 Solution

Expand a perturbed eigenvector and energy as $\psi ={\psi }_0+\lambda {\psi }_1+O({\lambda }^2)$ and $E(\lambda )=E+\lambda \delta +O({\lambda }^2)$. At first order, $(H-E){\psi }_1+(V-\delta ){\psi }_0=0$. Projecting onto the $E$ eigenspace eliminates the first term and gives $P_{E}V{P}_E{\psi }_0=\delta {\psi }_0$. Thus degenerate perturbation theory diagonalizes the restriction of $V$ to that eigenspace; in the nondegenerate case this reduces to its expectation value.

For the specified two-dimensional eigenspace, the restricted matrix is $\left(\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right)$. Its eigenvectors are $(|1⟩\pm |2⟩)/\sqrt{2}$ and its eigenvalues $\alpha \pm \beta$. Hence the perturbed energies are $E+\lambda (\alpha \pm \beta )+O({\lambda }^2)$.

The square-box unperturbed energies are $E_{pq}={\pi }^2{\hslash }^2(p^2+q^2)/(2m{a}^2)$. Every diagonal matrix element of $xy/a^2$ is $1/4$, because each coordinate has mean $a/2$. The ground state is $(p,q)=(1,1)$, giving $E_{11}+\lambda /4$. The first excited eigenspace is spanned by $(1,2)$ and $(2,1)$. Its off-diagonal entry factors into the square ofThus that entry is $256/(81{\pi }^4)=(4/(3\pi ){)}^4$, and the three levels areThe neglected energy corrections are $O({\lambda }^{2}m{a}^2/{\hslash }^2)$ in the given weak-field regime.