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?
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...
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...