Selection rules in one electron atoms are:

  1. $\Delta n=$ any

  2. $\Delta l=\pm1$

  3. $\Delta m_l=0,\pm1$

  4. $\Delta s=0$

  5. Parity must change

Under strong spin orbit interaction:

  1. $\Delta j=0,\pm 1$, but $j=0\nRightarrow\: j'=0 $

  2. $\Delta m_j=0,\pm1$

In my notes, it states that while considering a $\rm H$ atom in the $2p$ state with $m_l=0$, the only possible decay is to the $1s$ ground state with $m_l=0$. This implies that the $2p\rightarrow2s$ transition is not possible but upon looking at the selection rules, I can't find anything wrong with it. What am I missing?


The transition is possible, though it's important to note that the energy ordering is opposite to what you seem to think it is - the $2s$ energy is higher than the $2p$ energy. Moreover, this energy difference is absolutely tiny - either $4\:\rm \mu eV$ or $50 \:\rm \mu eV$, depending on the total angular momentum in the $2p$ state, corresponding to wavelengths of order $30\:\rm cm$ and $3\:\rm cm$ (and therefore frequencies of order $970\:\rm MHz$ and $12\:\rm GHz$), respectively. (For more details about this splitting see e.g. my answer to this question.)

Still if you have a good enough state-preparation procedure and a stable enough microwave source, probably together with a pretty fancy atomic-beam apparatus, you should be able to observe the transition.

The lines are listed in the NIST ASD database - it's a good exercise to learn to use it so that you can find the states, the transitions, and the listed references to experimental observations of all the lines involved.


The story is more complicated. There are three states, $^2S_{1/2}$, $^2P_{1/2}$ and $^2P_{3/2}$. These are degenerate in the Schrödinger solution of the hydrogen atom. The Dirac equation shifts the $^2P_{3/2}$ and leaves the $^2S_{1/2}$ and $^2P_{1/2}$ degenerate. QED radiative corrections lift this degeneracy by a small amount of about 1 GHz, the famous Lamb shift, which is an important precision test of QED. Willis Lamb received the Nobel prize in 1955 for its experimental determination in 1947.


For a one electron system, the energy only depends on the principle quantum number. Therefore, the sublevels are degenerate. When you have degenerate energy levels, the state you are in is a superposition of the levels. So, the transition from 2p to 2s is not a transition: the electron is in a superposition of them (as long as they are degenerate). The only level that is lower in energy than n=2 is the 1s state, and then your selection rules are at play as to why we only ever see a 2p-1s transition.

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    $\begingroup$ This is wrong. The 2p and 2s levels are only degenerate at the coarsest level of approximation, but they're non-degenerate once you take fine structure into account. The transition is perfectly real - see iopscience.iop.org/article/10.1088/0026-1394/22/1/003/meta for an example of an experimental observation. $\endgroup$ – Emilio Pisanty Mar 13 '19 at 21:31
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    $\begingroup$ As for "why we only ever see a 2p-1s transition" - are you sure you understand the selection rules for electric quadrupole, magnetic dipole, and higher-order selection rules (en.wikipedia.org/wiki/Selection_rule#Summary_table) to fully account for why they forbid the 2s-1s pathway in first-order perturbation theory? What about higher orders of perturbation theory? How would you account, say, for this experiment? $\endgroup$ – Emilio Pisanty Mar 13 '19 at 21:45
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    $\begingroup$ both helpful points, will read up on them, thanks. $\endgroup$ – jezzo Mar 14 '19 at 4:00

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