Homework Assignment 03: Function of Operators; Quantum
Measurements (Due: 02/11/2016)
Problem 1 (50%) In class we have shown that if [A, B] ≠ 0 , then
e A+B ≠ e Ae B ,
where A and B are two operators, not necessary Hermitian.
If A and B do not commute, but they both commute with C = [A, B] , show that
e
A+B
A B
=e e e
1
− C
2
.
[Hints: You can first show that
[B, e λA ] = e λA[B, A]λ ,
where λ is a variable (not an operator). You can use this basic quantum commutator formula
[A, BC ] = [A, B]C + B[A, C ] .
Next define
G(λ) ≡ e λAe λB ,
and you can show that
dG(λ)
= (A + B +[A, B]λ)G(λ) .
dλ
Then integrate this to obtain the desired result.]
Problem 2 (50%)
Consider the following Hermitian operators in a Hilbert space,
⎡
⎤
⎡
⎤
⎡ 1 0 0 ⎤
⎢
⎥
1 ⎢⎢ 0 1 0 ⎥⎥
1 ⎢⎢ 0 −i 0 ⎥⎥
⎢
Lx =
,
L
=
,
L
=
1 0 1 ⎥
i 0 −i ⎥
0 0 0 ⎥⎥ .
y
z
⎢
⎢
⎢
2⎢
2⎢
⎥
⎢ 0 0 −1 ⎥
0 ⎥⎦
⎣ 0 1 0 ⎦
⎣ 0 i
⎣
⎦
(Note: These are orbital angular momentum operators for l = 1 , or spin operators for s = 1 .
But here we omitted the constant  .)
(a) [3%] What are the possible values one can obtain if Lz is measured?
(b) [10%] Take the state in which Lz is measured to be 1. In this state evaluate (ΔLx )2 .
(c) [10%] If the particle is in the state with Lz measured to be −1 , and then Lx is measured.
What are the possible outcomes and their corresponding probabilities?
(d) [15%] Consider the state
⎡ 1/ 2 ⎤
⎢
⎥
⎢
⎥
ψ = ⎢ 1/ 2 ⎥ ,
⎢
⎥
⎢ 1/ 2 ⎥
⎢⎣
⎥⎦
1
in the Lz basis. If L2z is measured in this state and a result of +1 is obtained, what is the state
(denoted by φ ) after the measurement? How probable was this result? After this
measurement if Ly is measured, what are the outcomes and respective probabilities?
(e) [12%] A particle is in a state for which the probabilities are p(Lz = 1) = 1 / 4 ,
p(Lz = 0) = 1 / 2 , and p(Lz = −1) = 1 / 4 . The most general and normalized state with this
property is (with δ1 , δ2 and δ3 real numbers)
iδ
iδ
iδ
e 1
e 2
e 3
ψ =
Lz = 1 +
Lz = 0 +
L = −1 .
2
2 z
2
It was stated that if ψ is a normalized state then e iθ ψ is a physically equivalent
iδ
iδ
normalized state. Does this mean that the factors e 1 , e 2 , e
iδ3
are irrelevant?
[Note: The notations are non-trivial in physics. Bad notations confuse you severely. Let’s denote
Lz = 1 ≡ z;+ , Lz = 0 ≡ z;0 , Lz = −1 ≡ z;− , and we use similar notations for
eigenstates of Lx and Ly . In addition, we often (by default) use eigenstates of Lz as basis,
denoting them simply as z;± ≡ ± and z;0 ≡ 0 . Thus
iδ
ψ =
iδ
iδ
e 1
e 2
e 3
+ +
0 +
− .]
2
2
2
2