# 3D Quantum harmonic Oscillator

1. Sep 4, 2011

### c299792458

1. The problem statement, all variables and given/known data
What are the stationary states of an isotropic 3D quantum harmonic oscillator in a potential $$U(x,y,z) = {1\over2}m\omega^2 (x^2+y^2+z^2)$$ in the form $$\psi(x,y,z)=f(x)g(y)h(z)$$ and how many linearly independent states have energy $$E=({3\over 2}+n)\hbar\omega$$?

2. Relevant equations

See above.

3. The attempt at a solution
The solution of a HO in 1D, say the x-direction, is $$c H_n e^{\sqrt{m\omega\over\hbar}x}$$ where $$H_n$$ is the $$nth$$ Hermite polynomial. So I am guessing $$\psi = c^3 H_n^3 e^{{m\omega\over\hbar}(x^2+y^2+z^2)}$$. I don't quit understand how to count L.I. states, though. I am guessing the number of combinations of $$n_x,n_y,n_z$$ such that $$n_x+n_y+n_z=n$$? Please help! Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 4, 2011

### vela

Staff Emeritus
That's not quite right.
Decent guess. Try plugging $\psi(x,y,z)$ into the Schrodinger equation. You should find the equation separates, so you can solve it.
Yes, that's correct.

3. Sep 4, 2011

### c299792458

Thanks, vela. So is $$\psi=cH_{n_x}H_{n_y}H_{n_z}e^{{m\omega\over\hbar}{(x^2+y^2+z^2)}}$$, where c is some normalization constant?

4. Sep 4, 2011

### vela

Staff Emeritus
Close. You're missing a factor of -1/2 in the exponent. What's the argument of the Hermite polynomials?

5. Sep 4, 2011

### c299792458

Ah, thanks. Is the argument $$\sqrt{m\omega\over\hbar}x_i$$, where $$x_i\in\{x,y,z\}$$?

6. Sep 4, 2011

### vela

Staff Emeritus
Yup!

7. Sep 4, 2011

Thanks! :-)