Particle in harmonic oscillator potential

[gumpyworm]

Homework Statement

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?

Homework Equations

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.

 

[vela]

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

 

[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?

 

[cbetanco]

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

 

[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?

 

[vela]

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?

Yes.

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

Because the potential only depends on x. The particle isn't bound when moving in the y and z directions, so E_y and E_z can take on any value. E_x is quantized, but the total energy isn't.

 

[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 $n=n_x+n_y+n_z$

 

[cbetanco]

Because the potential only depends on x.

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 $r^2=x^2+y^2+z^2$