How Does the 3D Harmonic Oscillator Model Extend from 1D Solutions?

[cscott]

Homework Statement

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 Schrödinger.

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 Schrödinger would give me a different answer.

 

[Avodyne]

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

 

[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?

 

[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.

 

[cscott]

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

 

[Avodyne]

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

Nope! It all looks correct now.