130_notes.dvi

Now as usual, the coefficient for each power ofymust be zero for this sum to be zero for ally. Before
shifting terms, we must examine the first few terms of this sum to learn about conditions ona 0 and
a 1. The first term in the sum runs the risk of giving us a power ofywhich cannot be canceled by
the second term ifk <2. Fork= 0, there is no problem because the term is zero. Fork= 1 the
term is (2ℓ+ 2)yℓ−^1 which cannot be made zero unless

a 1 = 0.

This indicates that all the odd terms in the sum will be zero, as we will see from the recursion
relation.

Now we will do the usual shift of the first term of the sum so that everything has ayℓ+kin it.

k→k+ 2

∑∞

k=0

[

ak+2(k+ 2)(2ℓ+k+ 3)yℓ+k+ak

[

2 E

̄hω

−(2ℓ+ 2k+ 3)

]

yℓ+k

]

= 0

ak+2(k+ 2)(2ℓ+k+ 3) +ak

[

2 E

̄hω

−(2ℓ+ 2k+ 3)

]

= 0

ak+2(k+ 2)(2ℓ+k+ 3) =−ak

[

2 E

̄hω

−(2ℓ+ 2k+ 3)

]

ak+2=−

2 E

̄hω−(2ℓ+ 2k+ 3)

(k+ 2)(2ℓ+k+ 3)

ak

For largek,

ak+2≈

2

k

ak,

Which will cause the wave function to diverge. We must terminate theseries for somek=nr=
0 , 2 , 4 ..., by requiring

2 E

̄hω

−(2ℓ+ 2nr+ 3) = 0

E=

(

nr+ℓ+

3

2

)

̄hω

These are the same energies as we found in Cartesian coordinates. Lets plug this back into the
recursion relation.

ak+2=−

(2ℓ+ 2nr+ 3)−(2ℓ+ 2k+ 3)

(k+ 2)(2ℓ+k+ 3)

ak

ak+2=

2(k−nr)

(k+ 2)(2ℓ+k+ 3)

ak

To rewrite the series in terms ofy^2 and letktake on every integer value, we make the substitutions
nr→ 2 nrandk→ 2 kin the recursion relation forak+1in terms ofak.