1000 Solved Problems in Modern Physics

(Romina) #1

2.3 Solutions 127


The eigen vector associated withλ 1 =1is

|ψ 1 >=

∑^2

n= 1

Cn|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= 1

Cn|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
2

1
2

]

,matrix ofP 2 =

[ 1

2

i
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 iσz

2.81Proof : 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
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