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Here $V(x,y)$ is the potential energy and $E$ is the stationary energy eigenvalue. The Schrödinger equation is the eigenvalue equation for kinetic plus potential energy. Its boundary conditions select the allowed energies. For a normalized wavefunction, the Born rule assigns probability $|\psi (x,y){|}^{2}dxdy$ to a small area, and $\int |\psi {|}^{2}dxdy=1$. A stationary state's full time dependence is $\psi (x,y)e^{-iEt/\hslash }$; its probability density is time-independent.

The infinite walls impose $\psi =0$ on all four boundaries. For a separated solution $\psi =X(x)Y(y)$ in the zero-potential interior,The first term depends only on $x$ and the second only on $y$, so each is constant, $-k_x^2$ and $-k_y^2$. Vanishing at $x=0$ and $y=0$ removes the cosine terms. Vanishing at the other boundaries givesThe normalized two-dimensional infinite square well eigenbasis and energies areNormalization follows from the two sine-squared integrals $a/2$ and $b/2$. Zero quantum numbers would give an identically zero function. These separated products form a basis, not every possible wavefunction: a general state is their superposition, and a degenerate energy eigenspace can have nonseparable stationary combinations of products with that same energy.

For $a=1\mathrm{nm}$ and $b=2\mathrm{nm}$, the ground state is $(1,1)$ and the nearest excitation is $(1,2)$, along the longer side. The next quantum along the shorter side costs more energy. ThusThis transition has nonzero dipole matrix element for an electric-field component along $y$. Light polarized strictly along $x$ would not drive it, so the minimum assumes an allowed polarization rather than an arbitrary fixed one.

For the requested degeneracy, keep $b>a$. The first two states are necessarily $(1,1)$ and $(1,2)$, since $E_{21}-E_{12}=3{\pi }^2{\hslash }^2(a^{-2}-b^{-2})/(2m)>0$. The third level is the smaller of $E_{21}$ and $E_{13}$: states with both indices at least two lie above $E_{21}$, and higher states on the $n_x=1$ sequence lie above $E_{13}$. Make these two candidates equal:The third-level degeneracy in a rectangular quantum box then has lowest distinct energies $11/8$, $5/2$ and $35/8$ in units ${\pi }^2{\hslash }^2/(2m_e{a}^2)$, with multiplicities one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.

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 ${\pi }^2{\hslash }^2/(2M{a}^2)$ 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 $(1,3)$.