130_notes.dvi

u 0 (x) =

(mω

π ̄h

)^14

e−mωx

(^2) /2 ̄h
This is a Gaussian (minimum uncertainty) distribution. Since the HO potential has a parity sym-
metry, thesolutions either have even or odd parity. The ground state is even parity.
Thefirst excited stateis an odd parity state, with a first order polynomial multiplying the same
Gaussian.
u 1 (x) =
(mω
π ̄h
)^14 √ 2 mω
̄h
xe−mωx
(^2) /2 ̄h
Thesecond excited stateis even parity, with a second order polynomial multiplying the same
Gaussian.
u 2 (x) =C

(

1 − 2

mωx^2

̄h

)

e−mωx

(^2) /2 ̄h
Note thatnis equal to the number of zeros of the wavefunction. This is a common trend. With
more zeros, a wavefunction has more curvature and hence more kinetic energy.
The general solution can be written as
un(x) =

∑∞

k=0

akyke−y

(^2) / 2
with the coefficients determined by the recursion relation
ak+2=
2(k−n)
(k+ 1)(k+ 2)
ak
and the dimensionless variableygiven by.
y=

√

mω

̄h

x

The series terminates with the last nonzero term havingk=n.