2.3 Solutions 127
The eigen vector associated withλ 1 =1is|ψ 1 >=∑^2
n= 1Cn|n>with−C 1 −iC 2 = 0 ,C 2 =iC 1 ,|ψ 1 >=1
√
2
| 1 >+
i
√
2| 2 >
The eigen vector associated withλ 2 =−1is|ψ 2 >=∑^2
n= 1Cn|n>withC 1 −iC 2 = 0 ,C 2 =−iC 1 ,|ψ 2 >=1
√
2
| 1 >−
i
√
2| 2 >
(c) The projector onto|ψi>isPi=|ψi>< ψi|.Matrix ofP 1 =[ 1
2 −
i
2
i
21
2]
,matrix ofP 2 =[ 1
2i
2
−i 2 12]
P 1 †P 2
[
00
00
]
= 0 ,P 1 P 1 † +P 2 P 2 †=I
2.80 (i)σx
(^2) =
(
01
10
)(
01
10
)
=
(
10
01
)
(ii) [σx,σy]=(
01
10
)(
0 −i
i 0)
−
(
0 −i
i 0)(
01
10
)
=
(
i 0
0 −i)
−
(
−i 0
0 i)
=
(
2 i 0
0 − 2 i)
= 2 i(
10
0 − 1
)
= 2 iσz2.81Proof : AX=λ 1 X (1)
BX=λ 2 X (2)
whereλ 1 andλ 2 are the eigen values belonging to the same stateλ.
BAX=Bλ 1 X=λ 1 BX=λ 1 λ 2 X (3)
ABX=Aλ 2 X=λ 2 AX=λ 2 λ 1 X=λ 1 λ 2 X (4)
Subtracting (3) from (4)
(AB−BA)X= 0
ThereforeAB−BA=0, becauseX
= 0
Operate withBonAin (1) and withAandBin (2)
Or [A,B]= 0