〈ψ(t)|Lx|ψ(t)〉 =(
e+iμBBt/h ̄
2
√^1
2e−iμBBt/ ̄h
2) ̄h
√
2
0 1 0
1 0 1
0 1 0
e−iμBBt/ ̄h(^21)
√ 2
eiμBBt/h ̄
2
〈ψ(t)|Lx|ψ(t)〉 =
̄h
√
2(
e+iμBBt/ ̄h
21
√
2
+
1
√
2
(
e−iμBBt/ ̄h
2+
eiμBBt/ ̄h
2)
+
e−iμBBt/ ̄h
21
√
2
)
=
̄h
4(4 cos(
μBBt
̄h)) = ̄hcos(
μBBt
̄h)
Note that this agrees with what we expect att= 0 and is consistent with the angular momentum
precessing about the z axis. If we checked〈ψ|Ly|ψ〉, we would see a sine instead of a cosine,
confirming the precession.
18.10.9Expectation ofSxin General Spin^12 State
Letχ=
(
α+
α−)
, be some arbitrary spin^12 state. Then the expectation value of the operator〈Sx〉 = 〈χ|Sx|χ〉= (α∗+ α∗−)
̄h
2(
0 1
1 0
)(
α+
α−)
=
̄h
2(α∗+ α∗−)(
α−
α+)
=
̄h
2(α∗+α−+α∗−α+).18.10.10 Eigenvectors ofSxfor Spin^12
First the quick solution. Since there is no difference between x and z,we know the eigenvalues of
Sxmust be± ̄h 2. So, factoring out the constant, we have
(
0 1
1 0)(
a
b)
= ±
(
a
b)
(
b
a)
= ±
(
a
b)
a = ±bχ(+x) =( 1
√
2
√^1
2)
χ(−x) =( 1
√
2
−√ 1
2)
These are the eigenvectors ofSx. We see that if we are in an eigenstate ofSxthe spin measured in
the z direction is equally likely to be up and down since the absolute square of either amplitude is
1
2.
The remainder of this section goes into more detail on this calculationbut is currently notationally
challenged.