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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?

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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.

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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.

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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 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 at 21:45
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    $\begingroup$ both helpful points, will read up on them, thanks. $\endgroup$ – jezzo Mar 14 at 4:00

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