# The stark effect of Hydrogen

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.

• Hint: symmetry. Jul 10, 2019 at 7:25

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 $$|210\rangle$$.