# Is the $2p\rightarrow2s$ transition possible?

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.)
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.
I will narrow the scope of the question as follows: is the transition with spontaneous emission of a photon between the Lamb split levels (i.e. $$2S_{1/2}→2P_{1/2}$$) forbidden by the dipole selection rules? No, it is allowed. However the spontaneous emission rate, which is $$\sim f^3$$, where $$f$$ is the transition frequency ($$\sim 1\rm GHz$$ in this case), is much lower than would be for a typical visible spectrum transition ($$f\sim 10^{15}\rm Hz$$, emission rates$$\sim 10^8\rm s^{-1}$$). Since on the other hand $$2S_{1/2}→1S_{1/2}$$ is forbidden by the parity selection rule, the decay rate of $$2S_{1/2}$$ ends up being very low: $$8 \rm s^{-1}$$, leading to this state being considered metastable.