Is the $2p\rightarrow2s$ transition possible?

Question

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:

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

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

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.