e = i~ iω0 0
=−
=− =− =−
V eiω0 t − 1 ~ω0 iω0 t V iω0 t iω0 t e 2 e 2 − e− 2 ~ω0 2iV iω0 t ω0 t 2 e sin ~ω0 2 i(E2 −E1 )t 2iV (E2 − E1 )t 2~ e sin (E2 − E1 ) 2~
(5.1.60)
Taking the modulus squared of (5.1.60), 2
P1→2 = |c1 (t)| −→ P1→2
4V 2 2 (E2 − E1 )t = sin (E2 − E1 )2 2~
Problem 13 Consider a hydrogen atom in its ground state. An electric field, E(t), is applied in the z-direction. 0 t0
(5.1.61)
a)
Apply first-order time-dependent perturbation theory to calculate an expression for the probability that the hydrogen atom is in an excited state.
b)
What is the probability if the excited state is (i) 2s and (ii) 2p? Explain briefly the justification for your answer.
c)
How do your answers to (b) behave in the limits τ = 0 and τ = ∞? Explain physically whether this behavior is sensible.
Solution Before we get down to business, let’s remember a few things. For instance, ψn`m (r, θ, φ) =Rn`(r)Y`m (θ, φ) s ` 3 2 (n − ` − 1)! − r 2r 2r 2`+1 na = Ln−`−1 Y`m (θ, φ) 3 e na na na 2n [(n + `)!] And the first three radial wave functions are:
(5.1.62)
389
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
3 R10 = 2a− /2 exp − ar R20 = R21 =
3 √1 a− /2 1 − r exp 2a 2 3 √1 a− /2 r exp − r a 2a 24
r − 2a
Y00 = Y10 = Y1±1 =
1 1 /2 4π 1 3 /2 cos θ 4π 1 3 /2 ∓ 8π sin θe±iφ
So, the first excited state has n = 2 with ` = 0, 1, and admits the following states: ψ200 (r, θ, φ) =R20 (r)Y00 (θ, φ) h r i 1 3 1 r √ = √ a− /2 1 − exp − 2a 2a 2 4π h r i r 1 − 3/2 exp − =√ a 1− 2a 2a 8π ψ210 (r, θ, φ) =R21 (r)Y10 (θ, φ) ! h r i r 3 1 − 3/2 r = √ a cos θ exp − a 2a 4π 24 h r i 3 r 1 = √ a− /2 exp − cos θ a 2a 4 2π
(5.1.63)
(5.1.64)
We also recall the time-dependent coefficients, 1 0 −iω0 t H e cb (t) i~ ab 1 0 iω0 t c˙b (t) = Hba e ca (t) i~
c˙ a (t) =
And in the spirit of first order time-dependent perturbation theory, Z 1 t 0 0 iω0 t0 0 (1) cb = H (t )e dt i~ 0 ba
(5.1.65)
The matrix element in (5.1.65) is given by 0 Hba = hψb |H 0 |ψa i t
The perturbation in our case is E(t) = E0 e− /τ , and for dipole radiation the perturbation becomes t
H 0 = −pE = −qzE0 e− /τ If we recall that in spherical coordinates z = r cos θ, then the full matrix element becomes t 0 Hba (t) =hψb | −qE0 e− /τ r cos θ |ψai t
= − qE0 e− /τ hψb |r cos θ|ψa i
(5.1.66)
For the most general expression for the probability that hydrogen is in an excited state, we carry on calculating our matrix element from (5.1.66) using the full wave functions for hydrogen: t
0 Hba = − qE0 e− /τ hψn`m (r, θ, φ)|r cos θ|ψ100 (r, θ, φ)i 3
− /2 t a r = − qE0 e− /τ √ hRnl (r)Ylm (θ, φ)|r cos θ|e− /a i π
W. Erbsen
QUANTUM MECHANICS
3
− t/τ
= − qE0 e
a− /2 √ π − 3/2
t a = − qE0 e− /τ √
π
Z
2π
0
Z
Z
π
0
2π
0
Z
Z
∞
0
π
0
r (Rn`(r)Y`m (θ, φ)) (r cos θ) e− /a r 2 sin θ drdθdφ
Y`m (θ, φ) cos θ sin θ dθdφ
Z
∞
r
r 3 Rn`(r)e− /a dr
(5.1.67)
0
To find the time-dependent probability, we would take the matrix element (5.1.67) and substitute it into the integral (5.1.65) and take the modulus squared. Since We do not know what state we are calculating the probability of jumping to, things will get very messy very quickly. Although they do call me messy wessy, this does not have anything to do with messy equations. Yuck. a)
The dipole selection rules prevent the 1s → 2s transition (` ± 1), so let’s find the probability of exciting to the 2p state. The matrix element would be t
0 Hba (t) = − qE0 e− /τ hψ210 (r, θ, φ)|r cos(θ)|ψ100 (r, θ, φ)i 1 t 3 r r 3 r 1 √ a− /2 e− /2a cos θ r cos θ √ a− /2 e− /a = −qE0 e− /τ a π 4 2π r E D 3 r 1 1 1 = − qE0 e− /2 3 √ re− /2a cos(θ) r cos(θ) e− /a a 4π 2 a t r E qE0 e− /τ D − r/2a =− √ re cos(θ) r cos(θ) e− /a 4π 2 a4 Z 2π Z π Z ∞ r r =α re− /2a cos(θ) (r cos(θ)) e− /a r 2 sin(θ) drdθdφ 0
=α
Z
0
0
2π
dφ
Z
0 π
2
cos (θ) sin(θ) dθ
0
Z
∞
r 4 e−
3r
/4a
dr
(5.1.68)
0
Where I have temporarily clumped all constants infront of the integrals as α. To solve the θ integral, we require the following trig identities: 1 1 (1 − cos(2θ)) , cos2 (θ) = (1 + cos(2θ)) 2 2 1 sin(α) cos(β) = [sin (αβ ) + sin (α − β)] 2 sin2 (θ) =
So the θ integral becomes Z π Z cos2 θ sin(θ) dθ = 0
π
1 (1 + cos(2θ)) sin(θ) dθ 2 Z Z 1 π 1 π = sin(θ) dθ + sin(θ) cos(2θ) dθ 2 0 2 Z π0 1 1 π = [− cos(θ)| 0 + [sin(3θ) + sin(−θ) dθ] 2 4 0 1 11 1 = [1 + 1] + [1 + 1] − [1 + 1] 2 43 4 1 1 =1 + − 6 2 2 = 3 0
For the r integral, we recall the following integral
(5.1.69)
391
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Z
∞
xn e−ax dx =
0
n! an+1
Then the r integral becomes Z
∞
r 4 e−
3r
/4a
dr =
0
4! ( 3/4a)5
=
45 a5 4! 35
Combining (5.1.69) and (5.1.70) into (5.1.68) (let’s throw α back in there too), 5 5 t qE0 e− /τ 2 4 a 4! 45 4! 0 − t/τ √ = −qE e a Hba(t) = − √ (2π) 0 3 35 4π 2 a4 36 2
(5.1.70)
(5.1.71)
Substituting (5.1.71) into (5.1.65), (1) cb
−qE0 a 45 4! √ = i~ 36 2 =
−qE0 a 45 4! √ i~ 36 2
−qE0 a 45 4! √ = i~ 36 2 −qE0 a 45 4! √ = i~ 36 2 =
−qE0 a 45 4! √ i~ 36 2
=
−qE0 a 45 4! √ i~ 36 2
=
−qE0 a 211 4! √ ~ 36 2
Z
t
e−
t0
/τ iω0 t0
e
dt0
0
Z
t
et (iω0 − /τ ) dt0 0
1
0
"
t 0 t0 eiω0 t e− /τ (iω0 − 1/τ )
0 # 1 e(iω0 − /τ )t 1 − (iω0 − 1/τ ) (iω0 − 1/τ ) h i 1 (iω0 − 1/τ )t e − 1 (iω0 − 1/τ ) 1 t i 1 e(iω0 − /τ ) 2 h (iω0 − 1/τ ) t t 2 − e−(iω0 − /τ ) 2 e 1 (iω0 − /τ ) iω0 t t e 2 e 2τ (iω0 − 1/τ )t sin (iω0 − 1/τ ) 2
"
(5.1.72)
The probability of transition is then 2 −qE a 211 4! e iω20 t e 2τt (iω0 − 1/τ )t 0 √ P (1s → 2p) = sin ~ 36 2 (iω0 − 1/τ ) 2
Which finally leads to
P (1s → 2p) = b)
t 1 q 2 E02 a2 221 (4!)2 e /2τ 2 (iω0 − /τ )t sin ~2 312 ω02 + ( 1/τ )2 2
If we let τ = 0, then (5.1.73) leads to P (1s → 2p) =
q 2 E02 a2 221 (4!)2 1 2 iω0 t sin ~2 312 ω02 2
(5.1.73)
W. Erbsen
QUANTUM MECHANICS
Problem 14 Deuterons have spin-1 (bosons). a)
In general, what are the possible total spin S and total spin angular momentum J of two deuterons in an arbitrary (relative) orbital angular momentum state L? Assume the state vector is a product of a spin part and an orbital part. Determine just the allowed S, J, not the state vector.
b)
In particular, for L = 0 and L = 1, what S and J are allowed?
Solution a)
For bosons, ms = ±1, so S = {0, 1, 2}, and the possible arbitrary J values are just J = L + S = {|L − S|, ..., |L + S|}. The (arbitrary) total angular momentum quantum numbers are given by L L L L
b)
S 2 1 0
J {|L − 2|, ..., |L + 2|} {|L − 1|, ..., |L + 1|} {L}
Let’s make a table and find out! L 0 1 0 1 0 1
Problem 15
S 2 2 1 1 0 0
J 2 1, 2, 3 1 0, 1, 2 0 1
(Old Problem 16)
A particle of mass m is confined to a one-dimensional potential −γδ(x), γ > 0. Show that there is one bound state. Calculate its binding energy and its eigenfunction.
Solution We imagine our potential to consist of three regions: I is before the potential, II is at the potential, and III is after the potential. Plugging our given potential into the time-dependent Schr¨ odinger equation, ~2 ∂ 2 − − γδ(x) ψ(x) = Eψ(x) (5.1.74) 2m ∂x2 We now solve (5.1.74) for each of the regions. For the first region (region I), ~2 ∂ 2 ∂2 ψ(x) = Eψ(x) −→ ψ(x) − k 2 ψ(x) = 0, − 2m ∂x2 ∂x2
p 2m|E| k= ~
(5.1.75)
393
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
The sign changed because we know that E is negative. The solution to (5.1.75) is rather obvious, ψI (x) = Aekx + Be−kx −→ ψI (x) = Aekx
(5.1.76)
Where we found that B = 0 due to boundary conditions. The solution for region III is similar, ψIII (x) = Cekx + De−kx −→ ψIII (x) = De−kx
(5.1.77)
And boundary conditions dictated that C = 0, as before. We now move on to the trickier region, that of II: ~2 ∂ 2 ψ(x) + V (x)ψ(x) = Eψ(x) 2m ∂x2 Z Z 2 Z 2 ∂ ~ ψ(x) dx + − V (x)ψ(x) dx = E ψ(x) dx 2m − ∂x2 − − ! Z ~2 ∂ ∂ − lim − − δ(x)ψ(x) dx = 0 →0 2m ∂x ∂x − − ~2 ∂ ∂ − ψIII (x) − ψI (x) = γψ(0) 2m ∂x ∂x ∂ 2mγ ∂ De−kx − Aekx = − 2 D ∂x ∂x ~ 2mγ −kD − kA = − 2 D ~ −
At x = 0, ψI (0) = ψIII (0) −→ A = D, so: 2mγ mγ 2kA = 2 A −→ K = 2 = ~ ~
p 2m|E| ~
Rearranging and solving,
2m|E| =
m2 γ 2 mγ 2 mγ 2 −→ |E| = −→ E = − 2 2 2 ~ 2~ 2~
(5.1.78)
We can see that the only values in (5.1.78) are numbers; there are no indicies like n, that would infer a different eigenenergy. Therefore, (5.1.78) is the only bound state. For normalization, Z ∞ |ψIII (x)| dx =1 0 Z ∞ 2A2 e−2kx dx =1 0 √ mγ 1 −2kx 1 −2kx A2 2 A − e − − e =0 −→ = 1 −→ A = 2k 2k k ~ ∞ 0 So that we have
ψ(x) =
√
h mγ i mγ exp − 2 x ~ ~
W. Erbsen
QUANTUM MECHANICS
Problem 16 Consider the isotropic harmonic oscillator where the potential is given by V (r) = 1/2 mω2 r 2
(5.1.79)
a)
Show that the eigenfunctions can be expressed in the form Rn`(r)Y`m` .
b)
The problem is separable in Cartesian coordinates, show that the eigenenergies can be expressed as En = n + 3/2 ~ω (5.1.80)
c)
Find the degeneracy D(n) of En for the first 4 values of energies. You obtain the degeneracy from the solutions in Cartesian coordinates. For each energy, identify the value(s) of orbital angular momentum (or momenta if more than one) for each eigenenergy.
Solution a)
I don’t wanna
b)
The full Hamiltonian for the 3-D isotropic harmonic oscillator is 2 ~2 2 1 ~2 ∂ ∂2 ∂2 1 H=− ∇ + mω2 r 2 = − + + + mω2 x2 + y2 + z 2 2 2 2 2m 2 2m ∂x ∂y ∂z 2 We now insert (5.1.81) into the time-independent Schr¨ odinger equation, ~2 ∂2 ∂2 ∂2 mω2 2 2 2 − + + x + y + z ψ(x, y, z) = Eψ(x, y, z) + 2m ∂x2 ∂y2 ∂z 2 2
(5.1.81)
(5.1.82)
We will solve this partial differential equation by separation of variables, whereby we assume that a separable solution exists in the form ψ(x, y, z) = X(x)Y (y)Z(z). Substituting this into (5.1.82), 2 ∂ ∂2 ∂2 m2 ω2 2 2mE + + X(x)Y (y)Z(z) − x + y2 + z 2 X(x)Y (y)Z(z) = − 2 2 2 2 2 ∂x ∂y ∂z ~ ~ 1 ∂2 1 ∂2 1 ∂2 m2 ω2 2 2mE X(x) + + − x + y2 + z 2 X(x)Y (y)Z(z) = − 2 X(x) ∂x2 Y (y) ∂y2 Z(z) ∂z 2 m2 ~ mω 2 1 ∂ 2 mω 2 1 ∂ 2 mω 2 1 ∂2 2mE X(x) − x2 + Y (y) − y2 + Z(z) − z2 = − 2 X(x) ∂x2 ~ Y (y) ∂x2 ~ Z(z) ∂x2 ~ ~ (5.1.83) If we let E be the separation constant, where E = Ex + Ey + Ez , we find that (5.1.83) can be fully separated into three unique equations: mω 2 1 ∂2 2mEx X(x) − x2 + =0 (5.1.84a) 2 X(x) ∂x ~ ~2 mω 2 1 ∂2 2mEy Y (y) − y2 + =0 (5.1.84b) 2 Y (y) ∂y ~ ~2 mω 2 1 ∂2 2mEz Z(z) − z2 + =0 (5.1.84c) 2 Z(z) ∂z ~ ~2
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
It is clear that (5.1.84a)-(5.1.84c) represent the 1-D differential equations for the harmonic oscillator along the x, y, and z directions, respectively. The eigenvalues for the three 1-D solutions are well known, Ex = ~ω nx + 1/2 , Ey = ~ω ny + 1/2 , Ez = ~ω nz + 1/2 So that the total energy becomes
c)
Enx ,ny ,nz = nx + ny + nz + 3/2 −→ En = ~ω n + 3/2
We can discover the degeneracies by using a table as follows nx 0 1 0 0 1 1 0 1 2 0 0 2 2 0 0 1 1 3 0 0
ny 0 0 1 0 1 0 1 1 0 2 0 1 0 2 1 0 2 0 3 0
nz 0 0 0 1 0 1 1 1 0 0 2 0 1 1 2 2 0 0 0 3
E(~ω) 3 /2 5 /2 5 /2 5 /2 7 /2 7 /2 7 /2 9 /2 7 /2 7 /2 7 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2
This table includes all possible values for the lowest four energy levels: E0 = 3/2 ~ω, E1 = 5/2 ~ω, E2 = 7/2 ~ω, E3 = 9/2 ~ω. The degeneracies, as well as the possible values for the orbital angular momentum are: E E0 E1 E2 E3
Problem 17
d n 1 0 3 1 6 2 10 3
` 0 0 0, 1 0, 1, 2
W. Erbsen
QUANTUM MECHANICS
a)
Consider a helium ion. The two electrons are in the 2p and 3p states. The energy levels will have definite total angular momentum (J = L + S). What J-values can occur and how many degenerate energy levels for each J can there be?
b)
How do your answers change if the electrons are both in the 3p states?
Solution a)
Let’s call the electron in the 2p state “1”, and the electron in the 3p state “2.” Then `1 = 1, and `2 = 1. The total orbital angular momentum can then take the values L = {|`1 − `2 |, ..., |`1 + `2 |} = {0, 1, 2}
(5.1.85)
And for electrons, ms = ± 1/2 , so the possible values for the total spin angular momentum are S = {0, 1}
(5.1.86)
And we also know that the possible values for the total angular momentum is determined by J = L + S = {|L − S|, ..., |L + S|} And to find the possible values, we create a table: L 0 1 2 0 1 2
S 1 1 1 0 0 0
J 1 0, 1, 2 1, 2, 3 0 1 2
So, the possible values for J are J = {0, 1, 2, 3} , each with degeneracies 2J + 1: J 0 1 2 3 b)
D 1 3 5 7
If both electrons are in the 3p state, then L = {0, 1, 2} S = {0} And the possible total angular momentum quantum numbers can be found by L 0 1 2
S 0 0 0
J 0 1 2
So the possible values for J are J = {0, 1, 2} . The degeneracies are
(5.1.87)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
J 0 1 2
D 1 3 5
Problem 18 A particle of spin zero is confined in a two-dimensional infinite square well of dimension 2a on each side. a)
Find the energies and the wave functions of the first two states.
b)
If there are two indistinguishable noninteracting spin 1/2 particles inside the box, write down the total wave function of the first two states. Identify the degree of degeneray for each state.
c)
If there are five indistinguishable spin 1/2 noninteracting particles in the box, what is the total energy of the ground state?
d)
If there are five indistinguishable spin 1 noninteracting particles in the box, what is the total energy of the ground state?
Solution a)
In general, the eigenstates and energy eigenvalues for this problem are given by n πx n πy 1 x y ψ(x, y) = sin sin a 2a 2a π 2 ~2 2 Enx ,ny = (n + nxy) = E0 (n2x + n2y ) 8ma2 x
(5.1.88a) (5.1.88b)
The the ground state, nx = ny = 1, (5.1.88a) and (5.1.88b) become ψ(x, y) =
πx πy 1 sin sin , a 2a 2a
E11 = 2E0
πx πy 1 sin sin , a a 2a
E12 = 5E0
For the first excited state, nx = 2, and ny = 1, so we have ψ(x, y) =
So, the first excited state is doubly degenerate. b)
We first recall that the total wave function of a fermion (space and spin part) must be completely anti-symmetric under the exchange of any two particles in the system. For the case at hand, we may have to “symmetrize,” or “anti-symmetrize” the wave function such that the final total wave function is anti-symmetric. In other words, if the spin part is symmetric, then the space part must be anti-symmetric, and also vice versa. We recall that, in general,
W. Erbsen
QUANTUM MECHANICS
1 ψ(x1 , x2 ) = √ [ψn1 (x1 )ψn2 (x2 ) ± ψn2 (x1 )ψn1 (x2 )] 2
(5.1.89)
And also ψ(x1 , x2 ) =
n πx n πx 1 1 1 2 2 sin sin a 2a 2a
For the ground state, n1 = n2 = 1 and S = 0 (singlet) so the space part is just πx πx 1 1 2 sin ψ(x1 , x2 ) = sin a 2a 2a
(5.1.90)
(5.1.91)
Since the space part is symmetric, the spin part must be anti-symmetric: 1 χ00 = √ [|↑↓i − |↓↑i] 2
(5.1.92)
The total wave function for the ground state for two indistinguishable spin- 1/2 fermions is just the product of (5.1.91) and (5.1.92) πx πx 1 2 1 sin [|↑↓i − |↓↑i] ψ(x1 , x2 ) = √ sin 2a 2a a 2 The ground state has singlet degeneracy. For the first excited state, where are two different options: electron 1 can be excited or electron 2 can be excited. Either of the first excited states may have by a symmetric (S = 0) or an anti-symmetric (S = 1) spatial wave function, like (5.1.89). In our case, letting n1 = 1 and n2 = 2, πx πx πx i 1 h πx1 2 1 2 ψ(x1 , x2) = √ sin sin + sin sin (5.1.93a) 2a a a 2a a 2 h i 1 πx1 πx2 πx1 πx2 = √ sin sin − sin sin (5.1.93b) 2a a a 2a a 2 Where (5.1.93a) is symmetric and thus has an anti-symmetric spin part, while (5.1.93b) is antisymmetric , and must have a symmetric spin part. The former is familiar; the latter (S = 1) has three components, each corresponding to one of the available symmetric spin functions. So, this gives us four wave functions: 1 ψ(x1 , x2 ) = √ a 2 1 = √ a 2 1 = √ a 2 1 = √ a 2
h πx πx πx πx i 1 1 2 1 2 √ [|↑↓i − |↓↑i] sin sin + sin sin 2a a a 2a 2 h πx πx πx πx i 1 2 1 2 sin sin − sin sin |↑↑i 2a a a 2a h πx πx πx πx i 1 1 2 1 2 √ [|↑↓i + |↓↑i] sin sin − sin sin 2a a a 2a 2 h πx πx πx πx i 1 2 1 2 sin sin − sin sin |↓↓i 2a a a 2a
We arrive at four more wave functions if we choose n1 = 2 and n2 = 1. Therefore, the first excited state is eightfold degenerate, and E = 5E0
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
c)
If there are five indistinguishable spin 1/2 fermions in the box, the total energy of the ground state can be determined via a table: n1 1 1 2
n2 1 2 1
E 2E0 5E0 5E0
d 1 4 4
There can be two electrons per each set of quantum numbers, so we can fit five in the first two rows, so
E = 2 · 2E0 + 3 · 5E0 −→ E = 19E0 d)
For five indistinguishable spin-1 bosons, all the particles can be in the lowest level, so the ground state is just
E = 5 · 2E0 −→ E = 10E0
Problem 19 Ten electrons are confined in a box of dimensions (a, 2a, 2a) on each side. Calculate the total energy of the ground state of these ten electrons if we assume that the electrons do not interact with each other.
Solution The separated eigenfunctions and associated eigenenergies are r
n πx 2 x , ψ(x) = sin a a r n πy 1 y ψ(y) = sin , a 2a r n πz 1 z ψ(z) = sin , a 2a The total energy is then En x n y n z =
n2xπ 2 ~2 2ma2 n2y π 2 ~2 Ey = 8ma2 n2 π 2 ~2 Ez = z 2 8ma Ex =
π 2 ~2 4n2x + n2y + n2z 8ma2
(5.1.94a) (5.1.94b) (5.1.94c)
(5.1.95)
In order to find out which combination of atomic numbers yield the ground state, we create a table:
W. Erbsen
QUANTUM MECHANICS
nx 1 1 1 2 2 2 1 1 1 3
ny 1 1 2 1 1 2 2 1 3 1
nz 1 2 1 1 2 1 2 3 1 1
4n2x 4 4 4 16 16 16 4 4 4 36
n2y 1 1 4 1 1 4 4 1 9 1
n2z 1 4 1 1 4 1 4 9 1 1
Σ # 6 1 9 2 9 3 18 21 21 12 4 14 5 15 38
With fermions, two can have each energy (± 1/2 ), so we picked the five lowest energies and now multiply them by two and sum to find the total ground state energy:
E=
100π 2 ~2 π 2 ~2 [2 (6 + 9 + 9 + 12 + 14)] −→ E = 2 8ma 8ma2
Problem 23 Suppose a particle moves in one dimension under the potential V (x). V (x) is even, that is V (x) = V (−x). Also, all the eigenstates of the Hamiltonian H are known to be non-degenerate. a)
Prove that every eigenstate of H has either even or odd parity.
b)
Consider a completely free particle in one dimension. Show that there are solutions to the eigenvalue equation which do not have well-defined parity. Explain why this is not a contradiction to the theorem of part (a).
Solution a)
To show that every eigenstate of H has either even or odd parity, we apply the parity operator to the time-independent Schr¨ odinger equation: 2 2 2 p p p P Hψ(x) =P + V (x) ψ(x) = + V (−x) ψ(−x) = + V (x) ψ(−x) (5.1.96) 2m 2m 2m If we were to apply the parity operator twice, 2 2 p p 2 P Hψ(x) = P + V (x) ψ(−x) = + V (x) ψ(x) 2m 2m
(5.1.97)
So, we can see that P 2 has an eigenvalue of 1: P 2 = 1 −→ P = ±1. Furthermore, P and H commute, so they must share the same eigenvector. Since the eigenstates of P can have either even or odd parity (p = ±1), and H and P commute (have same eigenvector), then every eigenstate of H must have either even or odd parity.
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
A free particle is described by ψ+ (x) = Aeikx,
ψ− (x) = Ae−ikx
And if we apply the parity operator once, P ψ+ (x) =P Aeikx = Ae−ikx P ψ− (x) =P Ae−ikx = Aeikx Since ψ+ (x) and ψ− (x) have the same energy (they are degenerate), we cannot follow the same logic as in a).
Problem 24 Consider an isotropic 3D spherical square well. What are the conditions on the depth V0 and radius a such that a particle of mass m has exactly one bound state in this well? Show your reasoning and calculations.
Solution The radial equation that we wish to solve is in terms of a intermediary function U (r) = rR(r) One could just imagine it to be ψ and carry on, since it plays no important part here. ~2 ∂ ~2 `(` + 1) − +V + U (r) = EU (r) (5.1.98) 2m ∂r 2 2m r 2 However, for the ground state, we recognize that ` = 0 and Veff → V , and (5.1.98) becomes ~2 ∂ − + V U (r) = EU (r) 2m ∂r 2
(5.1.99)
For our isotropic spherical square well, we define two regions; I is inside the sphere, where V = −V0 , and II is outside, where V = 0. Our task is to first solve (5.1.99) for each of these two regions. Starting with region I, ~2 ∂ − − V U (r) = EU (r) 0 2m ∂r 2 ∂2 2m(E + V0 ) u(r) + U (r) = 0 2 ∂r ~2 p ∂2 2m(E + V0 ) 2 U (r) + k U (r) = 0, k= (5.1.100) 2 ∂r ~ The solution of (5.1.100) is UI (r) = A sin (kr) + B cos (kr) And now for region II,
(5.1.101)
W. Erbsen
QUANTUM MECHANICS
−
~2 ∂ 2 U (r) = EU (r) 2m ∂r 2
∂2 U (r) + κ2 U (r) = 0, ∂r 2
κ=
√
2mE ~
(5.1.102)
The solution of (5.1.102) we wish to put in exponential form, UII (r) = Ceiκr + De−iκr
(5.1.103)
Due to boundary conditions our solutions for regions I and II become UI (r) =A sin (kr) UII (r) =De−iκr 0 Our two boundary conditions are UI (a) = UII (a), and UI0 (a) = UII (a). These are, respectively:
A sin (ka) = De−iκa
(5.1.104a) −iκa
KI A cos (ka) = −iκDe
(5.1.104b)
If we take (5.1.104a) and divide it by (5.1.104b), tan (ka) = −
k iκ −→ cot (ka) = − iκ k
(5.1.105)
√ Let’s define z = ka and z0 = a 2mV0 /~. Then z02 =
a2 ~2 2 2mV0 −→ V0 a2 = z 2 ~ 2m 0
and (5.1.105) becomes − cot(z) =
r
z0 2 −1 z
(5.1.106)
If we graph (5.1.106), it is very clear what the boundaries are that determine what the radius and depth of the well should be. z0 will intersect the z-axis between π/2 and 3π/2 if there is to be exactly one bound state (less than π/2 means there is no bound state). So, π 3π < z0 < 2 2 2 ~2 π 2 ~2 3π 2 < V0 a < 2m 2 2m 2 So, the conditions on the width a and depth v0 of the well such that there is exactly one bound state is: ~2 π 2 9~2 π 2 < V0 a2 < 8m 8m
403
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 25 If the spin state of an electron is prepared such that it has a spin ~/2 along the x-direction, what is the probability of finding this spin to have eigenvalue ~/2 if the spin is measured along the y-direction? Show the steps in your calculation. The Pauli spin matrices are: 0 1 0 −i 1 0 σx = , σy = , σz = 1 0 i 0 0 −1
Solution In the spirit of showing all the steps in our calculations, our first step would be to find the eigenvalues and eigenvectors of Sx , Sy and Sz . We know that S = ~/2σ, where σ represents the Pauli spin matrices. Starting with Sx first, 2 −λ ~/2 = λ2 − ~ = 0 −→ λ = ± ~ ~/2 −λ 4 2
And now we find the eigenvectors for each of the eigenvalues: ~ ~ 1 −~/2 ~/2 α 0 1 (x) √ For λ = + ~/2 : = − α + β = 0 −→ χ+ = ~/2 −~/2 β 0 2 2 2 1 ~ 1 ~ ~/2 ~/2 α 0 1 (x) For λ = − ~/2 : = α + β = 0 −→ χ− = √ ~/2 ~/2 β 0 2 2 2 −1 Doing the same for Sy ,
Now the eigenvectors are For λ = + ~/2 : For λ = − ~/2 :
−λ i~/2
~ ~2 −i~/2 2 = λ − = 0 −→ λ = ± −λ 4 2
α 0 = β 0 ~/2 −i~/2 α 0 = i~/2 ~/2 β 0 −~/2 −i~/2 i~/2 −~/2
i~ α− 2 i~ α+ 2
~ 1 1 (y) β = 0 −→ χ+ = √ 2 2 i ~ 1 1 (y) β = 0 −→ χ− = √ 2 2 −i
And now for Sz
Whose eigenvectors are
~/2 − λ 0
For λ = + ~/2 : For λ = − ~/2 :
2 0 = − ~ + λ2 = 0 −→ λ = ± ~ −~/2 − λ 4 2
α 0 = β 0 0 α 0 = 0 β 0
0 0 0 −~
~ 0
1 (z) − ~β = 0 −→ χ+ = 0 0 (z) ~α = 0 −→ χ− = 1
If, for the sake of exercise, we have a particle that is prepared in the z-direction with eigenvalue ~/2 and we wish to know the probability of obtaining ±~/2 in the x-direction,
W. Erbsen
QUANTUM MECHANICS
† α 1 (x) 1 = χ+ χ(z) = √ 1 β 2 † α 1 (x) (z) = χ− χ = √ 1 −1 β 2
α + β 2 α+β √ −→ P = √ 2 2 α − β 2 α−β √ −→ P = √ 2 2
And to do the same for the y-direction,
(y) χ+ (y)
χ−
†
†
χ
(z)
1 =√ 1 2
1 χ(z) = √ 1 2
α − iβ 2 α α − iβ −i = √ −→ P = √ β 2 2 α + β 2 α α + iβ i = √ −→ P = √ β 2 2
However, if the system is initially prepared in the x-direction with an eigenvalue of ~/2, as is asked in the problem, the calculations are no different:
(y) χ+
χ−
(y)
1 1 1 1 = √ 1 −i = √ −→ P = 0 2 2 2
†
(x) χ+
†
1 (x) χ+ = √ 1 2
1 1 1 i = √ −→ P = 0 2 2
Problem 26 a)
A two-dimensional harmonic oscillator has the potential V (x, y) 21 mω2 (x2 + 4y2 ). Calculate the energies of the first three lowest states, and identify the degrees of degeneracy for each energy.
b)
If there is an additional small coupling term W (x, y) = ax2 y present, where a is a small constant, calculate the first-order correction to each of the three levels.
Write down the matrix elements that you need to calculate. If the matrix element is zero, you have to say so and explain why. If it is not zero, say so and represent each integral by a constant and proceed further.
