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I have read the stark effect of Hydrogen (calculating energy levels of the n=2 states of a Hydrogen atom placed in an external uniform electric field along the positive z-direction) from Quantum Mechanics by N. Zetilli. (chapter 9, example 9.3, page 498) Using the degenerate perturbation theory, we can see that initially there were four degenerate levels (|nlm>=|200>,|211>,|210>,|21-1>) and on adding the perturbation term in Hamiltonian, the degeneracy is partially removed. The states |211> and |21-1> still have the same energy while the other two have different energies. I have understood the mathematics but i want to know the physics behind this. As in if there is a physical basis by which we can determine which states will still remain degenerate and which will not.

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    $\begingroup$ Hint: symmetry. $\endgroup$ Jul 10 '19 at 7:25
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The levels $|nlm\rangle = |200\rangle, |210\rangle, |211\rangle, |21-1\rangle$ are eigenfunctions of the Hamiltonian

$$H=-\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\varepsilon_0 r}$$

so the four levels are clearly degenerate since the energy cannot depend on which direction we choose for the $\hat{z}$ axis. When we introduce an electric field along a certain direction we destroy the isotropy of the system. Let the direction of the field be the $\hat{z}$ direction, we will have a perturbation $H'=eEz$. We can observe that $[H', L_{z}]=eE[z, L_{z}]=0$, it means that the system is invariant under rotation about $\hat{z}$ axis, in fact it can be shown that a rotation of a finite angle $d\phi$ around $\hat{z}$ axis applied to an eigenfunction $|r(d\phi)\rangle$ of the position operator $r_{op}$ can be written as

$$|r(d\phi)\rangle=|r, \theta, \phi + d\phi \rangle = \exp\left( -\frac{i}{\hbar}L_{z}d\phi \right)|r\rangle$$

This shows that $L_{z}$ is the generator of the rotations around $\hat{z}$ axis. As a consequence of that, $H'$ only connects the levels with the same value of $m$, those levels are $|200\rangle$ and $|210\rangle$.

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