In quantum mechanics, a particle of mass $m$ in a 2D infinite square well has an energy spectrum of
$$E_{n_x,n_y} = \frac{n_x^2 \hbar^2 \pi^2}{2 m L_x^2} + \frac{n_y^2 \hbar^2 \pi^2}{2 m L_y^2}$$
where $n_x$ and $n_y$ are positive integers, and $L_x$ is the width of the well in the $x$ direction, while $L_y$ is the width of the well in the $y$ direction.
If $(L_x / L_y)^2$ is irrational, it is straightforward to show that there are no states with degenerate energy levels by assuming that one exists, and showing that $(L_x / L_y)^2$ can then be written as a ratio of integers, arriving at a contradiction.
If $L_x / L_y = p/q$ is rational, then choosing $n_x = Np$ and $n_y = q$ gives the same energy as $n_x' = p$ and $n_y' = Nq$ (with integer $N$), and so there do exist degenerate states.
I haven't been able to figure out what happens if $L_x / L_y$ is irrational, but $(L_x / L_y)^2$ is rational. Are degenerate states possible in such a case?
Best Answer
I don't know if you need to find a general rule, but here is a specific example showing that degeneracy is possible: If $L_x/L_y = \sqrt{2}$, then $\{n_x = 2, n_y = 5\}$ is degenerate with $\{n_x = 6, n_y = 3\}$.