Solution a)
We find a solution by employing separation of variables. The Hamiltonian for our system is H =−
~2 ∂ 2 ~2 ∂ 2 mω2 2 − + x + 2mω2 y2 2m ∂x2 2m ∂y2 2
Substituting (5.1.107) into the time-independent Schr¨ odinger equation, ~2 ∂ 2 ~2 ∂ 2 mω2 2 2 2 − − + x + 2mω y ψ(x, y) = Eψ(x, y) 2m ∂x2 2m ∂y2 2
(5.1.107)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
∂2 ∂2 m2 ω2 x2 4m2 ω2 y2 + 2 − − 2 2 ∂x ∂y ~ ~2
ψ(x, y) = −
2mE ψ(x, y) ~2
(5.1.108)
At this point we introduce the separable solutions ψ(x, y) = X(x)Y (y). Substituting this into (5.1.108), 2 ∂ ∂2 m2 ω2 x2 4m2 ω2 y2 2mE + − − X(x)Y (y) = − 2 X(x)Y (y) 2 2 2 2 ∂x ∂y ~ ~ ~ 2 2 ∂ m2 ω2 x2 ∂ 4m2 ω2 y2 2mE − X(x)Y (y) + − X(x)Y (y) = − 2 X(x)Y (y) ∂x2 ~2 ∂y2 ~2 ~ 1 ∂2 m2 ω2 x2 1 ∂2 4m2 ω2 y2 2mE X(x) − + Y (y) − =− 2 (5.1.109) 2 2 2 2 X(x) ∂x ~ Y (y) ∂y ~ ~ In our case the separation constant is E = E1 + E2 , and the separated equations become mω 2 2 2mEx ∂2 X(x) − x − X(x) = 0 −→ Ex = ~ω(nx + 1/2 ) ∂x2 ~ ~2 " # 2 ∂2 2mω 2mEy 2 Y (y) − x − X(x) = 0 −→ Ey = 2~ω(ny + 1/2 ) ∂y2 ~ ~2 And we know that E = Ex + Ey , so: En = ~ω(nx + 1/2 ) + ~ω(2ny + 1) −→ En = ~ω nx + 2ny + 3/2 To find the degree of degeneracy for the first three energies, let’s make a table: nx 0 1 0 1 2 0 2 2
ny 0 0 1 1 0 2 1 2
E(~ω) 3 /2 5 /2 7 /2 9 /2 7 /2 11 /2 11 /2 15 /2
order 1 2 3 4 3 5 5 6
To display the degeneracies for the energies, let’s make another table, E(~ω) 3 /2 5 /2 7 /2 b)
d 1 1 2
Now the game has changed; we add a perturbation to our oscillator of the form W (x, y) = ax2 y. Before we can calculate the first order energy shift, we must recall some stuff about harmonic oscillators: r r ~ ~ † x= (ax + ax ), y= (a† + ay ) 2mω 4mω y √ √ a† ψn (x, y) = n + 1 ψn+1 (x, y), aψn (x, y) = n ψn−1 (x, y) With this, our perturbation becomes:
W. Erbsen
QUANTUM MECHANICS
!2 r ! ~ ~ † † W =a (a + ax ) (a + ay ) 2mω x 4mω y r ~ ~ =a (a† + ax )2 (a†y + ay ) 2mω 4mω x r
(5.1.110)
In general, the first order shift in energy is given by En(1) =hnx |hny |W (x, y)|ny i|nxi r a~ ~ = hnx |hny |(a†x + ax )2 (a†y + ay )|ny i|nxi 2mω 4mω r a~ ~ = hnx |hny |(a†x + ax )2 a†y |ny i|nxi + hnx | hny |(a†x + ax )2 ay |ny i|nxi 2mω 4mω r a~ ~ p √ = ny + 1 hnx |hny |(a†x + ax )2 |ny + 1i|nxi + ny hnx |hny |(a†x + ax )2 |ny − 1i|nxi 2mω 4mω (5.1.111)
Problem 28 well. That is A particle is in the ground state of an infinite square 0 0≤x≤L V (x) = ∞ otherwise
(5.1.112)
At t = 0 the wall at x = L is suddenly moved to x = 2L - this happens very fast, approximately instantaneously. a)
Calculate the probability that long after t = 0 the system is in the ground state of the new potential.
b)
How fast must the change take place for this “instantaneous” assumption to be good?
Solution The eigenstates for the ground states before and arbitrary eigenstates after the transition are r πx 2 ψgs(x) = sin (5.1.113a) L L r nπx 1 ψn (x) = sin (5.1.113b) L 2L
a)
We wish to express our initial ground state wave function ψgs (x) in terms of a linear combination of ψn (x) states: r r πx X nπx 2 1 X ψgs = sin = cn ψn (x) = cn sin L L L n 2L n For the probability that we will be in the ground state, we calculate cn for n = 1:
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
cn =
Z
L
ψgs(x)ψn (x) dx
0
√ Z L πx πx 2 sin sin dx L 0 L 2L √ Z L h πx πx πx πx i 2 1 cos − − cos + dx = L 0 2 L 2L L 2L √ Z πx 2 1 L 3πx = cos − cos dx L 2 0 2L 2L √ L πx 2L 2 2L 3πx = sin − sin 2L π 2L 3π 2L 0 √ π 2L 3π 2 2L = sin − sin 2L π 2 3π 2 √ 2 2L 2L = + 2L π 3π √ 4 2 = 3π =
So, the probability that we will remain in the ground state is 2
|c1 | = b)
32 9π 2
To find out how fast this process must take place, we use the time-energy uncertainty relation: ∆t∆E ∝ ~. We find the change in energy first ∆E =
n2 π 2 ~2 3n2 π 2 ~2 n2 π 2 ~2 − = 2 2 8mL 2mL 8mL2
So, the necessary time is (approximately) ∆t =
8mL2 3n2 π 2 ~
Problem 29 The scattering amplitude for a proton on a target atom is f(θ). That is, if the target is at rT = 0 then far from the target the wavefunction is eikr ψ(r) = eik·r + f(θ) (5.1.114) r a)
Suppose that there are N (N 1) target atoms located at the point r1 = A, r2 = 2A, r3 = 3A and so on. What is ψ(r) for positions r far from the line? That is, find ψ(r) for |r| N |A|. It will have the form eikr ik·r ψ(r) = e + F (θ) (5.1.115) r Find F (θ) and dσ/dΩ = |F (θ)|2 . Is dσ/dΩ ∼ N ?
W. Erbsen
b)
QUANTUM MECHANICS
Suppose now that the N (N 1) target atoms are located at random positions, but all inside a small finite ball of radius R. Calculate F (θ) and dσ/dΩ. Is dσ/dΩ ∼ N ?
Solution a)
For some point P very far from the target, then we can make the approximation that r1 ≈ r2 ≈ r3 ≈ r. We start with the template provided, (5.1.115), but add to it an additional phase term: ψ(r) =eikr + f(θ) =eikr +
N X eikr−inkA r n=0
N X eikr f(θ) e−inkA r n=0
The sum in (5.1.117) is finite; to evaluate it, we let N X
P
e−inkr =
P
(5.1.116) an , where a = e−ikr . Then
an = 1 + a + a2 + a3 + ... + aN−1
n=0
1+
N X
an+1 = 1 + a + a2 + a3 + ... + aN−1 + aN
n=0
1+
N X
an+1 =
n=0
1 − aN =
n=0
an + aN
n=0
1 − aN =
N X
N X
an =
N X
n=0 N X
n=0
an −
N X
an+1
n=0
an (1 − a)
1 − aN 1−a
Using this, (5.1.117) becomes ψ(r) = eikr +
eikr 1 − e−iNkr f(θ) r 1 − e−ikr
From (5.1.117), it is clear that F (θ) is simply the term in the brackets,
F (θ) = And now to find the differential cross section, dσ 2 = |F (θ)| dΩ
1 − e−iNkr f(θ) 1 − e−ikr
(5.1.117)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
From which it is clear that
2 1 − e−iNkr = f(θ) −ikr 1−e −iNkr/2 iNkr/2 2 e e − e−iNkr/2 2 ) |f(θ)| = −ikr/2 e eikr/2 − e−ikr/2 2 −iNkr e sin (N kr/2) 2 |f(θ)| = −ikr e sin (kr/2) sin2 (N kr/2) dσ 2 = |f(θ)| dΩ sin2 (kr/2)
No, the differential cross section is not proportional to N . b)
For the case of gathered target atoms we start in much the same way, ψ(r) =eikr + f(θ) =eikr +
N X eikr−iXn k r n=0
N X eikr−iXn k f(θ) e−iXn k r n=0
(5.1.118)
Where Xn represents the displacement of the nth target particle. From (5.1.118) it is clear that
F (θ) = f(θ)
N X
e−iXn k
(5.1.119)
n=0
To find the differential cross section, dσ 2 = |F (θ)| dΩ 2 N X = f(θ) e−iXn k n=0
= |f(θ)|
2
= |f(θ)|
2
N X
iXn k
e
n=0 N X
!
N X
−iXm k
e
m=0
!
eik(Xn −Xm )
n=0 m=0
2 = |f(θ)|
2 = |f(θ)|
N X
eik(Xn −Xm ) +
n=m=1 N X
n=m=1
N X
n6=m=1
e0 +
N X
n6=m=1
eik(Xn −Xm )
eik(Xn −Xm )
W. Erbsen
ELECTRODYNAMICS
2 = |f(θ)| N +
N X
n6=m=1
eik(Xn −Xm )
(5.1.120)
The remaining sum is often referred to as random walk, where the trajectories are completely random and as an ensemble, average to zero. Thus, our differential cross section is dσ 2 = N |f(θ)| dΩ There is no escaping the fact that the differential cross section is proportional to the number of atoms in this particular configuration.
5.2
Electrodynamics
Problem 1 A conducting sphere of radius a is placed in a uniform electrostatic field E0 . If the sphere is well insulated, find the expressions for potential and electric fields V and E at the point P (r, θ, φ) lying outside the sphere. Also obtain the expression for the induced surface charge density σ and show that it forms a dipole distribution. Specifiy your boundary conditions clearly.
Solution The induced surface charge density is σ(θ) = ε0 E; the plan of action then is to calculate E, which can be readily done by intermediately solving for the potential, V : ∞ X B` V (r, θ) = A` r ` + `+1 P` (cos θ) (5.2.1) r `=0
Where (5.2.1) stems from (3.63) in Griffiths. The problem, then, is to take (5.2.1) and adjust it so that it describes our problem. We do this by applying boundary conditions. It should be mentioned that as a conductor, the sphere has a constant potential, but we are only interested in the potential outside of the sphere, so we can choose it to be whatever we want. The most convenient choice would be to let Vsphere = 0. Additionally, we recognize that in the field (absent from the sphere) the potential is V (r, θ) = −E0 r cos θ. We can now form our boundary conditions: 0 at r = R V (r, θ) = −E0 r cos θ at r → ∞ Applying our first boundary condition to (5.2.1), ∞ X B` B` V (R, θ) = A` R` + `+1 P` (cos θ) = 0 −→ A` R` + `+1 = 0 −→ B` = −A` R2`+1 R R `=0
(5.2.2)
411
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Substituting our new value for B` into (5.2.1), we have ∞ X R2`+1 V (r, θ) = A` r ` − A` `+1 P` (cos θ) r
(5.2.3)
`=0
If we apply the second boundary condition to (5.2.3), we find that the second term in parenthesis approaches 0 as r → ∞, and we are left with V (r, θ) =
∞ X `=0
A` r ` P` (cos θ) = −E0 r cos θ
(5.2.4)
From (5.2.4) we can see that the second boundary condition is satisfied only if ` = 1 and A` = A1 = −E0 . With these modifications, (5.2.3) is now R3 V (r, θ) = −E0 r + E0 2 cos θ (5.2.5) r From the potential, we can now find E, and in turn σ: ∂ V (r, θ) ∂r ∂ R3 = − ε0 −E0 r − E0 2 cos θ ∂r r R3 = − ε0 −E0 − 2E0 3 cos θ r r=R
σ(θ) = − ε0
From this it is easy to see that
σ(θ) = 3E0 ε0 cos θ
Problem 2 Two insulated square parallel conducting plates, of side L and separation d, are charged with surface densities +σ on the upper plate and −σ on the lower. Two dieletric slabs, each with thickness d/2 and area L × L, are inserted between the plates, one slab above the other. The dieletric constants are K1 and K2 . Assume that d L. Determine: a)
D everywhere between the plates.
b)
E everywhere between the plates.
c)
The bound surface charge densities σb on the three dialetric surfaces.
d)
The capacitance.
Solution
W. Erbsen
ELECTRODYNAMICS
a)
We first recall that Gauss’ Law for for dielectrics can be written as I D · dA = Q Where D = ε0 KE is the electric displacement. In the current case, the area is just L · L = L2 , and the surface charge density defines the total charge as Q = σL2 , so that we have DL2 = σL2 −→ D = −σ ˆ z for
b)
− d/2 < z < d/2
We again use the fact that D = ε0 KE, such that Gauss’ Law becomes: σ ε0 K1,2 EL2 = −σL2 ˆ z −→ E = − ˆ z ε0 K1,2
(5.2.6)
(5.2.7)
And we now have
c)
σ z −ε K ˆ 0 1 E= σ (−ˆ z) ε0 K2
for 0 < z < d/2 for − d/2 < z < 0
The bound surface charge density is related to the polarization as σb = P · n ˆ, where we remember that P = ε0 χE. We also recall that the dielectric constant is defined as K = (1 + χ). The (general) polarization is then given by σ 1 |P1,2 | = ε0 χ1,2 |E1,2 | = ε0 (K1,2 − 1) E = ε0 (K1,2 − 1) = 1− σ (5.2.8) ε0 K1,2 K1,2 The three surfaces the prompt is speaking about refers to the top surface, where slab 1 interfaces with the positive conducting plate (and analogously for slab 2 on the lower negative conducting plate), and the interface between the two dielectrics. The only tricky bit is in the interstitial region; but all we have to do is add the bound charge densities being contributed from both slabs. The result is
d)
1 − 1 − σ K1 1 1 σb = 1− σ− 1− σ K1 K2 1 1− σ K2
for surface 1 for surface 2 for surface 3
In order to calculate the capacitance, we first recall the relation Q = C|∆V |. So, it seems like all we would have to do is calculate ∆V and we are home-free. We must integrate over both dielectrics, going from the negative terminal at the bottom of slab 2 to the positive terminal at the top of slab 1: Z 0 Z d/2 ∆V = − E2 dz − E1 dz − d/2
σ =− − ε0 K2
Z
0
0
− d/2
Z d/2 σ dz − − dz ε0 K1 0
413
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
σ d σ d 0− − + −0 ε0 K2 2 ε0 K1 2 dσ dσ + = 2ε0 K2 2ε0 K1 dσ 1 1 = + 2ε0 K1 K2 =
(5.2.9)
Using (5.2.9) we can now try to find C: C=
Q = ∆V
Which leads us to C= d
dσ 2ε0
σL2 1 K1
+
1 K2
2ε0 L2 1 1 + K1 K2
Problem 3 A hollow dieletric sphere is centered on the origin. It has dielectric constant K 6= 1. Its inner radius is a1 and its outer radius is a2 . It is uncharged. Now a point charge q > 0 is placed at the origin.
a)
Find the electric field strength E for r < a1 , a1 < r < a2 , and a2 < r.
b)
Find the surface charge density (charge/area) on the inside (at r = a1 ) and on the outside (at r = a2 ).
c)
Now the situation is changed. A thin conducting coating is applied to the outside of the sphere, and this surface is maintained at +V0 volts relative to infinity. Find E(r) for r > a2 .
Solution
W. Erbsen
ELECTRODYNAMICS
a)
b)
The electric field in all three regions can be found from Gauss’ Law as 1 q 4πε0 r 2 1 q E= 2 4πε K r 0 1 1 q 4πε0 r 2
for
0 < r < a1
for a1 ≤ r ≤ a2 for
r > a2
To find the bound surface charge density we must use the following relation σb = P · n ˆ, where P = ε0 χE. The general solution can be found by, 1 q 1 q σb = ε0 χ1,2 E1,2 · n ˆ = ε0 (K1,2 − 1) ˆ r·n ˆ = (K1,2 − 1) ˆ r·n ˆ (5.2.10) 2 4πε0 K1,2 r 4πK1,2 r 2 Our solution from (5.2.10) is applicable as long as we recognize that the induced charges on the inner surface are negative, while those on the outer surface are positive. Therefore, for the inner surface we have (−ˆ r) · n ˆ = −1, and for the outer ˆ r·n ˆ = 1. With this, our solutions become
c)
1 q 1 − 1 K1 4π a21 σb = 1 1 q 1− K2 4π a22
at r = a1 at r = a2
Since we are only interested in the field outside the sphere, the particular configuration of all the interior charges is irrelevant; all that is important is the total charge. We know that there is one point charge at the center of the sphere, q, and that the conducting shell is held at V0 . What we don’t know is how much charge exists on the exterior of the conductor; we can find this out as follows: V (r = a2 ) = V0 =
1 (Q + q) −→ Q = 4πε0 a2 V0 − q 4πε0 a2
(5.2.11)
And now using (5.2.11) we can find the potential at a greater distance r > a2 : V (r > a2 ) =
1 (Q + q) 1 (4πε0 a2 V0 − q + q) V0 a2 = −→ V (r) = 4πε0 r 4πε0 r r
Recalling that E = −∇V , from (5.2.12) we have
1 ∂ V0 a2 1 E(r) = − ˆ r=− r ∂r r r
From which it is easy to see that the field is E(r) =
Problem 4
V0 a2 ˆ r r3
V0 a2 − 2 ˆ r r
(5.2.12)
415
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
A linear molecule with a permanent electric dipole moment p0 and moment of inertia I (for example, HCL) is placed in a uniform electric field E. a)
Make a sketch showing the charges that make up the dipole, the dipole moment and the electric field when the dipole is in the equilibrium orientation.
b)
Derive an expression for the frequency ω0 of small amplitude oscillations about the equilibrium orientation.
c)
Describe the polarization and power of the radiation produced by this oscillating dipole, assuming it has been initially displaced by an angle θ 90 degrees from its equilibrium orientation. Make a sketch of the angular distribution of this power.
d)
Over what time scale will it continue radiating appreciably, and therefore what range of frequencies are emitted?
Solution a)
Picture picture picture
b)
To find an expression for ω0 , we begin with Newton’s Second Law applied to rotational kinematics: ¨ We note that the torque is also defined by τ = r × F, where r in our case is the τ = Iα = I θ. distance between the two charges in the dipole, let’s call this distance d. With this, we have q F τ = r × F = d × F = d × F = qd × =⇒ τ = P0 × E q q
(5.2.13)
Where in (5.2.13) I have used the fact that P0 = qd and also that F = qE. The magnitude of τ is of course |τ | = P0 E sin θ. Using this with Newton’s Second Law, we have I θ¨ = P0 E sin θ −→ θ¨ =
P0 Eθ P0 E sin θ ≈ =⇒ θ¨ = −ω2 θ I I
(5.2.14)
Where I have used the small angle approximation sin θ ≈ θ. We know that (5.2.14) is the differential equation for the harmonic oscillator, where we typically define ω as:
ω0 = c)
r
P0 E I
(5.2.15)
µ0 P02 ω4 12πc
(5.2.16)
Equation (11.22) from Griffiths is: hP i =
Which represents the total time averaged power radiated from an oscillating dipole. If we imagine our molecule rotating periodically (pendulum-like), it has a combined energy equal to E=
1 2 1 2 mv + Iω 2 2
(5.2.17)
And also sources of potential energy, which we neglect. When the molecule is perfectly aligned with the field, that is, θ = 0, then (5.2.17) reduces to E = 1/2 Iω2 . We also note that power is defined as the amount of work per unit time, and therefore from (5.2.16) we can say
W. Erbsen
ELECTRODYNAMICS
hP i =
dE µ0 P02 4 ENERGY = ω = Cω4 = 12πc TIME dt | {z } C
Where the time derivative of the rotational kinetic energy is simply Iωω. ˙ Recalling that C 0 = C/I, we have Iωω˙ = Cω4 −→
dω 1 = C 0 ω3 −→ 3 dω = C 0 dt dt ω
Whose solution is t=
1 2Cω2
Problem 5 A current I flows into a parallel plate capacitor with circular plates of radius R separated by a distance d. The current was 0 before t = 0 and I 6= 0 after. a)
What is the charge on the plates as a function of time?
b)
What is the electric field between the plates?
c)
What is the displacement current density between the plates?
d)
What is the magnetic field between the plates at r = R/2 from the center of the plates?
e)
What is the Poynting vector between the plates at R/2?
Solution a)
We note that the fundamental definition for current is I=
dq(t) dt
We can solve this differential equation, noting that I(0) = 0: Z I dt = q(t) −→ It + A = q(t) −→ q(t) = It b)
(5.2.18)
To find the electric field between the plates, we note that our plates have a (time dependent) surface charge density σ(t). Applying Gauss’ Law, I qenc E · dA = (−ˆ z) ε0 A σ(t)A EA = ε0
417
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
E=
1 q(t) (−ˆ z) ε0 πR2
(5.2.19)
From (5.2.19) and using (5.2.18), the electric field is given by E= c)
It (−ˆ z) ε0 πR2
(5.2.20)
We know that the displacement current density is denoted by Jd and is given by Jd = ε0
∂E ∂P + ∂t ∂t
(5.2.21)
The displacement current density is obtained simply by taking the time derivative of the electric displacement field D = ε0 E+P. In our case, we have no dielectric media so the last term in (5.2.21) is neglected. Using (5.2.20), Jd becomes It ∂ Jd = (−ˆ z ) ∂t ε0 πR2 Which is simple enough to evaluate. The result is: Jd = d)
I (−ˆ z) ε0 πR2
(5.2.22)
The magnetic field at a position r = R/2 between the plates can be computed using Amp´ere’s Law: I B · d` =µ0 Ienc In we are allowed to switch Ienc with Id since the magnetic field within the plates is the same as between the plates. We also note that the displacement current density Jd is related to the displacement current Id by Jd = Id /πR2 (in the following case R is actually r). So, with (5.2.22), (d) becomes: B2πr =µ0 Jd πr 2 I B2πr =µ0 (−ˆ z) πr 2 πR2 µ0 Ir B =− ˆ z 2πR2
(5.2.23)
If we recall that r = R/2, it is easy to see that (5.2.23) can be written as: B=− e)
µ0 I ˆ φ 4πR
(5.2.24)
The fundamental formulation for the Poyinting Vector is S=
1 E×B µ0
(5.2.25)
Inserting our previously tabulated values for E from (5.2.20) and B from (5.2.24), (5.2.25) becomes: 1 It µ0 I ˆ S= − ˆ z × − φ µ0 ε0 πR2 4πR
W. Erbsen
ELECTRODYNAMICS
Which can be rewritten as S=
I2t ρ ˆ 4ε0 π 2 R3
Problem 6 A sphere of radius R carries charge Q distributed uniformly over its surface. The sphere is rotated at a constant angular velocity ω around the z-axis. Calculate the magnetic dipole moment. Find the magnetic field at the point (x, 0, 0) where x R.
Solution The integral form of the magnetic dipole moment is Z Z µ = I dA = A dI
(5.2.26)
We also recall that the definition of current is I = dQ/dt, but if we want the differential element of the amount of current over some specified length of time (the period, T ), this becomes dI = dQ/T . The differential charge element in our case since we have a surface charge density is dQ = σ dA. So, theoretically we just insert our values into (5.2.26) and integrate. we begin by looking for the charge element dQ for a thin strip of the sphere, rotating around at some radius r = a sin θ with some angular velocity ω. This rotating charge, of course, produces current, and in turn a magnetic dipole moment. The thickness of the stripe is a dθ, and so we have: Q 1 (5.2.27) dQ = σ dA = (a dθ) (2πa sin θ) = Q sin θ dθ 4πa2 2 Using (5.2.27) and remembering that the temporal period at some rotational frequency ω is T = 2π/ω, our differential current element becomes dI =
dQ ω 1 Qω = Q sin θ dθ = sin θ dθ 2π/ω 2π 2 4π
(5.2.28)
2
The loop area is A = πr 2 = π (a sin θ) , and the magnetic dipole moment becomes Z Z π Z Qω 4ωQa2 π 3 2 µ = AdI = π (a sin θ) ˆ z sin θ dθ sin θ = sin θ dθ ˆ z 4π 4 0 0 Straightforward integration yields Z 4ωQa2 π 1 − cos2 θ sin θ dθ ˆ z 4 0 Z π Z π 4ωQa2 = sin θ dθ − cos2 θ sin θ dθ ˆ z 4 0 0
µ=
(5.2.29)
419
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
4ωQa2 4 4ωQa2 = 4 4ωQa2 = 4 =
1 3 π cos θ 0 ˆ z 3 1 − (−1 − 1) + (−1 − 1) ˆ z 3 4 ˆ z 3 π
[− cos θ| 0 +
From which we conclude that µ=
ωQa2 ˆ z 3
(5.2.30)
To calculate the field B, we recount the following equation, the bastard love-child of the Biot-Sevart law: Bdip (r) =
µ0 1 [3 (µ · ˆ r) ˆ r − µ] 4π r 3
(5.2.31)
In our case we use ˆ r=x ˆ, and with (5.2.30), (5.2.31) becomes B(x) =
µ0 1 ωQa2 [3 (ˆ z·x ˆ) x ˆ−ˆ z] 4π x3 3
Where we note that ˆ z·x ˆ = 0, so we finally have B(x) = −
µ0 ωqa2 ˆ z 12πx3
Problem 8 Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a dielectric oil (dielectric constant ε, density ρ). The inner one is maintained at potential V and the outer one is grounded. To what height does the oil rise in the space between the tubes? (Hints: Find the electric field, then the electrical energy and then the force).
Solution To find the maximum height that the oil reaches, we must think about what forces are acting on the fluid. The force that is lifting the oil upwards is due to the capacitance between the two cylinders. The motion of the fluid stops when this force is exactly cancelled by the opposing force due to gravity. We can, then start from the end and work towards the beginning. Let’s say the fluid is currently at its maximum height, h. Then the mass density can be used as follows: ρ=
MASS m = −→ m = ρπ b2 − a2 h 2 2 VOLUME (πb − πa ) h
The force acting downward on the center of mass is then
W. Erbsen
ELECTRODYNAMICS
ρgπ b2 − a2 h ˆ z Fg = 2
(5.2.32)
Where the factor of 2 in the denominator stems from the fact that we are interested in the average height. We also know that (5.2.32) must be equal to the force pulling up on the fluid. To find out what this opposing (lifting) force is, recall that 1 dC F = − V2 ˆ z 2 dz At this point we take a break to calculate what the capacitance of our system is. In general, I Q 1 Q Qenc −→ E2πrL = −→ E = ˆ r E · dA = ε0 ε0 2πε0 rL A And from (5.2.34) we now calculate the potential: Z a Z 1 Q a1 1 Q V =− E ·ˆ rdr = − ˆ r·ˆ rdr = log b/a 2πε0 L b r 2πε0 L b
(5.2.33)
(5.2.34)
(5.2.35)
Again using the handy relation Q = CV , (5.2.35) becomes C=
Q 2πε0 L = V log ( b/a )
(5.2.36)
What we have just found in (5.2.36) is the general relation for the capacitance of a cylindrical capacitor. In our case, part of the capacitor is filled with the dielectric, and the other with oil, so the total capacitance would just be the sum of the two: 2πε0 (L − z) log ( b/a) 2πε0 z (1 + χ) 2πε0 zK1 = = b log ( /a ) log ( b/a )
Cair = Coil So that the total capacitance is C = Cair + Coil =
2πε0 (L − z) 2πε0 z (1 + χ) 2πε0 + = (L + zχ) b b log ( /a ) log ( /a ) log ( b/a )
At this point we substitute (5.2.37) into (5.2.33): 1 2 d 2πε0 V 2 χπε0 F=− V − (L + zχ) ˆ z = ˆ z 2 dz log ( b/a ) log ( b/a ) z=h And now we set (5.2.32) equal to (5.2.38):
ρgπ b2 − a2 h V 2 χπε0 ˆ z= ˆ z 2 log ( b/a ) If we solve for the height h we finally arrive at h=
2V 2 (r − 1) ε0 ρg (b2 − a2 ) log ( b/a)
(5.2.37)
(5.2.38)
421
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
The factor of 1/2 comes from the fact that h represents the average height of the oil, which is half-way up the vertical height.
Problem 9 A cylindrical capacitor consists of two long, concentric tubes of sheet metal of radii R1 and R2 , respectively. The space between the tubes is filled with a dielectric constant ε. a)
Find the capacitance of this capacitor.
b)
Suppose the potential difference between the shells is V0 , find the electrostatic energy.
Solution a)
The typical equation relating to capacitance is Q = CV → C = Q/V . We can firmly say that if the total height of the capacitor is L, and if the plates have a line charge density λ, then the total charge is Q = λL. We now calculate the potential, Z R1 V =− E · dr (5.2.39) R2
We now find E as follows E=
1 Q λL 1 = 4πε0 ε r 4πε0 ε r
(5.2.40)
Using (5.2.40), we can now calculate V from (5.2.39): Z R1 1 λL λL R2 λL R1 λL dr = − (log [R2 ] − log [R1 ]) = − log = log V =− 4πε0 ε R2 r 4πε0 ε 4πε0 ε R1 4πε0 ε R2 (5.2.41) From (5.2.41) we can finally say that C= b)
λL 2π0 L h i = V 2 log R R1
(5.2.42)
We recall that the equation for the electrostatic energy is U = 1/2 CV 2 . From (5.2.41) and (5.2.42), 2 1 4πε0 ε λL R2 h i U= log 2 log R2 4πε0 ε R1 R1
2 2πε0 ε λ2 L2 R2 h i = log 2 2 2 16π ε0 ε R1 2 log R R1
From which it is easy to arrive at
W. Erbsen
ELECTRODYNAMICS
U=
λ2 L2 R2 log 4πε0 ε R1
Problem 11 A long wire is bent into a hairpin-like shape shown in the figure. Find an exact expression for the magnetic field at the point P which lies at the center of the half circle.
Solution It is easy to show from Amp´ere’s Law that the magnetic fields produced by an infinite straight wire and an arc are, respectively: B=
µ0 I , 2πr
and
B=
µ0 Iφ 4πr
The two lengths of wire contribute each half of the magnetic field produced by an infinite straight wire, while the arc extends through an angle φ = π, so: µ0 I µ0 I µ0 I B= + −→ B = 2πR 4R 4R
2 +1 π
Problem 12 A copper wire with a diameter of 2 mm has a resistance of 1.6Ω for every 300 meters of length. It is carrying a current of 25 amps. Assuming that this DC current density is uniformly distributed over a cross section, a)
What is the numerical value of the resistivity ρ of the copper from the information given?
b)
Determine the electric field vector E, the magnetic induction vector B, and the Poynting vector S at the surface of the wire (be sure to state their directions and actual numerical values). µ0 /4π = 10−7 , 1/4πε0 = 9 × 109 in MKS units.
c)
How much energy (in Joules/meter) is stored in the electric field and in the magnetic field within the wire?
d)
Suppose the current is increasing at a rate of 1 Amp/s. What statement can be made about the radial dependence of the induced electric field that results within the wire?
Solution
423
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
The fundamental relation relating resistance and resistivity is: R=
RA ρL −→ ρ = A L
(5.2.43)
2 In our case, A = πr 2 = π 10−3 m , so that the resistivity is given by
(1.6Ω)(π(10−3m)2 ) −→ ρ = 1.67 × 10−8 Ω · m 300m To find the electric field vector E, we must employ Gauss’ Law: I 1 λ 1 λ 1 λ −→ E · 2πr = −→ E = E · d` = ε0 L ε0 L 2πε0 r ρ=
b)
(5.2.44)
(5.2.45)
We notice that the equaled part in (5.2.45) looks suspiciously similar to the definition of the scalar potential V . And we can go ahead and find the scalar potential using the given information: V = IR = (25A)(1.6Ω) = 40v. Using this information, and manipulating (5.2.45), we have 1 λL 2 2V = (5.2.46) E= 4πε0 r L L Substituting the given values, (5.2.46) becomes: E=
2 v ˆ z 15 m
(5.2.47)
The magnetic field may similarly be found using Amp´ere’s Law: I µ0 I 2 × 10−7 · 25A B · d` =µ0 Ienc −→ B = = 2πr 10−3 m Evaluating this, we arrive at ˆ B = 5 × 10−3 T φ
(5.2.48)
Using (5.2.47) and (5.2.48), we can calculate the Poynting Vector S: S= c)
1 1 E×B = µ0 µ0
4 v ˆ −→ S = −1061 W ρˆ ˆ z × 3 × 10−3 T φ 15 m m2
(5.2.49)
Finding the total energy stored in the both the electric and magnetic fields entails recalling the following relations: ηE =
energy 1 = ε0 |E|2 , volume 2
and
ηB =
energy 1 1 = |B|2 volume 2 µ0
Let’s say that we are interested in the amount of energy in a 1m length of wire. Then the volume is V = πr 2 = π(10−3 m)2 (1 m) = 3.14 × 10−6 m3 . Accordingly, the amount of energy stored in the electric and magnetic fields is given by 2 A · s 2 v J 8.887 × 10 3.14 × 10−6 m3 −→ Energy in E = 2.473 × 10−19 V · m 15 m m 2 1 1 A·m ˆ 3.14 × 10−6 m3 −→ Energy in B = 1.562 × 10−5 J 5 × 10−3 T φ −7 2 4π10 V·s m 1 2
−12
W. Erbsen
ELECTRODYNAMICS
d)
Qualitatively, when an time-varying current is present in a conductor (eg, AC current), the current has a greater tendency to exist near the surface of the conductor (or on the “skin”). This is called the skin effect.
Problem 13 A battery with emf V , a resistor with resistance R, and a capacitor with initial capacitance C have been connected in series as shown for a very long time. The dielectric, with permittivity ε, occupies 1/2 the gap in the parallel plate capacitor. Now suppose that the dielectric is removed very quickly, in a time short compared to any relevant time constants. The new capacitance is C 0 . a)
Give an explanation of what happens to each circuit element. After a very long time, a new equilibrium is reached. In terms of C, C 0, R and V , give expressions for: i)
The net change in energy in the capacitor
ii)
b)
The net change in energy in the battery
iii)
The total energy dissipated in the resistor
iv)
The amount of work that was done to remove the dielectric
In terms of ε, what is the ratio C/C 0?
Solution a)
We first recall that the capacitance for a parallel plate capacitor is given by C = εA/d, where A is the surface area of the plates and d is the distance between them, while ε corresponds to whatever material lies between. To find the capacitance of the capacitor before the dielectric is removed, we recognize that since the dielectric does not occupy the entire interstitial space between the plates we essentially have two capacitors in series, Cair and Cdia . Mathematically, these are (respectively) Cair =
ε0 A 2ε0 A = , d/ d 2
The total capacitance is then just
and
Cdia =
εA 2εA = d/ d 2
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
C=
1 1 CairCdia (2ε0 A/d)(2εA/d) 2εε0 A + = = = Cair Cdia Cair + Cdia 2ε0 A/d + 2εA/d d(ε + ε0 )
(5.2.50)
While after the dielectric is removed, the capacitance is simply C0 = i)
ε0 A d
The amount of energy in the capacitor decreases. Adding dielectrics between the plates of a capacitor act to decrease the magnitude of the electric field, but increase the capacitance (and therefore the energy stored) through the polarization of the static dielectric atoms. To calculate this change, we just have to remember that the amount of energy in a capacitor is just U = 1/2 CV 2 . So, ∆Uc = 1/2 C 0 V 2 − 1/2 CV 2 −→ ∆Uc = 1/2 V 2 (C 0 − C)
ii)
(5.2.52)
When the dielectric is removed, there is a drop in capacitance and charge flows from the plates of the capacitor towards the battery. The change in energy of the circuit occurs due to thermal losses in the resistor; the potential energy of a charge in a potential is U = QV , so the change in energy is then ∆Ub = Q0 V − QV = C 0 V 2 − CV 2 −→ ∆Ub = (C 0 − C) V 2
iii)
(5.2.51)
(5.2.53)
As mentioned previously, the total amount of energy dissipated in the circuit is done through the resistor. The basic equation to calculate this is P = I 2 R, however all we have to do is take the difference of (5.2.52) and (5.2.53): ∆Ur = ∆Uc − ∆Ub = 1/2 V 2 (C 0 − C) − (C 0 − C) V 2 −→ ∆Ur = 1/2 V 2 (−C 0 + C)
(5.2.54)
The ratio of the capacitances before and after the dielectric is removed is C C 2εε0 A d 2ε = · −→ = C0 d(ε + ε0 ) ε0 A C0 ε + ε0
Problem 14 A series LCR circuit with L= 2 H, C=2 µF, and R=20 Ω is driven by a generator with maximum emf of 100 V and variable frequency. Find the resonant frequency ω0 and the phase φ and maximum current Imax when the generator angular frequency is ω = 400 s−1 .
