1000 Solved Problems in Modern Physics

3.3 Solutions 221

=[(1−0)(1+ 0 +1)]^1 /^2 | 1 >=

√

2 | 1 >

J+

∣

∣

∣ 3 >=[[1−(−1)](1− 1 +1)]

(^12)


∣

∣

∣ 2 >=

√

2 | 2 >

J−| 1 >=[(j+m)(j−m+1)]^1 /^2 | 2 >

=[(1+1)(1− 1 +1)]^1 /^2 | 2 >=

√

2 | 2 >

J−| 2 >=

√

2 | 3 >

J−| 3 >= 0

Matrices forJxandJy:

(Jx) 11 =< 1 |Jx| 1 >=^1 / 2 < 1 |J++J−| 1 >

=^1 / 2 < 1 |J+| 1 >+

1

2

< 1 |J−| 1 >

=^1 / 2 [(j−m)(j+m+1)]^1 /^2 δm′m+ 1

+^1 / 2 [(j+m)(j−m+1)]^1 /^2 δm′,m− 1 = 0

Similarly; (Jx) 22 =(Jx) 33 = 0

(Jx) 12 =< 1 |Jx| 2 >= 1 / 2 < 1 |J++J−| 2 >= 1 / 2 < 1 , 1 |J++J−| 1 , 0 >

=^1 / 2 [(j−m)(j+m+1)]^1 /^2 δm′m+ 1 + 1 /2[(j+m)(j−m+1)]^1 /^2 δm′m− 1

The second delta is zero

∴ (Jx) 12 =

1

2

[(1+0)(1+ 0 +1)]^1 /^2 =



√

2

Similarly, (Jx) 21 =(Jx) 23 =(Jx) 32 =√ 2

By a similar procedure the matrix elements ofJycan be found out. Thus

Jx=



√

2

⎛

⎝

010

101

010

⎞

⎠;Jy=√

2

⎛

⎝

0 −i 0

i 0 −i

0 i 0

⎞

⎠;Jz=

⎛

⎝

10 0

00 0

00 − 1

⎞

⎠

(b) For the matrix elements ofJwe can use the relation

<j′m′|J^2 |jm>=j(j+1)^2 δj′jδm′m

Thus (J^2 ) 11 =(J^2 ) 22 =(J^2 ) 33 =1(1+1)^2 = 2 ^2

(J^2 ) 12 =(J^2 ) 21 =(J^2 ) 13 =(J^2 ) 31 =(J^2 ) 23 =(J^2 ) 32 = 0

J^2 = 2 ^2

⎛

⎝

100

010

001

⎞

⎠

Alternatively,J^2 =Jx^2 +Jy^2 +Jz^2

Using the matrices which have been derived the same result is obtained.

3.86 With the addition ofj 1 =1 andj 2 = 1 /2, one can getJ= 3 /2or^1 / 2 .Inthe
(J,M) notation in all one gets 6 states

ψ

(

3

2

,

3

2

)

,ψ

(

3

2

,

1

2

)

,ψ

(

3

2

,−

1

2

)

,ψ

(

3

2

,−

3

2

)

andψ

(

1

2

,

1

2

)

,ψ

(

1

2

,−

1

2

)