Physics 443, Solutions to PS 11

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Physics 443, Solutions to PS 11. 1.Griffiths 9.1. Hydrogen atom is placed in a time dependent electric field E = E(t)k. The perturbation is given by H = eE(t)z.
Physics 443, Solutions to PS 11 1.Griffiths 9.1 ˆ The Hydrogen atom is placed in a time dependent electric field E = E(t)k. perturbation is given by H = eE(t)z. s

1 −r , exp 3 πa a

s

1 1 −r r 1− exp , 2πa 2a 2a 2a

ψ100 = ψ200 =









s

ψ211





1 1 −r r exp sin θeiφ , = − 2 πa 8a 2a 

s

1 1 −r r exp cos θ, 2 2πa 4a 2a

s

−r 1 1 r exp sin θe−iφ . 2 πa 8a 2a

ψ210 = ψ21−1 =











By inspection or calculation, you will find that all the matrix elements are zero except ! 28 h100|H|210i = − √ 5 eEa. 23 2. Griffiths 9.11 We shall find the decay rates from the A coefficients which are A=

ω3 |P|2 . 3π0 h ¯ c3

Refer to the previous problem for explicit representation of the hydrogenic wavefunctions. Using that x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, you can convince yourself that all the matrix elements between ψ100 and ψ200 are zero. Therefore the state ψ200 has an infinite lifetime. For the state ψ210 we have that the x and y matrix elements are zero (think selection rules!) while (see previous problem) 28 a √ h100|z|210i = , 235 from which we can calculate the A coefficient as A=

ω 3 215 (qa)2 . 3π0 h ¯ c3 310 1

We now consider the case of ψ211 and ψ21−1 , where the z matrix element is zero, and up to a phase, the x and y components are equal. We have that |P|2 = 2(qa)2 214 3−10 . Notice that this is the same as what we calculated for the ψ210 state. Using that h ¯ ω = E2 − E1 = 0.75E1 , we have that τ = 1/A = 1.6 x 10−9 sec. 3.Griffiths 9.14 The allowed decay routes are |300i −→ |210i −→ |100i |300i −→ |211i −→ |100i |300i −→ |21 − 1i −→ |100i Recall the selection rules that ∆m = ±1, 0 and ∆l = ±1. Also note that on pg. 360 Griffiths derives the m-selection rules. We have if ∆m = ±1 then the z matrix element is zero, while the x and y terms are equal up to a phase. While if ∆m = 0 then the x and y matrix elements are zero. Using these rules, notice that |h210|r|300i|2 = |h210|z|300i|2 , and |h21 ± 1|r|300i|2 = 2|h21 ± 1|x|300i|2 . Comparing the integrals, √ you can notice that the ratio between h210|z|300i and h21 ± 1|x|300i is 2. We therefore have that |h210|r|300i|2 = |h21 ± 1|r|300i|2 . This means that all three transition rates are equal, and that each route has an equal probability of 1/3. To calculate the lifetime, we add the A coefficients to get that ω 3 e2 |hri|2 A=3 3π0 h ¯ c3 215 37 |hri|2 = 12 a2 5 5 h ¯ ω = E3 − E2 = − E1 . 36 1 −7 τ = = 1.58 x 10 sec. A !

4. Griffiths 9.20 (a) Manipulations with the Pauli matrices give that ~ = − γ¯h H = −~µ · B 2

B0 Brf eiωt −iωt Brf e −B0 2

!

h ¯ =− 2

ω0 Ωeiωt −iωt Ωe −ω0

!

.

(b) Using the time dependent Schrodinger equation for a spinor χ(t) = (a(t), b(t)), we can write: ω0 Ωeiωt −iωt Ωe −ω0

h ¯ − 2

!

a(t) ˙ ˙ b(t)

i = 2

!

a b

!

∂ = i¯ h ∂t

a(t)ω0 + b(t)Ωeiωt a(t)Ωe−iωt − b(t)ω0

a b

!

!

.

(c) Solving these coupled differential equations is done by first differenti˙ ating the b(t) equation, h i ¨b = i (a˙ − iωa)Ωe−iωt − bω ˙ 0 2

then substituting in for a(t). ˙ We have ¨b = i ( i (aω0 + bΩeiωt ) − iωa)Ωe−iωt − bω ˙ 0 2 2 

and

"

¨b = i (iaΩe−iωt ( ω0 − ω) + i b Ω2 − bω ˙ 0 2 2 2



#

Then substitute the expression for a(t) from the original equation ˙ + b(t)ω0 . a(t)Ωe−iωt = −2ib(t) and we get "

  ˙ 0 ¨b = i (i (−2ib˙ + bω0 )eiωt /Ω Ωe−iωt ( ω0 − ω) + i b Ω2 − bω 2 2 2 "

  ¨b = i (i −2ib˙ + bω0 ( ω0 − ω) + i b Ω2 − bω ˙ 0 2 2 2 "

#

  ¨b = i (−2bω ˙ + i b ω 2 − 2ω0 ω + Ω2 2 2 0   ¨b = −ibω ˙ − b ω 2 − 2ω0 ω + Ω2 4 0   ¨b = −ibω ˙ − b ω02 − ω2 4