Solution √ We first recall that for a resonant RLC circuit, we have Z = R, ω = 1/ LC , XC = XL , and φ = 0. This is not the case, however; the first thing we want to do is find the resonant frequency ω0 of the circuit: 1 1 rad ω0 = √ = p −→ ω0 = 500 −6 sec LC (2 H)(2 × 10 F)
(5.2.55)
W. Erbsen
ELECTRODYNAMICS
In order to find the phase angle, we must first recall that XC = 1/ωC, XL = ωL, and tan (φ) = (XL − XC ) /R. Using these, we have −1
φ = tan
ωL − R
1 ωC
−1
= tan
"
(400 s−1 )(2 H) −
1 (400 s−1 )(2×10−6 F)
20 Ω
The maximum current occurs at resonance, where R = Z = with this condition, and also the result from (5.2.55), Imax =
V V = q Z R2 + ωL −
−→ Imax = 0.222 A
1 ωC
2 = r
q
#
−→ φ = −87.46o 2
R2 + (XL − XC ) . Using Ohm’s Law 100 V
(20 Ω)2 + (400 s−1 ) (2 H) −
1 (400s−1 )(2×10−6
F)
2
Problem 15 A grounded conducting plane is connected to a conducting sphere with radius a through a battery of voltage V0 . The sphere is a distance d a above the plane. a)
Find the charge on the sphere.
b)
Find the force between the plane and the sphere.
Solution a)
Since we are told that d a, it is fair to go ahead and assume that the sphere can be thought of as isolated; and since it is connected to a battery of potential V0 , we could go ahead and assume that this is the potential of the sphere. The standard capacitance of a conducting sphere is C = 4πε0 r
(5.2.56)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Where in our case r = a. Furthermore, we know that Q = CV , so the total charge on our sphere is just Q = 4πε0 aV0 b)
(5.2.57)
The force between the sphere and the grounded conducting plane can be calculated via the method of images, where we imagine an identical, albeit oppositely charged sphere a distance 2d away. Using (5.2.57) and Coulomb’s law, we have 2
F=
1 (4πε0 aV0 ) 1 Q2 = 2 4πε0 (2d) 4πε0 4d2
Which can be readily simplified to F =
πε0 a2 V02 d2
Problem 17 A conducting sphere of radius a and total charge Q is surrounded by a spherical shell of dielectric material (with permittivity ε) of inner radius a and outer radius b. Find the electrostatic energy of the system.
Solution The general form for the energy stored in any electrostatic system is Z ε W = E2 dτ 2
(5.2.58)
In our case, the electric field of the conducting sphere is given by E=
1 Q ˆ r 4πε r 2
Inserting this value for E into (5.2.58), we have 2 ZZZ Z 2π Z π Z ∞ 1 Q 1 1 ε 2 W = ˆ r dV = Q · r 2 sin θ drdθdϕ 2 2 2 2 4πε r 2 16π ε r4 0 0 0
(5.2.59)
(5.2.60)
The radial integral in (5.2.60) must be divided according to each section: from 0 → a we are inside the conductor, so E = 0. However from a → b and b → ∞ we must evaluate the integral, and (5.2.60) becomes: " Z # Z 2π Z π Z ∞ Q2 ε20 b 1 1 W = dϕ sin θ dθ 2 dr + dr 32π 2 ε20 0 ε a r2 r2 0 b
W. Erbsen
ELECTRODYNAMICS
" b ∞ # Q2 ε20 1 1 = (2π)(2) 2 − + − 32π 2 ε20 ε r a r b 2 2 Q ε0 1 1 1 = − + + 2 2 8πε0 ε b a b From which is may be readily deduced that Q2 W = 8πε2
1 1 − a b
+
Q2 1 8πε20 b
Problem 18 The small loop of wire (radius a and resistance R) falls under gravity towards the larger loop (radius A), which has a constant current I. The small loop is contrained to move along the axis of the large loop and remains parallel to the large loop.
a)
Explain (in words) what happens to the small loop.
b)
For a particular height h and velocity v, what is the induced emf in the small loop? Assume a A.
c)
What is the electromagnetic force acting on the small loop?
d)
Does the system radiate? If so, describe the radiation pattern qualitatively (e.g., frequency, direction, polarization).
Solution a)
Words words words
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
The emf induced on the smaller loop will be due to the change of flux passing through it, changing as it falls: I dΦB d E=− =− B · dA (5.2.61) dt dt If we assume that the field is homogeneous throughout the smaller loop, we can expand on (5.2.61) as follows E=−
d B · πa2 dt
(5.2.62)
At this point all we have to do is find B. The best way to do this is with the Biot-Sevart Law, but before we do that we should define some needed quantities. d`, the length element of the source loop (the bigger one) is A dφ. Furthermore, the radial √ vector r is found by subtracting by boring trig stuff: r = z ˆ k − Aρ. ˆ The magnitude of r is just z 2 + A2 . Using all this in the Biot-Sevart Law: B= =
µ0 I d` × r 4π |r|3 ˆ × (z ˆ µ0 I A dφφ k − Aρ) ˆ 4π
(z 2 + A2 ) z ρˆ + Aˆ z
3/ 2
µ0 IA dφ 4π (z 2 + A2 ) 3/2 Z 2π Z 2π µ0 IA 1 = z ρ ˆ dφ + A ˆ z dφ 4π (z 2 + A2 ) 3/2 0 0 µ0 I A = [0 + 2πA] ˆ z] 4π (z 2 + A2 ) 3/2 =
=
µ0 I A2 ˆ z 2 (z 2 + A2 ) 3/2
We can now use (5.2.63) in (5.2.62) to find E: " # µ0 I A2 2 d E = − πa dt 2 (z 2 + A2 ) 3/2 " # 1 πµ0 Ia2 A2 d dt dz =− 2 dt dz dt (z 2 + A2 ) 3/2 " # πµ0 Ia2 A2 d 1 =− z˙ 2 dz (z 2 + A2 ) 3/2 " 3z πµ0 Ia2 A2 d =− z˙ − 5 /2 2 2 2 dz (z + A ) z=h
(5.2.63)
(5.2.64)
By evaluating (5.2.64) at z = h, and recognizing that z˙ is the downward velocity, we have
E=
3hµ0 Iπa2 A2 ·v 5 2(h2 + A2 ) /2
(5.2.65)
W. Erbsen
ELECTRODYNAMICS
c) d)
Not too sure on this one Yes , the system does radiate, in the ρ ˆ direction.
Problem 19 A cylindrically shaped conductor has length L, radius a, and conductivity σ. It carries a uniform time-independent current density J = Jˆ e0 , parallel to its long axis. a)
Determine the electric and magnetic fields within the conductor.
b)
Calculate the Poynting vector S within the conductor.
c)
Write out Poynting’s theorem for this situation. Verify that it is satisfied for any point within the conductor.
Solution a)
We first recall the fundamental relation for current density: J = σE. So, the electric field is given simply by
E=
J/σ ˆ z for 0≤ρ≤a 0 elsewhere
(5.2.66)
And to find the internal magnetic field B, we apply Amp´ere’s Law: I B · d` =µ0 Iinc = µ0 πρ2 J 2πρB =µπρ2 J
Such that we have
B= b)
ˆ for µ0 Jρ/2 φ 0≤ρ≤a 0 elsewhere
(5.2.67)
We note that the Poynting vector takes the form S=
1 E×B µ0
And since we know that E = J/σ ρˆ from (5.2.66) and B = µ0 Jρ/2 ρ ˆ from (5.2.67), then the Poynting vector is simply
S=
J 2ρ (−ρ) ˆ 2σ
(5.2.68)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
c)
Poynting’s theorem states that ∂u +∇·S =−J·E ∂t Where u is the energy density. We can express the energy density as: B2 1 2 ε0 E + u= 2 µ0 Substituting (5.2.66) and (5.2.67) into (5.2.70), J2 µ20 J2 ρ2 1 ε0 2 + µ0 u= 2 σ 4 2 2 2 1 J µ0 J ρ = ε0 2 + 2 σ 4
(5.2.69)
(5.2.70)
(5.2.71)
And we note that u is time independent, so ∂u/∂t = 0. We also note that the divergence in cylindrical coordinates takes the form ∇·F =
1 ∂ 1 ∂Fϕ ∂Fz (ρFρ ) + + ρ ∂ρ ρ ∂ϕ ∂z
So finally we can calculate 1 ∂Sϕ ∂Sz 1 ∂ (ρSρ ) + + ρ ∂ρ ρ ∂ϕ ∂z 1 ∂ = (ρSρ ) ρ ∂ρ 2 J ρ 1 ∂ = ρ (−ρ) ˆ ρ ∂ρ 2σ 1 J2 ρ = (−ρ) ˆ ρ σ J2 =− ρ ˆ σ
∇·S=
(5.2.72)
So, using (5.2.72), Poynting’s theorem (5.2.69) is now −
J J2 J2 ρˆ = −J · E = −J ρˆ = − ρˆ X σ σ σ
Problem 21 An electron beam of uniform charge density ρ0 inside the beam radius r0 is traveling on the axis of a metal tube with an inner radius Ri and an outer radius of Ro which is kept at a potential φτ . a)
Calculate the potential at the center of the electron beam.
b)
Calculate the net charge on the tube per unit length `.
W. Erbsen
c)
ELECTRODYNAMICS
Calculate the energy that is required to double the tube potential to 2 × φτ assuming that the electron beam charge density remains constant.
Solution a)
Calculating the potential at the center of the beam poses no major obstacles. We imagine to be dragging a unit charge (“test” charge) from ∞ to the center of the beam . The total potential is how much work it took. So we start from the outside, from ∞, and boundary by boundary calculate the potentials and at the end add them together. Z 0 φ(r = 0) = − E · ds ∞ Ro
=−
Z
|∞
E4 · ds − {z } φτ
Z
Ri
Ro
E3 · ds −
Z
Ra
Ri
E2 · ds −
Z
0
E1 · ds
(5.2.73)
a
We are told that the potential of the shell is kept at φτ , so we needn’t even calculate E4 . The rest are ρ0 (πr 2 ) ρ0 ˆ r= rˆ r 2πε0 2ε0 ρ0 (πr02 ) ρ0 r02 E2 = ˆ r= ˆ r 2πε0 r 2ε0 r E3 =0
E1 =
(5.2.74a) (5.2.74b) (5.2.74c)
The reason why E3 = 0 is because this region corresponds to within a conductor, where the internal electric field is guaranteed to be zero. Using (5.2.74a), (5.2.74b) and (5.2.74c), we can evaluate our potential from (5.2.73): Z Ra Z 0 ρ0 r02 ρ0 φ(0) =φτ − ˆ r · dr − rˆ r · dr 2ε r 2ε 0 0 Ri a Z Z 0 ρ0 r02 Ro 1 ρ0 =φτ − dr − r dr 2ε0 Ri r 2ε0 a 0 ρ0 r02 ρ0 r 2 Ro =φτ − [log r|Ri − 2ε0 2ε0 2 a 2 ρ0 r0 Ri ρ0 a2 =φτ − log + (5.2.75) 2ε0 Ro 2ε0 2 From (5.2.75) it is easy to see that the potential is ρ0 r02 Ri ρ0 a2 φ(0) = φτ + log + 2ε0 Ro 4ε0 b)
The net charge on the tube per unit length would just be the charge density times the area in question where we take the length to be equal to 1, λ = −ρ0 πa2
433
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 22 A charge 2q is located at (0, 0, 0), a second charge, −q, is located at (0, −a, 0), and a third charge, −q, is located at (0, 0, −a). a)
Calculate the monopole and dipole moments.
b)
Calculate all quadrupole moments.
c)
What is the potential for r a using multipole approximation?
Solution a)
The monopole moment is zero, since it is just the sum of the charges. The dipole moment for an array of charges is defined by p=
N X
ri q i
(5.2.76)
i=1
Applied to our geometric arrangement of charges, ˆ ˆ p = (−aˆj)(−q) + (0)(2q) + (−ak)(−q) −→ p = aq(ˆj + k) b)
(5.2.77)
The defining equation for the quadrupole moment is ∞
1X Q0 = qi 3xiα xiβ − ri2 δαβ 2 i=1
(5.2.78)
1 1 1 Q0 = − (a2 )(−q) + (0)(−2q) − (a2 )(−q) −→ Q0 = a2 q 2 2 2
(5.2.79)
Applying (5.2.78) to our case,
c)
The full equation for multipole expansion is Z ∞ 1 X 1 V (r) = (r 0 )n Pn (cos θ0 )ρ(r0 ) dτ 0 4πε0 n=0 r (n+1) Z Z Z 1 1 1 3 1 1 0 0 0 0 0 0 0 2 2 0 0 0 = ρ(r ) dτ + 2 r cos θ ρ(r ) dτ + 3 (r ) cos θ − ρ(r ) dτ + ... 4πε0 r r r 2 2 (5.2.80) The integrals represent the various -pole terms through quadrupole. We know that the monopole term is zero, so we simply substitute (5.2.77) and (5.2.79) into (5.2.80): 1 V (r) ≈ 4πε0
"
ˆ aq(ˆj + k) a2 q + 2 2 r r
#
W. Erbsen
ELECTRODYNAMICS
Problem 23 Consider a nonconducting sphere of radius R with a uniform charge density ρ, except for a cavity of radius a, a distance b > a from the center. a)
Find the electric field at the center of the cavity.
b)
Find the voltage potential (V (∞) = 0) at the center of the cavity.
Solution a)
We are given that the dielectric object has a charge density equal to ρ; and since the geometry of the problem is simple enough, from this we can readily calculate the total charge Q: Q Q CHARGE = 4 = 4 −→ Q = 4/3 πρ R3 − a3 (5.2.81) ρ= VOLUME /3 πR3 − 4/3 πa3 /3 π (R3 − a3 ) And now we can apply Gauss’ Law to find the field E: E4πb2 =
Qenc 1 4 −→ E = ρ πb3 2 ε0 4πb 0 3
Which allows us to say E=
ρb ˆ r 3ε0
In order to find the voltage at the center of the cavity, we must imagine that we are dragging a unit charge to the center of the cavity from infinity. We do this in steps, addressing each boundary individually: Z a Z R Z a+b Z a V (r) = − E · ds = − E1 · ds − E2 · ds − E3 · ds (5.2.82) ∞
∞
R
a+b
The respective electric fields are E1 =
ρ(r 3 − a3 ) ρ(R3 − a3 ) ˆ r = ˆ r 3ε0 r 2 3ε0 r 2
(5.2.83a)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
ρ(r 3 − a3 ) ˆ r 3ε0 r 2 ρb ˆ r E3 = 3ε0
E2 =
(5.2.83b) (5.2.83c)
We now substitute (5.2.83a)-(5.2.83c) into (5.2.82) and integrate. I will do each of the three integrals in turn and combine them later. The first is Z ρ(R3 − a3 ) R 1 dr V1 (r) = 2 3ε0 ∞ r R ρ(R3 − a3 ) =− r 3ε0 ρ(R3 − a3 ) =− 3ε0 R
∞
(5.2.84)
And now for the second, Z
a+b
r 3 − a3 dr r2 R "Z # Z a+b a+b 1 ρ 3 = rdr − a dr 3ε0 R r2 R " a+b a+b # ρ r 2 1 3 = −a − 3ε0 2 R r R ρ (a + b)2 R2 1 1 − + − = 3ε0 2 2 a+b R
V2 (r) =
And that last little bitch is
ρ 3ε0
Z a ρb dr 3ε0 a+b ρb = [a − (a + b)] 3ε0 ρb2 =− 3ε0
(5.2.85)
V3 (r) =
We now take (5.2.84) - (5.2.86) and substitute it into (5.2.82): ρ(R3 − a3 ) ρ (a + b)2 R2 1 1 ρb2 V (r) = + − + − + 3ε0 R 3ε0 2 2 a+b R 3ε0 Which finally leads us to ρ (R3 − a3 ) (a + b)2 R2 1 1 2 V (r = 0) = + − + − +b 3ε0 R 2 2 a+b R
Problem 24
(5.2.86)
W. Erbsen
ELECTRODYNAMICS
A particle of charge +q and mass m has a velocity of V = (0, vy , vz ) when at a position of (x, 0, 0). There is a magnetic field of B = (0, 0, B). a)
Find an expression for the cyclotron period, i.e. the time to go once around a circular orbit.
b)
Find the position of the charge after one cyclotron period.
Solution a)
The cyclotron period can be found by: FC = Fm −→ mω2 r = qvB −→ ω =
qB m
So the period is τ= b)
2π 2πm −→ τ = ω qB
Most generally, we would approach the problem like this F = m¨ r = qv × B Which leads us to (¨ xx ˆ + y¨y ˆ + z¨ˆ z) =
qB [x(−ˆ ˙ y ) + yˆ ˙ x] = ω [yˆ ˙ x − xˆ ˙ y] m
This equation may be separated very easily; the resulting coupled differential equations are x ¨ =ωy˙ y¨ = − ωx˙ z¨ =0 We begin by integrating (5.2.87a) with respect to t:
(5.2.87a) (5.2.87b) (5.2.87c)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
x˙ = ωy + C(vx )
(5.2.88)
y¨ = −ω2 y − ωC(vx )
(5.2.89)
Substituting (5.2.88) into (5.2.87b),
Our initial conditions state that at t = 0 then vx = 0, so C(vx ) in (5.2.89) vanishes. We are left with a second order differential equation, with solutions of the form y(t) = A cos ωt + B sin ωt If we again apply our initial conditions, we note that at t = 0 then y = 0, so we must require that A = 0. Our solution is then y(t) = B sin ωt
(5.2.90)
We now substitute (5.2.90) into (5.2.88): x˙ = ωyB sin ωt −→ x(t) = −B cos ωt + C(x) −→ x(t) = −B cos ωt + x0
(5.2.91)
Where we note that initial conditions imply that C(x) is none other than the initial position of our particle. Additionally, we can unleash the true identity of the integration constant B everpresent in both (5.2.90) and (5.2.91). We do this by taking the derivative of (5.2.90) and setting it equal to vy : vy y(t) ˙ = Bω cos 0 = vy −→ B = ω And now we can rewrite (5.2.90) and (5.2.91): x(t) = − y(t) =
vy cos ωt + x0 ω
vy sin ωt ω
(5.2.92a) (5.2.92b)
And now, after one full period, t → τ , which implies that ωτ = 2π, it can be easily (very easily) seen that (5.2.92a) and (5.2.92b) reveal the location of the particle to be x(τ ) = x0 −
vy , ω
y(τ ) = 0
Problem 25 A long wire lies along the z-axis. A square coil of wire, each side of length a, lies in the xz-plane. The side nearest the wire is parallel to the wire and located at x = a. a)
Find the vector potential A if the wire is carrying a current I and there is no current in the loop.
b)
What is the flux through the loop, using the results from part a)?
c)
What is the mutual inductance of the loop and on the wire?
W. Erbsen
ELECTRODYNAMICS
d)
If the current I is varying at a rate dI/dt = K, what is the induced emf in the coil?
e)
If there is a current in the loop varying at a rate dI/dt = C, what is the size of the voltage across the ends of the long wire?
Solution a)
In order to find the magnetic vector potential A, we this by first setting up the determinant, ρˆ ˆ φ 1 ∂ ∂ ∂ρ ρ ∂φ Aρ Aφ
must “uncurl” it from B = ∇ × A. We do ˆ z ∂ ∂z Az
Due to the geometry of this particular problem, since I flows along the ˆ z direction and the face of the loop is adjacent aligned to the wire, we can make some shortcuts. We know that A only flows in the ˆ z direction in our case, so the other terms are zero. Additionally, we can see that A will depend only on ρ, so any derivatives of A that aren’t with respect to ρ will be zero. ρ ˆ z ˆ φ ˆ ∂ ∂ ˆ 0 0 = − Az φ ∂ρ ∂ρ 0 0 Az So, we now have
B=∇×A=−
∂ ˆ = µ0 I Az φ ∂ρ 2πρ
(5.2.93)
The differential equation in (5.2.93) can be solved straightforwardly, ∂ µ0 I ˆ µ0 I a ˆ Az φ = φ −→ Az = log ∂ρ 2πρ 2π ρ b)
To find the flux, we use the typical definitions: Z Z Z Z Z a µ0 I 2a 1 µ0 I ΦB = B · dA = (∇ × A) · dA = dρdz = dρ dz 2πρ 2π a ρ 0
(5.2.94)
(5.2.95)
The integration in (5.2.95) is straightforward, and the result is ΦB = c)
µ0 Ia log [2] 2π
(5.2.96)
If we recall that the mutual inductance is qualitatively defined as M = ΦB /I, then the mutual inductance is simply: M=
µ0 a log [2] 2π
439
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
d)
Using (5.2.96), we can straightforwardly find the emf to be E=−
dΦB d µ0 Ia µ0 a dI =− log [2] = − log [2] dt dt 2π 2π dt
And using the handy substitution the prompt suggests, E=− e)
Kµ0 a log [2] 2π
With the given substitution, the voltage across the ends of the long wire is E=−
Cµ0 a log [2] 2π
Problem 26 A disc carrying a uniform surface charge density σ has radius a and rotates at an angular frequency ω. This disc lies in the xy-plane and rotates along the z-axis. A second identical disc rotating at the same frequency is located with its center on the z-axis at Z, but this one lies in the yz-plane about an axis parallel to the x-axis. Find the torque on the second disc. Indicate in a sketch the direction of the torque and the sense of rotation of the two discs.
Solution We begin by recanting a rather unfamiliar form of Biot-Sevart’s Law: dB =
µ0 KdS × r 4π |r|3
(5.2.97)
Where K is known as the surface charge density, and is described by K = σv. Similar to what we did in a previous problem, we define all the elements needed to compute the magnetic field from (5.2.97). The line element and its magnitude are, respectively p r = zˆ z − ρρ, ˆ and |r| = z 2 + ρ2 And we can also expand upon the surface charge density:
ˆ K = σv = σωρφ And we also recall that that the volume element for cylindrical coordinates is ds = ρ dρdφ. We can now calculate, Kds × r = (σωρρ) ˆ (ρ dρdφ) × (zˆ z − ρρ) ˆ 2 ˆ × (zˆ =σωρ dρdφφ z − ρρ) ˆ
W. Erbsen
ELECTRODYNAMICS
=σωρ2 dρdφ (z ρˆ + ρˆ z)
(5.2.98)
We can now substitute (5.2.98) into (5.2.97), µ0 σω ρ2 (z ρˆ + ρˆ z) dρdφ 3/ 4π (z 2 + ρ2 ) 2 ZZ 2 µ0 σω ρ (z ρˆ + ρˆ z) B= dρdφ 3/ 2 2 4π (z + ρ ) 2 " Z # Z a a µ0 σω ρ2 ρ3 z = ˆ dρ + z dρ 3/ ρ 3/ ˆ 2 2 2 0 (z 2 + ρ2 ) 0 (z 2 + ρ2 )
dB =
(5.2.99)
The first integral in (5.2.99) is neglected, due to symmetry. The second integral can be found in tables, and the result is µ0 σω a2 + 2z 2 √ B= − 2z ˆ z (5.2.100) 2 z 2 + a2 Now, if we can reconcile with the fact that K = σv, then we can find the magnetic moment of the upper loop as follows Z a Z a Z a Z a Z a πσωa4 µ= I · dρ = K(πρ2 ) dρ = σv(πρ2 ) dρ = σωρ(πρ2 ) dρ = πσω ρ3 dρ = 4 0 0 0 0 0 (5.2.101) And if we recall that within this context τ = µ × B, then by using (5.2.100) and (5.2.101) we can find the torque:
τ=
πµ0 σ 2 ω2 a4 a2 + 2z 2 √ − 2z (−ˆ y) 8 z 2 + a2
Problem 27 The KSU ultra-short pulse, ultra-high intensity laser in the JRM Lab has λ = 800 nm, pulse width of 20 femtoseconds, pulse energy of 6 mJ, and can be focused to a spot size of 100 µm. a)
What is the intensity of the light in the focus?
b)
What average electric field produces this intensity?
c)
To how many atomic units of the intensity and electric field does this correspond?
Solution
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
Intensity is defined as the amount of energy per unit area per unit time and has units W/m2. Mathematically, I=
b)
ENERGY = (AREA) (TIME)
6 × 10−3 J 2 −→ I ≈ 3.8197 W/m 2 100×10−3 m π (20 × 10−13 s) 2
(5.2.102)
The relationship between the field and intensity is
1 I = cε0 |E0 |2 −→ |E0| = 2
r
2I cε0
(5.2.103)
Where we can plug in our value from (5.2.102) and also for the various constants to arrive at a final value. It should be noted however that I ∝ |E|2 . c)
Donno
Problem 28 A parallel plate capacitor is being charged with a current I. The capacitor is circular and has an area of πa2 . a)
What is the displacement current between the plates?
b)
Referring to the drawing below, what is H in regions 1, 2, and 3 for ρ < a. (Neglect fringing fields)
c)
From the boundary condition on Ht , where H is evaluated at r = ρ(< a), what are the values on the surface currents K on the face of each of the capacitor plates?
d)
Make a sketch of all the current flows and label them.
Solution a)
Most formally, the displacement current is found by differentiating the electric displacement field
W. Erbsen
ELECTRODYNAMICS
D
=
ε0 E
↓ Jd
↓ ∂E ε0 ∂t
=
+
+
P ↓ ∂P ∂t
We have no dielectric so the polarization term is neglected in our case. One thing that should definitely be remembered is that the electric field between the plates of a parallel plate capacitor is E = σ/ε0 . We are now in a place to find the displacement current Jd : ∂E ∂ σ ∂σ ∂ Q(t) = ε0 = = (5.2.104) Jd = ε0 ∂t ∂t ε0 ∂t ∂t πa2 ˙ If we recall that I(t) = Q(t), then (5.2.104) is Jd = b)
Recall that Amp´ere’s Law for H fields is
I
I ˆ z πa2
(5.2.105)
H · d` = Id
Within region 2, we can use (5.2.105) to find H: H2 (2πρ) =
I ˆ I ˆ φ −→ H2 = φ πa2 2πρ
(5.2.106)
And in regions 1 and 3, the value for H is simply H1,3 =
I ˆ φ 2πρ
(5.2.107)
Problem 29 A spherical conductor A contains two spherical cavities. The total charge on A itself is zero. However, there is a point charge qb at the center of one cavity, and qc at the center of the other. A considerable distance r away is another charge qd . What force acts on each of the four objects A, qb , qc, and qd ? Which answers, if any, are only approximate? If qb is displaced slightly away from the center of the cavity, will it experience a resistance force? (If so, what is it?)
Solution The forces on charges qb and qc themselves are, of course, zero. To calculate the force of qd acting on A, F=
1 qd (qb + qc ) 4πε0 r2
(5.2.108)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Which is the same as the net force of A acting on qd . We must recognize, however that (5.2.108) is an approximation. To get the exact answer you must employ the method of images with an image charge inside the sphere. The exact result turns out to be F=
1 qd (qb + qc) 1 q 2 Rr − 4πε0 r2 4πε0 (r 2 − R2 )2
Problem 30 A uniformily charged rod of length a resides along the z-axis as drawn. The linear charge density is λ. The Z a voltage potential at (x, 0, 0) is 1 λ √ V = dz (5.2.109) 2 4πε0 0 z + x2 a)
Expand the denominator of this expression for x z to two nonzero terms.
b)
Substitute this expansion into the expression for V , work with two integrals.
c)
Identify the two results as monopole, dipole, quadrupole, octapole or whatever and give an expression for the two moments.
Solution
W. Erbsen
ELECTRODYNAMICS
a)
The series expansion around z is √
1 1 1 z2 3z 4 5z 6 = p = − 3+ 5− − ... x 2x 8x 16x7 z 2 + x2 x z 2 /x2 + 1
So, expanded to the first two nonzero terms:
1 1 z2 √ = − 3 x 2x z 2 + x2 b)
(5.2.110)
Substituting (5.2.110) into (5.2.109), we have Z a 1 1 z2 V = λ − dz 4πε0 0 x 2x3 Z a Z a 1 1 1 2 λ = dz − 3 z dz 4πε0 x 0 2x 0 1 a a3 = λ − 3 4πε0 x 6x So the result is V =
c)
λ a λ a3 − 4πε0 x 24πε0 x3
(5.2.111)
The first term in (5.2.111) is monopole, because of the 1/x factor, and the second is quadrupole, since it has a 1/x3 term.
Problem 31 a)
Find the voltage potential at point P a distance b from the right end of a uniform line charge with length ` and total charge Q as drawn.
b)
Using this voltage find the x-component of the electric field at point P .
Solution a)
We first note that our object has a charge given by dq = λ d`, and the total charge is Q, and the total length is `, so we know that λ = Q/`. Furthermore, we know that E = −∇V −→ dV = −Edx. We can take this further: Z 0 λ 1 V =− dx 4πε0 −` (b − x) 1 Q 0 =− [log (b − x)|−` 4πε0 `
445
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
1 4πε0 1 =− 4πε0 =−
Q [log (b) − log (b + `)] ` Q b log ` b+`
So that the potential is V = b)
1 Q log 4πε0 `
b+` b
(5.2.112)
E can be calculated rather straightforwardly as follows ∂ Vx ˆ ∂x 1 Q ∂ b+` log x ˆ =− 4πε0 ` ∂b b 1 Q 1 − (b+`)/b =− x ˆ 4πε0 ` b+` 1 Q1 b−b−` =− x ˆ 4πε0 ` b b+`
E=−
From which we arrive at 1 Q E= 4πε0 b
1 b+`
x ˆ
Problem 32 A dipole consistent of equal and opposite charges ±q separated by x resides a distance d/2 above a conducting, grounded, infinite plane as drawn. a)
On the drawing, accurately draw the image charge(s).
b)
Find the force between the dipole and the plane.
c)
Use the binomial expansion to expand this force to order (x/d)2 when d x. With what power does the force depend on the distance d?
W. Erbsen
ELECTRODYNAMICS
Solution a)
Two charges are imagined to be symmetrically placed about the grounded conducting plane; the magnitudes of opposing charges are opposite.
b)
We use the following equation: F=
3(P · ˆ r)P0 + 3(P · ˆ r)P + 3(P · P0 )ˆ r − 15(P · ˆ r)(P0 · ˆ r)ˆ r r4
(5.2.113)
Which should probably be memorized. The vectors P and P0 represent the dipole moments of the charged, and uncharged dipoles (respectively). Their values are P = qxˆ x and P0 = qx(−ˆ x). The vector ˆ r represents the distance between the two moments r = 2d(−ˆ y). We can see that the majority of the terms in (5.2.113) vanish, F=
3(P · P0 ) ˆ r 3q 2 x2 3(qxˆ x · qx(−ˆ x)) ˆ r = −→ F = y ˆ 4 4 r r 16d4
(5.2.114)
Problem 33 a)
Two spheres initially uncharged are connected by a battery of voltage V . After the switch is closed, what is the charge on the larger sphere?
b)
What is the capacitance of the larger sphere?
Solution a)
When the switch is closed, the amount of charge that flows into the larger sphere (of radius r = b) can be found by V =
1 Q −→ Q = 4πεbV 4πε b
447
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
To find the capacitance of the larger sphere we recall that Q = CV , so: C = 4πεb
Problem 34 A circular loop of wire with radius a and electrical resistance R lies in the xy-plane. A uniform mgnetic field is turned on at time t = 0; for t > 0 the field is B 0 B(t) = √ ˆj + kˆ 1 − e−λt (5.2.115) 2 a)
Determine the current I(t) induced in the loop.
b)
Sketch a graph of I(t) versus t.
Solution a)
When the magnetic field is turned on, flux passes through the loop which causes current to flow. The amount of current is determined by how much flux passes through. We start from Ohm’s law, = IR. Faraday’s Law says, E=−
dΦB dt
(5.2.116)
Where the magnetic flux, ΦB , is proportional to the amount of field lines that pass through a given area. In our case, this is I πa2 B0 ˆ ˆ ΦB = B · dA = √ j + k 1 − e−λt (5.2.117) 2 Substituting (5.2.117) into (5.2.116) we can find E: d πa2 B0 ˆ ˆ √ E=− j + k 1 − e−λt dt 2
W. Erbsen
ELECTRODYNAMICS
=−
λπa2 B0 ˆ ˆ −λt √ j+k e 2
(5.2.118)
Using (5.2.118) and also Ohm’s Law, we arrive at λπa2 B0 ˆ ˆ −λt I(t) = − √ j+k e 2R b)
(5.2.119)
Looks like an exponentially decaying signal. The negative sign denotes the direction, the coefficients the magnitude.
Problem 35 The solenoid pictured below is long, and has n turns per unit length and cross section A. a)
If the current in the solenoid changes from I1 to I2 , how much charge passes through the resistor?
b)
If the current in the solenoid, as a function of timet t, is t I(t) = I1 e− /τ + I2 1 − e− /τ
(5.2.120)
what is the current IR (t) in the resistor?
c)
Sketch separately the I1 and I2 terms and the total I vs. t/τ for I1 = 1A and I2 = 2A. On a separate graph sketch IR (t) vs. t/τ .
