Physics 710-711-712 November 16, 2009Problem Set 5

Problem 1: Do exercise 4.2.1 of the text.

(1)

Solution: The possible outcomes are Lz = {1, 0,−1}, which are the eigenvalues of Lz.

(2)

Solution: Lz|ψ〉 = 1 · |ψ〉 implies

|ψ〉 =

100

.

(Note that I have normalized |ψ〉!) Then

〈Lx〉 = 〈ψ|Lx|ψ〉 =(1 0 0

) 1√2

0 1 01 0 10 1 0

100

=1√2

(1 0 0

)010

= 0.

〈L2x〉 = 〈ψ|L2

x|ψ〉 =(1 0 0

) 12

0 1 01 0 10 1 0

0 1 01 0 10 1 0

100

=12(0 1 0

)010

=12.

∆Lx =√〈L2

x〉 − 〈Lx〉2 =

√(12

)− 02 =

1√2.

(3)

Solution: The characteristic equation for Lx is

0 = det(Lx − λ) = det

−λ1√2

01√2−λ 1√

2

0 1√2−λ

= λ− λ3, ⇒ λ ∈ {1, 0,−1}.

The corresponding eigenvectors, |λ〉, then satisfy

0 = (Lx − λ)|λ〉 =

−λ1√2

01√2−λ 1√

2

0 1√2−λ

abc

=

−λa+ b√2

a√2− λb+ c√

2b√2− λa

where we have parameterized the components of |λ〉 by (a b c). For λ = 1, we can solve

for b and c in terms of a, giving b =√

2a and c = a. We then determine a by normalizing

|λ = 1〉:

|λ = 1〉 =

a√2aa

, ⇒ 1 = 〈λ = 1|λ = 1〉 =(a∗√

2a∗ a∗) a√

2aa

= 4|a|2, ⇒ a =12

(where I have chosen the arbitrary phase to be 1). Thus, and doing the same thing for

λ = 0 and λ = −1, gives

|λ = 1〉 =12

1√2

1

, |λ = 0〉 =1√2

10−1

, |λ = −1〉 =12

1−√

21

.

(4)

Solution: The possible outcomes are Lx = {1, 0,−1}, which are the eigenvalues of Lx.

|ψ〉 is the normalized eigenstate of Lz with eigenvalue Lz = −1, which is

|ψ〉 =

001

.

So (here P stands for "probability of"):

P(Lx = 1) = |〈λ = 1|ψ〉|2 =

∣∣∣∣∣∣12 (1 √2 1

)001

∣∣∣∣∣∣2

=14,

P(Lx = 0) = |〈λ = 0|ψ〉|2 =

∣∣∣∣∣∣12 (1 0 −1)0

01

∣∣∣∣∣∣2

=12,

P(Lx = −1) = |〈λ = −1|ψ〉|2 =

∣∣∣∣∣∣12 (1 −√

2 1)0

01

∣∣∣∣∣∣2

=14.

(5)

Solution:

L2z =

10

1

, ⇒ the possible outcomes are L2z = {0, 1}.

An eigenbasis of the L2z = 1 eigenspace is {|a〉, |b〉} with

|a〉 =

100

, |b〉 =

001

.

Therefore, upon measuring L2z = 1, the state collapses to

|ψ〉 −→ |ψ′〉 =(|a〉〈a|+ |b〉〈b|)|ψ〉|(|a〉〈a|+ |b〉〈b|)|ψ〉|

.

But

[|a〉〈a|+ |b〉〈b|] |ψ〉 =

100

(1 0 0)

+

001

(0 0 1) 1

2

11√2

=

100

12

+

001

1√2

=12

10√2

,

has norm √√√√√12(1 0

√2) 1

2

10√2

=√

32,

so

|ψ′〉 =1√3/2

12

10√2

=1√3

10√2

.

The probability of L2z = +1 is

P(L2z = 1) = 〈ψ| (|a〉〈a|+ |b〉〈b|) |ψ〉 = |〈a|ψ〉|2 + |〈b|ψ〉|2

=

∣∣∣∣∣∣12 (1 0 0) 1

1√2

∣∣∣∣∣∣2

+

∣∣∣∣∣∣12 (0 0 1) 1

1√2

∣∣∣∣∣∣2

=14

+12

=34.

If we measured Lz the posible outcomes are the eigenvalues Lz, {0,±1}, with probabilities

P(Lz = 1) =∣∣(1 0 0

)|ψ′〉

∣∣2 =

∣∣∣∣∣∣ 1√3

(1 0 0

) 10√2

∣∣∣∣∣∣2

=13.

