In Quantum Mechanics we can deal with the one-dimensional harmonic oscilator by using the trick of the ladder operators. In that case, the original Hamiltonian is
$$H = \dfrac{1}{2m}P^2+\dfrac{m\omega^2}{2}X^2$$
and upon defining the operators
$$a = \dfrac{1}{\sqrt{2}}(\hat{X}^2+i\hat{P}), \quad a^\dagger = \dfrac{1}{\sqrt{2}}(\hat{X}^2-i\hat{P}^2)$$
with $\hat{X} = \sqrt{\frac{m\omega}{\hbar}}X$ and $\hat{P} = \frac{1}{\sqrt{m\hbar\omega}}P$ we can write
$$H = \hbar\omega\left(a^\dagger a + \frac{1}{2}\right)$$
and the problem can be dealt with using just the operators $a$ and $a^\dagger$ and their product. Now, on one exercise the author asks to find the eigenvectors and eigenvalues of the Hamiltonian for the anisotropic harmonic oscilator in three dimensions, where the potential is
$$V(X,Y,Z) = \dfrac{m\omega^2}{2}\left[\left(1+\dfrac{2\lambda}{3}\right)(X^2+Y^2)+\left(1-\dfrac{4\lambda}{3}\right)Z^2\right].$$
The Hamiltonian for this seems quite complicated, but I imagine there is some trick like the one dimensional case which simplifies the problem a lot. Is there some way to reduce this to the one-dimensional case?
How can we treat the three dimensional anisotropic harmonic oscilator in Quantum Mechanics without complicating the problem a lot?
Best Answer
The potential $V(X,Y,Z)$ corresponds to three uncoupled harmonic oscillators with \begin{equation} \omega_X = \omega_Y = \omega \sqrt{1+\frac{2\lambda}{3}}, \qquad \omega_Z = \omega \sqrt{1-\frac{4\lambda}{3}}. \end{equation} Because the oscillators are uncoupled, the Schroedinger equation is separable and the eigenkets can therefore be written as $\left| \psi \right> = \left| n_x \right> \left| n_y \right> \left| n_z \right>$ as mentioned by dmckee in the comments. The energy spectrum can now be found straightforwardly by defining three sets of creation and annihilation operators that each act on one of the $\left| n_i \right>$.
Note that $-\frac{3}{2} < \lambda < \frac{3}{4}$, otherwise at least one of the oscillator potentials is inverted and there are no bound states in that case.