# $\Delta \ell = 0, \ell = 0 \rightarrow \ell = 0$ transition in atomic ground state

Assume I am in the $S_{1/2}$ state of, say, Rubidium:

Let's say I am in the higher $>F\rangle$ state ($3$ for $^{85}\textrm{Rb}$, $2$ for $^{87}\textrm{Rb}$), how do I decay to the lowest ground state (green arrows)?

This process would need a $\Delta \ell = 0$, $\ell = 0 \rightarrow \ell = 0$ transition, which is forbidden by selection rule. But a photon must be involved, which takes away an integer number of angular momentum. Is there a light assisted collision in the process?

The selection rule forbids photon transitions that go $0\rightarrow 0$. The reason is that the photon must carry away angular momentum. So if the initial state had $0$ angular momentum, the final state must have exactly opposite the photon's angular momentum (which is necessarily nonzero).
Other kinds of transitions, such as collisional deexcitations, are not subject to the no $0\rightarrow 0$ rule.
• In principle, this transition can also relax radiatively via a magnetic dipole transition: en.wikipedia.org/wiki/Magnetic_dipole_transition . However, the sponetaneous decay lifetime is, I believe, extremely long. This does involve emission of a photon. Since the total angular momentum $F$ changes by one, there is no problem with conservation. – Rococo Dec 15 '17 at 1:38
• @Rococo The matrix elements for $0\rightarrow 0$ magnetic dipole decays are very small indeed; they are typically nonzero only because of relativistic corrections. In many cases, the magnetic dipole process is slower than a two-photon decay process. – Buzz Dec 15 '17 at 1:45