1000 Solved Problems in Modern Physics

3.3 Solutions 219

Apart from the factor 1/r^2 , the angular part is seen to be

∂^2 ψ

∂θ^2

+cotθ

∂ψ

∂θ

+

1

sin^2

∂^2 ψ

∂φ^2

3.84 (a) The (i,j)th matrix element of an operatorOis defined by

Oij=<i|o|j> (1)

Forj= 1 / 2 ,m= 1 /2 and− 1 /2. The two states are

| 1 >=|

1

2

,

1

2

>and| 2 >=|

1

2

,−

1

2

> (2)

With the notation|j,m>

Now

<j′m′|Jz|jm>=mδjj′δmm′ (3)

Thus

(Jz) 11 =< 1 |Jz| 1 >=

1

2

 (4)

(Jz) 22 =< 2 |Jz| 2 >=−

1

2

 (5)

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

〈

1

2

,

1

2

|Jz|

1

2

,

1

2

〉

=0(6)

because of (3).

Similarly

(Jz) 21 =0(7)

Therefore

Jz=



2

(

10

0 − 1

)

(8)

ForJxandJy, we use the relations

Jx=^1 / 2 (J++J−) andJy=−

(

1

2 i

)

(J+−J−)

<j,m|Jx|j,m>=<j,m|

1

2

(J++J−)|j,m>

=^1 / 2 [(j+m+1)(j−m)]^1 /^2 <j,m′|j,m+^1 >

+^1 / 2 [(j−m+1)(j+m)]^1 /^2 <j,m′|j,m− 1 >

=^1 / 2 [(j+m+1)(j−m)]^1 /^2 δm′,m+ 1

+^1 / 2 [(j−m+1)(j+m)]

(^1) / 2
δm′,m− 1
That is the matrix element is zero unlessm′=m+1orm′=m−1.
The first delta factor survives