3

#

#

Define γ = iω, and α = 4b (ω 0 2 − ω 2 ). Now our dcoupled second order differential equation looks like the equation of motion for a damped harmonic oscillator. ¨b = −γ b˙ − αb The general solution has the form b = Aeiθt Substitution into the differential equation gives −θ2 = −iθγ − α and q q 1 1 ω ω02 iγ ± −γ 2 + 4α = −ω ± ω 2 + (ω 0 2 − ω 2 ) = − ± 2 2 2 2 



θ= so







0

i ω2 t

b = Ce

0

+ De

−i ω2 t



ω

e−i 2 t

or even better ω

b = [A cos(ω 0 t/2) + B sin(ω 0 t/2)] e−i 2 t The boundary conditions give us b(0) = A = b0 and i ˙ b(0) = (−iω/2)A + (ω 0 /2)B = (Ωa0 − ω0 b0 ) 2 → ωb0 + iω 0 B = (ω0 b0 − Ωa0 ) i → B = 0 [(ω − ω0 )b0 + Ωa0 ] ω Then 

b = b0 cos(ω 0 t/2) +

i 0 −i ω t 2 [(ω − ω )b + Ωa ] sin(ω t/2) e 0 0 0 ω0 

4

1 0.9 0.8 0.7 0.6 P(ω) 0.5 0.4 0.3 0.2 0.1 0 -10

-5

0 (ω/ω0 )

5

10

Figure 1: Transition probability as a function of driving frequency

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(d) If a0 = 1 and b0 = 0 then P (t) = |b(t)|2 = |

Ω 2 2 0 ω2 | sin (ω t/2) = sin2 (ω 0 /2)t 0 2 2 ω (ω − ω0 ) + Ω

(e) The full width at half maximum ∆ω = 2Ω. (f) ω0 = γB0 =

(5.59)(1.6 × 10−19 )C) ge B0 = (1T ) = 2.68 × 108 Hz 2mp 2(1.67 × 10−27 kg)

Ω = γBrf =

Brf ω0 = 10−6 ω0 = 2.68 × 102 Hz B0

5. Griffiths 9.21 (a) The interaction hamiltonian H 0 = −qE(r, t) · r = −qE0 (cos ωt + (k · r) sin ωt) · r = −qˆE0 (cos ωt + (k · r) sin ωt) · r The first term on the right gives us the dipole approximation and the corresponding spontaneous decay rate is R=

ω 3 |q hψf | ˆ · r | ψi i |2 π0 h ¯c

The second term on the right gives the magnetic dipole or electric quadrupole and the decay rate R = =

ω 3 |q hψf | (ˆ · r)(k · r) | ψi i |2 π0 h ¯c D E 3 ˆ · r) | ψi |2 ω |q|k| ψf | (ˆ · r)(k π0 h ¯c E ˆ · r) | ψi |2 ω 5 |q ψf | (ˆ · r)(k D

=

π0 h ¯ c3

where we use |k| = ω/c 6

(b) For the 1-dimensional oscillator, we can choose r to be in the z-direction. The photon that is emitted can go in any direction k and with any polarization ˆ. Suppose that k is in the x-y plane at an angle θ with respect to r. kˆ · r = r cos θ. The polarization is perpendicular to k. If the polarization is in the x-y plane then ˆ · r = r sin θ. If it is at an angle φ with respect to the x-y plane then its projection into the plane is cos φ. So in general ˆ · r = r sin θ cos φ and putting it all together (ˆ · r)(kˆ · r) = r2 cos θ sin θ cos φ and E q2ω5 D 2 | f | r cos θ sin θ cos φ | i |2 π¯hc5 E q2ω5 D 2 = | f | r | i |2 (cos θ sin θ cos φ)2 π¯hc5 The calculation of the expectation value does not depend on the angles since it is a 1-dimensional oscillator. And since it is 1-dimensional we might as let r = x. Then

R =

R =

E q2ω5 D 2 | f | x | i |2 (cos θ sin θ cos φ)2 5 π¯hc 2

s

D

E

f | x2 | i

E h ¯  D f | (a+ + a− )2 | i =  2mω0

=

! D E h ¯ f | (a2+ − a+ a− − a− a+ + a2− )2 | i 2mω0

ω0 is the natural frequency of the oscillator. Since the final state has less energy then the initial state, the only non-zero contribution comes q 2 2 from a− . If the initial state is | ni, then a− | ni = n(n − 1)| n − 2i and E D q2ω5 h ¯ 2 R = | n − 2 | a | n |2 (cos θ sin θ cos φ)2 − π¯hc5 2mω0 !q q2ω5 h ¯ = | n(n − 1)|2 (cos θ sin θ cos φ)2 5 π¯hc 2mω0 2 3 q ω h ¯ = n(n − 1)(cos θ sin θ cos φ)2 2 πm c5 !

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where we use the fact that ω0 = ω/2. Finally, average over all angles for photon direction and polarization. 1 Z Z 1 (cos θ sin θ cos φ)2 sin θdθdφ = 4π 15 so q2 ω3h ¯ n(n − 1) R = 15πm2 c5 (c) To compute the rate for 2S → 1S in hydrogen D

2

E

D

2

1S | r cos θ sin θ cos φ | 2S = R(r)1S | r | R(r)2S

EZ Z

cos θ sin θ cos φ sin θdθdφ

The angular integral is zero. Note that the 1-dimensional harmonic oscillator has a preferred direction in space. The 1S and 2S states in hydrogen are spherically symmetric.

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