Suppose you have a particle in a square box $[0,L]\times[0,L]$. As the box is a square, the (2,1) and (1,2) eigenfunctions will have the same energy. If you were to apply an oscillating electric field in the $x$-direction so that the Hamiltonian would become $$ H \psi=-\frac{\hbar^2}{2m}\Delta \psi -e\,\vec{x}\cdot \vec{E}(t)\psi $$
(where, again, the field is in the $x$-direction--say $\vec{E}(t) = (A cos(\omega t),0)$ ), the second energy level would split in two.
I'd like to know what the two lowest energy levels of the perturbed Hamiltonian are, and, more importantly, whether the second energy level corresponds to (2,1) or (1,2).
