130_notes.dvi

= ̄h^2

(

χ 00 −

1

2

√

2

(

χ(1)+χ(2)− −χ(1)−χ(2)+ +χ(1)+χ(2)− −χ(1)−χ(2)+

))

= ̄h^2 (1−1)χ 00 = 0 ̄h^2 χ 00

21.8.4 Adding anyℓplus spin

We wish to write the states of total angular momentumjin terms of the product statesYℓmχ±. We
will do this by operating with theJ^2 operator and setting the coefficients so that we have eigenstates.

J^2 ψjmj=j(j+ 1) ̄h^2 ψjmj

We choose to write the the quantum numbermjasm+^12. This is really just the defintion of the
dummy variablem. (Other choices would have been possible.)

The z component of the total angular momentum is just the sum of the z components from the
orbital and the spin.
mj=ml+ms

There are only two product states which have the rightmj=m+^12. If the spin is up we needYℓm
and if the spin is down,Yℓ(m+1).

ψj(m+ (^12) )=αYℓmχ++βYℓ(m+1)χ−
audio
We will find the coefficientsαandβso thatψwill be an eigenstate of
J^2 = (~L+S~)^2 =L^2 +S^2 + 2LzSz+L+S−+L−S+.
So operate on the right hand side withJ^2.
J^2 ψj,m+ 12 = α ̄h^2

[

ℓ(ℓ+ 1)Ylmχ++

3

4

Ylmχ++ 2m

1

2

Ylmχ+

+

√

ℓ(ℓ+ 1)−m(m+ 1)

√

1 Yl(m+1)χ−

]

+ β ̄h^2

[

ℓ(ℓ+ 1)Yℓ,m+1χ−+

3

4

Yℓ,m+1χ−+ 2(m+ 1)

(

− 1

2

)

Yℓ,m+1χ−

+

√

ℓ(ℓ+ 1)−(m+ 1)m

√

(1)Ylmχ+

]

And operate on the left hand side.

J^2 ψj,m+^12 =j(j+ 1) ̄h^2 ψj,m+^12 =j(j+ 1) ̄h^2

(

αYlmχ++βYℓ,(m+1)χ−

)

Since the two terms are orthogonal, we can equate the coefficientsfor each term, giving us two
equations. TheYℓmχ+term gives

αj(j+ 1) =α

(

ℓ(ℓ+ 1) +

3

4

+m

)

+β

√

ℓ(ℓ+ 1)−m(m+ 1).