= Solution
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 $|\psi(x,y)|^2dx\,dy$ to a small area, and $\int|\psi|^2dx\,dy=1$. A stationary state's full time dependence is $\psi(x,y)e^{-iEt/\hbar}$; 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,
$$
\frac{X''}{X}+\frac{Y''}{Y}=-\frac{2mE}{\hbar^2}.
$$
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 gives
$$
\boxed{k_x=\frac{n_x\pi}{a},\qquad k_y=\frac{n_y\pi}{b},\qquad n_x,n_y=1,2,\ldots.}
$$
The normalized <two-dimensional infinite square well> <eigenbasis> and energies are
$$
\boxed{\psi_{n_xn_y}(x,y)=\frac2{\sqrt{ab}}\sin\frac{n_x\pi x}{a}\sin\frac{n_y\pi y}{b},\qquad E_{n_xn_y}=\frac{\pi^2\hbar^2}{2m}\left(\frac{n_x^2}{a^2}+\frac{n_y^2}{b^2}\right).}
$$
Normalization 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>. Thus
$$
\boxed{\epsilon_{\min}=E_{12}-E_{11}=\frac{3\pi^2\hbar^2}{2m_eb^2}=4.52\times10^{-20}\,\mathrm J=0.282\,\mathrm{eV}.}
$$
This 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 $E_{21}-E_{12}=3\pi^2\hbar^2(a^{-2}-b^{-2})/(2m)>0$. 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:
$$
\frac4{a^2}+\frac1{b^2}=\frac1{a^2}+\frac9{b^2},\qquad \boxed{b=\sqrt{\frac83}\,a=1.633\,\mathrm{nm}.}
$$
The <third-level degeneracy in a rectangular quantum box> then has lowest distinct energies $11/8$, $5/2$ and $35/8$ in units $\pi^2\hbar^2/(2m_ea^2)$, with multiplicities one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.
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