130_notes.dvi

TheYℓ(m+1)χ−term gives

βj(j+ 1) =β

(

ℓ(ℓ+ 1) +

3

4

−(m+ 1)

)

+α

√

ℓ(ℓ+ 1)−m(m+ 1).

Collectingαterms on the LHS andβterms on the RHS, we get two equations.

(

j(j+ 1)−ℓ(ℓ+ 1)−

3

4

−m

)

α =

√

(ℓ−m)(ℓ+m+ 1)β

√

(ℓ−m)(ℓ+m+ 1)α =

[

j(j+ 1)−ℓ(ℓ+ 1)−

3

4

+ (m+ 1)

]

β

Now we just cross multiply so we have one equation with a common factor ofαβ.

(ℓ−m)(ℓ+m+ 1) =

[

j(j+ 1)−ℓ(ℓ+ 1)−

3

4

−m

][

j(j+ 1)−ℓ(ℓ+ 1)−

3

4

+ (m+ 1)

]

While this equation looks like a mess to solve, if we notice the similarity between the LHS and RHS,
we can solve it if

ℓ=j(j+ 1)−ℓ(ℓ+ 1)−

3

4

.

If we look a little more carefully at the LHS, we can see that another solution (which just interchanges
the two terms in parentheses) is to replaceℓby−ℓ−1.

−ℓ−1 =j(j+ 1)−ℓ(ℓ+ 1)−

3

4

.

These are now simple to solve

j(j+ 1) =ℓ(ℓ+ 1) +ℓ+

3

4

⇒ j=ℓ+

1

2

j(j+ 1) =ℓ(ℓ+ 1)−ℓ−1 +

3

4

⇒ j=ℓ−

1

2

So these are (again) the two possible values forj. We now need to go ahead and findαandβ.

Pluggingj=ℓ+^12 into our first equation,

(ℓ−m)α=

√

(ℓ−m)(ℓ+m+ 1)β

we get the ratio betweenβandα. We will normalize the wave function by settingα^2 +β^2 = 1. So
lets get the squares.

β^2 =

(ℓ−m)^2

(ℓ−m)(ℓ+m+ 1)

α^2 =

(ℓ−m)

(ℓ+m+ 1)

α^2

α^2 +β^2 = 1⇒

ℓ+m+ 1 +ℓ−m

ℓ+m+ 1

α^2 = 1

α=

√

ℓ+m+ 1

2 ℓ+ 1

β=

√

ℓ−m

ℓ+m+ 1

√

ℓ+m+ 1

2 ℓ+ 1

=

√

ℓ−m

2 ℓ+ 1