Consider a Hamiltonian $H$ with known eigenstates and eigenvalues (possibly degenerate). Derive a general method for calculating the energies of a new Hamiltonian $H+\lambda V$ to first order in the parameter $\lambda$. Apply this method to find approximate expressions for the new energies close to an eigenvalue $E$ of $H$, given that there are just two orthonormal eigenstates $|1\rangle$ and $|2\rangle$ corresponding to $E$ and that

$$\langle 1|V| 1\rangle=\langle 2|V| 2\rangle=\alpha, \quad\langle 1|V| 2\rangle=\langle 2|V| 1\rangle=\beta$$

A charged particle of mass $m$ moves in two-dimensional space but is confined to a square box $0 \leqslant x, y \leqslant a$. In the absence of any potential within this region the allowed wavefunctions are

$$\psi_{p q}(x, y)=\frac{2}{a} \sin \frac{p \pi x}{a} \sin \frac{q \pi y}{a}, \quad p, q=1,2, \ldots$$

inside the box, and zero outside. A weak electric field is now applied, modifying the Hamiltonian by a term $\lambda x y / a^{2}$, where $\lambda m a^{2} / \hbar^{2}$ is small. Show that the three lowest new energy levels for the particle are approximately

$$\frac{\hbar^{2} \pi^{2}}{m a^{2}}+\frac{\lambda}{4}, \quad \frac{5 \hbar^{2} \pi^{2}}{2 m a^{2}}+\lambda\left(\frac{1}{4} \pm\left(\frac{4}{3 \pi}\right)^{4}\right)$$

[It may help to recall that $2 \sin \theta \sin \varphi=\cos (\theta-\varphi)-\cos (\theta+\varphi)$.]