= Third-level degeneracy in a rectangular quantum box
{title2=$b/a=\sqrt{8/3}$}
For $b>a$ in a <two-dimensional infinite square well>, the lowest two states are $(1,1)$ and $(1,2)$, both nondegenerate. The next energy is the smaller of $E_{21}$ and $E_{13}$: every other candidate lies above at least one of these. Requiring equality gives $3/a^2=8/b^2$, or $b/a=\sqrt{8/3}$. In units $\pi^2\hbar^2/(2Ma^2)$ the lowest distinct energies are then $11/8$, $5/2$ and $35/8$, with degeneracies one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.
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