1000 Solved Problems in Modern Physics

(Tina Meador) #1

220 3 Quantum Mechanics – II


(Jx) 12 =< 1 |Jx| 2 >=


1

2

,

1

2

|Jx|

1

2

,−

1

2


=

1

2

[(

1

2


1

2

+ 1

)(

1

2

+

1

2

)] 1 / 2

=^1 / 2 

(Jx) 21 =


1

2

,−

1

2

|J|

1

2

,

1

2


=^1 / 2 

because the second delta factor survives
(Jx) 11 =< 1 |Jx| 1 >=

〈 1

2 ,

1
2 |Jx|

1
2 ,

1
2


=0 because of delta factors.
Similarly, (Jx) 22 = 0
ThusJx= 2

(

01

10

)

;Jy= 2

(

0 −i
i 0

)

Jz=



2

(

10

0 − 1

)

(9)

These three matrices are known as Pauli matrices.
(b)J^2 =Jx^2 +Jy^2 +Jz^2
Using the matrices given in (6), squaring them and adding we get

J^2 =^3 / 4 ^2

(

10

01

)

3.85 (a) For j = 1 ,m = 1 ,0 and−1, the three base states are denoted by
| 1 >,| 2 >and| 3 >.Inthe|j,m>notation| 1 >=| 1 , 1 >,| 2 >=
| 1 , 0 >,| 3 >=| 1 ,− 1 >


Jz| 1 >=m| 1 >=| 1 >
Jz| 2 >= 0 .| 2 >= 0
Jz| 3 >=−| 3 >
(Jz) 11 =<^1 |Jz|^1 >=<^1 ,^1 |Jz|^1 ,^1 >=
(Jz) 22 =< 2 |Jz| 2 >=< 1 , 0 |Jz| 1 , 0 >= 0
(Jz) 33 =< 3 |Jz| 3 >=< 1 ,− 1 |Jz| 1 ,− 1 >=−
(Jz) 12 =(Jz) 21 =(Jz) 13 =(Jz) 31 =(Jz) 23 =(Jz) 32 = 0
because ofδ– factorδmm′

Jz=



10 0

00 0

00 − 1



For the calculation ofJxandJy, we need to work outJ+andJ−.
J+| 1 >= 0
J+| 2 >=[(j−m)(j+m+1)]^1 /^2 | 1 >
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