# A 3D Harmonic Oscillator

1. Jun 1, 2009

### cscott

1. The problem statement, all variables and given/known data

Consider a particle of mass $m$ moving in a 3D potential

$$V(\vec{r}) = 1/2m\omega^2z^2,~0<x<a,~0<y<a$$.

$$V(\vec{r}) = \inf$$, elsewhere.

2. The attempt at a solution

Given that I know the solutions already for a 1D harmonic oscillator and 1D infinite potential well I'm going to combine them $E_x + E_y + E_z = E$, and $\psi=\psi(z)\psi(x)\psi(y)$ as for separation of variables of 3D Schrodinger.

Therefore,

$$E = (n_z+1/2)\hbar\omega + \frac{\pi^2\hbar^2}{2ma^2}(n_x^2+n_y^2)$$

$$\psi=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n_z}n_z!}}H_{n_z}(\zeta)e^{-\zeta^2/2}\sqrt{\frac{2}{a}}\sin\left(\frac{n_x\pi}{a}x\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n_y\pi}{a}y\right)$$

where $H$ are the Hermite polynomials and $$\zeta=\sqrt{m\omega/\hbar}z$$.

Is this a correct approach? I couldn't see how going all through the separation of variables for 3D Schrodinger would give me a different answer.

Last edited: Jun 2, 2009
2. Jun 1, 2009

### Avodyne

You are correct (modulo some minor typos in your expression for psi).

3. Jun 2, 2009

### cscott

I believe I fixed the typo. Did I miss anything?

My OP was because the second part asks: Assuming $$\hbar\omega>3\pi^2\hbar^2/(2ma^2)$$ find the energies of the ground state and the first excited state, labeling by each state by its three quantum numbers.

And I don't see why this assumption needs to be made. Wouldn't I just sub into $E$,

$$(n_x, n_y, n_z) = (1, 1, 0)$$ for the ground state and,

$$(n_x, n_y, n_z) = (1, 1, 1)$$ for the first excited state?

4. Jun 2, 2009

### Hao

For the ground state, the assumption makes no difference. However, it has an effect for the first excited state.

The first excited state is defined as the state that is second lowest in energy.

If $$\hbar \omega$$ were to be very small, $$(n_x, n_y, n_z) = (1, 1, 1)$$ would be second lowest in energy, and hence the first excited state.
However, if $$\hbar \omega$$ were to be very large, it can be possible for $$(n_x, n_y, n_z) = (2, 1, 0)$$ to be lower in energy than $$(n_x, n_y, n_z) = (1, 1, 1)$$ instead.

5. Jun 2, 2009

### cscott

Thanks, this makes sense now. I was too caught up thinking it had something to do with keeping the energy positive/negative.

6. Jun 2, 2009

### Avodyne

Nope! It all looks correct now.