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If the lamb shift causes the $^2$S$_{\frac{1}{2}}$ state to be higher in energy than $^2$P$_{\frac{1}{2}}$, why does the $^2$S$_{\frac{1}{2}}$? state not decay to $^2$P$_{\frac{1}{2}}$ Is there a selection rule that forbids this or is there a more subtle effect at work?

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This is incorrect. The Lamb shift makes the $2\,{}^2 P_{1/2}$ state lower than the $2\,{}^2 S_{1/2}$; as such, the former will not spontaneously decay into the latter. In any case, these are states within the $n=2$ shell, so that their main decay channel is to the ground state, $1\,{}^2S_{1/2}$; this can and will happen on a ~nanosecond timescale (i.e. slow compared to the radiation, but fast as transitions go).

If the atom starts off in the $2\,{}^2 S_{1/2}$ state, then it will first decay to $2\,{}^2 P_{1/2}$, and from there to the $1\,{}^2 S_{1/2}$ ground state. This is because dipole transitions are forbidden between $S$ states, so the direct jump would require a quadrupole transition, which is a lot less likely than two dipole jumps.

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  • $\begingroup$ Yeah I got that completely backwards... $\endgroup$ Sep 9 '14 at 11:26

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