P(Lz = 0) =∣∣(0 1 0

)|ψ′〉

∣∣2 =

∣∣∣∣∣∣ 1√3

(0 1 0

) 10√2

∣∣∣∣∣∣2

= 0.

P(Lz = −1) =∣∣(0 0 1

)|ψ′〉

∣∣2 =

∣∣∣∣∣∣ 1√3

(0 0 1

) 10√2

∣∣∣∣∣∣2

=23.

(6)

Solution: In the Lz eigenbasis

|Lz = 1〉 =

100

, |Lz = 0〉 =

010

, |Lz = −1〉 =

001

,

write the unknown state as

|ψ〉 =

abc

.

Then

P(Lz = 1) =14

= |〈Lz = 1|ψ〉|2 =

∣∣∣∣∣∣(1 0 0)ab

c

∣∣∣∣∣∣2

= |a|2,

P(Lz = 1) =12

= |〈Lz = 0|ψ〉|2 =

∣∣∣∣∣∣(0 1 0)ab

c

∣∣∣∣∣∣2

= |b|2,

P(Lz = 1) =14

= |〈Lz = −1|ψ〉|2 =

∣∣∣∣∣∣(0 0 1)ab

c

∣∣∣∣∣∣2

= |c|2.

The most general solution to these three equations is then

a =12eiδ1 , b =

1√2eiδ2 , c =

12eiδ3 ,

for some arbitrary phases δi, which gives the desired answer.

The δi phase factors are not irrelevant. For example

P(Lx = 0) = |〈λ = 0|ψ〉|2 =

∣∣∣∣∣∣ 1√2

(1 0 −1

) 12

eiδ1√2eiδ2

eiδ3

∣∣∣∣∣∣2

=∣∣∣∣ eiδ12√

2− eiδ3

2√

2

∣∣∣∣2

=18(eiδ1 − eiδ3

) (e−iδ1 − e−iδ3

)=

18

(1− ei(δ3−δ1) − e−i(δ3−δ1) + 1

)=

14

(1− cos(δ3 − δ1)) ,

so something measurable (a probability) depends on the difference of the phases.

Problem 2: Do exercise 4.2.2 of the text.Solution:

〈P 〉 = 〈ψ|P |ψ〉 =∫ ∞−∞dx〈ψ|x〉〈x|P |ψ〉

=∫ ∞−∞dxψ∗(x)

(−i~ d

dx

)ψ(x) = −i~

∫ ∞−∞dxψ(x)

dψ(x)dx

= − i~2

∫ ∞−∞dx

d

dx

(ψ(x)2

)= − i~

2ψ2∣∣∞−∞ = 0

if ψ → 0 as |x| → ∞.

Alternatively, use the k-basis:

〈P 〉 = 〈ψ|P |ψ〉 =∫ ∞−∞dk〈ψ|k〉〈k|P |ψ〉 =

∫ ∞−∞dk ~k 〈ψ|k〉〈k|ψ〉 =

∫ ∞−∞dk ~k ψ∗(k)ψ(k).

But

ψ(k) = 〈k|ψ〉 =∫ ∞−∞dx 〈k|x〉〈x|ψ〉 =

1√2π

∫ ∞−∞dx eikxψ(x),

therefore

ψ(−k) =1√2π

∫ ∞−∞dx e−ikxψ(x) = ψ∗(k)

since ψ(x) is real. So

〈P 〉 =∫ ∞−∞dk ~kψ∗(k)ψ(k) =

∫ ∞−∞dk ~kψ(−k)ψ(k).

and under the change of variables k → −k, this becomes

〈P 〉 =∫ ∞−∞dk ~(−k)ψ(k)ψ(−k) = −〈P 〉,

and so 〈P 〉 = 0.

Problem 3: Do exercise 2.4.3 of the text.Solution:⟨eip0x/~ψ

∣∣∣P ∣∣∣eip0x/~ψ⟩

=∫ ∞−∞dx 〈eip0x/~ψ|x〉〈x|P

∣∣∣eip0x/~ψ⟩

=∫ ∞−∞dx(eip0x/~ψ(x)

)∗(−i~)

d

dx

(eip0x/~ψ(x)

)= −i~

∫ ∞−∞dxψ∗(x)e−ip0x/~

[ip0

~eip0x/~ψ(x) + eip0x/~ dψ

dx

]=

∫ ∞−∞dxψ∗(x) p0 ψ(x)− i~

∫ ∞−∞dxψ∗(x)

dψ

dx

= p0

[∫ ∞−∞dx 〈ψ|x〉〈x|ψ〉

]+[∫ ∞−∞dx 〈ψ|x〉〈x|P |ψ〉

]= p0〈ψ|ψ〉+ 〈ψ|P |ψ〉 = p0 + 〈P 〉.