Solution a)
For discrete changes in quantities, there are no need for integrals or derivatives, eg dt → ∆t. Keeping this in mind, E=−
∆ΦB ∆ ∆B µ0 n(I2 − I1 )A = − BA = − A=− ∆t ∆t ∆t ∆t
(5.2.121)
449
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
We also know Ohm’s Law, which states that E = IR, where I = ∆Q/(∆t). With this knowledge, (5.2.121) becomes ∆Q µ0 n(I2 − I1 )A R=− ∆t ∆t From which it is clear that the ∆t on the denominator cancels, and we are left with ∆Q = − b)
µ0 n(I2 − I1 )A R
To calculate the current, E=−
∂ ∂ ∂I(t) BA = − µ0 nIA = −µ0 nA ∂t ∂t ∂t
(5.2.122)
Recalling again Ohm’s Law, IR (t) can be found by combining (5.2.122) and (5.2.120): i t E µ0 nA ∂ h − t/τ µ0 nA I1 t I2 t IR (t) = =− I1 e =− − e− /τ + e− /τ + I2 1 − e− /τ R R ∂t R τ τ From which it is easily seen that IR (t) = −
5.3
µ0 nA − t/τ e [I2 − I1 ] Rτ
Modern Physics
Problem 1 13 Recently, 2000 rubidium atoms (87 atoms/cm3 , were ob37 Rb), which had been compressed to a density of 10 served to undergo a Bose-Einstein condensation as a function of temperature. This occurs when the deBroglie wavelengths overlap and the system forms a single macroscopic quantum state. Estimate the temperature at which Bose-Einstein condensation occurs for this system.
Solution We first recall that the deBroglie wavelength is given by λ=
h p
Where we are also told that N =2000 atoms ρ =1013 atoms · cm3
(5.3.1)
W. Erbsen
MODERN PHYSICS
So, we can say that the spacing is proportional to the deBroglie wavelength, and the spacing can be found by 2000 atoms = 1013 atoms · cm3 −→ L = L3
2000 atoms 101 3 atoms · cm3
1/3
We now wish to set (5.3.2) equal to the deBroglie wavelength. But first, we recall that √ p p = 2mE → p ∼ 2mkB T E ∼ kB T
(5.3.2)
(5.3.3)
Where putting (5.3.3) into (5.3.1) of course yields h λ∼ √ 2mkB T
(5.3.4)
Setting (5.3.4) equal to (5.3.2) yields
2000 atoms 13 10 atoms · cm3
1/3
h = √ 2mkB T
Solving for T leaves us with 1
h /2 T ∼ 2mkB
2000 atoms 1013 atoms · cm3
1/6
Problem 2 Rutile, TiO2 , is a tetragonal crystal a = 4.4923˚ A. The position of the Ti atoms in the unit cell are 0, 0, 0; 1 /2 , 1/2 , 1/2 and the position of the oxygen atoms are 0.31, 0.31, 0; 0.81, 0.19, 0.5; 0.69, 0.69, 0; 0.19, 0.81, 0.5. a)
What would be the d-spacing of the 310 planes of rutile?
b)
Using CuK x-radiation, λ = 1.54 ˚ A, what would be the diffraction angle from the (310) planes of Rutile?
c)
Assuming atomic scattering factors of fTi for the 2 titanium atoms and fO for the 4 oxygen atoms, what is the crystalline scattering factor from the (310) planes of TiO2 ?
Solution We first note that: a =4.4923 ˚ A Ti =
0, 0, 0 1 /2 , 1/2 , 1/2
451
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
0.31, 0.31, 0 0.81, 0.19, 0.5 O= 0.69, 0.69, 0 0.19, 0.81, 0.5 a)
In order to find the d-plane spacing, we recall that 1 h2 k2 `2 = 2 + 2 + 2 2 d a b c
(5.3.5)
Where h, k and ` correspond to the the indices of the plane (eg h = 3, k = 1, and ` = 0). Furthermore, the numbers a, b, and c correspond to the sides of the crystal. In our case a = b = c → a. Taking all this into account, (5.3.5) becomes 1 h2 k2 1 1 = + −→ d = a + (5.3.6) d 2 a2 a2 h k Plugging into (5.3.6) the appropriate values, we have 1 1 d = 4.4923 ˚ A + −→ d = 1.42 ˚ A 32 1 b)
(5.3.7)
We recall that the diffraction equation from a periodic structure is given by nλ = 2d sin θ
(5.3.8)
Where integer multiples of λ yield maxima. Since we are interested in the first maxima, then n = 1 in our case, and after applying this to (5.3.8) and solving for θ, we have λ −1 θ = sin (5.3.9) 2d
c)
Putting in the correct numbers into (5.3.9) leaves us with " # 1.54 ˚ A −1 −→ θ = 32.8o θ = sin 2 1.42 ˚ A
In order to find the crystalline scattering factor for our lattice, we employ the following equation for reasons unknown: X Fhk` = fj exp [iφhk` (j)] (5.3.10) j
Where φhk` (j) = 2π (hxj + kyj + `zj ) Substituting (5.3.11) into (5.3.10) yields X Fhk` = fj exp [i2π (hxj + kyj + `zj )]
(5.3.11)
(5.3.12)
j
In our case, ` is always zero, so we can make our lives easier by ignoring the last term in (5.3.14): X Fhk` = fj exp [i2π (hxj + kyj )] j
W. Erbsen
MODERN PHYSICS
And we also know what h and k are, so let’s go ahead and put them in too: X F310 = fj exp [i2π (3xj + yj )]
(5.3.13)
j
Completing (5.3.10) and (5.3.11) for Titanium is the easiest, so let’s do that sum first: Ti F310 =fTi exp [i2π (3 · 0 + 0)] + fTi exp i2π 3 · 1/2 + 1/2 =fTi [1 + exp [i4π]]
(5.3.14)
Now, let’s do the same for Oxygen: O F310 = fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] {z } | {z } | {z } | {z } | I
II
III
IV
So let’s do these one at a time..
I =fO exp [i2π (3 · 0.31 + 0.31)] ⇒ fO exp [2.48πi] II =fO exp [i2π (3 · 0.81 + 0.19)] ⇒ fO exp [5.24πi] III =fO exp [i2π (3 · 0.69 + 0.69)] ⇒ fO exp [5.52πi] IV =fO exp [i2π (3 · 0.19 + 0.81)] ⇒ fO exp [2.76πi]
(5.3.15a) (5.3.15b) (5.3.15c) (5.3.15d)
Combining (5.3.15a)-(5.3.15d) back into our original equation, O F310 =fO [exp [2.48πi] + exp [5.24πi] + exp [5.52πi] + exp [2.76πi]]
(5.3.16)
And now, combining (5.3.14) with (5.3.16), Ti O F310 = F310 + F310
−→ F310 = fTi [1 + exp [i4π]] + fO [exp [2.48πi] + exp [5.24πi] + exp [5.52πi] + exp [2.76πi]]
Problem 3 Copper is a monovalent metal of density 8 g/cm3 and atomic weight of 64 g/mole. Using the free electron approximation, and considering a cubic system of side L, a)
Find a formula for the number of states dN within an energy interval dε.
b)
Use your expression to evaluate the Fermi energy for copper in eV, at 0 K.
Solution The Free Electron Approximation assumes that the valence electrons are essentially detached from their parent ions. Subsequently, the free electrons are thought of as “conduction” electrons, and are able to move freely throughout the crystal. This approximation completely ignores all electron-electron and ion-electron interactions.
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
In this part we are looking for the density of states for a single valence electron. Using the free electron approximation, we imagine the electron to be bound within a cube, each side of length L. The TISE equation says that ~2 2 ~2 ∂ 2 ∂2 ∂2 − ∇ ψ = Eψ −→ − + + ψ = Eψ (5.3.17) 2m 2m ∂x2 ∂y2 ∂z 2 Whose corresponding eigenvalues are E=
~2 π 2 n2x + n2y + n2z 2 2mL
(5.3.18)
If we imagine the energy levels to be continuous rather than discrete, then the total number of 1/2 states can be approximated by the volume of a sphere of radius r = n2x + n2y + n2z . The total number of states, N , can then be found by integrating this radius within n-space: N =2·
1 4 3 · πr 8 3
(5.3.19)
Where in (5.3.19), the factor of 2 is due to spin degeneracy (two different spins with the same energy), while the factor of 1/8 is due to the fact that we are only interested in the quadrant where all n’s are positive (nx , ny , nz > 0), which happens only within one quadrant, which corresponds to 1 /8 of the total volume of the sphere. Substituting r in (5.3.19) for its equivalent form in terms of n gives 3/2 1 4 (5.3.20) N = 2 · · π n2x + n2y + n2z 8 3 Now, we solve (5.3.18) as follows
n2x + n2y + n2z =
2mL2 E ~2 π 2
(5.3.21)
Substituting (5.3.21) back into (5.3.20), 3/ 2mL2 E 2 1 4 N =2 · · π 8 3 ~2 π 2 3
/
π (2m) 2 L3 3/2 = · E 3 ~3 π 3 3
/
(2m) 2 L3 3/2 = E 3~3 π 2
(5.3.22)
We now recall that the density of states g(N ), is defined as the first derivative of N with respect to energy. Using (5.3.22), this becomes g(N ) =
dN dE "
# 3 / (2m) 2 L3 3/2 E 3~3 π 2 3 (2m) /2 L3 d 3/2 = E 3~3 π 2 dE d = dE
Carrying out the simple derivative in (5.3.23), we have
(5.3.23)
W. Erbsen
MODERN PHYSICS
3
/
g(N ) = b)
(2m) 2 L3 1/2 E 2~3 π 2
(5.3.24)
In order to find the Fermi Energy, EF , we first recall that the number of particles is given by Z EF N= g(E) · f(E) dE (5.3.25) 0
Where g(E) is the density of states, and f(E) is the Fermi-Dirac distribution function, which is given by f(E) =
1 exp [(E − µ) /kBT ] + 1
(5.3.26)
At T = 0 K, then µ(T = 0) = EF , where EF is the Fermi Energy. If E < µ, then (5.3.26) becomes f(E) =
1 −→ 1 exp [(E − µ) /kB T ] + 1
With this, (5.3.25) becomes N =
Z
EF
g(E) dE # 3 Z EF " (2m) /2 L3 1/2 E = dE 2~3 π 2 0 3 Z / (2m) 2 L3 EF 1/2 = E dE 2~3 π 2 0 0
3
/
=
(2m) 2 L3 2 3/2 · EF 2~3 π 2 3
=
3 (2m) 2 L3 / · EF2 3 2 3~ π
3
/
(5.3.27)
From the values given initially, we know that N N = 3 V L 3
=8
g 1 mol (100 cm) · ·· · NA 3 cm3 64 g (1 m)
g 1 mol (100 cm)3 atoms · ·· · 6.022 × 1022 3 3 cm 64 g mol (1 m) atoms =7.528 × 1028 m3 =8
Now, we wish to solve (5.3.27) for EF , which yields 2/ ~2 N 3 EF = 3π 2 3 2m L Substituting in (5.3.28) and the other appropriate values into (5.3.29) yields
(5.3.28)
(5.3.29)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
"
# 2/3 3 3 3 6.582 × 10−16 eV · s π 2 8 28 atoms EF = · 3 × 10 m/s · 7.528 × 10 −→ EF = 6.4 eV (2 · 0.511 × 106 eV) m3
Problem 4 A white dwarf is a star which has used up its nuclear fuel and contracted under its own weight to a state of high density and arbitrarily low temperature. It is held up by the kinetic energy of its degenerate electrons. Assume that all electrons in the star behave like free electrons in a metal, that all the electron states are full up to the Fermi energy, that the star is spherical and made of helium, and that the electrons are non-relativistic. a)
Discuss briefly the physical principles needed to find the equilibrium radius of the star. Write down a set of equations which, when combined, will determine the radius. Don’t worry too much about the factors of 2.
b)
Show that this radius is given approximately by R=C
1 h2 M − /3 (Gme mp ) 5/3
(5.3.30)
Where C is a purely numerical constant, G is the gravitational constant, me and mp are the electron and proton masses, and M is the mass of the star.
Solution a)
The equilibrium radius is defined as the radius which minimizes the total energy of the star. In our case the energy comes from the gravitational energy, which acts to contract the star, and also the degenerate energy, which is attributed to the tendency of Fermions to repel one another. Mathematically, Etot = Egrav + Edeg
b)
(5.3.31)
Let’s find the gravitational energy in (5.3.31) first. In order to find the gravitational self-energy of some spherical mass with a density ρ, we recall that ρ = m/V → m = ρV . In other words, 4 4 m = πr 3 ρ recalling that V = πr 3 (5.3.32) 3 3 From which we can say that dm = 4πr 2 ρ dr
(5.3.33)
We also recall that the potential energy is given by Gm dm r R Substituting (5.3.33) into (5.3.34), and recalling that U = Egrav, we have Z R Gm Egrav = − dm r 0 dU = −
(5.3.34)
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MODERN PHYSICS
=−G
Z
R
0
1 r
4πr 3 ρ 3
4πr 2 ρ dr
(5.3.35)
Where I have substituted m for the value from (5.3.32) and likewise for dm from (5.3.33). Continuing, Z 16π 2 Gρ2 R 4 Egrav = − r dr 3 0 16π 2 Gρ2 R5 =− 3 5 2 2 16π Gρ R5 =− 15 16π 2 Gρ2 3M =− 15 4πR3 2 3 GM (5.3.36) =− 5 R We now recall that the total energy of a degenerate electron gas is given by∗ 3 Edeg = N EF 5
(5.3.37)
Where the Fermi Energy, EF , is given by ~2 EF = 2me Substituting (5.3.38) into (5.3.37), Edeg
3π
2N
V
2/3
(5.3.38)
" 2/3 # 3 ~2 2N = N 3π 5 2me V 2/3 3~2 2N N 3π = 10me V 2 5 3~2 1 / 3 3π 2 · 2/ · N /3 = 3 10me V
(5.3.39)
Recalling that V = 4/3πR3 , (5.3.39) becomes Edeg
2/3 2 5 3~2 3 2 /3 = 3π · · N /3 3 10me 4πR 2/3 2 5 3~ 9π = · N /3 3 10me 4R 2/ 5 3~2 1 9π 3 = · N /3 2 10me R 4
(5.3.40)
Substituting (5.3.36) and (5.3.40) into (5.3.31), Etot ∗ Recall
that E =
R
E · d (ε) dε
3 GM 2 3~2 1 =− + 5 R 10me R2
9π 4
2/3
N
5
/3
(5.3.41)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
And now minimizing the total energy with respect to R, we take the first derivative of (5.3.41) with respect to R and set it equal to zero: " # 2/ ∂Etot ∂ 3 GM 2 3~2 1 9π 3 5/3 = − + N ∂R ∂R 5 R 10me R2 4 2/3 5 3 1 3~2 9π 1 2 ∂ /3 ∂ = − GM + N 5 ∂R R 10me 4 ∂R R2 2/3 5 3 1 3~2 9π 2 = − GM 2 − 2 + N /3 − 3 5 R 10me 4 R 2 / 3 1 6~2 9π 3 5/3 1 = GM 2 − N (5.3.42) 2 5 R 10me 4 R3 Setting (5.3.42) equal to zero and solving for R, we find that 2/3 5 6~2 9π 3 2 GM R = N /3 5 10me 4 2/3 5 ~2 9π R= N /3 Gme M 2 4 Recalling that N = M/mp , (5.3.43) becomes 2/ ~2 9π 3 R= G 4 2/ ~2 9π 3 = G 4 2/ ~2 9π 3 = G 4 2/ ~2 9π 3 = G 4
1
5
2
me (mp · N )
N /3
1 1 me · m2p N 1/3 m 1/3 1 p
me · m2p
M
1
1 M 1/3
5/ 3
me · mp
And now recalling that ~ = h/2π, (5.3.44) becomes 2/3 h2 9π 1 1 R= 2 5/ 1 3 M /3 4π G 4 me · mp 2/3 2 1 9π h 1 1 = 2 4π 4 G me · mp5/3 M 1/3 And we are finally left with R=C
h
1
5/ 3
Gme mp
M − /3
Where C=
(5.3.43)
1 4π 2
9π 4
2/3
(5.3.44)
(5.3.45)
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MODERN PHYSICS
Problem 5 A particular atom has two energy levels with a transition wavelength of 1064 nm. At 297 K there are 2.5 × 1021 atoms in the lower state. a)
How many atoms are in the upper state?
b)
Suppose that 1.8 × 1021 of the atoms in the lower state are pumped to the upper state. How much energy could this system release in a single laser pulse? How many photons are emitted in this pulse?
Solution a)
We first recall that the number of atoms in a particular state is 1 E N = exp − Z kB T
(5.3.46)
Taking the ratio of the two populations, and using (5.3.46), we have NU (EU − EL) = exp − NL kB T
(5.3.47)
Where we note that ∆E =EU − EL hc = λ 4.136 × 10−5 eV · s 3 × 108 m · s−1 = (1064 × 10−9 m) =1.166 eV
(5.3.48)
Substituting (5.3.48) back into (5.3.47) and solving for NU yields " # (1.166 eV) NU = 2.5 × 1021 atoms · exp − 8.617 × 10−5 eV · K−1 (300 K) −→ NU = 40.87 atoms
b)
(5.3.49)
If 1.8 × 1021 atoms are pumped to the upper state, and we are interested in how much energy they could release in a single laser pulse, this is the same as asking what happens if all atoms simultaneously fall down to the lower state from the upper state. We know that the energy difference between the two levels is ∆E = 1.166 eV from (5.3.48), so the total energy that would be released is eV Epulse = 1.8 × 1021 (1.166 eV) 1.602 × 10−19 J · eV−1 −→ Epulse = 2.099 × 1021 Pulse −→ Epulse = 336.292
Joules Pulse
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 6 A helium-neon laser emits light with a wavelength of 6328 ˚ A in a transition between two states of neon atoms. What would be the ratio of population of the upper state relative to the lower state of the neon atoms if they were in thermal equalibrium at 300 K?
Solution At equilibrium (T = 300 K in this case), the population of the two states are equal. That is, N1 = N2 And the ratio would be N2 −(E2 − E1 ) = exp N1 kB T
(5.3.50)
Where ∆E =E2 − E1 hc = λ 4.136 × 10−15 eV · s 3 × 108 m · s−1 = (632.8 × 10−9 m) =1.961 eV Applying this to (5.3.50) at T = 300 K yields " # N2 (1.961 eV) N2 = exp − −→ = 1.136 × 10−33 −2 5 N1 N1 8.617 × 10 eV · K (300 K)
Problem 7 The atomic number of Na is 11. a)
Write down the electronic configuration for the ground state of the Na atom showing in standard notation the assignment of all electrons to various one-electron states.
b)
Give the standard spectroscopic notation for the ground state of the Na atom.
c)
The lowest frequency line in the absorption spectrum of Na is a doublet. What are the spectroscopic designations of the pair of energy levels to which the atom is excited as a result of this absorption process?
d)
Discuss the mechanism responsible for the splitting between this pair of energy levels.
e)
Which of these levels lies the lowest? Discuss the basis on which you base your choice.
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MODERN PHYSICS
Solution a)
The ground state electronic configuration of Sodium in standard notation is: 1s2 2s2 2p6 3s1
b)
Only the unpaired 3s electron contributes to the total angular momentum J: L = 0, S = 1/2 −→ J = 1/2
So, the spectroscopic notation is:
2
c)
S 1/2
The first excited state is the 3p state, which means that L = 1, S = 1/2 −→ J = 1/2 , 3/2
And therefore, the spectroscopic notation for the lowest excited states is 2
d)
2
P 3/2
The levels 32 P 1/2 and 32 P 3/2 are split when the spin-orbit interaction is taken into account. In the electron’s rest frame, it sees the nucleus orbiting it. This causes the electron to see a magnetic field caused by the nucleus “orbiting” the electron. This is seen as a current loop, and is proportional to the electron’s orbital angular momentum, Li . This field interacts with the electron’s magnetic dipole moment, µi , which is proportional to its spin angular momentum, si . This creates a perturbation term which is proportional to Li , Si for each electron. The perturbation term is given by H=
e)
P 1/2 ,
1 1 dV(ri ) Li · S i 2m2 c2 ri dri
The 32 P 1/2 state lies lower than the 32 P 3/2 when spin-orbit is taken into account. The reason is because of Hund’s rules, which states that for any two states with identical L and S values, the one with the lowest J is more tightly bound with the lowest energy shells that are less than half-full (as is the case with a single electron in a P -state).
Problem 8 A slab of lead shielding 1.0 cm thick reduces the intensity of 15 MeV γ rays by a factor of 2. a)
By what factor will 5.0 cm of lead reduce the beam?
b)
What is the assumption cross-section for 15 MeV γ’s in lead nuclei? Lead has a density of 11.4 g/cm3 , and an atomic weight of 207 g/mol.
Solution
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
The decrease in intensity of radiation passing through matter is mathematically given by I = I0 exp [−µd]
(5.3.51)
Where µ = nσ, and n is the number of atoms per cm3 , and σ is the absorption cross section. a)
First, we find µ for the initial case: 0.5I0 = I0 exp [−µ] −→ µ = log [2] So, for d = 5.0 cm, (5.3.51) becomes I = I0 exp [− log [2] · 5] −→ I =
b)
I0 32
Recalling that µ = nσ, and that for lead we found that µ = log [2], we have: log [2] = nσ −→ σ =
log [2] n
(5.3.52)
Where for lead, we find that g 1 mol NA atoms · · 3 cm 207 g 1mol 11.4 · NA atoms = 207 cm 3
n =11.4
(5.3.53)
Substituting (5.3.53) into (5.3.52), σ = log [2] ·
207 cm3 −→ σ = 2.090 × 10−23 cm3 11.4 · 6.022 × 1023
Problem 9 The potential energies of two diatomic molecules of the same reduced mass are shown. From the graph, determine which molecule has the larger a)
Inter-nuclear distance
b)
Rotational inertia (moment of inertia)
c)
Separation between rotational energy levels
d)
Binding energy
e)
Zero-point energy
f)
Separation between low-lying vibrational states
Solution
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MODERN PHYSICS
a) b) c)
1 has a larger inter-nuclear distance 1 has a larger rotational intertia , because I = µR2 and since we have the same µ and 1 has a larger inter-nuclear distance, than 1 must also have a higher rotational inertia. 2 has a larger separation between rotational energy levels , because Erot ∝
` (` + 1) ~2 2I
And from part b), I1 > I2 so we must conclude that E1 < E2 . d) e)
2 has a larger binding energy , since the well is deeper. 2 has the largest zero-point energy . We recall that the zero-point energy is defined as the lowest energy possible e.g. the lowest vibrational state. We can approximate, using the simple harmonic oscillator, that r 1/2 2m Ki E1 = 1/2 ~ω1 T = (Vi − E) , ωi = E2 = 1/2 ~ω2 ~2 m And since V2 > V1 , then K2 > K1 and ω2 > ω1 , and therefore E2 > E1 .
f)
2 has the largest separation between the lowest-lying vibrational states , since ∆Evib ∝ ~ω −→ ω2 > ω1
Problem 10 A K0 meson (mass 497.7 MeV/c2 ) decays into a π + , π − meson pair with a mean life of 0.89−10 s. a)
Which one of the fundamental interactions is responsible for this decay?
463
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
Suppose the K0 has a kinetic energy of 276 MeV when it decays, and that the two π Mesons emerge at equal angles to the original K0 direction. Find the kinetic energy of each π meson and the opening angle between them. The mass of a π meson is 139.6 MeV/c2 .
Solution a)
This decay is caused by the weak interaction . This is know because K o has strange quarks, while π-particles do not. Only in the weak interaction strangeness is not conserved.
b)
At this point, we must employ the wonders of conservation of momentum, which says pk = pπ + + pπ −
(5.3.54)
We now recall the handy formula p2 c2 = E 2 − mc2
2
(5.3.55)
First it would be nice to combine the two momenta from the Pions in (5.3.54): pk = 2pπ
(5.3.56)
However, we must be careful! The Pions do not both go in the same directions – they fly off at equal angles. Therefore, the momentum we concern ourselves with from with is pk = 2pπ cos θ
(5.3.57)
At this point we recognize that pπ in (5.3.57) can be rewritten using our good ol pal (5.3.55): q 2 2 (5.3.58) p2π c2 = Eπ2 − mπ c2 −→ pπ c = Eπ2 − (mπ c2 )
Now multiplying both sides of (5.3.58) by c,
pk c = 2pπ c cos θ
(5.3.59)
q 2 pk c = 2 Eπ2 − (mπ c2 ) cos θ
(5.3.60)
h 2 i p2k c2 = 4 Eπ2 − mπ c2 cos2 θ
(5.3.61)
And substituting (5.3.58) into (5.3.59),
And squaring both sides,
And substituting (5.3.55) into (5.3.61) for the LHS, h 2 2 i Ek2 − mk c2 = 4 cos2 θ Eπ2 − mπ c2
We know all the constants in this equation, except for Eπ . From conservation of energy, Ek = 2Eπ −→ Eπ = 1/2 Ek Substituting this back into (5.3.65), Ek2 − mk c2
2
=4 cos2 θ
2 Ek2 − mπ c2 4
(5.3.62)
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MODERN PHYSICS
We can now solve (5.3.63) for θ, −1
θ = cos
h 2 i = cos2 θ Ek2 − 2mπ c2
(5.3.63)
"s
(5.3.64)
2
Ek2 − (mk c2 )
2
Ek2 − (2mπ c2 )
#
Before we go any further, we must recognize that Ek represents that total energy of the K o , so it includes both the mass energy term and the kinetic term. Using all the constants applied, (5.3.64) becomes "s # 2 2 (497.7 + 276 MeV) − (497.7 MeV) θ = cos−1 (5.3.65) 2 2 (497.7 + 276 MeV) − (2 · 139.6 MeV) Evaluating (5.3.65), we are left with θ = 34.82o HOWEVER, if we want the total angle between the two Pions, we must double this result, which yields θ = 69.34o The kinetic energy can just be found via Conservation of Energy, which says that Tk = 2Tπ −→ Tπ =
Tk 276 eV −→ Tπ = −→ Tπ = 138 MeV 2 2
Problem 11 ˚ spacing between the crystal planes. The A beam of 50 eV electrons is scattered from a cubic crystal with 1.2A electron beam is perpendicular to the crystal plane. At what angle will the first diffraction maximum be detected?
Solution We first recall that the deBroglie wavelength is given by h p
(5.3.66)
√ p2 −→ p = 2mT 2m
(5.3.67)
λ= And the momentum is T = Substituting (5.3.67) into (5.3.66) yields
465
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
λ= √
h 2mT
(5.3.68)
We also recall that the equation for diffraction from a 3D period structure is nλ = 2d sin θ
(5.3.69)
Where we recognize that the path length must be in integer multiples of λ for completely constructive interference. For the first maxima, we have n = 1, and with (5.3.68), we have h h 1 −1 √ √ = 2d sin θ −→ θ = sin (5.3.70) 2d 2mT 2mT Substituting in the appropriate values into (5.3.70) yields −34 6.626 × 10 J·s 1 q θ = sin−1 2 (1.2 × 10−10 m) 2 (9.109 × 10−31 Kg) (50 eV) 1.602 × 10−19 J · eV−1 −→ θ = 46.3o
Problem 12 One of the strongest emission lines observed from distant galaxies comes from Hydrogen and has a wavelength of 122 nm (in the ultraviolet region). a)
How fast must a galaxy be moving away from us in order for that line to be observed in the visible region of 366 nm?
b)
What would be the wavelength of the line if that galaxy were moving towards us at the same speed?
Solution a)
If the source and receiver are moving away from each other, the Doppler effect says that p 1 − u2 /c2 0 f =p fo 1 + u2 /c2 p 1 − u2 /c2 = fo 1 + u2 /c2 Recalling that f = c/λ, we can rewrite (5.3.71) in the following form s 1 1 − u/c 1 = λ0 1 + u/c λo
(5.3.71)
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MODERN PHYSICS
λo λ0
2
1 − u/c 1 + u/c [1 − u/c] (c/c) = [1 + u/c] (c/c) c−u = c+u =
So, with (5.3.73) and the values given, we have 2 2 2 λ0 122 nm 1 1 c−u = = = = 0 c+u λ 366 nm 3 9
(5.3.72)
(5.3.73)
(5.3.74)
Rewriting, (5.3.74) becomes c−u 1 = c+u 9 9 (c − µ) =c + u
9c − 9u =c + u u + 9u =9c − c
From which it is easy to see that u= b)
8 c 10
If the galaxy is moving towards Earth at the same speed, we simply change the sign of u in (5.3.72): s 1 1 + u/c 1 = 0 λ 1 − u/c λ0 s 1 − u/c λ0 = λ0 1 + u/c r 1 − 0.8 = 122 nm 1 + 0.8 122 nm = 3 Which finally leads us to λ0 = 40.67 nm
Problem 13 Hall effect: Consider a free electron gas with an electric field E along the x-direction. The electrons are constantly undergoing collisions resulting in some average velocity E is applied. a)
The conductivity is defined using the current density j = Eσ. Find an expression for σ in terms of the relaxation time T which is half of the time between successive collisions. Let n be the density of electrons in
467
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
the material and m the electron mass. b)
Now assume the electron gas is confined to a wire lying along the x-direction (E is still applied). Under these circumstances the equation of motion of an electron is m
v dv = −e (E + v × B) − m dt T
(5.3.75)
Where the first term in the Lorentz force and the second term takes account of collisions. It is found that a voltage develops along the y-direction (across the wire). A measure of this voltage is given by the Hall constant R = Ey /(j · Bz ). Show that R = −1/(ne). c)
Will the Hall constant be larger in a metal or a semiconductor and why?
Solution a)
To find an expression for the conductivity, σ in terms of the relaxation time, T , we must employ several not-so-tedius substitutions. First, we write down the (given) Ohm’s Law: J = σE
(5.3.76)
We also recall that the drift velocity is defined by v=
J nq
(5.3.77)
Where v is the velocity vector. We also recall that v = a·T
(5.3.78)
Where a is the acceleration of course. Now, from Newton’s second law, F = ma
(5.3.79)
And we must also recall that the force on a charged particle is given by F = −qE
(5.3.80)
Setting (5.3.79) equal to (5.3.80), ma = −qE −→ a = −
qE m
We now take (5.3.81) and substitute it back into (5.3.78): qE v= − ·T m
(5.3.81)
(5.3.82)
We now set (5.3.82) equal to (5.3.77), which yields J qE nq 2 ET =− · T −→ J = nq m m And finally, we set (5.3.83) and (5.3.83) equal to one another:
(5.3.83)
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MODERN PHYSICS
σE = b)
nq 2 ET nq 2 T −→ σ = m m
(5.3.84)
The very first thing we need to do in this problem is to solve (5.3.75), which reads (for convienence) m
dv mv = −q (E + v × B) − dt T
(5.3.85)
Let’s evaluate the cross product first. We are told that the magnetic field is along the z-direction, and so ˆi v × B = vx 0
ˆj vy 0
ˆ z vz = vy Bz ˆi − vx Bz ˆj Bz
Substituting this in to (5.3.85) and breaking down the components, m(v + v + v ) d(vx + vy + vz ) x y z m = −q Ex + Ey + Ez + vy Bz ˆi − vx Bz ˆj − dt T
(5.3.86)
(5.3.87)
A voltage only exists along the y-direction, so lets get rid of everything not in the y-direction, and (5.3.88) becomes mv dvy y = −q Ey − vx Bz ˆj − m dt T And now,
Ey ˆ 0 = −q Ey − vx Bz ˆj −→ vx = − j Bz
(5.3.88)
Recalling that J · R = vx , (5.3.88) becomes
R=
Ey ˆ j JBz
(5.3.89)
And now, recall that J · R = vx ,
J =q·v·n
Combining these, Rqvx n = −vx −→ R = − c)
1 nq
The hall constant will be larger .
Problem 14 The density of states g(E) is defined as the number of electronic states per unit energy range.
(5.3.90)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
Write down an expression for the density of the electron states in metal.
b)
Find an expression for the Fermi energy from a).
c)
Find an expression for the fraction of the conduction electrons which are within kT of the Fermi energy.
d)
Evaluate this fraction for Copper at T = 300K. [Copper is a monovalent metal with ρ = 8.92g/cm3 and atomic weight M = 63.5g/mole.]
