130_notes.dvi

9.7.4 Solving the HO Differential Equation*

The differential equation for the 1D Harmonic Oscillator is.

− ̄h^2

2 m

d^2 u

dx^2

+

1

2

mω^2 x^2 u=Eu.

By working with dimensionless variables and constants, we can see the basic equation and minimize
the clutter. We use the energy in terms of ̄hω.

ǫ=

2 E

̄hω

We define a dimensionless coordinate.

y=

√

mω

̄h

x

The equation becomes.

d^2 u

dx^2

+

2 m

̄h^2

(E−

1

2

mω^2 x^2 )u= 0

d^2 u

dy^2

+ (ǫ−y^2 )u= 0

(Its probably easiest to just check the above equation by substituting as below.

̄h

mω

d^2 u

dx^2

+

(

2 E

̄hω

−

mω

̄h

x^2

)

= 0

d^2 u

dx^2

+

2 m

̄h^2

(E−

1

2

mω^2 x^2 )u= 0

It works.)

Now we want to find the solution fory→∞.

d^2 u

dy^2

+ (ǫ−y^2 )u= 0

becomes
d^2 u
dy^2

−y^2 u= 0

which has the solution (in the largeylimit)

u=e−y

(^2) / 2
.
This exponential will dominate a polynomial asy→∞so we can write our general solution as
u(y) =h(y)e−y
(^2) / 2
whereh(y) is a polynomial.