### 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