= Solution
The <Hamiltonian operator> separates into two commuting one-dimensional <quantum harmonic oscillators>. Write
$$
\xi_j=\sqrt{\frac{m\omega_j}{\hbar}}x_j,\qquad
G=e^{-(\xi_1^2+\xi_2^2)/2},\qquad E_{00}=\frac{\hbar}{2}(\omega_1+\omega_2).
$$
Products of <harmonic oscillator eigenfunctions> give
$$
\psi_{n_1,n_2}\propto H_{n_1}(\xi_1)H_{n_2}(\xi_2)G,\qquad
E_{n_1,n_2}=E_{00}+\hbar(n_1\omega_1+n_2\omega_2).
$$
Here $H_n$ is the physicists' <Hermite polynomial>, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six <energy eigenstates>, ordered with multiplicity, are
$$
\boxed{\begin{array}{c|c|c}
(n_1,n_2)&E_{n_1,n_2}&\psi_{n_1,n_2}\ \text{up to a nonzero constant}\\\hline
(0,0)&E_{00}&G\\
(1,0)&E_{00}+\hbar\omega_1&\xi_1G\\
(0,1)&E_{00}+\hbar\omega_2&\xi_2G\\
(2,0)&E_{00}+2\hbar\omega_1&(4\xi_1^2-2)G\\
(1,1)&E_{00}+\hbar(\omega_1+\omega_2)&\xi_1\xi_2G\\
(0,2)&E_{00}+2\hbar\omega_2&(4\xi_2^2-2)G
\end{array}}
$$
For this <two-dimensional oscillator level ordering>, the excitation <energies> obey
$$
0<\omega_1\leq\omega_2<2\omega_1\leq\omega_1+\omega_2\leq2\omega_2<3\omega_1.
$$
Every state with $n_1+n_2\geq3$ costs at least $3\hbar\omega_1$, so none has been missed. Equalities within the list occur when $\omega_2=\omega_1=\omega$. Then the <two-dimensional isotropic harmonic oscillator> has
$$
\boxed{E_N=\hbar\omega(N+1),\qquad g_N=N+1,\qquad N=0,1,2,\ldots.}
$$
The <quantum degeneracy> counts $n_1=0,\ldots,N$ with $n_2=N-n_1$. The six listed states occupy three distinct <energy levels>, of <quantum degeneracies> one, two and three.
Back to article page