Solution a)
In order to find an expression for the density of states for a 3D electron gas, we start with the assumption that the metal is in the shape of a cube, each side having a length of L. Assuming for the time being hat we have one free electron and infinite potential at the cube boundaries, the Eigenenergies are given by E=
~2 π 2 2 n + n2y + n2z 2mL2 x
(5.3.91)
For macroscopic systems, we assume that L is large enough that we can assume that the energy spectrum is continuous. In this approximation, the total number of states (up to an energy E) 1/2 is proportional to the volume of a sphere of radius r = n2x + n2y + n2z . Since we are only interested in one quadrant, we concern ourselves with only 1/8th of the total volume of the sphere. Furthermore, we also have a 2-fold degeneracy, so we must also include a factor of 2. So, we have 2/2 1 4 N =2 · · π n2x + n2y + n2z 8 3 3/2 π 2 = nx + n2y + n2z (5.3.92) 3 From (5.3.91) we have
Substituting (5.3.93) into (5.3.92),
2 2mL2 nx + n2y + n2z = 2 2 E ~ π 3/2 π 2mL2 N= E 3 ~2 π 2
(5.3.93)
(5.3.94)
Recalling that the density of states is defined by g(E) =
dN dE
(5.3.95)
We then substitute (5.3.94) into (5.3.95), yielding " 3/2 # d π 2mL2 g(E) = E dE 3 ~2 π 2 3
(2m) 2 L3 d h 3/2 i = E 3~3 π 2 dE 3 (2m) /2 L3 3 1/2 = E 3~3 π 2 2 /
(5.3.96)
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MODERN PHYSICS
Simplifying, (5.3.96) leads us to 3
(2m) /2 L3 1/2 E g(E) = 2~3 π 2 b)
We first recall that the number of particles in a system is given by Z EF N= g(E) · f(E) dE
(5.3.97)
(5.3.98)
0
Where g(E) is the density of states, and f(E) is the relevant distribution function, which in our case is the Fermi-Dirac function: 1 (5.3.99) f(E) = exp [(E − µ)/kB T ] + 1 For T = 0, this distribution gives the average number of particles in a state of energy E to be 1 for E < µ and 0 for E > µ. This maximum occupied energy (µ(T = 0)) is known as the Fermi Energy. So, letting f(E) → 1, and substituting (5.3.97) into (5.3.98), we have # 3 Z EF " / (2m) 2 L3 1/2 N = E dE 2~3 π 2 0 3 Z / (2m) 2 L3 EF 1/2 E dE = 2~3 π 2 0 3 / (2m) 2 L3 2 3/2 = E 2~3 π 2 3 F 3
/
(3m) 2 L3 3/2 = EF 2~3 π 2
(5.3.100)
Solving (5.3.100) for EF , EF =
"
3~3 π 2 (2m)
3 /2
L3
N
# 2/3
Which can be rewritten slightly nicer as 2/ ~2 3N π 2 3 EF = 2m L3 c)
(5.3.101)
In order to find the fraction of conduction electrons within kB T of the conduction band, we need to make a few assumptions. First, we assume that the density of states is constant over the entire kB T band, and secondly that the Fermi-Dirac distribution is still on the order of 1 for particles in this region. Then the number of particles within the kBT band is given by NkB T =g(EF ) · kB T Substituting (5.3.97) into (5.3.102), NkB T
"
# 3 / (2m) 2 L3 1/2 = E · kB T 2~3 π 2
(5.3.102)
(5.3.103)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
And also from (5.3.94) we know that N=
3/2 π 2mL2 E 3 ~2 π 2
(5.3.104)
Taking the ratio of (5.3.103) and (5.3.104) leads us to NkB T 3 1 = kB T N 2 EF d)
(5.3.105)
We now wish to evaluate (5.3.105) for Copper. The first thing that we want to do is find the fraction N/V , which is: 3 N 8.92 mol 100 cm atoms = · (5.3.106) · 6.022 × 1023 3 V 63.5 cm 1m mol The Fermi Energy as found in (5.3.101) is " EF =
3~3 π 2
# 2/3
N 3 (2m) /2 L3 " # 2/3 2 / 3~3 π 2 N 3 = 3/ V (2m) 2
Where I have set L3 → V . Substituting (5.3.107) into (5.3.105), " # 2/3 2 −1 / 3 2 NkB T 3 3~ π N 3 = kB T 3 / N 2 V (2m) 2 " 2 3 # /3 2/3 /2 3 (2m) V = kB T 2 3~3 π 2 N
(5.3.107)
(5.3.108)
Substituting in to (5.3.108) the appropriate values, 2 2/3 /3 2/3 −31 2 · 9.109 × 10 Kg NkB T 3 63.5 · = 1.381 × 10−23 J · K−1 (300 K) N 2 3 · 1.055 × 10−34 J · s 8.92 · 103 · 6.022 × 1023
Evaluating this, we are left with
NkB T = 5.5 × 10−3 N
Problem 15 A hypothetical atom has only two excited states, at 4.0 and 7.0eV, and has a ground-state ionization energy
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MODERN PHYSICS
of 9.0eV. If we used a vapor of such atoms from the Franck-Hertz experiment, for what voltages would we expect to see decreases in the current? List all voltages up to 20V.
Solution We first recall the principles behind the Franck-Hertz experiment. This seminal experiment demonstrated the quantization of atomic energy states by means of accelerating negatively charged electrons through a positively charged grid, surrounded by Mercury vapor. The resulting data takes the form of an increases oscillating signal, when plotting the net electron current vs the potential difference of the grid. The hypothetical atom described in the prompt has three energy levels: V3 =9 eV
[Continuum]
V2 =7 eV V1 =4 eV The electron current will drop whenever any multiple of V1 , V2 , or V3 is reached. Therefore, 1V1 =4 eV 2V1 3V1 4V1 5V1
=8 eV =12 eV =16 eV =20 eV
1V2 2V2 1V3 2V3
=7 eV =14 eV =9 eV =18 eV
Also, allowing for combinations of V1 , V2 and V3 , V1 + V2 =11 eV V1 + V3 =13 eV V2 + V3 =16 eV 2V1 + V2 =15 eV 2V1 + V3 =17 eV 3V1 + V2 =19 eV So, we expect the current to drop at the following values: V = 4, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20 eV
Problem 16
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Consider two protons of mass mp and electric charge e, separated by a distance r. a)
Compute the ratio of the electromagnetic to gravitational forces acting on the protons.
b)
Given your every day experiences with bulk neutral bodies, what do you conclude from the numerical values that you computed in a)? In particular, what force is more important for small bodies? For very large bodies?
c)
Consider a large spherical body, of constant density, made up of N hydrogen atoms. Determine the critical mass of this body at which the binding energy of the body just balances the gravitational potential energy of teh body. Use the fact that the atom size is approximately the Bohr radius [= 1/(ame )] and the atomic binding energy is twice the Rydberg energy [=a2 me /2], where a is the fine structure constant and me is the electron mass. Also assume that the mass of the hydrogen atom is equal to mp .
d)
Determine the critical radius of the body, corresponding to the critical mass derived in c).
Solution a)
The electrostatic force between the two protons is Fe =
1 e2 e2 ⇒K 2 2 4πε0 r r
(5.3.109)
While the gravitational force acting on the bodies is FG = G
Mp2 r2
(5.3.110)
So that if we were to find the ratio of electromagnetic and gravitational forces acting on the protons, we would divide (5.3.109) by (5.3.110), which leads us to Fe Ke2 /r 2 Ke2 = = 2 2 Fg Gmp /r Gm2p
b)
Plugging into (5.3.111) the appropriate values, 2 8.988 × 109 N · m2 · C−2 1.6022 × 10−19 C Fe Fe 36 = 2 −→ F ≈ 10 −2 −11 2 −27 Fg g 6.674 × 10 N · m Kg (1.673 × 10 Kg)
(5.3.111)
(5.3.112)
Rewriting (5.3.111),
2 e mp 2 e 20 ≈10 mp
Fe K = Fg g
(5.3.113)
So, unless the mass of the particle is more than ten orders of magnitude larger than the charge, the electrostatic force will be more prevalent. However, in our everyday experiences, we have mostly bulk substances, being neutral in nature, so for really large objects (stars, planets, etc), the gravitational force wins, whereas on the atomic level (atoms, molecules), the electrostatic force still dominates.
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MODERN PHYSICS
c)
Here we must find the critical mass, in which the binding energy just balances the gravitational energy. In the following calculations, we will assume that the Hydrogen atom radius is on the order of the Bohr radius, and that the binding energy of Hydrogen is twice the Rydberg energy. We first recall that ρ=
m 4 −→ m = V · ρ = πr 3 ρ V 3
(5.3.114)
Taking the differential limit of (5.3.114), dm = 4πr 2 ρ dr
(5.3.115)
We also recall that the gravitational self energy, in differential form is Ug = −
GM dm r
Substituting (5.3.115) into (5.3.116) and carrying out the integration, Z R Gm Ug = − dm r 0 Z R G 4 3 =− πr ρ 4πr 2 ρ dr r 3 0 Z 2 16π G 2 R 4 =− ρ r dr 3 0 16π 2 G 2 R5 ρ =− 3 5 2 2 16π Gρ 5 R =− 15
(5.3.116)
(5.3.117)
And the density is (recall that M = mp N ): ρ=
M 3mp N = 4/3πR3 4πR3
And according to the prompt, the binding energy of one Hydrogen atom is 2 a m1 Ub = 2 2
(5.3.118)
(5.3.119)
For the entire mass, (5.3.119) becomes Ub = N A2 me Putting (5.3.118) into (5.3.117), 16π 2 GR5 9m2p N 2 · 15 16π 2 R6 2 5 2 G9mp R N =− 15R G9m2p N 2 =− 15R G3m2p N 2 =− 5R
UG = −
(5.3.120)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
3 m2p 2 =− G N 5 R
(5.3.121)
Recalling now that Ub = N a2 me
(5.3.122)
Setting (5.3.121) and (5.3.122) equal to one another, 3 m2p 2 G N 5 R 5 a2 me mp N = R 3 Gmp
N a2 me =
(5.3.123)
We now recall that M = mp · N , so that (5.3.123) becomes M= d)
5 a2 me R 3 Gmp
(5.3.124)
The critical radius is related to the critical mass as∗ R=N
1
/3
RBohr
Applying this to (5.3.124), we are left with†
5 a2 me 2 R= R 3 Gm2p Bohr
1/3
Problem 17 Use the Bohr theory to answer the following question: a)
Derive the energy levels of the hydrogen atom.
b)
The orbiting speed of the electron in the ground state.
c)
If an excitation laser pulse has a duration of one picosecond, what is the principal quantum number n for which the orbiting period of the electron is about one picosecond?
d)
If the electron is replaced by a negatively charged muon, what is the ground state energy of this muon? Use these facts to get your numerical answers: the ground state of atomic hydrogen has a binding energy of 13.6 eV, the mass of the proton is 1837 times that of en electron and teh mass of a muon is 206 times that of an electron. What is the photon energy for the n = 2 to n = 1 transition in this case?
e)
Use the uncertainty principle to estimate the ground state energy of atomic hydrogen and compare the results from the Bohr model.
∗ Hint: †R
volume of sphere = N · volume of hydrogen.
=N
1
/3
r0
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MODERN PHYSICS
Solution a)
The potential (energy) of a Hydrogen atom is given in its most general form by V =−
1 e2 4πε0 r
(5.3.125)
While the total energy of the system (kinetic plus potential) is E =T + V =
me v2 1 e2 − 2 4πε0 r
(5.3.126)
Assuming that we have a circular orbit, we may use the following formula for centripetal force: mv2 = r
F =
1 e2 4πε0 r 2 | {z }
(5.3.127)
Coulomb Force
Solving (5.3.127) for v2 , v2 =
1 e2 4πε0 me r
And now substituting (5.3.128) back into (5.3.126), me 1 e2 1 − E= 2 4πε0 me r 4πε0 1 1 e2 2 1 = − 2 4πε0 r 2 4πε0 1 e2 =− 8πε0 r
(5.3.128)
e2 r e2 r (5.3.129)
One of Bohr’s key assumptions is that circular orbit had quantized angular momentum. Mathematically, L = mvr = n~
(5.3.130)
Solving (5.3.130) for r, r=
n~ n2 ~2 −→ r 2 = 2 2 me v me v
(5.3.131)
Substituting our value for v2 from (5.3.128) back into (5.3.131), we have n2 ~2 me r 4πε0 2 m2e e 2 2 n ~ 4πε0 r= me e2
r2 =
(5.3.132)
Now substituting (5.3.132) into (5.3.129), E=−
1 2 me e2 e · 2 2 8πε0 n ~ 4πε0
(5.3.133)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
We can rearrange (5.3.133), which finally leads us to E =− b)
me e4 2
2 (4πε0 ) n2 ~2
(5.3.134)
For the ground state, n = 1, and using (5.3.128), and (5.3.132), we have 1/2 1 e2 me e2 v= 4πε0 me ~2 4πε0 " # 1/2 1 e4 = 2 (4πε0 ) ~2 =
1 e2 4πε0 ~
(5.3.135)
Plugging in the appropriate values into (5.3.135), 9
2
v = 98.987 × 10 N · m · C c)
−2
2 1.6023 × 10−19 C −→ v = 2.187 × 106 m · s−1 (1.055 × 10−34 J · s)
(5.3.136)
By default, the orbital period is given by Tn =
2πrn vn
(5.3.137)
Substituting our previous functional value of v into (5.3.137), as well as our value for r and also (5.3.137), we have 2 2 n ~ 4πε0 n~ Tn =2π 4πε · 0 me e2 e2 2
=
2π (4πε0 ) n3 ~3 me e4
(5.3.138)
Solving (5.3.138) for n, n=
Tn me e4
! 1/3
2π (4πε0 )3 ~3 1/ Tn me e4 K 2 3 = 2π~3
Substituting in the appropriate values into (5.3.139), 4 ! 10−12 s 9.109 × 10−31 Kg 1.602 × 10−19 C 8.987 × 109 N · m2 · Kg−2 n= 3 2π (1.055 × 10−34 J · s)
(5.3.139)
(5.3.140)
Evaluating (5.3.144), we are left with, n = 18.73 −→ n = 19 d)
The energy of a Muon instead of an electron is found by replacing me in (5.3.136) by µ, which is the reduced mass of the atom, and is mathematically described by
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MODERN PHYSICS
µ=
mp · mµ mp + mµ
(5.3.141)
The relative mass of a proton to an electron is mp = 1838 me , so (5.3.141) becomes µ=
(1836 me ) (207 me ) = 186.0 me 1836 me + 207 me
(5.3.142)
With this new value of reduced mass, we must re-evaluate our results from (5.3.134), which leads us to E =− =− =
me e4 2 k 2n2 ~2 4 9.109 × 10−31 Kg 1.602 × 10−19 C 8.897 × 109 N · m2 · Kg−2 2n2 (1.055 × 10−34; J · s)
−13.6 eV n2
2
(5.3.143)
Applying our reduced mass from (5.3.145) to (5.3.146), −14.6 eVP n2 186.0 me me −13.6 eV · 186.0 = n2
E=
(5.3.144)
For the ground state, (5.3.144) becomes Egs(µ) = 2629 eV
(5.3.145)
And the photon energy for this case would just be E2 − E1 evaluated using (5.3.144), which yields 1 ∆E = E2 − E1 = −2529 eV − 1 −→ ∆E = 1896 eV 4 e)
From the uncertainty relation, we recall that ∆p∆r ∼ ~ −→ ∆p ∼
~ ∆r
Using these results, the total energy is E= =
p2 e2 −k 2m r ~2 2
2m (∆r)
−
ke2 r
(5.3.146)
We must minimize (5.3.146) in the usual fashion, ∂E(r) =0 ∂r ~2 ∂ 1 ∂ 1 − ke2 =0 2m ∂r r 2 ∂r r (5.3.147) Rearranging things a bit, (5.3.147) becomes
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
1 ~2 ·2· 2m Ke2 ~2 = kme2
r=
Substituting (5.3.148) back in to (5.3.146), Kme2 ~2 K 2 m2 e4 2 − ke E= 2m ~4 ~2 me4 k 2 me4 k 2 = − 2 2~ ~2 4 2 me k = ~2
(5.3.148)
(5.3.149)
From (5.3.149), we can easily deduce that E=
me4 2 k 2~2
(5.3.150)
Which we can see is identical to (5.3.144).
Problem 18 The resistivity ρ of a metal due to scattering of the quasi-free electrons as per the Drude model is given by me vav (5.3.151) ρ= ne2 λ a)
Derive the expression above. What are the quantities n and λ, and explain qualitatively is ρ inversely proportional to them?
b)
The average electron velocity vav is very different in the classical and quantum-mechanical treatments. Explain the origin of the difference (why the quantum-mechanical velocity is much larger at room temperature, and is nearly independent of temperature).
c)
How is λ related to the cross-section σ for scattering of conduction electrons in the metal?
d)
Naively, one might take σ to be the geometrical area of the lattice ions. A more correct treatment gives a much smaller cross section which depends on temperature. Explain how this arises.
Solution a)
We first note that n is the number of electrons per unit volume, while λ is the mean free path. The electrons will move with a drift velocity, vd . In some time interval ∆t all charge that passes through cross sectional area A will be ∆Q = neAvd ∆t. We can find the current by taking the derivative, I=
dQ(t) d = [neAvd ∆t] dt dt
(5.3.152)
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MODERN PHYSICS
If we assume that the time interval is appropriately constant, (5.3.152) becomes I = neAvd
(5.3.153)
At this point, we must recall a number of seemingly arbitrary definitions. We of course recall Ohm’s law as V = IR, and the resistivity is given as R=
ρL A
(5.3.154)
We also recall that the electric field within the material is given by E=
V −→ V = EL L
(5.3.155)
Substituting (5.3.155) into Ohm’s Law, we have EL = IR We now substitute our expression for resistivity from (5.3.154) into (5.3.156), ρL EA EL = I −→ I = A ρ
(5.3.156)
(5.3.157)
We now set (5.3.153) equal to (5.3.157), neAvd =
E EA −→ ρ = ρ nevd
(5.3.158)
In the same mantra as before, we recall that Newton’s second law says F = ma, F = eE, and v = aτ , so we can solve for v and arrive at eEτ m
(5.3.159)
m m E · −→ ρ = 2 ne eEτ ne τ
(5.3.160)
vd = Putting (5.3.159) back into (5.3.158), ρ=
But hold on there jack, τ is really the average time the particle goes without a collision, and we want our expression in terms of the mean free path. The formula which connects these two entities is simple: τ = λ/vavg . Substituting this back into (5.3.160), ρ= b)
m vavg mvavg · −→ ρ = ne2 λ ne2 λ
(5.3.161)
The average electron velocity, vavg is a very different animal when investigated in either the classical or quantum-mechanical regime. Classically, T =
mv2 3 = kB T 2 2
Playing with (5.3.162) a little bit, v=
r
3kB T m
(5.3.162)
(5.3.163)
Quantum-mechanically, only electrons that have energies within the kBT band contribute, so the average energy is the Fermi energy: Eavg = EF :
481
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
T = EF = Playing some more, vavg =
r
mv2 2
2EF m
(5.3.164)
(5.3.165)
Where we note that EF does not depend on T , so vavg does not depend on it either. Freaky! A qualitative summary, Ef = kB TF kB Troom c)
−→
vavg (Quantum) vavg (Classical)
To answer this we need only consider two elements. Firstly, we recognize that an electron entering on the left of the cylinder will be scattered before it reaches the right side of the same cylinder. We know that the volume of the cylinder is λσ, and that the density of scatterers is n = 1/λσ, so, our answer is then λ=
d)
1 nσ
The area swept out by the atomic vibrations perpendicular to the direction of propagating electrons is given by the cross section, σ. Increasing the temperature, T increases the vibrations. Therefore, as they both increase without limit, we have no choice but to say that σ, T → ∞ at the same rate.
Problem 19 The band theory of solids can be used to explain how a pure, intrinsic semiconductor (such as germanium) can become a p-type semiconductor. a)
How do conduction bands arise in a solid?
b)
Sketch an energy level diagram for electrons in a pure germanium crystal. Indicate both the conduction and valence bands and give order of magnitude estimates of the energies in your diagram.
Solution a)
This is a tricky question to answer, in the sense that the full explanation is a science in and of itself. Irregardless, we start with the idealized scenario of a repeating series of potentials (a “Dirac Comb”) that has a fixed height (V0 ) and each barrier is a fixed distance from one another a. The Barrier itself has a width, which we call b. These series of potentials go on forever, as to neglect any edging effects. At this point, we introduce our old friend the TISE in 1-D: ~2 ∂ − + V (x) ψ(x) = Eψ(x) (5.3.166) 2m ∂x Whose realizable solutions end up begin
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MODERN PHYSICS
ikx
ψ(x) = Ae
,
k=
p
2m (E − V (x)) ~
(5.3.167)
Our periodicity boundary condition also satisfies the TISE, but the solution will be shifted by some phase factor: ψ (x + N (a + b)) = eik(a+b) ψ(x)
(5.3.168)
Where (5.3.168) is lovingly referred to Bloch’s Theorem. Solving the TISE for repeating lattices using boundary conditions (eventually yields a transcidental equation, which can be used to visually determine the bands and band-gaps within a particular solid. The size, length, and location of the bands and bandgaps depends on the distance between the potentials, a, the width of the potentials, b, and the height of the potentials, V0 . From these three variables, we can plot the energy versus the wave number, and depending on the combination of the supplied variables, the plots may indicate a conductor, insulator, or a semiconductor.
b)
For a conductor, the Fermi Energy lies within a band and accordingly, the valence bands and conduction bands overlap. This implies that at T = 0, some of the electrons will be “free” to “conduct.” For an insulator, the Fermi energy lies either at the beginning, or the end of the gap (or somewhere inbetween). Subsequently, at T = 0 all orbitals below EF are filled, leaving no ”free“ electrons to “conduct.” For the case of a semi-conductor, the gaps are small enough for some electrons to jump across and become free. These particular characteristics can be controlled by way of introducing impurities to the pure semiconductor (known as “doping”). I can’t sketch here, but it should be fairly straightforward. Anyway, as far as an order of magnitude estimate for the gap spacing between the valence and conduction bands in a semi-conductor, the only way I can think to solve this problem is by experience, which tells me that the gap should be ∼ 1 eV
Problem 20 When sodium metal is illuminated with light of wavelength 4.20 × 102 nm, the stopping potential is found to be 0.64 V; when the wavelength is changed to 3.10 × 102 nm, the stopping potential becomes 1.69 V. Using only this data and the values of the speed of light and electronic charge, find the work function of sodium and the value of Planck’s constant.
Solution We first recall that the key equation for the photoelectric effect is Emax = hf − φ
(5.3.169)
The stopping potential is the potential that is required to stop an electron with some maximum kinetic energy Emax . Since the potential energy created by a potential difference V is U = qV , (5.3.169) becomes qV =hf − φ
483
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
=
hc −φ λ
(5.3.170)
Given that λ = 4.20 × 102 nm, and also that the stopping potential is V − 0.65 V, (5.3.170) becomes hc V1 = 0.65 V qV1 = −φ (5.3.171) λ1 = 420 × 10−9 m λ1 And similarly, for λ = 3.10 × 102 nm and V = 1.69 V, we have hc V2 = 1.69 V −φ qV2 = λ2 = 310 × 10−9 m λ2
(5.3.172)
Solving (5.3.171) and (5.3.172) for φ and setting them equal to one another leads us to hc hc − qV1 = − qV2 λ1 λ2 1 1 − hc =q (V1 − V2 ) λ1 λ2 q h = (V1 − V2 ) c
1 1 − λ1 λ2
−1
(5.3.173)
Substituting in the correct values into (5.3.173) leads us to −1 1 1 1.602 × 10−19 C (0.65 V − 1.69 V) − h= 2.998 × 108 m · s−1 420 × 10−9 m 310 × 10−9 m −→ h = 6.578 × 10−34 J · s
(5.3.174)
Now, using either (5.3.171) or (5.3.172), we can calculate the work function φ. Using (5.3.172), φ=
hc − qV1 λ1
Substituting in the appropriate values, 6.578 × 10−34 J · s 2.998 × 108 m · s−1 φ= −→ (420 × 10−9 m)
φ = 3.661 × 10−19 J φ = 2.285 eV
Problem 21 Two identical photons are produced when a proton and an antiproton, both at rest, annihilate each other. What are the frequencies and corresponding wavelengths of the photons? Is it possible to produce a single photon in proton-antiproton annihilation? Why or why not?
Solution
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MODERN PHYSICS
We first recall the following energy/momentum relations: Ep2 =m2 c4 + p2 c2
(5.3.175a)
Ep2
(5.3.175b)
2 4
=m c
We also recall that Ep = mc2 = 938.3 MeV. Now, recall from Einstein’s relation that Eγ h 938.3 × 106 eV = 4.136 × 10−15 eV · s =2.296 × 1023 Hz
Eγ = hf −→ f =
(5.3.176)
Now, we can say that λ=
c 3 × 108 m · s−1 = −→ λ = 1.322 × 10−15 m f 2.269 × 1023 Hz
It is not possible to produce a single photon in this particular interaction, since it would violate conservation of momentum (recall that the protons start off at rest )
Problem 22 A dust particle at rest relative to the sun is in equilibrium between the radiation pressure and gravitational attraction. What is the size of the dust particle? (For simplicity, you can assume that the dust particle in spherical)
Solution We first recall that the radiation pressure is defined by Prad =
hSi c
(5.3.177)
Where hSi is the time-averaged Poynting vector, aka the intensity. The power emitted by the sun is around P ≈ 3.8 × 1026 W, and so the intensity is I = hSi =
3.8 × 1026 W 4πr 2
(5.3.178)
Applying (5.3.178) to (5.3.177), Prad =
3.8 × 1026 W 4πr 2 c
Since pressure is defined as P = F/A, then (5.3.179) becomes
(5.3.179)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Frad
3.8 × 1026 W 4πR2 = 4πr 2 3.8 × 1026 W 2 = R r2
Now, if we assume that the dust particles are perfectly reflective, then (5.3.180) becomes 3.8 × 1026 W 2 Frad =2 R r2
(5.3.180)
(5.3.181)
And we also recall that the gravitational force is given by Fgrav = G
mM r2
(5.3.182)
Equating (5.3.181) and (5.3.182), 3.8 × 1026 W 2 mM 2 R =G 2 r2 r Solving for R, this becomes R=
s
GmM 2 (3.8 × 1026 W)
(5.3.183)
Problem 23 What is the density of a neutron star? Evaluate the size (radius) of a neutron star as massive as the sun, −14 given that the radius 197 m, mass of the sun is 2 × 1030 kg. 79 Au nucleus is 7 × 10
Solution Density is of course given by ρ = m/v, and given the supplied constants, we have ρ=
197 × 1.67 × 10−27 Kg 3
4/3π (7 × 10−14 m)
−→ ρ = 2.194 × 1014 Kg · m−3
To find the radius of the neutron star, we recall that from the density, we can say that 1/ M 3M 3 ρ= −→ R = 4/3πR3 4πρ Plugging into (5.3.185) the appropriate values, R=
! 1/3 (3) 2 × 1030 Kg −→ R = 129589 m 4π (2.194 × 1014 Kg · m3 )
(5.3.184)
(5.3.185)
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MODERN PHYSICS
Problem 24 A typical particle production process is
p + p −→ p + p + π 0
where mp = 938MeV/c2 , mass of π 0 = 135MeV/c2 . a)
In a standard accelerator one of the initial protons would be at rest. In this case what is the minimum laboratory energy for the moving particle so that the reaction listed above occurs?
b)
In colliding beam accelerators, such as the one proposed for the superconducting supercollider, the two initial particles are moving towards each other with equal speeds. Why do colliding beam accelerators have an energetic advantage over the type of accelerator from a)?
c)
What energy would be required of each proton in a colliding beams accelerator to produce a pion?
Solution We first recall the following: E 2 =p2 c2 + m0 c2 E =mc
2
2
(5.3.186a) (5.3.186b)
And that the invariant quantity is Center of mass frame invariant after collision
2 2
z }| { E 2 − p 2 c2
=
m0 c | {z }
Lab frame invariant before collision
Where in the center of mass frame (COM) the momentum is zero, so (5.3.187) becomes 2 m0 c2 = E 2
(5.3.187)
(5.3.188)
Let’s make a table:
Before • −→ • Σm a)
After •→
•→
2mp
•→
(5.3.189)
2mp + mπ
If we choose not to use the COM frame, then the momentum is conserved (let’s do it this way). We first apply conservation of energy to our system: Ei = Ef
(5.3.190)
Squaring (5.3.190) and substituting in (5.3.186a), p2i c2 + m2i c4 = p2f c2 + m2f c4 Conservation of momentum says that pi = pf , and so (5.3.191) becomes m2i c4 =m2f c4
(5.3.191)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
2mp c2
2
2 = (2mp + mπ ) c2
4m2p c4 =4m2p c4 + 2mp mπ c4 + 2mp mπ c4 + m2π c4 0 =4mp mπ c4 + m2π c4 mπ c2 + 2 0 =mπ c2 2mp c2 mπ 0 =mπ c2 +2 2mp
(5.3.192)
We can now say that this is the threshold energy, which is the minimum possible energy that will allow this reaction to happen. The RHS of (5.3.192) is this so-called threshold energy: mπ c2 ET = mπ c2 + 2 2mp c2 Substituting in the appropriate values, we have 135 MeV ET = 135 MeV + 1 −→ ET = 279.7 MeV 2 · 938 MeV
(5.3.193)
And to find the total energy, we combine the results of (5.3.193) plus the rest mass energy of the proton: Etot = 279.7 MeV + 938 MeV −→ Etot = 1217.7 MeV b)
Since in the case of a), all the energy needed to create the π o is carried by the first proton, however if both were to move towards one another, then the energy would need to be much less in order to reach the threshold energy.
c)
If they are moving towards one another, we use the same logic as before - we want the threshold energy, that is the energy that is just barely enough to complete the reaction. In this case, we care only about the creation of three particles: two Protons, and one Pion. Therefore, each incident proton must have an initial energy equal to Ep =
1 (2mp + mπ ) 2
Plugging in the appropriate number, Ep =
1 [2 (938 MeV) + 135 MeV] −→ Ep = 1005.5 MeV 2
Problem 25 In a muonic hydrogen atom, the electron is replaced by a muon, which is 207 times heavier. a)
What is the wavelength of the n = 2 to n = 1 transition in Muonic hydrogen? In what region of the electromagnetic spectrum does this wavelength belong?
b)
How would the fine structure and Zeeman effects differ from those in ordinary atomic hydrogen?
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MODERN PHYSICS
Solution a)
We first recall the Rydberg formula, which reads 1 1 − 2 n2f ni
1 =R λ
!
(5.3.194)
While the Rydberg constant is given by me R= 4πc~3 R ∼C · me
e2 4πε0
2
= 1.097 × 107 m−1
(5.3.195)
Where me is the electron mass. For Muonic Hydrogen, we replace me with µ/me , where µ is the reduced mass, where mµ = 207 me ,
mp = 1836 me
So, the reduced mass is µ=
(207 me ) (1836 me ) = 186.0 me (207 me ) + (1836 me )
(5.3.196)
Substituting (5.3.196) into (5.3.194), 1 µ =R· λ me
1 1 − 2 n2f ni
!
Substituting in the appropriate values, (5.3.197) becomes 1 me 1 = 1.097 × 107 m−1 186.0 −1 λ me 4 =1.530 × 109 m−1
(5.3.197)
(5.3.198)
From which it is easy to see that λ = 6.535 × 10−10 m Which are x-rays. b)
The only appreciable difference between Hydrogen and Muonic Hydrogen is the difference in the masses of the electron and the Muon, respectively. Therefore, we must find the splitting as a function of mass for both fine structure and Zeeman effects.
The fine structure is of order: Efs ∼ α4 me2 −→ Efs ∼ m While the Zeeman effect is of order: Ez ∼ me ·
e~ 1 · B −→ Ez ∼ 2m m
489
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 26 For an intrinsic semiconductor, the number of electrons per unit volume in the conduction band and the number of holes in the valence band are given by (EC − EF ) n =NC exp − (5.3.199a) kB T (EF − EV ) (5.3.199b) p =NV exp − kB T Where 3
2(2πmkB T ) /2 NC = NV = h3
(5.3.200)
a)
Find the position of the Fermi energy in an intrinsic semiconductor.
b)
Show that the product of the number of electrons in the conduction band and the number of holes in the valence band depends only on temperature and the gap energy, Eg .
c)
Discuss how one can use the temperature (T ) dependence of the conductivity σ to approximately measure the gap energy provided that T is not too high.
Solution We first recall that the Fermi Energy is defined as the highest occupied state at T = 0 K. We also recall that the Intrinsic Fermi Energy (Ei ) is defined as the Fermi Energy in the absence of doping. a)
In an intrinsic semiconductor, the number of electrons is equal to the number of holes (n = p), since the thermal excitation of an electron leaves behind a hole in the valence band. The very first thing that we wish to do is multiply (5.3.199a) and (5.3.199b): (Ec − Ev ) (5.3.201) np = Nc Nv exp − kB T And furthermore, from (5.3.200), we may deduce that 3/ 2πme kB T 2 NC = 2 h2 3/ 2πmh kB T 2 NV = 2 h2
(5.3.202a) (5.3.202b)
Where mh is the electron effective mass near the top of the valence band. Using (5.3.202a)(5.3.202b), we can now rewrite (5.3.199a) and (5.3.199b), respectively: 3/2 2πme kB T (Ec − EF ) n =2 · exp − (5.3.203a) h2 kB T 3/ 2πmh kB T 2 (EF − Ev ) p =2 · exp − (5.3.203b) h2 kB T Multiplying (5.3.203a) and (5.3.203b) leads us to
W. Erbsen
MODERN PHYSICS
np = 4
2πkBT h2
3
(me mh )
3
/2
(Ec − Ev ) exp − kB T
(5.3.204)
The intrinsic carrier concentration depends exponentially on (Ec − EV ) /2kBT . At the Fermi Energy, n = p because the number of electrons and the number of holes are created in pairs for intrinsic semiconductors, as is what we have here. So, (Ec − Ef ) (EF − Ev ) Nc exp − =Nv exp − kB T kB T Since Nc and Nv are equal in this case, this becomes (EF − Ev ) (Ec − Ef ) =− kB T kB T Ec − Ef = − EF − Ev From which it is easy to gather that EF =
Ev + Ec 2
In other words, the Fermi Energy is located exactly inbetween the valence band and the conduction band. b)
From the results of the previous problem, we found that np = 4
c)
2πkBT h2
3
(mh me )
3
/2
Eg exp − kB T
(5.3.205)
Where Eg = Ec − Ev is the gap energy.
We recall from Ohm’s law that
σ=
j , E
Vd =
j nq
Combining these, we find that σ=
nqv E
(5.3.206)
nq 2 v F
(5.3.207)
Also recalling that F = qE, (5.3.206) becomes σ=
Now, we recall that v = aτ , where τ is the time between collisions. Combining this with Newton’s second law, F = ma, we have F = vm/τ , and (5.3.207) becomes σ= Recalling from before that σ =σc + σv
nq 2 τ m
(5.3.208)
491
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
q2τ = m
( 3/ 3/ ) 2πme kB T 2 (Ec − EF ) 2πmh kB T 2 (EF − Ev ) 2 · exp − +2 · exp − h2 kB T h2 kB T
From this we can easily say that
(Ev + Ec ) σ ∼ C exp − kB T
So, as long as the temperature is low, the gap will dominate this equation, and we can solve for the gap as needed.
Problem 27 The potential energy between two atoms in a molecule can often be described rather well by the Lennard-Jones potential, which can be written a 6 a 12 U(r) = U0 −2 (5.3.209) r r
Where U0 and a are constants. a)
Find the interatomic separation r0 in terms of a for which the potential energy is minimum.
b)
Find the corresponding value of Umin .
c)
Use the figure below to obtain numerical values for r0 and U0 for the H2 molecule. Express your answer in nanometers and electron volts.
d)
Make a plot of the potential energy U (r) versus the internuclear separation r for the H2 molecules. Plot each term separately, together with the total U (r).
