2.3 Solutions
127
The eigen vector associated with λ 1 = 1 is
|ψ 1 > =
2
n=1
C n |n > with − C 1 − iC 2 = 0, C 2 = iC 1 ,
|ψ 1 > =
1
√
2
|1 > +
i
√
2
|2 >
The eigen vector associated with λ 2 = −1 is
|ψ 2 > =
2
n=1
C n |n > with C 1 − iC 2 = 0, C 2 = −iC 1 ,
|ψ 2 > =
1
√
2
|1 > −
i
√
2
|2 >
(c) The projector onto |ψ i > is P i = |ψ i >< ψ i |.
Matrix of P 1 =
1
2
−
i
2
i
2
1
2
, matrix of P 2 =
1
2
i
2
−
i
2
1
2
P 1
† P 2
0 0
0 0
= 0, P 1 P 1
†
+ P 2 P 2
†
= I
2.80 (i) σ x
2
=
0 1
1 0
0 1
1 0
=
1 0
0 1
(ii) [σ x , σ y ] =
0 1
1 0
0 −i
i 0
−
0 −i
i 0
0 1
1 0
=
i 0
0 −i
−
−i 0
0 i
=
2i 0
0 −2i
= 2i
1 0
0 −1
= 2iσ z
2.81 Proof : A X = λ 1 X
(1)
B X = λ 2 X
(2)
where λ 1 and λ 2 are the eigen values belonging to the same state λ.
B AX = Bλ 1 X = λ 1 B X = λ 1 λ 2 X
(3)
AB X = Aλ 2 X = λ 2 AX = λ 2 λ 1 X = λ 1 λ 2 X
(4)
Subtracting (3) from (4)
(A B − B A)X = 0
Therefore A B − B A = 0, because X = 0
Operate with B on A in (1) and with A and B in (2)
Or [A, B] = 0