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Solving the HO Differential Equation *****

The differential equation for the 1D Harmonic Oscillator is.

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

We define a dimensionless coordinate.

The equation becomes.

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

It works.)
Now we want to find the solution for
.

becomes

which has the solution (in the large
limit)

This exponential will dominate a polynomial as
so we can write our general solution as

where
is a polynomial.
Take the differential equation

and plug

into it to get

This is our differential equation for the polynomial
.
Write
as a sum of terms.

Plug it into the differential equation.

We now want ot shift terms in the sum so that we see the coefficient of
.
To do this, we will shift the term
down two steps in the sum.
It will now show up as
.

*(Note that in doing this shift the first term for
and for
get shifted out of the sum.
This is OK since
is zero for
or
.)*
For the sum to be zero for all
, each coefficient of
must be zero.

Solve for

and we have a recursion relation giving us our polynomial.
But, lets see what we have. For large
,

The series for

has the coefficient of
equal to
and
the coefficient of
equal to
.
If
,

So our polynomial solution will approach
and our overall solution will not
be normalizable.
(Remember
.)
We must avoid this.
We can avoid the problem if the series terminates and does not go on to infinite
.

The series will terminate if

for some value of
.
Then the last term in the series will be of order
.

The acceptable solutions then satisfy the requirement

Again, we get quantized energies when we satisfy the boundary conditions at infinity.
The ground state wavefunction is particularly simple, having only one term.

Lets find
by normalizing the wavefunction.

Jim Branson
2013-04-22