# Particle in harmonic oscillator potential

1. Dec 4, 2011

### gumpyworm

1. The problem statement, all variables and given/known data
A particle with mass m moves in 3-dimensions in the potential $$V(x,y,z)=\frac{1}{2}m\omega^{2}x^{2}$$. What are the allowed energy eigenvalues?

2. Relevant equations

3. The attempt at a solution
The Hamiltonian is given by $$H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2}$$ where P is the momentum operator in three dimensions. Projecting this into the coordinate basis, we have $$\left( -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\right)\psi = \left(E + \frac{\hbar^{2}}{2m}\left( \frac{d^{2}}{dy^{2}}+\frac{d^{2}}{dz^{2}}\right) \right)\psi$$
The left side is now in the form of a simple harmonic oscillator. However, I am stuck after this point.

2. Dec 4, 2011

### vela

Staff Emeritus
Try assuming a solution of the form $\psi(x,y,z)=X(x)Y(y)Z(z)$.

3. Dec 5, 2011

### gumpyworm

Hmm so if we assume that, then we find that $E=E_{x}+E_{y}+E_{z}$. But $E_{y}$ and $E_{z}$ correspond to the eigenvalues of a free particle, since the particle is free in the y and z directions. So does this just mean that the eigenvalues are continuous, since the energy is only quantized along one direction?

4. Dec 5, 2011

### cbetanco

Why do you say it's only quantized in one direction? And no, the energy is not continuous in this case.

5. Dec 5, 2011

### gumpyworm

Don't we just have $$E=E_{x}+E_{y}+E_{z}=(n+\frac{1}{2})\hbar\omega+\frac{p_{y}^{2}}{2m}+\frac{p_{z}^{2}}{2m}$$Then, since p_y and p_z are continuous, wouldn't we just have that the ground state energy is $\frac{\hbar\omega}{2}$ and that above that, the eigenvalues are continuous?

6. Dec 5, 2011

### vela

Staff Emeritus
Yes.
Because the potential only depends on x. The particle isn't bound when moving in the y and z directions, so Ey and Ez can take on any value. Ex is quantized, but the total energy isn't.

7. Dec 5, 2011

### cbetanco

Why do you treat y and z differently from x? your solution should be the same for X(x), Y(y) and Z(z). I would say $n=n_x+n_y+n_z$

8. Dec 5, 2011

### cbetanco

OMG, I'm sorry. I didn't read the OP carefully enough. Yes, then you are right. I was thinking of $\frac{1}{2}m\omega r^2$ where $r^2=x^2+y^2+z^2$