I am trying to solve the hydrogen atom in parabolic coordinates and find the first level correction of the Stark effect.
The Hamiltonian is (free part + interaction): $$ H^0 = -\frac{1}{2}\nabla^2 - \frac{1}{r} = -\frac{2}{\xi + \eta}\left[\frac{\partial}{\partial\xi}\left(\xi\frac{\partial}{\partial\xi}\right) + \frac{\partial}{\partial\eta}\left(\eta\frac{\partial}{\partial\eta}\right)\right] - \frac{1}{2\xi\eta}\frac{\partial^2}{\partial\phi^2} - \frac{2}{\xi + \eta} $$ $$ H^1 = -Ez = -\frac{1}{2}E(\xi - \eta) $$ During separation of variables I found that the perturbed Hamiltonian $H^0 + H^1$ is also separable: $$ \left\{ \begin{aligned} &4\frac{\partial}{\partial\xi}\left(\xi\frac{\partial\Xi}{\partial\xi}\right) + \left(E\xi^2 + 2\varepsilon\xi + 2 + \lambda -\dfrac{m^2}{\xi}\right)\Xi = 0\\ &4\frac{\partial}{\partial\eta}\left(\eta\frac{\partial H}{\partial\eta}\right) + \left(-E\eta^2 + 2\varepsilon\eta + 2 - \lambda -\dfrac{m^2}{\eta}\right)H = 0\\ \end{aligned} \right. $$ Now I've gotten the eigenstates (bound states) of the unperturbed Hamiltonian: $$ |n,k,m\rangle = C_{nkm}\xi^{\frac{\left|m\right|}{2}}\eta^{\frac{\left|m\right|}{2}}\mathrm{e}^{\frac{\xi + \eta}{2n}}\mathcal{L}^{\left|m\right|}_{(n-\left|m\right|-1+k)/2}(\xi/n)\mathcal{L}^{\left|m\right|}_{(n-\left|m\right|-1-k)/2}(\eta/n)\mathrm{e}^{\mathrm{i}m\phi} $$ Where the L's are Associated Laguerre Polynomials.
Through direct integration I found that these states are already the "good" states which diagonalize the degenerate part of $H^1$, i.e.$\langle n,k,m|H^1|n,k',m'\rangle \propto \delta_{k,k'}\delta_{m,m'}$
My questions are:
1.How does the fact that the Stark Hamiltonian is separable in parabolic coordinates relate to $|n,k,m\rangle$ being the "good" states?
2. Usually when treating a degenerate problem, we find some other operators $A_i$ which commute with both $H^1$ and $H^0$. It's obvious that $L_z = -\mathrm{i}\frac{\partial}{\partial\phi}$ is the one with eigenvalue $m$, what's the one with eigenvalue $k$ then?
