quantum
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![Page 1: quantum](https://reader038.fdocuments.in/reader038/viewer/2022100802/577ccf0e1a28ab9e788ec577/html5/thumbnails/1.jpg)
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
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(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
,
![Page 3: quantum](https://reader038.fdocuments.in/reader038/viewer/2022100802/577ccf0e1a28ab9e788ec577/html5/thumbnails/3.jpg)
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.
![Page 4: quantum](https://reader038.fdocuments.in/reader038/viewer/2022100802/577ccf0e1a28ab9e788ec577/html5/thumbnails/4.jpg)
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.
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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 〉.