Solution a)
To find the value of the interatomic separation r0 when the potential energy is at a minimum, we must minimize (5.3.209). The general prescription for minimizing any function is to take the derivative with respect to the minimization variable and set it equal to zero, solving for the appropriate minimized value. In our case, ∂U(r) =0 (5.3.210) ∂r r=r0 So, applying (5.3.210) to (5.3.209), we have ∂ a 12
a 6 −2 =0 ∂r r r ∂ 1 ∂ 1 6 a12 − 2a =0 ∂r r 12 ∂r r 6
W. Erbsen
MODERN PHYSICS
12 6 a12 − 13 − 2a6 − 7 = 0 r r 12 a6 a12 13 = 12 6 r r a6 =r 6
(5.3.211)
Applying r = r0 at (5.3.211), we have r0 = a b)
The corresponding value of Umin can be found by applying (5.3.212) to (5.3.209): " 6 # 12 a a U(r0 ) =U0 −2 r0 r0 a 6 a 12 =U0 −2 a a
(5.3.212)
(5.3.213)
Which leads us to Umin = −U0 c)
(5.3.214)
The values for r0 and U0 can be obtained quite simply from the figure. The value of r0 is the horizontal position of the deepest point in the potential, while the minimal energy is simply the depth from the bottom:
493
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
r0 = 0.074 nm Umin = −|27.2 eV − 4.5eV| −→ Umin = −31.7 eV
Problem 28 The measured conductivity of copper at room temperature is 5.88 × 107 Ω−1 · m−1 , its Fermi energy is 7.03eV, density is 8.96g m/cm3 and molar mass is 63.5gm/mole. a)
Calculate the density of free electrons in copper.
b)
Calculate the average time between collisions of the conduction electrons.
c)
Calculate the mean free path of electrons in copper.
d)
Calculate the smallest possible deBroglie wavelength of electrons in copper at T = 0K. How does this value compare with the atomic separation in copper?
Solution a)
Copper has only one free (valence) electron. According to the Free Electron Approximation, all the conductivity is due to this one electron per copper atom. So, if we wanted to calculate the density of free electrons in copper, we recall that ρ = M/V , and since we have only one free electron per atom, we find the density as ρ Cu n= · NA m g g −1 atoms =8.96 · 63.5 · · 6.022 × 1023 3 cm mol mol g 1 mol 23 atoms =8.96 · · · 6.022 × 10 (5.3.215) cm3 63.5 g mol From (5.3.215), it is easy to see that we are left simply with n = 8.497 × 1022
b)
atoms cm3
(5.3.216)
In order to calculate the average time between collisions of the conduction electrons, we first must consider the kinetic energy of these electrons, and from that find the average velocity: r mv2 2E E= −→ v = (5.3.217) 2 m Where we take E in (5.3.217) to be the Fermi Energy, EF , which is given in the prompt. Accordingly, we can evaluate v in (5.3.217) as s 2 (7.03 eV) m 2 v= · (3 × 108 m/s) −→ v = 1.574 × 106 (5.3.218) (0.511 × 106 eV) s
W. Erbsen
MODERN PHYSICS
We now recall that the conductivity is defined as σ=
ne2 d mv
(5.3.219)
Where we recognize that the relationship between velocity, the time between collisions and the distance traveled between collisions is v = d/τ → d/v = τ . Substituting this into (5.3.219), ne2 d σ= · m v ne2 ·τ (5.3.220) = m Solving (5.3.220) for τ yeilds mσ τ= 2 ne 9.109 × 10−31 Kg 5.880 × 107 Ω−1 · m−1 = 3 2 (8.497 × 1022 atoms · cm−3 ) (100 cm · m−1 ) (1.602 × 10−19 C)
(5.3.221)
Evaluating all the goodness in (5.3.221) leaves us with τ = 2.462 × 10−14 s c)
The mean free path is found from the velocity found in part b) as d =v · τ
= 1.574 × 106 m · s−1
d)
(5.3.222)
2.462 × 10−14 s −→ d = 38.75 nm
(5.3.223)
The deBroglie wavelength can be found rather straightforwardly using (5.3.218) as λ=
h 6.626 × 10−34 J · s = p (9.109 × 10−31 Kg) (1.574 × 106 m · s−1 )
(5.3.224)
From (5.3.224) we see that λ = 4.621 × 10−10 m
(5.3.225)
The atomic separation of copper can be found from the density, ρ, which is given in the problem: 1/3 m m m ρ= = 3 −→ L = (5.3.226) v L ρ Substituting in the appropriate values into (5.3.226) yields " # 63.5 g · mol−1 6.022 × 1022 atoms · mol−1 L= −→ L = 2.275 × 10−10 m 3 (8.96 g · cm3 ) (100 cm · m−1 )
Problem 29 The proper mean lifetime of π Mesons (Pions) is 2.6 × 10−8 s. If a beam of such particles has a speed of 0.9 c:
495
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
What would their mean life be as measured in the laboratory?
b)
How far would they travel (on the average) before they decay?
c)
What would your answer be to part b) if you neglect time dilation?
d)
What is the interval in space-time between creation of a typical pion and its decay?
Solution We first recall that the equation for time-dilation is T = γT0
(5.3.227)
Where the Lorentz factor, γ, is equal to
a)
1 γ= p 1 − v2 /c2
In the lab frame, using (5.3.227)-(5.3.228) we have
2.6 × 10−8 s T = p −→ T = 5.965 × 10−8 s 1 − (0.9)2
c)
In the lab frame, we know that ∆x = v · T , and using (5.3.229), we have ∆x = (0.9) 3 × 108 m · s−1 5.965 × 10−8 s −→ ∆x = 16.105 m
d)
The Space-time interval is defined as
b)
(5.3.228)
(5.3.229)
(5.3.230)
If time-dilation is neglected, we can find ∆x as ∆x = (0.9) 3 × 108 m · s−1 2.6 × 10−8 s −→ ∆x = 7.020 m ∆s2 = −∆x2 + c2 ∆T 2
(5.3.231)
Substituting in the appropriate values into (5.3.231), 2 2 ∆s2 = 3 × 108 m · s−1 5.965 × 10−8 s − (16.105 m) −→ ∆s2 = 60.86 m2
So, the space-time interval is
∆s =
√ 60.86 m2 −→ ∆s = 7.801 m
Problem 30 Four identical, non-interacting particles are placed in a one dimensional box of length L. Find the lowest total energy of the system and list the quantum numbers of all occupied states (including ms ) if: a)
The particles are electrons
W. Erbsen
MODERN PHYSICS
b)
The particles are neutral Pions, π 0 .
c)
Two of the particles are protons and the other two are neutrons.
Solution We first recall that the 1-D, time independent SE reads −
~2 ∂ 2 ψ(x) = Eψ(x) 2m ∂x2
(5.3.232)
Solving (5.3.232) with the appropriate boundary conditions, we find that the corresponding Eigenfunction and Eigenvalues are, respectively r nπx 2 ψ(x) = sin (5.3.233a) L L n2 ~2 π 2 (5.3.233b) En = 2mL2 a)
Since electrons are Fermions, no two electrons can be described by the same set of quantum numbers, as necessitated by the Pauli Exclusion Principle. Therefore, the lowest occupied states are: |n, ms i = |1, 1/2 i, |1, −1/2 i, |2, 1/2 i, |2, −1/2 i And from (5.3.233b), the lowest energy corresponding to this state is Egs =
b)
5~2 π 2 ~2 π 2 [1 + 1 + 4 + 4] −→ Egs = 2 2mL me L2
The neutral pion has a spin of zero, making it a Boson. Since the Pauli Exclusion principle does not apply to Bosons, all the particles may lie in the lowest state: |n, ms i = |1, 0i, |1, 0i, |1, 0i, |1, 0i While the energy is, following (5.3.233b), Egs =
c)
~2 π 2 2~2 π 2 [1 + 1 + 1 + 1] −→ Egs = 2 2mL mπ L2
Both Protons and Neutrons are Fermions, but they are distinguishable from one another. Therefore, the lowest state is then |n, ms i = |1, 1/2 i, |1, −1/2 i, |1, 1/2 i, |1, −1/2 i And the Eigenenergies are Egs
~2 π 2 ~2 π 2 ~2 π 2 E = = [1 + 1] + [1 + 1] −→ gs 2mp L2 2mn L2 L2
1 1 + mp mn
497
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
5.4
Statistical Mechanics
Problem 1 a)
Define the chemical potential µ. Show that two systems are in diffusive equilibrium if µ1 = µ2 . You may start with F = F1 + F2 (free energy) and use the fact that µ1 = µ2 should minimize F .
b)
Now consider a 1 − D gas of length L and number of distinguishable and non-interacting particles N . Assume that temperature T is high enough that we are in the classical regime. Find the chemical potential µ in terms of T and other parameters of the problem.
c)
Now find the free energy F and the specific heat at constant volume CV .
Solution a)
The chemical potential, µ, can be thought of as a number that allows us to quantify whether or not two (or more) systems are in diffusive equilibrium. If they are equal, then there will be no (net) flux of particles moving from one system to another. If they are not equal, then particles will tend to flow from a region of high chemical potential to a region of low chemical potential. We also recall that the Helmholtz Free Energy (F ) measures the amount of “useful” work that is extractable from a system at constant temperature and volume. The Free Energy has several important characteristics: i) ii)
At equilibrium, the Free Energy is at a minimum An important form of the second law of thermodynamics is U = F + TS Where U is the internal energy, and S is the final entropy. At diffusive equilibrium, we know that dF = 0, and the number of particles in each system remains constant. From this, we may also deduce that the total number of particles remains constant. Quantitatively, N = N1 + N2 In fact, whether we are in diffusive equilibrium or not, the total number of particles must remain constant! That is, 0 = dN1 + dN2 −→ dN1 = −dN2
(5.4.1)
We can also think of the free energy of each system, as we did with the particle number previously. The total free energy would be described by F = F1 + F2
(5.4.2)
And when we are in diffusive equilibrium, according to our first stipulation, F must be minimized with respect to the particle number:
W. Erbsen
STATISTICAL MECHANICS
dF =0 dN
(5.4.3)
Combining (5.4.1) and (5.4.2), dF =
∂F1 ∂F2 dN1 + dN2 ∂N1 ∂N2
(5.4.4)
Setting (5.4.4) equal to zero, as demonstrated in (5.4.3), we have ∂F2 ∂F1 dN1 = − dN2 ∂N1 ∂N2
(5.4.5)
Applying (5.4.1) once again to (5.4.5), we are left with ∂F2 ∂F1 ∂F2 ∂F1 dN1 = − (−dN1 ) −→ = ∂N1 ∂N2 ∂N1 ∂N2 We now recall that the chemical potential is defined by ∂F µ= ∂N V,T
(5.4.6)
(5.4.7)
Applying the definition of chemical potential from (5.4.7) to our result (5.4.6), it is easy to see that at equilibrium, µ1 = µ2 b)
We are tasked with finding the chemical potential for distinguishable, non-interacting particles in the classical regime for a 1-D gas. As we do in all of statistical mechanics, we begin by finding the (single particle) partition function: X Ei Z1 = exp − (5.4.8) kB T i
For a classical gas, the energy will come in the form of the kinetic energy, which is E = p2 /(2m). Furthermore, within the classical regime, we may take our energy distribution to be continuous, and as such may be integrated in the following way: Z Z 1 p2 Z1 = 3 exp − d3 p d3 x (5.4.9) h 2mkB T For a 1-D gas, as is the case here, (5.4.9) becomes Z Z L 1 ∞ p2 Z1 = exp − dp dx h −∞ 2mkB T 0 i 1 hp = · 2πmkB T · [L] h Lp = 2πmkB T h
Within the classical approximation, the full partition function can be approximated by ZN ∼ = Z1N Applying this to the single particle partition function of (5.4.10), we have
(5.4.10)
499
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
ZN =
p N L 2πmkB T h
(5.4.11)
We now recall that the Free energy takes the form
F = −kB T log [ZN ]
(5.4.12)
Substituting (5.4.11) into (5.4.12), ( N ) Lp F = − kB T log 2πmkB T h p L = − N kBT log 2πmkB T h
(5.4.13)
Now, we recall that the chemical potential, given by (5.4.7), can be calculated from (5.4.13) as p ∂ L −N kB T log µ= 2πmkB T (5.4.14) ∂N h The answer is then
p L µ = −kB T log 2πmkB T h c)
From (5.4.13), F was found to be F = −N kB T log
p L 2πmkB T h
To find the specific heat capacity at constant volume, we have to get back to work. Recall that this quantity is defined as ∂U CV = (5.4.15) ∂T V We also recall that the internal energy is given by U =−
∂ log (ZN ) ∂β
Now substituting (5.4.11) into (5.4.16), ( N ) Lp 2πmkB T h p L log 2πmkB T h s " # 2πmL2 log βh2 ( 1/ ) 2πmL2 2 log βh2
∂ U =− log ∂β =−N
∂ ∂β
=−N
∂ ∂β
=−N
∂ ∂β
(5.4.16)
W. Erbsen
STATISTICAL MECHANICS
N 2 N =− 2 =−
∂ 2πmL2 log ∂β βh2 ∂ 2πmL2 1 log 2 ∂β {z } β | h
(5.4.17)
Garbage
The derivative of (5.4.17) comes out being very simple, N 1 · 2 β N kB T = 2
U=
(5.4.18)
Putting our cleaned up expression back into (5.4.15), we find CV : ∂ N kB T N kB CV = −→ CV = ∂T 2 2
Problem 2 Consider a particle in a 1-dimensional box of side L with available energy levels (in the presence of a magnetic field B along ˆ z) ~2 π 2 εn = γ 2 n2 + m · B; γ2 = (5.4.19) 2m L ˆ Where n is an integer greater than zero and m = ±mz k. a)
What is the partition function for this particle?
b)
What is the partition function for N classical, identical, noninteracting particles in the same box in the presence of a magnetic field? Hence find an expression for the internal energy U , and the Helmholtz free energy F .
c)
What is the equilibrium pressure and magnetization of this gas?
Solution From the information supplied, En = γ 2 n2 + m · B ~ π 2 γ2 = 2m L 2
a)
→ En =
n2 ~2 π 2 +m·B 2mL2
(5.4.20)
The first step in this problem, as is in most problems in this field, is to find the partition function. Let’s ride this pony: X X En Z1 = exp − = exp [−En β] (5.4.21) kB T n n
501
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Substituting (5.4.20) into (5.4.21), 2 2 2 n ~ π Z1 = exp − +m·B β 2mL2 n 2 2 2 X n ~ π β − m · Bβ = exp − 2mL2 n 2 2 2 X n ~ π = exp − β − ±m B β z 0 2mL2 n X
(5.4.22)
I changed m · B → ±mz B0 . Because of the ± in front of the spin quantum number, we need to split (5.4.22) into two distinct sums, each addressing its respective parity. With this, (5.4.22) is now 2 2 2 2 2 2 X n ~ π n ~ π β − m B β + exp − β + m B β (5.4.23) Z1 = exp − z 0 z 0 2mL2 2mL2 n At this point, we turn to our calculus loving friends and with a tote on their pipe I decide to go the way of the integral. And it was so, the sum was transferred into an integral, and (5.4.23) becomes 2 2 2 2 2 2 Z ∞ n ~ π n ~ π Z1 = exp − β − mz B0 β + exp − β + mz B0 β dn 2mL2 2mL2 0 2 2 2 2 2 2 Z ∞ n ~ π n ~ π β exp [−m B β] + exp − β exp [m B β] dn = exp − z 0 z 0 2mL2 2mL2 0 2 2 2 2 2 2 Z ∞ Z ∞ n ~ π n ~ π β exp − β dn = exp [−mz B0 β] exp − dn + exp [m B β] z 0 2 2mL 2mL2 0 0 Z ∞ Z ∞ β~2 π 2 2 β~2 π 2 2 = exp [−mz B0 β] exp − n dn + exp [m B β] exp − n dn z 0 2mL2 2mL2 0 0 Z ∞ β~2 π 2 2 = {exp [−mz B0 β] + exp [mz B0 β]} exp − n dn 2mL2 0 Z ∞ β~2 π 2 2 n dn =2 cosh [mz B0 β] exp − 2mL2 0 " s # 1 2πmL2 =2 cosh [mz B0 β] (5.4.24) 2 β~2 From (5.4.24), it is easy to see that the partition function becomes s 2πmL2 Z1 = cosh [mz B0 β] β~2 b)
(5.4.25)
But what if the partition function is to be applied to a classical, indistinguishable, non-interacting system? From this we must find the internal energy U , as well as the free energy F . We first recall that since we have indistinguishable particles, the full partition function can be found from the single-particle partition function as ZN ZN ∼ = 1 N! Rearranging the single particle partition function (5.4.25) and applying (5.4.26), we have
(5.4.26)
W. Erbsen
STATISTICAL MECHANICS
ZN
s " #N 2πmL2 1 cosh [mz B0 β] = N! β~2
(5.4.27)
Now, the free energy is F = −kB T log [ZN ]
(5.4.28)
And plugging (5.4.27) into (5.4.28), " s #N 2 2πmL − log [N !] F = −kB T log cosh [mz B0 β] β~2
(5.4.29)
At this point, we recall Stirling’s Approximation:
log [N !] ∼ = N log [N ] − N Applying this to (5.4.29),
#N 2 2πmL − (N log [N ] − N ) F = − kB T log cosh [mz B0 β] β~2 s ( " # ) 2πmL2 = − kB T N log cosh [mz B0 β] − N log [N ] + N β~2 s ( " # ) 2πmL2 = − N kB T log cosh [mz B0 β] − log [N ] + 1 β~2 "
s
Simplifying (5.4.30) further leads us to ( F = −N kB T
"
cosh [mz B0 β] log N
s
2πmL2 β~2
#
+1
)
(5.4.30)
(5.4.31)
To find the internal energy, U , we first recall that U =−
∂ log [ZN ] ∂β
Substituting the partition function from (5.4.27) into (5.4.32), s " #N 2 ∂ 1 2πmL U =− log cosh [mz B0 β] ∂β N! β~2 s " " ## 1 ∂ 2πmL2 =−N log cosh [mz B0 β] 1 ∂β β~2 N ! /N r ∂ 1 2πmL2 cosh [mz B0 β] √ =−N log 1 2 /N ∂β β |N ! {z ~ } Garbage
Proceeding,
(5.4.32)
503
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
∂ cosh [mz B0 β] √ U =−N log ∂β β n hp io ∂ =−N log [cosh [mz B0 β]] − log β ∂β ∂ 1 ∂ =−N log [cosh [mz B0 β]] − log [β] ∂β 2 ∂β Playing around with (5.4.33) a bit we can see that we have our answer: 1 U = −N mz B0 tanh [mz B0 β] − 2β c)
(5.4.33)
(5.4.34)
In order to find the equilibrium pressure, we first recall that P =−
∂F ∂F ⇒− ∂V ∂L
(5.4.35)
Where I took V → L since we are only in 1-D. Anyway, we can substitute our value for the Free Energy from (5.4.31) into (5.4.35): s " ( " # )# ∂ cosh [mz B0 β] 2πmL2 P =− −N kB T log +1 ∂L N β~2 s # # " " ∂ cosh [mz B0 β] 2πmL2 =N kB T +1 log ∂L N β~2 r ∂ cosh [mz B0 β] 2πm =N kB T (5.4.36) log L + 1 ∂L N β~2 | {z } Garbage
Eww smelly garbage yuck yuck. Anyway, ignoring the icky garbage, (5.4.36) becomes P =N kB T
N kB T ∂ [L] −→ P = ∂L L
(5.4.37)
The Magnetization, however, is a different story entirely. The relationship we are interested in is M =−
∂F ∂B
Substituting our previous value of F into (5.4.38) yields s " ( " #)# ∂ cosh [mz B0 β] 2πmL2 M =− −N kB T log +1 ∂B N β~2 s " # ∂ cosh [mz B0 β] 2πmL2 =N kB T log +1 ∂B N β~2 s ∂ 1 2πmL2 =N kB T log cosh [mz B0 β] +1 ∂B N β~2 ∂ M =N kB T log cosh [mz B0 β] ∂B
(5.4.38)
W. Erbsen
STATISTICAL MECHANICS
=N kB T [mβ tanh [mz B0 β]] Which finally leaves us with M = N mz tanh [mz B0 β]
Problem 4 a)
From the condition that two coexisting phases of a given chemical substance must have the same chemical ptential µ1 = µ2 , derive the Claussius-Clayperon equation S2 − S1 dP = dT V2 − V1
b)
(5.4.39)
Drawn below is the phase diagram for He4 . Carefully explain why the slope of the liquid-solid phase boundary is zero at T = 0.
Solution a)
We first recall that a common form of the first law of thermodynamics is dU = T dS − P dV
(5.4.40)
While the Helmholtz and Gibbs free energy are given, respectively, by F =U − T S G =F + P V
(5.4.41a) (5.4.41b)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
If we make the substitution (5.4.41a) into (5.4.41b), we have G = U − TS + PV
(5.4.42)
Taking (5.4.42) to the differential limit, we have dG =dU − T dS − S dT + P dV + V dP =dU − (T dS − P dV ) − s dT + V dP =dU − dU − S dT + V dP = − S dT + V dP
(5.4.43)
Where I substituted in (5.4.40) in the second to last step arriving at (5.4.43). We now recall that the chemical potential is given by ∂G ∂F µ= ⇒ (5.4.44) ∂N V,T ∂N We can now play around with (5.4.44) as follows µ=
∂G G −→ dµ = −→ dG = N dµ ∂N N
(5.4.45)
Applying (5.4.45) to (5.4.43), n dµ =V dP − S dT V S dµ = dP − dT N N
(5.4.46)
We now temporarily recall that since we are at diffusive equilibrium, then µ1 = µ2 and furthermore, dµ1 = dµ2 . Ergo, (5.4.46) is transformed into V1 S1 V2 S2 dP − dT = dP − dT N N N N V1 V2 S1 S2 dP − dP = dT − dT N N N N dP (V1 − V2 ) =dT (S1 − S2 )
(5.4.47)
From (5.4.47), it is most easily seen that dP S1 − S2 = dT V1 − V2 b)
(5.4.48)
We are given a phase diagram for He4 , which we immediately recognize as a Boson. This is especially interesting because Bosons do not obey the Pauli Exclusion Principle, and can therefore all occupy the same state at an arbitrarily low temperature. The multiplicity, that is, the number of states available to a system without changing anything, is only 1. e.g., at low temperatures, there is only 1 way that the system can be configured (all in the ground state), and subsequently the multiplicity is Ω = 1. This is particularly interesting, since He4 has a multiplicity of 1, then the entropy is 0: S = kBT log [Ω] −→ S = kB T log [1] −→ S = 0 The Claussius-Clayperon equation that we derived earlier in (5.4.48) describes the slope of the phase boundaries. In the case that we have low temperatures (approaching zero), where the multiplicity,
W. Erbsen
STATISTICAL MECHANICS
Ω, is 1, while the entropy, S, is 0. Plugging this in to (5.4.48), we get dP 0−0 dP = −→ =0 dT V1 − V2 dT The Claussius-Clayperon equation can be thought of as the “slope equation” for phase diagrams. When the result is zero, that implies that there is no slope, and so you get a straight line.
Problem 5 Consider an ideal gas of diatomic molecules. The rotational motion is quantized according to εj = j(j + 1)ε0 , j = 0, 1, 2, ... The multiplicity of each rotational level is gj = 2j + 1. a)
Find the rotational part of the partition function Z for one molecule.
b)
Convert the sum (for Z) to an integral and evaluate the integral for kT ε0 . From this find an expression for the specific heat capacity at constant volume at low temperatures.
Solution a)
For a system consisting of degeneracy, the form of the partition function that we will employ is X Ej Z1 = gj exp − (5.4.49) kB T j
In our case, we have gj = 2j + 1 ,
εj = j(j + 1) ε0
Applying both of these to our single-particle partition function in (5.4.49), we have X j(j + 1) ε0 Z1 = (2j + 1) exp − kB T
(5.4.50)
j
b)
Converting the sum in (5.4.50) into an integral: Z ∞ j(j + 1) ε0 Z1 = (2j + 1) exp − dj kB T 0 u = j(j + 1) = j 2 + j du = 2j + 1 dj And using this du substitution, we can transform (5.4.51) into Z ∞ uε0 Z1 = exp − du kB T 0 ∞ kB T uε0 =− exp − ε0 k B T 0
(5.4.51)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
From which it is easy to deduce that Z1 =
kB T ε0
(5.4.52)
We need to move away from the single-particle partition function, and since we are dealing indistinguishable entities: ZN ZN ∼ = 1 N!
(5.4.53)
Substituting the single particle partition function from (5.4.52) into (5.4.53), N 1 kB T ZN = N! ε0 Now hold on to yer britches, and recall that ∂U , CV = ∂T V
U =−
(5.4.54)
∂ log [ZN ] ∂β
First, let’s evaluate the internal energy U , by substituting into (5.4.54) this expression: " N # ∂ 1 1 U =− log ∂β N ! βε0 ∂ 1 1 =−N log ∂β N ! 1/N βε0 1 1 1 ∂ =−N log 1/ N ∂β ε0 β N! {z } |
(5.4.55)
And (5.4.55) becomes
(5.4.56)
N −→ u = N kB T β
(5.4.57)
U =−N
∂ log ∂β
1 β
Carrying out the simple derivative in (5.4.56) leads us to U =
To find the specific heat, we substitute the internal energy (5.4.57) into the appropriate formula CV =
∂ (nkB T ) −→ CV = N kB ∂T
For an ideal gas with three degrees of freedom, we find that U=
3 N kB T 2
And so the specific heat at constant volume becomes CV =
5 N kB 2
W. Erbsen
STATISTICAL MECHANICS
Problem 6 a)
Write down the Grand Partition Function (Gibbs sum), Ω. Define all quantities used and explain what the summations are over.
b)
Show that the average number of particles in a Grand Canonical Ensemble is given by hN i = τ
∂ log Ω ∂µ
(5.4.58)
where τ = kT and µ is the chemical potential. c)
Now consider a dilute gas of hydrogen atoms. Assume each atom may have the following states: State Ground Positive ion Negative ion
No. of Electrons 1 0 2
Energy 0 −∆ +δ
Find the condition that the average number of electrons per atom is 1.
Solution a)
We recall that within the framework of the Grand Canonical Ensemble, the system in question is in both thermal and diffusive equilibrium. The sum itself takes the form ∞ X X N µ − Es,N Ω (µ, T ) = exp (5.4.59) kB T N=0 s(N)
Ω = The Gibbs Sum µ = The Chemical Potential N = The Number of Particles Es,N = The energy of N-particles in S-states b)
We recall that, in general, the average number of particles is X hN i = N · P (N, Es,N )
(5.4.60)
s,N
Where the probability function is given by
1 N µ − Es,N P (N, Es,N ) = · exp Ω (µ, T ) kB T
(5.4.61)
And substituting the probability function (5.4.61) back into (5.4.60) to find the average number of particles, X 1 N U − Es,N hN i = N· · exp (5.4.62) Ω (µ, T ) kB T s,N
For the time being, let’s take the following (seemingly) arbitrary derivative:
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
∂Ω (µ, T ) ∂ X N µ − Es,N = exp ∂µ ∂µ kB T N,s(N) X N N µ − Es,N exp = kB T kB T N,s(N) X 1 N µ − Es,N = N exp kB T kB T
(5.4.63)
N,s(N)
Solving (5.4.63) for the portion that resembles a part of (5.4.62), X N µ − Es,N ∂ Ω (µ, T ) = N exp kB T · ∂µ kB T
(5.4.64)
We can substitute the RHS of (5.4.64) into (5.4.62): X kB T ∂ hN i = · Ω (µ, T ) Ω (µ, T ) ∂µ
(5.4.65)
N,s(N)
s,N
Let’s rewrite (5.4.65): hN i =
X
kB T
s,N
1 ∂ · Ω (µ, T ) Ω (µ, T ) ∂µ {z } |
(5.4.66)
Let’s look a little closer at the Gibbs function from (5.4.59) (ignore sums momentarily): N µ − Es,N Ω (µ, T ) = exp kB T Let’s take the first derivative with respect to µ:
∂ ∂ N µ − Es,N Ω (µ, T ) = exp ∂µ ∂µ kB T N N µ − Es,N = exp kB T kB T
(5.4.67)
If we take the first derivative of the Gibbs function with respect to µ, we get some coefficient times the original function. Now, there’s two ways of getting rid of the original function: assuming that you just want to play around with the coefficients for some reason. You could a) simply divide your answer by the original function, or b), you could take the same derivative you took before, but had the natural log of the function: 1 ∂ ∂ · Ω (µ, T ) = log [Ω (µ, T )] Ω (µ, T ) ∂µ ∂µ
(5.4.68)
We note that the LHS of (5.4.68) is identical to the underbraced term in (5.4.66). Substituting the RHS of (5.4.68) into (5.4.66), replacing the underbraced term, we have: X ∂ hN i = kB T log [Ω (µ, T )] (5.4.69) ∂µ s,N
Rearranging, (5.4.69) becomes
W. Erbsen
STATISTICAL MECHANICS
hN i = kB T
∂ log [Ω (µ, T )] ∂µ
(5.4.70)
Where we note that the summation is implicit in the Gibbs function. c)
The first thing that we have to do is find the Gibbs sum. Using the probability function (5.4.61) as our outline, the sum itself takes the form ∞ X X N µ − Es,N (5.4.71) Ω (µ, T ) = exp kB T N=0 s(N)
There are only three unique states, so the second sum in (5.4.71) only goes to 3. We also know that we have three states of electrons, of occupancy 0, 1, or 2 as given in the provided table. Accordingly, (5.4.71) becomes (1)µ − 0 (0)µ − (−∆) (2)µ − (δ) Ω (µ, T ) = exp + exp + exp kB T kB T kB T µ ∆ 2µ − δ = exp + exp + exp kB T kB T kB T = exp [µβ] + exp [∆β] + exp [(2µ − δ) β] (5.4.72) Substituting (5.4.72) into the equation for average number of particles from (5.4.70), 1 ∂ log [exp [µβ] + exp [∆β] + exp [(2µ − δ) β]] β ∂µ 1 β exp [µβ] + 2β exp [(2µ − δ)β] = β exp [µβ] + exp [∆β] + exp [(2µ − δ)β] exp [µβ] + 2 exp [(2µ − δ)β] = exp [µβ] + exp [∆β] + exp [(2µ − δ)β]
hN i =
(5.4.73)
We are actually looking for a condition where hN i = 1, so setting this in (5.4.73) yields exp [µβ] + exp [∆β] + exp [(2µ − δ)β] = exp [µβ] + 2 exp [(2µ − δ)β] exp [∆β] = exp [(2µ − δ)β] ∆β = (2µ − δ)β
(5.4.74)
From (5.4.74), one of the conditions that might be met is in terms of the chemical potential, which reads µ=
∆+δ 2
Problem 7 A system of N distinguishable noninteracting particles has one-particle energy levels of En = nε, where the degeneracy of the nth level is equal to n, and n = 1, 2, 3, ...
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
Show that the partition function of the system is" Z=
−1
where β = (kB T )
eβε 2
(eβε − 1)
#N
(5.4.75)
.
b)
Calculate the entropy of the system of particles.
c)
Determine the behavior of the entropy at high temperature.
d)
Determine the behavior of the specific heat at high temperatures.
Solution We first recognize that since we have distinguishable particles, then the partition function as a function of the single-particle partition function would be ZN ∼ = Z1N . a)
The standard form of the partition function (with degeneracy) is X En Z1 = n · exp − kB T n We now substitute En = nε into (5.4.76), Z1 =
(5.4.76)
X
nε n · exp − kB T
(5.4.77)
X
n
(5.4.78)
n
We can rewrite (5.4.77) slightly, Z1 =
n
ε n · exp − kB T
And now, we see that if we were to take some arbitrary derivative of some function then the answer could be (5.4.78). One such form is 1 ∂ X n Z1 = − n · exp [−εβ] (5.4.79) ε ∂β n Where I have changed form kB T to β −1 . We now recall the following summation identity: X r rn = 1−r
Applying this to (5.4.79), 1 ∂ exp [−εβ] Z1 = − ε ∂β 1 − exp [−εβ] 1 ∂ exp [−εβ] =− ε ∂β 1 − exp [−εβ] 1 ∂ 1 1 ∂ =− exp [−εβ] + exp [−εβ] ε ∂β 1 − exp [−εβ] 1 − exp [−εβ] ∂β
W. Erbsen
STATISTICAL MECHANICS
1 =− ε = = = =
(
exp [−εβ]
exp [−2εβ]
"
(1 − exp [−εβ])2
2
+
2
+
(1 − exp [−εβ]) exp [−2εβ]
#
−ε exp [−εβ]
1 + [−ε exp [−εβ]] 1 − exp [−εβ]
)
exp [−εβ] 1 − exp [−εβ]
exp [−εβ] 1 − exp [−εβ] · 1 − exp [−εβ] 1 − exp [−εβ]
(1 − exp [−εβ]) exp [−2εβ] + exp [−εβ] − exp [−2εβ] 2
(1 − exp [−εβ]) exp [−εβ] 2
(1 − exp [−εβ]) exp [−εβ] exp [2εβ]] = · 1 − 2 exp [−εβ] + exp [−2εβ] exp [2εβ]] exp [εβ] = exp [2εβ] − 2 exp [εβ] + 1
(5.4.80)
We an finally say that (5.4.80) can be reduced to exp [εβ]
Z1 =
(exp[εβ] − 1)2
(5.4.81)
Converting (5.4.81) into the full partition function, ZN = b)
"
exp [εβ] 2
(exp[εβ] − 1)
#N
(5.4.82)
To calculate the entropy for a system of particles, we first recall that S=−
∂F ∂T
(5.4.83)
So, we first need to find F : F = −kB T log [ZN ] Substituting the partition function (5.4.82) into (5.4.84), " #N exp [εβ] F = − kBT log (exp[εβ] − 1)2 " # exp [εβ] = − N kBT log 2 (exp[εβ] − 1) n h io 2 = − N kBT log [exp [εβ]] − log (exp [εβ] − 1) = − N kBT {log [exp [εβ]] − 2 log [(exp [εβ] − 1)]}
Substituting (5.4.85) back into (5.4.83), we can find S, S =−
∂ [−N kB T {log [exp [εβ]] − 2 log [(exp [εβ] − 1)]}] ∂T
(5.4.84)
(5.4.85)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
∂ ε ε T log exp + 2 log exp −1 ∂T kB T kB T ∂ ε ε +2 log exp −1 =N kB |{z} ∂T kB kB T |{z} Don’t matter =N kB
(5.4.86)
Garbage
Rewriting (5.4.86):
∂ ε log exp −1 S =2N kB ∂T kB T h i exp kBεT ε h i =2N kB − kB T exp ε − 1
(5.4.87)
kB T
Rearranging (5.4.87),
h i ε exp 2N kB T h i S =− ε T exp kB T − 1
(5.4.88)
Problem 8 In a mono-atomic crystalline solid each atom can occupy either a regular lattice site or an interstitial site. The energy of an atom at an interstitial site exceeds the energy of an atom at a lattice site by ε. Assume that the number of atoms, lattice sites, and interstitial sites are all equal in number. a)
Calculate the entropy of the crystal in the state where exactly n of the N atoms are at interstitial sites.
b)
What is the temperature of the crystal in this state, if the crystal is at thermal equilibrium?
c)
If ε = 0.5 eV and teh temperature of the crystal is 273 K, what is the fraction of atoms at interstitial sites?
