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\begin{document}

Quantize a square well. \( H_{0}=0\in [0,L] \) and \( H_{0}=\infty \in (-\infty ,0)or(L,\infty ) \).

\( H_{1}=\beta xy \).

solution for \( H_{0} \)is \( \psi _{n_{x}n_{y}}(x,y)=\frac{2}{L}sin(n_{x}\frac{\pi x}{L})sin(n_{y}\frac{\pi y}{L}) \) with \( n_{i}\in [1,\infty ) \) 

\( E_{n_{x}n_{y}}^{(0)}=\frac{\pi ^{2}}{2mL^{2}}(n_{x}^{2}+n_{y}^{2}) \) so \( E_{11}=\frac{\pi ^{2}}{ml^{2}} \), \( E_{12}=E_{21}=\frac{5\pi ^{2}}{2mL^{2}} \).

\( \triangle _{11}^{(1)}=<11|H_{1}|11>=\beta <11|xy|11>=\frac{\beta 4}{L^{2}}\int dx\int dy=\frac{1}{4}\beta L^{2} \).

On first excited state, you have:

\( H_{0}^{'}=H_{0}+PH_{1}P=\begin{array}{cc}
E_{12}+<12|H_{1}|12> & <12|H_{1}|21>\\
<21|H_{1}|12> & E_{21}+<21|H_{1}|21>
\end{array}
 \)

\( <12|H_{1}|12>=<21|H|21>=(\alpha =\frac{1}{4})\beta L^{2} \) and \( <12|H_{1}|21>=<21|H|12>=(\gamma =\frac{32}{9\pi ^{2}})\beta L^{2} \) This gives eigenvalues: \( \lambda =E_{12}+\alpha \beta L^{2}\pm \gamma \beta L^{2} \). So eigenstates are \( \frac{1}{\sqrt{2}}(|12>\pm |21>) \).

If symmetry under x,y interchange then diagonal basis is \( \frac{1}{\sqrt{2}}(|a>\pm |b>) \).

Parity (interchange of x,-x), then parity of perturbation splits parity
odd systems.

diagonal in parity eigenstates. Ex: \( e^{\pm ipx/\hbar } \) goes to \( sin(px/\hbar ) \) parity odd or \( cos(px/\hbar ) \) for
parity even.

\end{document}
