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!

  • $\begingroup$ Please can you give a little more context? What do you mean by "collapse operators", specifically? $\endgroup$ Commented Sep 5, 2018 at 13:02
  • $\begingroup$ @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. $\endgroup$
    – Jason Tao
    Commented Sep 6, 2018 at 16:52

1 Answer 1


If you are modeling something like spontaneous emission from an excited state into multiple ground states, you should keep in mind that the photon emitted from the atom typically carries information about which state the atom decayed to. You described a situation in which the polarization alone of the emitted photon does not distinguish between the atom having decayed into multiple different states. However, in this case, the frequency of the photon generally will carry this information.

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.

Always look for what information the photon has about the atom, either in its frequency, polarization, timing, or anything else. When the photon is lost to the environment this information about your atom will be revealed. So yes, you should model the decay from an excited state to each distinguishable ground state with a separate collapse operator.

  • $\begingroup$ Thank you. What I'm not clear is how much the measurement of the emitted photon gives. If the states are Zeeman shifted, which is often the case, we should also distinguish photons by their frequency. Then we'll have a bunch of collapse operators decaying to different Zeeman sublevels. $\endgroup$
    – Jason Tao
    Commented Sep 12, 2018 at 3:19
  • $\begingroup$ And is this information limited by our imagined detector quality, e.g. the frequency resolution.. $\endgroup$
    – Jason Tao
    Commented Sep 12, 2018 at 3:21
  • $\begingroup$ That's a great question. You are correct that for a small frequency separation, the emitted photon would not fully distinguish between the two states. This is characterized by the frequency separation relative to the bandwidth of the photon; in the case of spontaneous emission, the bandwidth of the photon is the linewidth of the transition (6 MHz for the D2 line of Rubidium). $\endgroup$ Commented Sep 12, 2018 at 3:53
  • $\begingroup$ Hi @HarryLevine, do you know of a textbook or lecture notes where this is addressed? In particular I assume as a magnetic field is turned on there's some transition of distinguishability between Zeeman levels, which would be interesting to see written down. $\endgroup$
    – Will C
    Commented Jun 5, 2023 at 14:12
  • $\begingroup$ @WillC Here's a reference for the idea that the polarization of the emitted photon becomes entangled with the final state of the atom: nature.com/articles/nature09256. I think this would be the primary effect for decay to different Zeeman states which would be polarization-selective transitions. But there might be examples where one could tune the distinguishability with a magnetic field, as you suggest… interesting to think about, I'm not sure! $\endgroup$ Commented Aug 15, 2023 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.