The series for

y^2 ey

(^2) / 2

∑y^2 n+2
2 nn!
has the coefficient ofy^2 n+2equal to 2 n^1 n!and the coefficient ofy^2 nequal to 2 n− (^1) (^1 n−1)!. Ifm= 2n,
am+2=

1

2 n

am=

1

m

am.

So our polynomial solution will approachy^2 ey

(^2) / 2
and our overall solution will not be normalizable.
(Rememberu(y) =h(y)e−y
(^2) / 2
.) We must avoid this.
We can avoid the problem if the series terminates and does not go on to infinitem.
am+2=
2 m+ 1−ǫ
(m+ 1)(m+ 2)
am
The series will terminate if
ǫ= 2n+ 1
for some value ofn. Then the last term in the series will be of ordern.
an+2=

0

(n+ 1)(n+ 2)

an= 0

The acceptable solutions then satisfy the requirement

ǫ=

2 E

̄hω

= 2n+ 1

E=

(2n+ 1)

2

̄hω=

(

n+

1

2

)

̄hω

Again, we get quantized energies when we satisfy the boundary conditions at infinity.

The ground state wavefunction is particularly simple, having only oneterm.

u 0 (x) =a 0 e

−y 22

=a 0 e−mωx

(^2) /2 ̄h
Lets finda 0 by normalizing the wavefunction.
∫∞
−∞
|a 0 |^2 e−mωx
(^2) / ̄h
dy= 1
|a 0 |^2

√

π ̄h

mω

= 1

u 0 (x) =

(mω

π ̄h

)^14

e−mωx

(^2) /2 ̄h