Homework Solutions #4 (McIntyre Chapter 5)

2 This is straightforward.

(a) 〈ψ|ψ〉 = A∗A(1 + 1 + 1) = 1. Choose A = 1√3.

(b) E1, E2, and E3 are possible, with equal probabilities of 13.

(c) 〈E〉 = 13(E1 + E2 + E3) = 14

3E1

(d) |ψ(t)〉 = A(e−iω1t|ϕ1〉 − e−4iω1t|ϕ2〉+ ie−9iω1t|ϕ3〉), where ω1 = E1/h.

(e) The same as (b), since H does not depend on t.

5 Obviously 〈x〉 = L2, but let’s calculate everything.

〈x〉 = 〈ϕn|x|ϕn〉 =2

L

∫ L

0dx x sin2

(nπ

Lx)

=L

2

〈x2〉 = 〈ϕn|x2|ϕn〉 =2

L

∫ L

0dx x2 sin2

(nπ

Lx)

=(

1

3− 1

2n2π2

)L2

∆x =(〈x2〉 − 〈x〉2

) 12 =

√1

12− 1

2n2π2L

〈p〉 = 〈ϕn|p|ϕn〉 = −ih 2

L

∫ L

0dx sin

(nπ

Lx)∂

∂xsin

(nπ

Lx)

=

−ih2nπ

L2

∫ L

0dx sin

(nπ

Lx)

cos(nπ

Lx)

= 0

〈p2〉 = 2m〈E〉 = 2m〈ϕn|H|ϕn〉 = 2mEn =n2π2h2

L2

∆p =(〈p2〉 − 〈p〉2

) 12 =

nπh

L

8 Normalization means

∫ L

0dxψ∗ψ = |A|2

∫ L

0dx x2(L− x)2 = 1 ⇒ |A| =

√30

L5

Phys 580 HW#4 Solutions 1

For the time evolution, we need to find the components of the state inthe energy basis.

〈ϕn|ψ(0)〉 =

√2

L

∫ L

0dx sin

(nπ

Lx)Ax(L− x) =

4√

15

(nπ)3[1− (−1)n]

Since ψ(x) is symmetric with respect to reflections about x = L2, we would

expect that all the even n components are zero, which is the case.The time evolution is

ψ(x, t) = e−ithHψ(x, 0) = e−i

thH∑n

ψnϕn(x) =∑n

ψne−in2ω1tϕn(x)

where ψn = 〈ϕn|ψ(0)〉 and ω1 = E1/h. This doesn’t look like it has anyeasier functional form.

The expectation value 〈x〉 = L2

always. Consider the operator Q, whicheffects a reflection of a function about x = L

2. (Clearly Q is closely related to

parity.) ψ(x, 0) is a Q eigenstate, and [H,Q] = 0. Therefore ψ(x, t) is also aQ eigenstate, and all Q eigenstates give 〈x〉 = L

2.

But that might not be obvious, so here’s a brute force calculation:

〈x〉 =∫ L

0dx

(∑n

ψ∗nein2ω1tϕ∗n

)x

(∑m

ψme−im2ω1tϕm

)=

∑nm

ψ∗nψmei(n2−m2)ω1t

∫ L

0dxϕ∗n(x)xϕm(x)

If you do the integral,∫ L

0dxϕ∗n(x)xϕm(x) =

2

L

∫ L

0dx x sin

(nπ

Lx)

sin(mπ

Lx)

=L

2δnm

Putting this in,

〈x〉 =L

2

∑n

ψ∗nψn =L

2

since the sum above is the sum of all energy probabilities, which is normalizedto 1.

17 The probabilities for x > a and x < −a are the same. Therefore

P|x|>a = 2A2∫ ∞a

dx e−qx =A2e−2qa

q=aA2e−2

√z20−z2√

z20 − z2

Phys 580 HW#4 Solutions 2

We now need to normalize the ground state to find A2.

P|x|<a = D2∫ a

−adx cos2(kx) = aD2

[sin(2ka)

2ka+ 1

]= aD2

[sin(2z)

2z+ 1

]

Since the probabilities sum to 1,

aA2e−2√

z20−z2√

z20 − z2+ aD2

[sin(2z)

2z+ 1

]= 1

From the quantization conditions for even parity (the ground state is even),we also get,Ae−qa = D cos ka and qAe−qa = kD sin ka. Using these, we get,after some algebra not worth reproducing,

P|x|>a =cos2 z

1 +√z20 − z2

Phys 580 HW#4 Solutions 3