SHO energy eigenvalues

[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?

[christianjb]

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

[indigojoker]

how can you arbitrarily say that though?

[christianjb]

how can you arbitrarily say that though?

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

[indigojoker]

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

[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.

[christianjb]

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

Yes.