The 3D Harmonic Oscillator

The 3D harmonic oscillator can also be separated in Cartesian coordinates. For the case of a central potential, \bgroup\color{black}$V={1\over 2}m\omega^2 r^2$\egroup, this problem can also be solved nicely in spherical coordinates using rotational symmetry. The cartesian solution is easier and better for counting states though.

Lets assume the central potential so we can compare to our later solution. We could have three different spring constants and the solution would be as simple. The Hamiltonian is

H&=&{p^2\over 2m}+{1\over 2}m\omega^2 r^2 \\
H&=&{p_x^2\over ...
...+{p_z^2\over 2m}+{1\over 2}m\omega^2 z^2 \\
H&=&H_x+H_y+H_z \\

The problem separates nicely, giving us three independent harmonic oscillators.

\bgroup\color{black}$\displaystyle E=\left(n_x+n_y+n_z+{3\over2}\right)\hbar\omega $\egroup
\bgroup\color{black}$\displaystyle \psi_{nx,ny,nz}(x,y,z)=u_{nx}(x)u_{ny}(y)u_{nz}(z) $\egroup

This was really easy.

This problem has a different Fermi surface in \bgroup\color{black}$n$\egroup-space than did the particle in a box. The boundary between filled and unfilled energy levels is a plane defined by

\begin{displaymath}\bgroup\color{black} n_x+n_y+n_z={E_F\over\hbar\omega}-{3\over 2} \egroup\end{displaymath}

Jim Branson 2013-04-22