PHYS 115A: Homework #3 Solution Set

PHYS 115A: Homework #3 Solution Set Charlotte Mason [email protected] Problem 1

Griffiths 2.3

Show that there is no acceptable solution to the time independent Schrdinger equation (TISE) for a particle in an infinite square well with E = 0 or E < 0. Following example 2.2 in the book we know that: 0 if 0 ≤ x ≤ a V (x) = ∞ otherwise 2

2

~ d ψ And the TISE is − 2m = Eψ. This applies inside the well, outside the well dx2 ψ(x) = 0. As the wavefunction is zero outside of the well this constrains our boundary conditions to be ψ(0) = ψ(a) = 0.

Let’s use the TISE to examine the two cases: E=0 −

~2 d2 ψ =0 2m dx2

⇒ ⇒

d2 ψ =0 dx2 ψ(x) = Ax + B

Consider the boundary conditions: ψ(0) = 0 → B = 0 ⇒ ψ(x) = 0 throughout the well ψ(a) = 0 → A = 0 E0 ~

Consider the boundary conditions: ψ(0) = 0 → C + D = 0 → C = −D ⇒ ψ(x) = 0 throughout the well ψ(a) = 0 → C(eωx − e−ωx ) = 0 → C = 0 or ω = 0

Problem 2

2

In both cases, the boundary conditions force ψ = 0 in the well, which is not an acceptable solution.

———— Problem 2

Griffiths 2.4

This problem involves the properties of the nth stationary state of the infinite square well. From equation (2.28) in Griffiths we know this is: r nπx 2 ψn (x) = sin a a (a) Z nπx 2 a x sin2 hxi = x|ψ| dx = dx a 0 a Z nπx 2 a 2 nπ Substitute y = , = y sin2 ydy a a nπ Z0 1 a 2 nπ y(1 − cos 2y)dy = a nπ 0 nπ 1 2 a 2 y 2 y − sin 2y − cos 2y Integrate by parts = a nπ 2 2 4 0 2 2 a 2 (nπ) 1 = − 0 − (1 − 1) a nπ 2 4 a ⇒ hxi = 2 Z

PHYS 115A

2

HW #3

Problem 2

3

(b) Z

2

hx i = Substitute

2

2

x |ψ| dx = = =

Integrate by parts

= = = =

Z 2 a 2 2 nπx x sin dx a 0 a Z 2 a 3 nπ 2 2 y sin ydy a nπ 0 Z 1 a 3 nπ 2 y (1 − cos 2y)dy) a nπ 0 Z nπ 2 a 3 y 3 nπ 2 y cos 2ydy | − a nπ 3 0 0 nπ 2 a 3 y 3 y 1 − cos 2y + sin 2y a nπ 3 2 4 0 3 3 2 a (nπ) nπ − a nπ 3 2 1 1 − a2 3 2(nπ)2

(c) hpi = m

dhxi via Ehrenfest’s theorem. hxi = 6 f (t) dt

⇒ hpi = 0

(d) 2

hp i =

PHYS 115A

Z

2 Z 2 ∂ 2 ∗ ∂ ψn ψn dx = −~ ψn dx i~ ∂x ∂x2 Z ∂ 2 ψn 2mE 2 = − 2 ψn by TISE ⇒ hp i = 2mE ψ ∗n ψn dx ∂x2 ~ = 2mE 2 nπ~ = a

ψn∗

HW #3

Problem 3

4

(e) Check the uncertainty principle, σx σp ≥ ~2 . nπ~2 a s a 1 2 ⇒ σx = − 2 3 (nπ)2 2 nπ~ 2 2 2 −0 σp = hp i − (hpi) = a nπ~ ⇒ σp = a s r 2 nπ~ 1 2 ~ (nπ) ⇒ σx σp = − −2 = 2 2 3 (nπ) 2 3 σx2 = hx2 i − (hxi)2 =

σx σp is closest to the uncertainty limit when n = 1: r ~ π2 ~ ~ σx σp = − 2 ≈ 1.136 > 2 3 2 2 Therefore the uncertainty principle is satisfied.