Hint: You have two choices to make in this problem - which atoms to put at interstitial sites, and which interstitial sites to put them in.
Solution a)
To calculate the entropy, we recall that S=−
∂F ∂T
(5.4.89)
And also that the Free energy is F = −kB T log [ZN ] While the single-particle partition function in the case of degeneracy is
(5.4.90)
W. Erbsen
STATISTICAL MECHANICS
Z1 =
X i
Ei gi · exp − kB T
(5.4.91)
Now, there are two different sources of degeneracy in this problem. There is one source coming from the n interstitial lattices, and then there is N for the regular lattices. So, combining these into (5.4.91), yields X X En EN Z1 = n exp − + N exp − (5.4.92) kB T kB T n N
Depending on where we are on the lattice we have a different energy, such that we should rewrite (5.4.92) as X X (E0 + ε) E0 Z1 = n exp − + N exp − kB T kB T n N (X X ) E0 ε = exp − n exp − + N (5.4.93) kB T k BT n N
At this point, we say that the degeneracy for this particular two-level case can be expressed in the following more compact form: g(N, n) =
N! (N − n)!n!
Using this equation, we can completely transform (5.4.93), which now reads N! E0 ε ZN = · exp − exp − +1 (N − n)!n! kB T kB T Using (5.4.90), we can now find the Free Energy by way of (5.4.95) as follows N! E0 ε F = − kB T log · exp − exp − +1 (N − n)!n! kB T kB T N! E0 ε = − kB T log + log exp − + log exp − +1 (N − n)!n! kB T kB T N! E0 ε = − kB T log − + log exp − +1 (N − n)!n! kB T kB T And to calculate the entropy, we now substitute (5.4.96) back in to (5.4.89): ∂ N! E0 ε S =− −kB T log − + log exp − +1 ∂T (N − n)!n! kB T kB T ∂ N! E0 ε =kB T log − + T log exp − +1 ∂T (N − n)!n! kB kB T N! ε ∂ ε =kB log + log exp − +1 +T log exp − +1 (N − n)!n! kB T ∂T kB T N! ε ε exp [−ε/(kB T )] =kB log + log exp − +1 +T (N − n)!n! kB T T 2 exp [−ε/(kB T )] + 1 We can simplify (5.4.97) again:
(5.4.94)
(5.4.95)
(5.4.96)
(5.4.97)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
N! ε ε exp [−ε/(kB T )] S = kB log + kB log exp − + 1 + kB (N − n)!n! kB T T exp [−ε/(kB T )] + 1 b)
(5.4.98)
We recall that from the first law of thermodynamics, dU = T dS − P dV
(5.4.99)
At constant volume, (5.4.99) becomes dU = T dS −→
dS 1 = dU T
(5.4.100)
Applying the chain rule to (5.4.100), dS dS dU dn dS dn 1 = · · = · = dU dU dn dU dn dU T And now, U =n (E + ε) + (N − n) E =nE + nε + N E − nE =nε + N E Taking the differential limit, dU = ε dn −→
dn 1 = dε ε
Substituting this back in to our chain rule, 1 dS ε dS 1 · = −→ = dn ε T dn T
(5.4.101)
At this point, we apply (5.4.101) to (5.4.98): d N! ε ε exp [−ε/(kB T )] dS = kB log + kB log exp − + 1 + kB dn dn (N − n)!n! kB T T exp [−ε/(kB T )] + 1 d N! =kB log (5.4.102) dn (N − n)!n! We can further rewrite (5.4.102), dS d =kB {log [N !] − log [(N − n)!] + log [n!]} dn dn
(5.4.103)
log [N !] = N log [N ] − N
(5.4.104)
Recall the identity
Using (5.4.104), we can rewrite (5.4.103) as: dS d =kB {log [(N − n)!] + n log [n] − n} dn dn We can apply the same logic to the mixed term in (5.4.105): log [(N − n)!] = (N − n) log [N − n] − (N − n)
(5.4.105)
W. Erbsen
STATISTICAL MECHANICS
=N log [N − n] − n log [N − n] − N + n
(5.4.106)
Substituting (5.4.106) back into (5.4.105), d dS =kB {N log [N − n] − n log [N − n] − N + n + n log [n] − n} dn dn d =kB {N log [N − n] − n log [N − n] + n log [n]} dn N n =kB − log [N − n] − + log [n] + 1 N −n N −n N −n N n + − +1 =kB log n N −n N −n N −n =kB log +2 n We now set (5.4.107) to ε/T :
ε N −n kB log +2 = n T
(5.4.107)
(5.4.108)
Solving (5.4.108) for T , T = c)
ε kB
−1 N −n log +2 n
(5.4.109)
To find the fraction of atoms in the interstitial sites, we simply solve (5.4.109) for N/n: ε N −n = log +2 kB T n ε N −n exp −1 = (5.4.110) kB T n From (5.4.110) we can see that N ε = exp −1 +1 n kB T
(5.4.111)
Problem 9 Calculate the magnetic susceptibility χ=
∂M ∂H H=0
(5.4.112)
Where M is the magnetic moment of the sample and H is the applied field. Calculate χ as a function of T of a gas of N permanent dipoles, each of moment µ, a)
If any direction is allowed (classical spines).
b)
If the dipole is only allowed to assume 2 directions, parallel and opposite to the applied field (Ising spins).
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Solution a)
The first thing that we recall is the energy of a dipole in an external field, which is given by U = −µ · H
(5.4.113)
And the magnetization is given by M =−
∂F ∂H
(5.4.114)
Now, if we have a whole bunch of dipoles, then they are not all perfectly aligned (or antialigned), but rather at some random angle θi . Then the total energy from the entirety of the molecules can be found as X U =−µ·H cos θi i
= − µH
X
cos θi
(5.4.115)
i
Since the particles in this case are distinguishable, the full partition function can be found from the single particle partition function as ZN ∼ = Z1N . The single-particle partition function is X µH cos θi (5.4.116) Z1 = exp kB T i Turning the sum in (5.4.116) into an integral, we have Z 2π Z π µH cos θ Z1 = exp sin θ dθdφ kB T 0 0 Letting u = cos θ and du = − sin θ dθ, (5.4.117) becomes Z −1 µHu Z1 = − 2π exp du kB T 1 −1 µHu kB T exp = − 2π · µH k B T 1 kB T −µH µH = − 2π · exp − exp µH kB T kB T kB T −µH =4π · sinh µH kB T The full partition function can be found from (5.4.118) easily as N kB T −µH ZN = 4π · sinh µH kB T We now use (5.4.119) to find the Free Energy, F : F = − kB T log [ZN ] ( N ) kB T µH = − kB T log −4π · sinh µH kB T
(5.4.117)
(5.4.118)
(5.4.119)
W. Erbsen
STATISTICAL MECHANICS
kB T µH = − N kB T log −4π · sinh µH kB T
(5.4.120)
We now substitute F from (5.4.120) into (5.4.114), the equation for the Magnetization, M : ∂ kB T µH M =− −N kB T log −4π · sinh ∂H µH kB T ∂ kB T µH =N kB T log −4π · sinh ∂H µH kB T 4πkB T µH ∂ log − log [H] + log − sinh =N kB T ∂H µ kB T 1 ∂ µH =N kB T − + log − sinh H ∂H kB T 1 µ µH =N kB T − + cosh H kB T kB T kB T µH + µ cosh =N − (5.4.121) H kB T Now, according to the prompt, to find the susceptibility χ, all we must do is differentiate M with respect to H. Doing this, we take the derivative of (5.4.121) with respect to H: ∂M ∂H ∂ kB T µH + µ cosh = N − ∂H H kB T ∂ 1 ∂ µH +µ cosh =N −kB T ∂H H ∂H kB T 1 µ 2 µH =N −kB T − 2 − µ − csch H kB T kB T 2 kB T µ µH =N + csch2 H2 kB T kB T
χ=
(5.4.122)
While (5.4.122) is the exact answer, and I believe that we were supposed to use an approximation. We make this approximation in (5.4.121), where we expand the cosh term in a series expansion: ∞ X kB T µH µ3 H 3 µH cosh = + − 3 T 3 ) + ... kB T µH 3kBT 45 (kB H=0
If we assume that the field is weak, then we can approximate this function by its first two expansion terms. Substituting them back into the Magnetization from (5.4.121), kB T kB T µH N µ2 H M =N − +µ + −→ M = H µH 3kB T 3kBT It is trivial then, to find the susceptibility in the weak-field regime: ∂ N µ2 H N µ2 χ= −→ χ = ∂H 3kB T 3kBT b)
We now need to see how things change if the dipoles are only allowed to assume 2 directions, parallel and anti-parallel. The only difference is that here our energies are discrete, so our problem
519
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
is actually easier. I will fly through it rather quickly. Firstly, the single particle partition function is µH µH + exp Z1 = exp − kB T kB T Whose full counterpart is then ZN
N µH µH = exp − + exp kB T kB T
And (5.4.123), we can find the Fermi energy ( N ) µH µH F = − kB T log exp − + exp kB T kB T µH µH + exp = − N kB T log exp − kB T kB T µH = − N kB T log 2 cosh kB T
(5.4.123)
(5.4.124)
Finding the magnetization of (5.4.124), µH ∂ log 2 cosh M =N kB T ∂H kB T µ µH =N kB T tanh kB T kB T
(5.4.125)
Approximating the first two terms in tanh in very much the same way we did before, ∞ X µH µH µ3 H 3 2µ5 H 5 tanh = − + 3 5 T 5 + ... 3 kB T kB T 3 (kB T ) 15kB H=0
Substituting in the first two terms back into the Magnetization equation (5.4.125), µH µ3 H 3 µ M =N kB T − 3 T 3) kB T kB T 3 (kB µH µ3 H 3 =N µ − 3 T 3) kB T 3 (kB And finally, to find the susceptibility we differentiate (5.4.126) with respect to H: ∂ µH µ3 H 3 χ= Nµ − 3 3 ∂H kB T 3 (kB T ) Playing around with this a little, we are finally left with χ = Nµ
µ µ3 H 2 − 3 3 kB T (kB T )
(5.4.126)
W. Erbsen
STATISTICAL MECHANICS
Problem 10 Similar to the van der Waals equation of state is the Dieterici equation of state, p(V − b) = RT e−a/RT V
(5.4.127)
Find the critical constants pc, V c and T c in this model of a weakly interacting gas. This equation of state was proposed to account for the interaction of gas atoms with walls.
Solution We first recall what a typical phase diagram looks like, which is of P vs. T . The critical point is where the phase boundaries cease to exist, and all phases coexist simultaneously. This point has the coordinates PC and TC . We can also plot a phase diagram of P vs. V , plotting a variety of different isotherms (lines of constant temperature). We find that in plotting one particular isotherm, that has a point which coincides with Pc and VC from the P T diagram plot. At this point, the slope is zero. The isotherms that lie inside the shaded area are in stable equilibrium of gas and liquid. At the critical point, on the isotherm at the critical temperature, the first (and therefore, the second) derivative with respect to V must be zero: 2 ∂P ∂ P = =0 ∂V Tc ∂V 2 Tc Solving the equation of state given from (5.4.127) for P yields h RT a i P = · exp − V −b RT V Taking the first derivative of (5.4.128) yields h ∂P ∂ RT a i = · exp − ∂V Tc ∂V V − b RT V h h RT ∂ a i a i ∂ RT = exp − + exp − V − b ∂V RT V RT V ∂V V − b h h RT h a a ii a i RT = exp − + exp − − V − b RT V 2 RT V RT V (V − b)2 h i h i a a RT a = 2 exp − − exp − V (V − b) RT V (V − b)2 RT V exp − RTa V a RT = − V −b V2 V −b And setting (5.4.129) equal to zero, ∂P a RT =0 −→ 2 = −→ V 2 RT − aV + ab = 0 ∂V Tc V V −b
(5.4.128)
(5.4.129)
(5.4.130)
Taking the second derivative of P is the same as taking the first derivative of (5.4.129), so let’s do that:
521
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
∂2P ∂V 2
Tc
# exp − RTa V a RT − V −b V2 V −b 2 exp − RTa V a (b − V )2 − 2aRT V (b − 2V )(b − V ) + 2R2 T 2 V 4 = 4 3 RT V (V − b) ∂ = ∂V
"
Setting (5.4.131) equal to zero, 2 ∂ P = 0 −→ a2 (b − V )2 = 2aRT V (b − 2V )(b − V ) + 2R2 T 2 V 4 ∂V 2 Tc
(5.4.131)
(5.4.132)
We now have two equations, and two unknowns. Let’s now solve (5.4.130) for T T =
a(V − b) RV 2
(5.4.133)
And substituting this into our second equation (5.4.132), 2 a(V − b) 2 a(V − b) a (b − V ) =2aRV (b − 2V )(b − V ) + 2R V4 RV 2 RV 2 (b − 2V ) + 2a2 (V − b)2 −a2 (V − b)2 =2a2 (V − b)2 V (b − 2V ) −1 =2 +2 V 2(2V − b) 3= V 2
2
From (5.4.134), we can see that Vc = 2b . Plugging in this value into (5.4.133) yields T =
a a(2b − b) −→ Tc = 2 R(2b) 4bR
Substituting both Vc and Tc into P from (5.4.128), R h a i a 4bR 1 a P = · exp − · · −→ Pc = 2 · exp [−2] 2b − b 4bR R a 2b 4b
Problem 11 For a gas of molecules with diameter d, number density n and at temperature T , find a)
The mean free path
b)
Their average speed
c)
The pressure of the gas using kinetic arguments
(5.4.134)
W. Erbsen
STATISTICAL MECHANICS
Solution a)
The mean free path is defined as the average distance that a gas particle transverses between collisions. If we imagine two gas particles, each of diameter d, passing close to one another, the question is: what is the effective collisional diameter? In other words, if two individual particles happen to collide at one point in space, then it is easy to see that if any particle approaches any other particle, then the net new particle diameter would have to be the sum of the two collided molecules, or 2d. This illustration works particularly well if we make a number of critical assumptions: i)
Only one particle is moving, while all other molecules (target molecules) are stationary. If the moving particle comes within a distance of 2d of a stationary particle, then a collision occurs. → The effective collision area is then A = πd2 .
If the particle travels through space some distance, and in its wake it sweeps out a volume proportional to its cross sectional area and the distance that it traveled. The volume of such a sphere is V = `A = `πd2 , where ` is the length of the swept out cylinder. We must also recognize that we must harness the time that passes while the particle snakes its path through space. Recalling that velocity is defined as v = ∆x/∆t, and in our case ∆x = ` and ∆t is just plain old t. Then we can say that v is the average velocity, and t is the time between successive collisions.
When the volume of this imaginary cylinder equals the average volume per particle, then we are likely to get a collision. → The mean free path (`) is the length of the cylinder when this condition is met Quantitatively, this condition is Volume of Cylinder =Avg Volume per Particle V πd2 ` = N
(5.4.135)
We can solve (5.4.135) for the mean free path, `: `=
1 V · πd2 N
(5.4.136)
It is important that we recognize that (5.4.136) is only an approximation, since according to stipulation i), we have assumed that the target molecules are stationary, which is most certainly never true. To remedy this, we can rewrite ` from (5.4.136) in terms of the average velocity, v: `=v·t=
1 V · 2 πd N
(5.4.137)
In truth, what we are interested in is the relative velocity between the various particles, not the average velocity. Luckily, this question has been asked before, and a quantitative link has been defined relating the two quantities. The savior comes in the form of the Maxwell-Boltzmann Speed Distribution Function, which I shall derive later. The punch-line is √ 1 v rel = 2 v −→ v = √ vrel (5.4.138) 2
523
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Substituting (5.4.138) into (5.4.137), 1 1 V ` = √ v rel · t −→ ` = √ · 2 2 2 πd N b)
(5.4.139)
We recall that any probability can be obtained by integrating the probability distribution function over a particular range: Z v2 P (v) = f(v) dv v1
Our goal is to find some distribution function that allows us to integrate and find a value for speed. We also note that there are many different speeds that can correspond to any particular velocity. In general, Probability of having Number of velocity vectors f(v) ∝ · (5.4.140) a velocity v corresponding to the same speed The first factor in (5.4.140) is simply the Boltzmann factor, so we can begin building our distribution function: mv2 f(v) ∼ exp − (5.4.141) 2mkB T And for the second term in (5.4.140), we need to get a little more abstract.. we imagine a 3-D coordinate system in which the axes correspond to different velocity components vx , vy , and vz . Some speed then, will act to mark an invisible sphere around the origin of our coordinate system, such that if one end of the vector is planted at the origin, it can trace out the sphere in any direction it pleases. This “velocity space” analogy convinces us that the second factor in (5.4.140) is only a proportionality factor, equal to the area that is swept out in v-space. This area is A = 4πv2 . Therefore, from (5.4.141), our distribution function is now mv2 f(v) ∼ exp − · 4πv2 · C (5.4.142) 2mkB T Where C normalizes our distribution function. Now, Z ∞ f(v) dv = 1 0 Z ∞ mv2 2 4πC v exp − dv = 1 2mkB T 0 " 3 # 1/2 1 2kB T 4πC π =1 4 m
3/ 2πkB T 2 C· =1 m 3/2 m C= 2πkBT
Substituting (5.4.143) back into our function (5.4.142)
(5.4.143)
W. Erbsen
STATISTICAL MECHANICS
m f(v) = 4πv · 2πkB T 2
3/2
mv2 · exp − 2mkB T
(5.4.144)
This, (5.4.144), is the famous Maxwell-Boltzmann Distribution Function. We will now use it. But first, going back to the fundamentals: an average quantity can be found by Z x hxi = x · f(x) dx 0
Following this example, we now wish to find the average speed, v, using our distribution function from (5.4.144): Z ∞ v= v · f(v) dv 0
m =4π 2πkB T
3/2 Z
0
∞
mv2 v · exp − 2mkB T 3
dv
(5.4.145)
The integral in (5.4.145) is tedious , and I would recommend doing it in Mathematica or possibly a table. Anyway, the result is r 8kB T v= (5.4.146) πm c)
What is pressure? Well, kinetically speaking, pressure is the average force exerted on the walls of some container. Mathematically, Favg = N ·
mv 2x L
(5.4.147)
Now, if the speed is the same in all directions, then (5.4.147) becomes Favg = 3N ·
mv2 L
(5.4.148)
And we recall that the pressure is defined as P = F/A, in our case, the force is the average force applied to the sides of the container. Using (5.4.148), the pressure becomes P =
Favg 3mv2 −→ P = N · Area L·A
Problem 12 Consider a large reservoir energy U0 − ε in thermal contact with a system with energy ε. a)
Give an argument based on the multiplicity (number of distinguishable ways a state with a given energy may ε be obtained) which leads to the result P (ε) ∼ exp − (5.4.149) kT
Where P (ε) is the probability of finding the system in a particular state with energy ε. Hint: Recall that the entropy S = k log (multiplicity).
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
A system of N magnetic moments (¯ µ) may P orient themselves parallel or antiparallel to an applied field. Deduce from a) the total magnetization M ≡ 1 µ ¯i as a function of H and T .
Solution a)
Multiplicity refers to the number of configurations that a system has available to it while staying essentially the same. The entropy can be linked to the multiplicity as S = kB log [Ω]
(5.4.150)
And the probability of being in one particular state is P =
1 Ω
(5.4.151)
And we recall that the first law says that dU = T dS − P dV
(5.4.152)
Since the volume in our case is constant, (5.4.152) reduces to dU = T dS −→
dS 1 = dU T
(5.4.153)
Applying the chain rule to (5.4.153), dS dU dΩ dS dΩ 1 · · = · = dU dΩ dU dΩ dU T
(5.4.154)
Sweet. Now, substitute our relation for S from (5.4.150) into (5.4.154) and let’s see what happens: dS dΩ d dΩ · = kB log [Ω] · dΩ dU dΩ dU kB dΩ · (5.4.155) = Ω dU And setting (5.4.155) equal to 1/T , as originally intended, kB dΩ 1 1 1 · = −→ dΩ = dU Ω dU T Ω kB T
(5.4.156)
Integrating both sides of (5.4.156), we find that log [Ω] =
E kB T
E Ω = exp kB T
(5.4.157)
And applying (5.4.151) to (5.4.157), we have
E P = exp − kB T b)
We first recall that for a two-state paramagnet, we can either be in the direction of the field, or anti-parallel to it:
W. Erbsen
STATISTICAL MECHANICS
E = µ·B
(5.4.158)
While the total number of domains is given by N = N↑ + N↓ Each with respective energies E↑ =µB
E↓ = − µB We also recall that the magnetization is defined as the density of magnetic moments. In our case, M = µ (N↑ + N↑ )
(5.4.159)
The single-particle partition function is then given by µB µB Z1 = exp + exp − kB T kB T µB =2 cosh kB T And assuming that our particles are distinguishable, then the full partition function can be found easily: N µB ZN = 2 cosh (5.4.160) kB T Now finding the Helmholtz Free Energy, we use the partition function from (5.4.160), and so: ( N ) µB F = − kBT log 2 cosh kB T µB = − N kBT log 2 cosh (5.4.161) kB T The magnetization, M , is then found using (5.4.161) as ∂ µB M =− −N kB T log 2 cosh ∂B kB T ∂ µB =N kB T log 2 cosh ∂B kB T µB µ =N kB T tanh kB T kB T Simplifying things a bit, the magnetization becomes M = N µ tanh
µB kB T
527
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 13 Consider a two-state system with energies 0 and ε. a)
Calculate the partition function
b)
Find the average energy at temperature T
c)
Find the heat capacity and graph it versus temperature
Solution a)
We first recall that the plain old partition function is given in its most basic form as X Es Z= exp − kB T s
(5.4.162)
We only have a two-level system, so it is easy to expand (5.4.162) to suit our needs: ε Z = exp [0] + exp − kB T Which can be rewritten as: ε Z = 1 + exp − kB T b)
In order to find the average energy, we recall that 1 X U= Es exp [−Es /(kB T )] Z s
(5.4.163)
(5.4.164)
Evaluating the internal energy from (5.4.164) out to the two allowed terms, much as we did before: U=
0 · exp [0] + ε exp [−ε/(kB T )] 1 + exp [−ε/(kB T )]
(5.4.165)
Getting rid of the null terms in (5.4.165) leads us to U=
ε exp [−ε/(kB T )] 1 + exp [−ε/(kB T )]
(5.4.166)
We can actually nicen (5.4.166) up a bit (for reasons that will become clear momentarily) by letting β = 1/(kBT ): U= c)
ε exp [−εβ] 1 + exp [−εβ]
To find the heat capacity at constant volume, we first recall: ∂U CV = ∂β V Substituting the internal energy from (5.4.167) into (5.4.168),
(5.4.167)
(5.4.168)
W. Erbsen
STATISTICAL MECHANICS
∂ ε exp [−εβ] CV = ∂β 1 + exp [−εβ] ∂ 1 1 ∂ + [ε exp [−εβ]] =ε exp [−εβ] ∂β 1 + exp [−εβ] 1 + exp [−εβ] ∂β " # 2 εβ exp [−εβ] 1 =ε exp [−εβ] + −ε β exp [−εβ] 2 1 + exp [−εβ] (1 + exp [−εβ]) # " εβ exp [−εβ] −ε2 β exp [−εβ] =ε exp [−εβ] 2 + 1 + exp [−εβ] (1 + exp [−εβ]) (" # ) exp [−2εβ] (− exp [−εβ]) (1 + exp [−εβ]) 2 =ε β 2 + (1 + exp [−εβ]) (1 + exp [−εβ]) (1 + exp [−εβ]) ) ( exp [−2εβ] − exp [−εβ] − exp [−2εβ] =ε2 β (1 + exp [−εβ])2 We can now deduce that 2
CV = −ε β
"
exp [−εβ] (1 + exp [−εβ])
2
#
Problem 14 An ideal diatomic gas has rotational energy levels given by Ej = a)
h2 j(j + 1), 8π 2 I
with degeneracies
(5.4.169)
For Oxygen, what fraction of the molecules is in the lowest rotational energy state at T = 50K? Recall that θrot =
b)
gj = 2j + 1
h2 = 2K 8π 2 IkB
(5.4.170)
Repeat this for Hydrogen, with θrot = 85K.
Solution a)
The single particle partition function (with degeneracy) is by default given by X Ej Z1 = gj · exp − kB T
(5.4.171)
Substituting in the appropriate values into (5.4.171), X 1 h2 Z1 = (2j + 1) · exp − · 2 j(j + 1) kB T 8π I j
(5.4.172)
j
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
At this point, we can rewrite Ej and Θrot as follows kB h2 · j(j + 1) 8π 2 I kB =Θrot · kB · j(j + 1)
Ej =
Substituting (5.4.173) back into our partition function from (5.4.172), X 1 · Θrot · kB · j(j + 1) Z1 = (2j + 1) · exp − kB T j X Θrot = (2j + 1) · exp − j(j + 1) T j We now wish to convert the sum in (5.4.174) into an integral: Z Θrot Z1 = (2j + 1) · exp − j(j + 1) dj T Using u = j(j + 1) = j 2 + 1 and du = (2j + 1) du, (5.4.175) becomes Z ∞ Θrot Z1 = exp − u du T 0 ∞ Θrot T u exp − =− Θrot T 0 T = Θrot
(5.4.173)
(5.4.174)
(5.4.175)
(5.4.176)
From the provided information, we have T = 50 K, and (5.4.176) becomes Z1 = b)
T 1 Θrot 1 1 −→ = −→ = Θrot Z1 T Z1 25
(5.4.177)
We recall the single particle partition function from (5.4.175) and using the fact that Θrot = 85 K, and T = 50 K, (5.4.175) becomes X 85 Z1 = (2j + 1) · exp − j(j + 1) (5.4.178) 50 j The sum in (5.4.178) seems to converge very quickly. Let’s look at the first few terms: Z1 (0) =1
85 Z1 (1) =3 · exp − · 2 = 0.11 50 85 Z1 (2) =5 · exp − · 6 = 0.00 50 So, taking only the first two terms, we have Z1 ≈ 1.100, and so the inverse of this yields 1 ≈ 0.909 Z1
W. Erbsen
STATISTICAL MECHANICS
Problem 15 On a glass surface, water molecules can hydrogen bond to either 1 or 2 OH groups as shown below with energies ε or 2ε, respectively. The molecule may also be weakly bound to the surface with energy ε1 ( ε) (not shown).
a)
Find an expression for the partition function for a water molecule on the surface.
b)
Find the limiting behavior of the internal energy U and the specific heat C at i) high temperatures and ii) low temperatures.
c)
Hence, sketch the behavior of U and C as a function of temperature.
Solution a)
The single-particle partition function is easily found to be Z1 = exp [εβ] + exp [−2εβ] + exp [−ε1 β]
b)
(5.4.179)
The relevant expression for the internal energy is U =kB T 2
∂ ∂ log [Zn ] = − log [ZN ] ∂T ∂β
(5.4.180)
Since our particles in this case are distinguishable, the partition function in (5.4.179) becomes ZN = Z1N = [exp [εβ] + exp [−2εβ] + exp [−ε1 β]]N Substituting (5.4.181) into (5.4.180), n o ∂ N U =− log [exp [εβ] + exp [−2εβ] + exp [−ε1 β]] ∂β ∂ =−N log [exp [εβ] + exp [−2εβ] + exp [−ε1 β]] ∂β ε exp [εβ] − 2ε exp [−2εβ] − ε1 exp [−ε1 β] =−N exp [εβ] + exp [−2εβ] + exp [−ε1 β] We now recall that the specific heat at constant volume is given by
(5.4.181)
(5.4.182)
531
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
CV =
∂U ∂T
(5.4.183) V
Rewriting (5.4.182) and substituting into (5.4.183), h i h i h i ε1 2ε ε − 2ε exp − − ε exp − ε exp 1 kB T kB T kB T ∂ h i h i h i CV = − N ∂T exp ε + exp − 2ε + exp − ε1 kB T kB T kB T 2 h ih h h i h ii h h i i h h iii 3ε ε+ε1 3ε 3ε 2ε+ε1 2 2 ε 4 + exp kB T + 9 exp kB T + 2εε1 exp kB T − 1 + ε1 exp kB T exp kB T N = h i h i h i2 kB T 2 ε1 3ε+ε exp k2ε + exp + exp T k T k T B B B (5.4.184)
Here, (5.4.184) represents the full expression for the specific heat. See attached plots of U vs. T as well as CV vs. T below. Cv
U 0.35
0.5 0.4
0.30
0.3 0.25 0.2 0.1 T 20
40
60
80
100 T 2
4
6
8
10
Problem 16 The Diesel engine cycle consists of several processes Suppose the cycle is applied to n moles of an ideal diatomic gas. In terms of the given pressures and volumes and the gas constant R, give expressions for a)
The heat QH given to the gas during combustion.
b)
The heat QC removed from the gas during the cooling process.
c)
The entropy change of the gas in the process BC.
d)
The entropy change of the gas in process CD.
AB:
Adiabatic compression from volume V0 to volume V1 , where pressure = p1 . The compression ratio rc = V0 /V1 is high enough that it causes ignition of the air-fuel mixture without needing a spark.
BC:
Expansion at constant pressure p1 from volume V1 to V2 < V0 , during the burning of the fuel.
CD:
Adiabatic expansion from V2 to V0 , with expansion ration = V0 /V2 .
W. Erbsen
DA:
STATISTICAL MECHANICS
Constant volume (V0 ) cooling back to the original temperature at point A.
Treat the gas as an ideal gas, and suppose that the constant pressure (CP ) and constant volume (CV ) heat capacities are known. a)
Calculate the efficiency of the cycle, defined as the ratio of the work output to the heat absorbed per cycle, ε = W/QH .
b)
Express ε as a function of γ = CP /CV , re , and rc.
c)
Calculate the entropy changes in each process of the cycle, and sketch the cycle in a S-T diagram.
Solution a)
To find the heat given off during ignition, QH, we are looking at the B→C isobar. We recall that the heat is given by Q = nC∆T
(5.4.185)
QBC = nCP [TC − TB ]
(5.4.186)
Applying this to the B→C isobar,
We now recall that the ideal gas law says that P V = nRT −→ T =
PV nR
Substituting this equation for T into (5.4.186), P1 V1 P1 V2 CP P1 QBC =nCP − −→ QBC = [V1 − V2 ] nR nR R
(5.4.187)
(5.4.188)
533
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
We now wish to find the heat QC removed during the cooling process, which along the D→A isochor. The process is very much the same as in the last part, adopting (5.4.186) for our purposes, QDA = nCV [TA − TD ]
(5.4.189)
And substituting (5.4.187) into (5.4.189), PD V0 PA V0 CV V0 QDA = nCV − −→ QDA = [PD − PA ] nR nR R c)
(5.4.190)
Now we want to calculate the change in entropy along B→C. We first recall that the entropy is defined by Q T
(5.4.191)
CP P1 [V1 − V2 ] RT
(5.4.192)
∆S = Substituting (5.4.188) into (5.4.191), ∆S =
And also that the specific heat at constant pressure is CP = T
∂S ∂T
(5.4.193)
Treating (5.4.193) like a differential equation,
1 1 S V2 dS = dT −→ = log CP T CP V1
(5.4.194)
Solving (5.4.194) for S, S = CP log
V2 V1
(5.4.195)
d)
To find the change in entropy along the isochore D→A, we can go ahead and use (5.4.195) and say that V0 S = CV log −→ S = 0 (5.4.196) V0
e)
We are now asked to find the efficiency of the entire cycle, which is given as ε = W/QH . We also recall that W = Qin − Qout , so that in our case, the efficiency is given by ε= =
QBC − QDA QBC CP P1 R [V1 − V2 ] − CP P1 R
CV V 0 R
[V1 − V2 ]
[PD − PA]
Which leads us to ε=
CP P1 [V1 − V2 ] − CV V0 [PD − PA ] CP P1 [V1 − V2 ]
W. Erbsen
STATISTICAL MECHANICS
f)
We now wish to rewrite ε in terms of some new variables, which are γ=
CP , CV
re =
V0 , V2
rc =
V0 V1
The result is ε=1− g)
1 P1 γ
PD − PA 1/ − 1/ rc re
−→ ε = 1 −
1 PD − PA rc re P1 γ rc − re
The entropy for both cycles are ∆SBC = ∆SDA
V2 = CP log V1
,
∆SAB = ∆SCD = 0
And the net change in entropy is of course ∆Snet = 0
Problem 17 A classical monatomic ideal gas of N particles (each of mass m) is confined to a cylinder of radius r and infinite height. A gravitational field points along the axis of the cylinder downwards. a)
Determine the Helmholtz energy A.
b)
Find the internal energy U and the specific heat CV .
c)
Why is CV 6= 3/2 N k?
