SHO energy eigenvalues

1. Oct 21, 2007

indigojoker

We know the eigenvalue relation for the Hamiltonian of a SHO (in QM) though relating the raising and lowering operators we get:

$$H= \hbar \omega (N+1/2)$$

This is true for $$H=\frac{p^2}{2m}+\frac{m \omega^2 x^2}{2}$$

I would like to solve for another case where $$V=a\frac{m \omega^2 x^2}{2}$$

where a is some constant

We now have $$H=\frac{p^2}{2m}+\frac{ a m \omega^2 x^2}{2}$$

I'm not sure how to go about this. When relating the creation and annihilation operators, we get: $$a^{\dagger} a = \frac{m \omega}{2 \hbar} x^2 + \frac{1}{2m \omega \hbar} p^2 -\frac{1}{2}$$

I'm not sure how to incorporate a constant into the potential, any ideas?

2. Oct 21, 2007

christianjb

This is equivalent to the substitution w'=sqrt(a)w, or am I missing something?

3. Oct 21, 2007

indigojoker

how can you arbitrarily say that though?

4. Oct 21, 2007

christianjb

It's mathematically true that you can make that substitution. Maybe I'm missing some subtlety here!

5. Oct 21, 2007

indigojoker

so you're saying that the energy eigenvalues will be:
$$H= \hbar \sqrt{a}\omega (N+1/2)$$

6. Oct 21, 2007

malawi_glenn

a is just a constant, now if you look at the harmonic potential, the $$\omega$$ is the "ground"(classical) angular frequency of the potential. So if you draw the potential as a function of x, i.e V(x) you see that the energy eigenvalues are $$\hbar (\omega \sqrt a)(n + 1/2)$$. because you simple do the change of variable that christianjb pointed out, so you get new annihilation operators and so on. Introducing this a, just implies that we change to the same 1-dim SHO but with another angular frequency.

7. Oct 21, 2007

Yes.