Griffiths 2.5

A particle in the infinite square well has initial wavefunction: ψ(x, 0) = A [ψ1 (x) + ψ2 (x)] (a) |ψ|2 = |A|2 (ψ1∗ + ψ2∗ ) (ψ1 + ψ2 ) = |A|2 |ψ1 |2 + ψ1∗ ψ2 + ψ2 ψ1∗ + |ψ2 |2 Z Z 2 2 |ψ1 |2 + |ψ2 |2 dx (ψ1,2 are orthonormal, cross-terms integrate to zero.) 1 = |ψ| dx = |A| = 2|A|2

PHYS 115A

1 ⇒ A= √ 2

HW #3

Problem 3

5 iEt

(b) Time evolution of states: ψ(x, t) ∼ ψ(x, 0)e− ~ . For our state: iE1 t iE2 t 1 ψ(x, t) = √ ψ1 e− ~ + ψ2 e− ~ 2 ! r r 1 2 πx −iωt 2 2πx −i4ωt π2~ =√ sin e + sin e , where ω = . a a a a 2ma2 2 πx 1 −iωt 2πx −i3ωt sin =√ e + sin e a a a 1 πx 2πx +i3ωt πx 2πx −i3ωt 2 |ψ(x, t)| = sin + sin e sin + sin e a a a a a 1 πx 2πx +i3ωt 2 πx 2 2πx −i3ωt = sin + sin + sin sin e +e a a a a a 1 πx 2πx 2 2πx 2 πx + sin + 2 sin sin cos 3ωt = sin a a a a a (c) Z

x|ψ(x, t)|2 dx Z 1 a πx 2πx 2 πx 2 2πx x sin = + sin + 2 sin sin cos 3ωt dx a 0 a a a a

hxi =

y2 y 1 y sin ydy = − sin 2y − cos 2y 2 2 4 Z a 2πx a2 a2 πx + sin2 + ⇒ x sin2 dx = a a 4 4 0 Z

Z 0

a

2

a

(see Q1.a above)

πx 3πx x cos − cos cos 3ωtdx a a 0 a 2 a πx ax πx a2 3πx ax 3πx = 2 cos + sin − 2 cos + sin cos 3ωt π a a a 9π a 3a a 0 a2 1 16a2 =− 2 1− cos 3ωt = − 2 cos 3ωt π 9 9π a 1 a2 a2 16a2 32 ⇒ hxi = + − cos 3ωt = 1 − 2 cos 3ωt a 4 4 9π 2 2 9π

πx 2πx 2x sin sin cos 3ωtdx = a a

Z

hxi oscillates in time with angular frequency 3ω = oscillation is PHYS 115A

3π 2 ~ . The amplitude of the 2ma2

32 a . 9π 2 2 HW #3

Problem 4

6

(d) dhxi dt 32 a − 2 (−3ω) sin 3ωt =m 2 9π

hpi = m

=

8~ sin 3ωt 3a

(e) Acting with H on the particle in state ψ(z, t) we see we measure energies E1 = 2 ~2 π 2 ~2 and E2 = 2π . The probability measuring either is just the probability of 2ma2 2ma2 being in those states, so just the normalisation coefficent (probability amplitude) squared. i.e. P = P1 = P2 = 21 . The expectation of the energy is: Z Z 1 ∗ hHi = ψ Hψdx = ψ1∗ Hψ1 + ψ2∗ Hψ2 2 1 5π 2 ~2 hHi = (E1 + E2 ) = 2 4ma2


Griffiths 2.6 Ψ(x, 0) = A ψ1 (x) + eiφ ψ2 (x)

From question 3.(b) above we can write down the time evolution of Ψ: i h πx 1 + sin 2πxae−3iωt eiφ Ψ(x, t) = √ e−iωt sin a a

where ω =

E1 π2~ = . ~ 2ma2

i h πx i 1 h πx sin + sin 2πxae−3iωt eiφ × sin + sin 2πxae+3iωt e−iφ a a a 1 πx 2πx i(3ωt−φ) −i(3ωt−φ) 2 πx 2 2πx = sin + sin + sin sin e +e a a a a a 1 πx 2πx 2 πx 2 2πx = sin + sin + 2 sin sin cos (3ωt − φ) a a a a a

|Ψ(x, t)|2 =

Following Q3.(c) above: a ⇒ hxi = 2

32 1 − 2 cos (3ωt − φ) 9π

The phase shift effectively starts the clock at a different time, i.e. t = 0 → t =