Solution a)
There are two components of energy in this system: kinetic (p2 /2m) and potential (mgh). The Hamiltonian for this system is then p2 + mgz 2m p2x + p2y + p2z = + mgz 2m
H=
While the single-particle partition function takes the form Z Z 1 H Z1 = 3 exp − d3 pd3 r h kB T
(5.4.197)
(5.4.198)
Substituting our Hamiltonian from (5.4.197) into the partition function from (5.4.198) yields
535
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
# p2x + p2y + p2z mgz exp − − d3 pd3 r 2mkB T kB T " # Z ∞ Z ∞ p2x + p2y + p2z mgz 1 exp − dpx dpy dpz exp − dz = 3 h −∞ 2mkB T kB T 0 " # Z ∞ Z ∞ Z ∞ Z ∞ p2y 1 p2x p2z mgz = 3 exp − dpx exp − dpy exp − dpz exp − dz h −∞ 2mkB T 2mkB T 2mkB T kB T −∞ −∞ 0 i hp i hp iZ ∞ 1 hp mgz = 3· 2πmkB T · 2πmkB T · 2πmkB T exp − dz h kB T 0 3 / Z [2πmkB T ] 2 ∞ mgz exp − dz (5.4.199) = h3 kB T {z } |0
1 Z1 = 3 h
Z Z
"
We actually want to solve this last integral in polar coordinates: Z ∞ Z ∞ Z 2π Z R mgz mgz exp − dz = exp − rdrdθdz kB T kB T 0 0 0 0 Z ∞ Z 2π Z R mgz = exp − dz dθ r dr kB T 0 0 0 ∞ 2 kB T mgz R =− exp − · [2π] · mg kB T 0 2 R2 kB T · 2π · mg 2 R2 πkB T = mg =
(5.4.200)
Substituting (5.4.200) back into the original integral of our function (5.4.199), 3
[2πmkB T ] /2 R2 πkB T · Z1 = h3 mg
(5.4.201)
And for classical particles, we recall that ZN ∼ = Z1N , so applying this to the single particle partition function (5.4.201) yields " #N 3 / [2πmkB T ] 2 R2 πkB T ZN = · h3 mg " #N 3 / 2πmkB T 2 R2 πkB T = · h2 mg " # 3 N 2πmk T R2 πk T 2/3 /2 B B = · (5.4.202) h2 mg
We now recall that the internal energy is defined by U =−
∂ log [Zn ] ∂β
(5.4.203)
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STATISTICAL MECHANICS
Substituting the partition function (5.4.202) into the internal energy (5.4.203), 3 N " 2 2/3 # /2 ∂ 2πmkB T R πkB T U =− log · ∂β h2 mg 2 2/3 # R π · βmg ! 2/3 3 1 /2 /2 5/2 3N ∂ 2 m π log =− 2 ∂β β 5/2 gh3
3N ∂ log =− 2 ∂β
=−N
"
2πm βh2
3/2 1/2 5/2 ∂ 2 m π 1 log ∂β gh3 β 5/2 | {z }
(5.4.204)
Garbage
Carrying out the derivative in (5.4.204), U =N
h 5 i 5N ∂ 5N ∂ log β /2 = log [β] −→ U = kB T ∂β 2 ∂β 2
(5.4.205)
Recalling that the specific heat at constant volume is none other than the first derivative of internal energy with respect to temperature, we have from (5.4.205) ∂ 5N 5N CV = kB T −→ CV = kB ∂T 2 2 b)
We recognize that CV 6= 3/2 N kB because we have more degrees of freedom in this particular system, which stems from our original Hamiltonian.
Problem 18 A thin-walled vessel of volume V is kept at a constant temperature T . A gas leaks slowly out of the vessel through a small hole of area A into surrounding vacuum. Find the time required for the pressure in the vessel to drop to half of its original value.
Solution We first note that Pi = P0 Pf = 1/2 P0
Ni = N Nf = 1/2 N
The number of particles leaving the container over a time interval dt is dN = −nAv dt
(5.4.206)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Where we recall that n = N/V . Furthermore, the average velocity is given by r 8kBT v= πm Substituting v from (5.4.207) into (5.4.206), and solving for dt, yields r 8kB T dN = − nA dt πm r V 1 πm dt = − · dN N A 8kB T
(5.4.207)
(5.4.208)
Integrating (5.4.208), r Z N/2 1 πm V ; dN A 8kB T N N r V πm N =− log /2 − log [N ] A 8kB T
t=−
(5.4.209)
Simplifying (5.4.209) gives us a neat expression for t: r V πm t= log [2] A 8kB T
Problem 19 Consider a paramagnetic substance with the equation of state M = AH/(T − T0 ). Here M is the magnetization, H is the applied magnetic field, A and T0 are constants, and T is the temperature. The equation of state is valid only for T > T0 . Show that CM , the heat capacity at constant magnetization, is independent of M .
Solution We first recall that specific heat (in general) is defined as ∂Q ∂T ∂S C =T ∂T C=
(5.4.210a) (5.4.210b)
From the 2nd Law, if the heat capacity at constant M is indeed independent of M itself, then taking the derivative with respect to it should be equal to zero: ∂CM =0 ∂M
(5.4.211)
This is the condition that we are trying to prove. One possible way to do this is to work backwards, hoping to arrive at the same result. Plugging (5.4.210b) into (5.4.211), we have
W. Erbsen
STATISTICAL MECHANICS
∂CM ∂ ∂S = T ∂M ∂M ∂T 2 ∂ S =T ∂M ∂T
(5.4.212)
Now, we switch the order of the derivatives in (5.4.212), yielding ∂CM ∂ ∂S =T ∂M ∂M ∂T ∂ ∂S =T · ∂T ∂M
(5.4.213)
And now, we recall a few seemingly unrelated identities: dU =T dS + HM
(5.4.214a)
dU =T dS + dF dF = − SdT + HdM
(5.4.214b) (5.4.214c)
Now, don’t forget the function that was given in the prompt: M=
AH M −→ H = (T − T0 ) · T − T0 A
(5.4.215)
We now recall one of the least-studied areas within the realm of statistical mechanics. One of them is particularly useful for this problem, and that is ∂S ∂H =− (5.4.216) ∂M T ∂T M We now substitute H from (5.4.215) into the RHS of (5.4.216) ∂S ∂ M =− (T − T0 ) · ∂M T ∂T A M = A
(5.4.217)
We now wish to substitute the LHS of (5.4.216) into the RHS of (5.4.213): ∂CM ∂ ∂S =T · ∂M ∂T ∂M ∂ M =T · ∂T A
(5.4.218)
And since neither M nor A has any time dependence, (5.4.218) is clearly null, and the result could not be more clear: ∂CM =0 ∂M
539
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 20 A sample of ideal gas is taken through the cyclic process abca shown in the figure. At point a, T = 300K.
a)
What are the temperatures of the gas at points b and c?
b)
Complete the table by inserting a plus sign, a minus sign, or a zero in each indicated cell. Note that Q is positive when heat is absorbed by the gas and W is positive when work is done by the gas. ∆E is the change in internal energy of the gas. Q
W
∆E
a −→ b b −→ c c −→ a
Solution a)
We recall from the ideal gas law that P V = N kB T
(5.4.219)
From the figure, we are able to deduce that both P and V strictly by visual inspection, while T is given only at point a. We can now solve for N kB in (5.4.219): N kB =
PV T
(5.4.220)
Throughout the cycle, the fraction in (5.4.220) is constant, so to find the temperature at some other point when the other variables are given, PaVa Pb Vb Pb Vb Ta = −→ Tb = Ta Tb Pa Va
(5.4.221)
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STATISTICAL MECHANICS
Plugging in the appropriate values into (5.4.221), 7.5 × 103 N · m2 3.0 m3 (300 K) Tb = −→ Tb = 2700 K (2.5 × 103 N · m2 ) (1.0 m3 ) We can repeat the same process to find Tc by making a change of variables from b → c: Tc =
Pc Vc Ta Pa Va
(5.4.222)
And substituting in the correct values into (5.4.222), we have 2.5 × 103 N · m2 3.0 m3 (300 K) −→ Tc = 900 K Tc = (2.5 × 103 N · m2 ) (1.0 m3 ) b)
We note that Q is positive when heat is absorbed, while W is positive when work is done by the gas. Accordingly, From a→b b→c c→a
∆t ↑ ↓ ↓
Q ↑ ↓ ↑
W ↑ = ↓
Problem 21 A sample of helium gas inside a cylinder terminated with a piston doubles its volume from Vi = 1m3 to Vf = 2m3 . 6 During this process the pressure and volume are related by P V /5 = A = constant. Assume that the product P V 2 always equals /3 U , where U is the internal energy. a)
What is the change in energy of the gas?
b)
What is the change in entropy of the gas?
c)
How much heat was added to or removed from the gas?
Solution a)
Given that the original formula P V
6
/5
= A, we can solve for P before and after this expansion:
Pi =
A Vi
6/ 5
,
Pf =
A 6/ 5
Vf
And the change in energy can be found using the second given expression, PV =
2 3 U −→ U = P V 3 2
And so the change in energy, ∆U , can be found by 3 ∆U = (Pf Vf − Pi Vi ) 2
(5.4.223)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
3 = 2
A
Vf −
6 /5
Vf
1
3 = A 2
1 /5
Vf
−
A Vi
6/ 5
1 Vi
1/ 5
Vi
!
! (5.4.224)
Substituting in the appropriate values into (5.4.224), 3 ∆U = A 2 b)
1 (2 m3 )
1/ 5
−
1 (1 m3 )
1/ 5
!
(5.4.225)
To find the change in entropy of the gas, we first recall that a common form of the first law of thermodynamics is dU = T dS − P dV
(5.4.226)
1 P dU + dV T T
(5.4.227)
Solving (5.4.226) for dS, dS =
We now recall from the ideal gas law, and also from the information given in the problem, that P V = N kB T =
2 U 3
3 N kB T 2 2 U −→ T = 3 N kB 2 U N kB T = −→ P = V 3 V −→ U =
Substituting (5.4.228a)-(5.4.228c) into (5.4.227), 3 N kB 1 N kB T dS = dU + dV 2 U T V 3 1 1 = N kB dU + nkB dV 2 U V
(5.4.228a) (5.4.228b) (5.4.228c)
(5.4.229)
Integrating, (5.4.229) becomes 3 U Vf S = N kB [log U |Ufi + nkB [log V |Vi 2 3 Uf Vf = N kB log + N kB log 2 Ui Vi
(5.4.230)
And we now innocently recall that 3 1 3 1 A = A 1/ 1 / 5 2 (2 m3 ) 2 Vf 5 3 1 3 1 = A 1/ Ui = A 1 / 5 3 2 (1 m ) 2 V 5
Uf =
i
And substituting (5.4.231a) and (5.4.231b) into S from (5.4.230),
(5.4.231a) (5.4.231b)
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STATISTICAL MECHANICS
− 1/ Vf 5 3 Vf S = N kB log − 1/ + N kB log 5 2 Vi Vi 3 Vi Vf + N kB log = N kB log 10 Vf Vi 7 Vf = N kB log 10 Vi
(5.4.232)
Substituting the appropriate values into (5.4.232), we are left with S= c)
7 N kB log [2] 10
We first recall that dQ = dW + dU
(5.4.233)
To find out what W is in (5.4.233), we recall that work is defined as W = 6 6 that P V /5 = A → P = AV − /5 , so we have Z Vf 6 W =A V − /5 dV Vi
V 1 f =−A·5 1 V /5
=−A·5
Vi
1
1/ 5
Vf
−
1 Vi
1/ 5
R
P dV . Also remembering
!
(5.4.234)
Now, substituting ∆U from (5.4.225) and W from (5.4.234) into (5.4.233), we have ! ! 1 1 3 1 1 − 1/ + · A − 1/ Q=−A·5 1 1 / / 2 Vf 5 Vi 5 Vf 5 Vi 5 ! 7 1 1 =− A − 1/ 1 /5 2 V V 5 f
i
And with the given values, this becomes 7 Q= A 2
1 (1 m3 )
1/ 5
−
1 (2 m3 )
1/ 5
!
Problem 24 An arbitrary network of resistors, capacitors, and inductors is in thermal equilibrium at temperature T . There are no sources in the network.
543
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
By considering a capacitor C as a subsystem, find the probability density P (V ) that a voltage V exists across it. Find the mean and the root-mean-square voltages on a 10 pF capacitor when T = 300 K and when T = 77 K.
b)
Similarly, find the probability density P (I) that a current I flows through an inductor L at temperature T . What is the rms current in a 1 mH inductor when T = 300 K and when T = 77 K?
Solution a)
We recall that the energy stored in a capacitor is given by E(V ) =
CV 2 2
(5.4.235)
While the probability function is most generally given by εs 1 P (εs ) = exp − Z kB T
(5.4.236)
Where the partition function in this case is εs exp − kB T
(5.4.237)
And substituting (5.4.235) into (5.4.237) yields X CV 2 Z= exp − 2kB T
(5.4.238)
Z=
X s
V
Passing this to the integral limit,(5.4.238) becomes r Z ∞ CV 2 2πkB T dV −→ Z = Z= exp − 2k T C B −∞
(5.4.239)
Substituting (5.4.235) and (5.4.239) into (5.4.236), r C CV 2 P (V ) = exp − 2πkBT 2kB T To find the mean voltage, hV i, we recall that a common way to define this quantity is Z 1 E(x) hxi = x · exp − dx Z kB T In our case, (5.4.240) becomes r hV i =
C 2πkB T
Z
∞
CV 2 V · exp − 2kBT −∞
(5.4.240)
dV −→ hV i = 0
This is because we were trying to integrate an even function multiplied by an odd function over a symmetric domain. The RMS, however, does not fit this criteria, and as such we would expect a non-zero result:
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STATISTICAL MECHANICS
2
r
CV 2 V 2 · exp − dV 2kBT −∞ s r 3 C 2k T 1 B = π· 2πkBT 2 C s C 8k 3 T 3 π · · B3 = 4 2πkBT C r 2T2 kB = C2
hV i =
Z
C 2πkBT
∞
(5.4.241)
From (5.4.241) it is easy to deduce that hV 2 i = b)
kB T C
Finding the probability density for current flowing through an inductor is identical to the process just underwent, therefore I will only summarize the results: r I IL2 P (I) = · exp − 2πkB T 2kB T hIi = 0 hI 2 i =
kB T L
Problem 25
a)
Compare the 4 level systems shown below. More than one particle may occupy a level. Which system has the Highest temperature Lowest specific heat
Lowest temperature Highest entropy
545
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
b)
Consider the system shown below. Explain why it can be thought of as having a negative temperature. (Hint: Consider the Boltzmann factor e−E/kT ).
c)
If this system (from part b)) is brought into contact with a large reservoir at temperature T R (positive) draw a graph indicating how its temperature will change as a function of time as it comes to equilibrium with the reservoir. (Take the initial temperature of the system to be T S a negative number).
d)
Explain why no problems with the third law are encountered in part c).
Solution a)
Highest temp: (ii) Lowest temp: (iii) Lowest specific heat: (ii) Highest entropy: (ii)
b)
If the temperature is increased slightly, then it is possible that the total energy will decrease, which is the opposite of the typical behavior.
c)
The third law states that as the entropy in a perfect crystal approaches zero, then the temperature approaches zero as well.
This example does not violate the third law of thermodynamics because system at a negative temperature is further away from absolute zero than a system at positive temperature (from the point of view of energy, anyway).
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STATISTICAL MECHANICS
Problem 26 Equipartition. A classical harmonic oscillator H=
p2 Kq 2 + 2m 2
(5.4.242)
is in thermal contact with a heat bath at temperature T . Calculate the partition function for the oscillator in the canonical ensemble and show explicitly that hEi = kB T,
and
2
2
h(E − hEi) i = (kB T )
(5.4.243)
Solution We first recall that the single-particle partition function is given by Z Z 1 Z1 = 3 exp [−βH] d3 pd3 q h
(5.4.244)
However, in our case we are assuming that the Hamiltonian is only in 1-D, so that substituting our Hamiltonian from (5.4.242) into (5.4.244), we have 2 Z Z 1 p kq 2 Z1 = + exp −β dpdq h 2m 2 Z Z ∞ 1 ∞ p2 Kq 2 = exp − dp exp dq h −∞ 2mkB T 2 −∞ ! r 2πk T 1 p B = 2mπkB T h K r 2πkBT m = (5.4.245) h K We now recall that the internal energy, hEi, is defined by hEi = −
∂ log Z1 ∂β
Rearranging (5.4.245) and substituting into (5.4.246), we have r 2π m ∂ log hEi = − −→ hEi = kB T ∂β βh K
(5.4.246)
(5.4.247)
Now, we recall that 2
h(E − hEi) i = hE 2 i − hEi2 And we can also express hEi in yet another way Z Z 1 hEi = E · exp [−βH] dpdq −→ hEi = kB T Z1
(5.4.248)
(5.4.249)
547
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
As before, We can extend this to find hE 2 i, however, and (5.4.249) becomes Z Z 1 hE 2 i = E 2 exp [−βH] dpdq Z1
(5.4.250)
And now, seemingly arbitrarily, we take the first derivative of hEi with respect to β in (5.4.247): ∂ ∂ 1 1 hEi = =− 2 ∂β ∂β β β
(5.4.251)
And now, stay with me here, we wish to do the same thing for the primary expression in (5.4.249): Z Z ∂ ∂ 1 hEi = E · exp [−βH] dpdq (5.4.252) ∂β ∂β Z1 Recognizing that Z1 has a dependence on β, we continue with (5.4.252), yielding Z Z Z Z ∂ 1 ∂ ∂ 1 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β Z1 ∂β ∂β Z1 Z Z Z Z 1 ∂ 1 2 E · exp [−βH] dpdq + E · exp [−βH] dpdq = Z1 ∂β Z | {z 1}
(5.4.253)
Let’s take a closer look at the underbraced term in (5.4.253), using the value of Z1 found in (5.4.245): " r # r ∂ 1 ∂ hβ K h K = ⇒ (5.4.254) ∂β Z1 ∂β 2π m 2π m Substituting (5.4.254) back in to (5.4.253), r Z Z Z Z ∂ 1 h K 2 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β Z1 2π m Inserting in to (5.4.255) our expression for Z1 found in (5.4.245), r r Z Z Z Z ∂ h K h K 2 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β 2πkB T m 2π m | {z }
Let us know rewrite the underbraced term in (5.4.256): r r Z Z h K h K E · exp [−βH] dpdq = · Z1 hEi 2π m 2π m r r h K 2πkB T m = · hEi 2π m h K 1 = hEi β
(5.4.255)
(5.4.256)
(5.4.257)
Plugging (5.4.257) back into (5.4.256), ∂ 1 hEi = ∂β Z1 |
Z Z
1 E 2 · exp [−βH] dpdq + hEi β {z }
(5.4.258)
W. Erbsen
STATISTICAL MECHANICS
We immediately recognize the underbraced term to be none other than our good friend hE 2 i. Using this knowledge, we can now rewrite (5.4.258), ∂ 1 hEi =hE 2 i + hEi ∂β β Rearranging this a bit, ∂ 1 hEi hE 2 i = hEi − β ∂β 1 1 ∂ 1 = − β β ∂β β 1 1 = 2+ 2 β β 2 = 2 β
(5.4.259)
Substituting (5.4.247) and (5.4.259) into (5.4.248), hE 2 i − hEi2 =
2 1 − 2 −→ hE 2 i − hEi2 = (kB T )2 β2 β
Problem 27 Consider two single-particle states and two particles. Calculate the entropy of this system when the particles have the following statistics: a)
Maxwell-Boltzmann (i.e. classical)
b)
Bose-Einstein
c)
Fermi-Dirac
d)
If two particles of mass m are confined to a box of volume V , approximately what temperature do classical statistics lose their validity?
Solution a)
We first recall how the entropy depends on the multiplicity S = kB log [Ω]
(5.4.260)
We first need to know what the multiplicity, Ω is before we can use this tool though. The multiplicity refers to the number of available states in the system. It’s as simple as that. For Maxwell-Boltzmann statistics, we recognize that this is a classical system, and both particles may occupy any of the states available with no restrictions what-so-ever:
549
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
• •
•
•
•
•
•
X
•
So, the entropy in this case is
(Ω) = 4
S = kB log [4] b)
Within the framework of hte Bose-Einsein universe, we recognize that the particles may still occupy the same orbital, however being indistinguishable particles, there is no difference between states with similar occupancies. Eg, • • • X (Ω) = 3 • • • And the entropy ends up being
S = kB log [3] c)
For the Fermi-Direac distribution, we recognize that we are constrained clenched fist of Pauli’s Exclusion Principle. According to Mr Pauli, there is only one possible configuration: • X (Ω) = 1 • And, as expected, the entropy, which is given as
S = kB log [1] −→ S = 0 d)
Within the classical regime, the single-particle partition function is given by " # Z Z p2x + p2y + p2z 1 Z1 = exp − d3 p d3 r h 2mkB T " # Z p2x + p2y + p2z V = exp − d3 p h 2mkB T =
3 V (2πmkB T ) /2 h
(5.4.261)
And the classical limit is defined as Z1 1 N
(5.4.262)
Substituting (5.4.261) into (5.4.262), V h
(2πmkB T ) N
(2πmkB T ) 2πmkB T
3
/2
3
/2
1
Nh V
Nh V 2/3
(5.4.263)
W. Erbsen
STATISTICAL MECHANICS
From (5.4.263), it is easy to see that the condition is 1 T 2πmkB
Nh V
2/3
Problem 28 A zipper has N links; each link has a state in which it is closed with energy 0 and a state in which it is open with energy ε. We require, however, that the zipper can only unzip from the left end, and that the link number, s, can only open if all links to the left (1, 2, ..., s − 1) are already open. a)
Show that the partition function can be summed in the form: QN =
b)
1 − exp [−(N + 1)βε] 1 − exp[−βε]
(5.4.264)
In the limit ε kT , find the average number of open links.
The above model is a very simplified model of the unwinding of two-stranded DNA molecules.
Solution Ooooh the zipper problem. I zip you so much. a)
The standard form of the single-particle partition function is ∞ X εs Z1 = exp − kB T s=0
(5.4.265)
Because the “particles” are distinguishable, then we can gain the full partition from (5.4.265) quite easily s ∞ X εs Zs = exp − (5.4.266) kB T s=0 We now take a field trip. Let’s take a close look at the following series: N X
xs = 1 + x + x2 + x3 + ... + xN
(5.4.267)
s=0
We now multiply both sides of (5.4.267) by x. This looks like x
N X
xs = x + x2 + x3 + x4 + ... + xN+1
s=0
We now take (5.4.267) and subtract from it (5.4.268). This yields
(5.4.268)
551
CHAPTER 5: DEPARTMENTAL EXAMINATIONS
(1 − x)
N X s=0
xs = 1 − xN+1
(5.4.269)
If we were to rearrange (5.4.269), then it would look like N X
xs =
s=0
1 − xN+1 1−x
(5.4.270)
This exact sum is what we need in order to evaluate (5.4.266)! Zs = b)
1 − exp [− (N + 1) εβ] 1 − exp [−εβ]
(5.4.271)
In general, the way to find the average number of open links, we use the form hsi =
N 1 X s · exp[−sεβ] Zs s=0
(5.4.272)
Now, notice that we can arrive at (5.4.272) if we are very tricky and take a derivative of the following form: hsi =
N ∂ X 1 exp [−sεβ] Zs ∂ (εβ) s=0 | {z }
(5.4.273)
Zs
Rewriting (5.4.273) once more,
hsi =
1 ∂ Zs Zs ∂x
(5.4.274)
Where I have made the temporary substitution x = εβ for convenience. Carrying out the first derivative in (5.4.274), ∂ ∂ 1 − exp [− (N + 1) x] Zs = ∂x ∂x 1 − exp [−x] 1 ∂ ∂ 1 = [1 − exp [− (N + 1) x]] + [1 − exp [− (N + 1) x]] 1 − exp [−x] ∂x ∂x 1 − exp [−x] " # 1 exp [−x] = [(N + 1) exp [−(N + 1)x]] − [1 − exp [−(N + 1)x]] 2 1 − exp [−x] (1 − exp [−x]) " # exp [−(N + 1)x] 1 − exp [−(N + 1)x] =(N + 1) − exp [−x] 1 − exp [−x] (1 − exp [−x])2 exp [−(N + 1)x] exp [−x] exp [−x] exp [−(N + 1)x] − 2 + 2 1 − exp [−x] (1 − exp [−x]) (1 − exp [−x]) exp [−(N + 1)x] 1 exp [−x] exp [−(N + 1)x] exp [−x] =(N + 1) − − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] 1 exp [−x] exp [−(N + 1)x] exp [−x] = (N + 1) exp [−(N + 1)x] − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] =(N + 1)
W. Erbsen
STATISTICAL MECHANICS
=
1 1 − exp [−(N + 1)x] (N + 1) exp [−(N + 1)x] − exp [−x] 1 − exp [−x] 1 − exp [−x]
(5.4.275)
Where we note that the expression in brackets in (5.4.275) is none other than Zs . Rewriting, 1 ∂ Zs = [(N + 1) exp [−(N + 1)x] − exp [−x] Zs ] ∂x 1 − exp [−x]
(5.4.276)
At this point we are ready to substitute (5.4.276) and (5.4.271) into (5.4.274): 1 1 · [(N + 1) exp [−(N + 1)x] − exp [−x] Zs ] Zs 1 − exp [−x] (N + 1) exp [−(N + 1)x] − exp [−x] = 1 − exp [−x] (N + 1) exp [−(N + 1)x] exp [−x] = − 1 − exp [−x] 1 − exp [−x]
hsi =
(5.4.277)
Putting the numbers back in to (5.4.277) leads us to hsi =
exp [−εβ] (N + 1) exp [−(N + 1)εβ] − 1 − exp [−εβ] 1 − exp [−εβ]
(5.4.278)
We are asked to evaluate the average number of open links from (5.4.278) given the following stipulation: ε kBT −→ εβ 1 If εβ 1, then all the exponential terms approach zero, and so the average number of open links is hsi = 0
Problem 29 a)
Obtain the Van der Waals equation of state N 2a p+ (V − N b) = N kT V
(5.4.279)
by making the following two corrections to the ideal gas: i) ii) b)
Use an “effective volume,” instead of the usual volume Include interactions in a mean field form. State clearly where the mean field assumptions come in.
What is the work done by the gas if it expands isothermally from V1 to V2 at some temperature T ?
Solution
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
a)
The ideal gas law is, of course, given by P V = N kB T
(5.4.280)
Let’s apply ii) first to (5.4.280), where we are including interactions from the “mass field form.” Mean field theory essentially consists of replacing all interactions attributed to any one body, with an average, or “effective” interaction. → This reduces a complicated multi-body problem to an effectively single-body problem, whose solutions are easier to find.
But what is the probability that two particles will be in the same place at the same time? → The density (ρ = m/v) is sort of like a probability, in the sense that if the density is large, the probability that two particles will try to be in the same space at the same time increases (this probability actually goes like P ∝ ρ2 ). → Think of two coins: if the probability that one is heads up is ρ, then the probability that they will both be heads up is ρ2 . This adds an “effective pressure” term to (5.4.280). The addition of a new term indicates that due to the chance of spatially co-existing molecules, the pressure is actually more, proportional to several constants. Anyway, with this correction, (5.4.280) becomes P + aρ2 V = N kB T (5.4.281)
Where the constant a is a proportionality constant, which depends on how much the particles interact with one another. We now recall how we define the particle density: ρ=
N V
Substituting this in to (5.4.281) yields N2 P + a 2 V = N kB T V
(5.4.282)
Next, we wish to address i), which is attributed to the “effective volume,” which basically means that the two particles cannot coexist at the same place at the same time. → Therefore, the amount of space available for these molecules flying around is actually less then the total volume of the container, by an amount proportional to the number of molecules present: Veff = V − N b
(5.4.283)
Where b is the volume of the individual gas particle, and N is the number of particles in the volume. Therefore, substituting (5.4.283) into (5.4.282), we have N2 P + a 2 (V − N b) = N kB T (5.4.284) V
W. Erbsen
STATISTICAL MECHANICS
b)
To find the work done by a gas which expands isothermally (constant temperature), from some initial volume Vi to some final volume Vf , we must first recall that Z Vf W = P dV (5.4.285) Vi
At this point, we solve (5.4.284) for P , which will subsequently be substituted into (5.4.285). But first things first: P =
N kB T aN 2 − 2 V − Nb V
And now substituting (5.4.286) back into (5.4.285), Z Vf N kB T aN 2 W = − 2 dV V − Nb V Vi Z Vf Z Vf 1 1 dV − aN 2 dV =N kB T 2 V − N b V Vi Vi V 1 f Vf 2 =N kB T [log [V − N b]|Vi − aN − V Vi
=N kB T [log [Vf − N b] − log [Vi − N b]] + aN 2
(5.4.286)
1 1 − Vf Vi
(5.4.287)
We can simplify (5.4.287) slightly, leaving us with Vf − N b 1 1 W = N kB T log + aN 2 − Vi − N b Vf Vi
Problem 30 A quantum harmonic oscillator has energy levels En = n + 1/2 ~ω0 , where n = 0, 1, 2, ... Treat this single oscillator to be a small system coupled to a heat bath at temperature T . What is the probability of finding the oscillator in its nth quantum state?
Solution The Quantum Harmonic Oscillator (or simply QHO, as we lovingly call it) has the following eigenenergies: En = n + 1/2 ~ω
(5.4.288)
And if we dutifully recall, the partition function is given by Z=
En exp − kB T n=0 ∞ X
(5.4.289)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Now, we are interested in finding the QHO in some other state, let’s call it n0 . How do we find this out? Well, we first recall that P (En) =
exp [−En /(kB T )] Z
(5.4.290)
exp [−En0 /(kB T )] Z
(5.4.291)
Now let’s tailor it to our system in need: P (En0 ) = Now, we must find Z: "
# n + 1/2 ~ω0 Z= exp − kB T n=0 ∞ X
(5.4.292)
Now, the nth excited state is n0 , and (5.4.291) becomes exp − n0 + 1/2 ~ω0 /(kB t) 0 P P (En) = ∞ 1 n=0 exp [− (n + /2 ) ~ω0 /(kB T )] 0 1 exp − n + /2 ~ω0 β = P∞ 1 n=0 exp [− (n + /2 ) ~ω0 β] ∞ X 0 1 = exp − n + /2 ~ω0 β · exp n + 1/2 ~ω0 β n=0
∞ X = exp −n0 ~ω0 β − 1/2 ~ω0 β · exp 1/2 ~ω0 β exp [n~ω0 β] n=0
∞ X = exp −n0 ~ω0 β − 1/2 ~ω0 β + 1/2 ~ω0 β · exp [n~ω0 β] n=0
= exp [−n0 ~ω0 β] ·
∞ X
exp [n~ω0 β]
(5.4.293)
n=0
At this point we recall the following arcane summation identity ∞ X
ean =
n=0
1 1 − ea
Applying this to (5.4.293), P (En0 ) =
exp [−n0 ~ω0 β] 1 − exp [~ω0 β]
Problem 31 The surface temperature of the sun is T0 (= 5800o K); its radius is R(= 7 × 108 m) while the radius of the
W. Erbsen
STATISTICAL MECHANICS
Earth is r(= 6.37 × 106 m). The mean distance between the sun and Earth is L(= 1.5 × 1013 m). In first approximation, one can assume that both the sun and the Earth absorb all electromagnetic radiation incident upon them. Assume that Earth has reached a steady state at some temperature T . a)
Find an approximate expression for the temperature T of Earth in terms of the astronomical parameters mentioned above.
b)
Calculate the temperature T numerically.
Solution a)
The Stefan-Boltzmann Law gives the total power that the Sun is emitting, and is given by PS = 4πR2 σT04
(5.4.294)
Since the sun emits radiation in all directions, the Earth only absorbs a small fraction of the total output, and depends on the cross sectional area of the Earth and the distance it is from the emitter. Quantitatively, A 4πL2 πr 2 =PS 4πL2 r2 =PS 4L2
PE =PS
(5.4.295)
As suggested in the prompt, if we assume in the first approximation that the Earth is a perfect absorber, then (5.4.295) holds, and we can use (5.4.294) to come up with another expression which describes the power absorbed by the Earth. We start by slightly rewriting (5.4.294) PE = 4πr 2 σT 4
(5.4.296)
At this point, we substitute both (5.4.294) and (5.4.296) into (5.4.295): 4πr 2 σT 4 = 4πR2 σT04 ·
r2 1 −→ T 4 = T04 · R2 · 4L2 4L2
(5.4.297)
From (5.4.297) it is easy to see that T = T0 b)
r
R 2L
Substituting in the appropriate values into (5.4.298), we solve s 7 × 108 m T = 5800 K −→ T = 28.017 K 2 (1.5 × 1015 ) m
(5.4.298)
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CHAPTER 5: DEPARTMENTAL EXAMINATIONS
Problem 32 A 1.00 gram drop of water is supercooled to −5.00o C (remains in liquid state). Then suddenly and irreversibly it freezes and becomes solid ice at the temperature of the surrounding air, 5.00o C. The specific heat of ice is 2.220 J · g−1 · K, and of water is 4.136 J · g−1 · K, and the heat of fusion of water is 333 J · g−1 . a)
How much heat leaves the drop as it freezes?
b)
What is the change in entropy of the drop as it freezes?
Solution m = 1.00 g T1 = −5.00o T2 = −5.00o a)
b)
Cw = 2.220 J · g−1 · K Ci = 4.186 J · g−1 · K
Lf = 3.333 J · g−1
Two equations that have been since long forgotten are Q =mC∆T
(5.4.299a)
Q =mLf
(5.4.299b)
Where we recall that the heat, Q is the key quantity in all this. If a cup of ice is sitting in room temperature, then heat is entering the vessel, causing it to melt. If, however you have the same cup of ice cooled to exactly 32o F (remember, ice doesn’t normally freeze lower than this), and it starts to turn to water of the same temperature. The heat required to make this phase transition may not contribute towards a change in ∆T , however energy is being applied because it takes energy to undergo a phase change. In our case, the sample does not undergo any change in temperature, and the amount of heat that leaves the drop is due only to the change of phase. Therefore, we only need to use (5.4.299b), and applying the appropriate constants, yields Q = (1.00 g) 333 J · g−1 −→ Q = 333 J (5.4.300) A common form of entropy is
dS =
∆Q T
And substituting in the appropriate values into (5.4.301), we have S=
333 J −→ S = 1.243 J · K−1 273 − 5 K
(5.4.301)
