Quadratic versus linear Stark shift I'm trying to understand why the Stark shift changes from quadratic to linear as the applied electric field increases. I think there is some kind of connection to whether the induced dipole moment is also proportional to the field, or whether it saturates to some constant value (and therefore makes the shift linear). But how do I see this mathematically? 
Also, why for the hydrogen n=2 case is the shift linear from the very start? I know the linearity is due to the fact that we can take superpositions of n=2 states that have a permanent electric dipole moment, but why is that not possible for more complicated atoms?
 A: The linearity in hydrogen is due to the fact that you can take energy eigenstates that have net dipole moments.  This requires a superposition of states with different values of $\ell$.  It is only because of the accidental degeneracy of the Coulomb problem that there are $n=2$ energy levels with different values of $\ell$.  The states $|2S\rangle\pm|2P_{z}\rangle$ have dipole moments (or order $ea_{0}$) along the $z$-direction. The expectation values of the electric operator $-e{\cal E}z$ in these states give the first order Stark shifts in hydrogen.
However, in atoms with more than one electron $|2S\rangle$ and $|2P\rangle$ are not degenerate, so it is not possible to form a superposition state that is also an energy eigenstate.  (In fact, in hydrogen the states are not degenerate either, because of the combine effects of spin-orbit coupling and the Lamb shift resulting from quantum fluctuations in the electric field.)  This means the perturbative energy shifts for small fields must be quadratic in ${\cal E}$  However, when the electric field ${\cal E}$ becomes large enough the energy splittings between the different $\ell$ states become negligible.  Then the effects are very similar to what they would be if the different $\ell$ states were truly degenerate.  However, this is, strictly speaking, beyond the perturbative regime, and the energies cannot be read off directly from the perturbative formulas (although on can get the limiting linear expressions by using first-order perturbation theory and neglecting the energy spliting between the different $\ell$ states, just as the relativistic and Lamb shift corrections were ignored in the above discussion of hydrogen).