PHYS 115A

φ . 3ω

HW #3

Problem 5

7

For φ = π2 → Ψ(x, t) = √1a e−iωt sin πx − i sin 2πxae−3iωt . a cos(3ωt − π2 ) = sin 3ωt. At t = 0, hxi = a2 . − sin 2πxae−3iωt . For φ = π → Ψ(x, t) = √1a e−iωt sin πx a cos(3ωt − π) = − cos 3ωt. 32 At t = 0, hxi = a2 1 + 9π 2 .


Griffiths 2.7

Particle in infinite square well. Ψ(x,0)

Aa/2

a/2

a

(a)

1 = A2

Z

a/2

x2 dx +

0

Z

!

a

(a − x)2 dx

a/2

! 3 a/2 3 a x (a − x) − = A2 3 0 3 a/2 3 3 2 3 a a Aa = A2 + = 8 8 16 r ⇒ A=2

3 a3

(b) From equation (2.17) in the book: Ψ(x, t) =

∞ X n=1

PHYS 115A

r −iEn t/~

cn ψn (x)e

, where ψn (x) =

2 nπx n2 π 2 ~2 sin( ) and En = . a a 2ma2 HW #3

Problem 5

8

From equation (2.37): r Z a 2 nπx )Ψ(x, 0)dx cn = sin( a 0 a ! r r Z a Z a/2 nπx 2 3 nπx (a − x) sin( = 2 )dx + )dx x sin( a a3 a a a/2 0 √ a/2 2 6 a 2 ax nπx nπx = 2 )− cos( ) sin( a nπ a nπ a 0 a )! a 2 nπx ax nπx nπx a2 cos( )− )+ cos( ) sin( + − nπ a nπ a nπ a a/2 √ 2 2 6 a 2 nπ a nπ = 2 sin( ) − cos( ) a nπ 2 2nπ 2 2 a 2 a cos(nπ) + sin(nπ) − nπ nπ a2 a2 nπ + cos(nπ) + cos( ) nπ nπ 2 a 2 nπ a2 nπ + sin( ) − cos( ) nπ 2 2nπ 2 √ √ nπ nπ 2 6 2a2 4 6 = 2 sin = sin a (nπ)2 2 (nπ)2 2 ( ⇒ cn =

0√

n−1 4 6 (−1) 2 (nπ)2

if n is odd if n is even

√ r ∞ nπx n−1 1 4 6 2 X 2 Ψ(x, t) = (−1) sin e−iEn t/~ (nπ)2 a n odd n2 a (c) Probability of measuring energy E1 is |c1 |2 (see Q4.(e)): √ 2 4 6 96 2 P1 = |c1 | = i 2 = 4 (π) π

PHYS 115A

HW #3

Problem 6

9

(d) The expectation of the energy is hHi, from equation (2.39) we have: X hHi = |c2 |2 En ∞ 96 π 2 ~2 X 1 2 n = 4 π 2ma2 n odd n4

=

∞ 96 π 2 ~2 X 1 π 4 2ma2 n odd n2

96 π 2 ~2 π 2 π 4 2ma2 8 48~2 π 2 6~2 = 2 2 = π ma 8 ma2 =


Griffiths 2.8

Particles in an infinite square well (width a). Starts in left region (0 < x < a/2), equally likely at t = 0 to be at any point in that region. (a) Initial wavefunction given above information: A 0 < x < a/2 Ψ(x, 0) = 0 otherwise Normalise it: Z 1= 0

a

a /2A2 dx = A2 ⇒ A = 2

r

2 a

(b) Probability of having energy E1 is |c1 |2 , from equation (2.37), : r Z a/2 2 πx c1 = A sin dx a 0 a ia/2 2h a = − cosπxa a 2 0 2 = π 2 2 P1 = |c1 | = π 2

————

PHYS 115A

HW #3