# Collapse operators in multi-level atomic systems?

I'm trying to understand the case of multiple ground states and excited states. Due to selection rules, usually we say there should be 3 collapse operators, namely C_+1, C_-1, C_pi, corresponding to sigma+, sigma- and pi spontaneous photon emissions. But if there are multiple ground states in one single kind of transition, say C_-1, the state after the collapse would be a superposition of all possible ground states. I'm not very clear on how to explain the physics idea? I saw in some examples (laser cooling simulation of 87Rb) people used 6 collapse operators. There were two sets of collapse operators for collapsing into F=1 and F=2 respectively. I find this more confusing and am not sure if it is right. Thank you for your answer!

• Please can you give a little more context? What do you mean by "collapse operators", specifically? – Mark Mitchison Sep 5 '18 at 13:02
• @MarkMitchison It happens when you try to simulate the atomic internal states evolution using master equation or quantum Monte Carlo method. In the latter case they are also called quantum jump operators. – Jason Tao Sep 6 '18 at 16:52

For example, since you brought up Rubidium: consider a Rubidium atom in the excited state $|5P_{3/2}, F=2, mF=1\rangle$. It can decay via a $\sigma^+$ transition to $|5S_{1/2}, F=2, mF=0\rangle$, or to $|5S_{1/2}, F=1, mF=0\rangle$ (the only difference being the $F$ number). However, the two decay channels are separated in frequency by $6.8$ GHz, so the emitted photon will indicate to the environment which state your atom ended up in and destroy the superposition.