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I Photon absorption and speed of wave function propagation...

  1. Oct 13, 2016 #1
    Hi all,

    Whew, last question for a while: I think I already know the answer, but want to confirm (e..g, I think this thread basically answers the question, https://www.physicsforums.com/threads/propagation-of-wavefunction.152053/)

    As an example, let's say I have an electron (in free space or bound to an atom) that absorbs a photon and transitions to a higher energy state. I understand that in QFT, this transition is not instantaneous (although very fast) - in terms of the electron probability distribution (square of wave function) for, e.g., the bound electron case, there should be a decrease in the probability of being found in one orbital over time, with an increase in the probability of being found in a higher orbital over time (again, with the time scale of this transition being very short).

    Now, the question: the electron wave function will change as a result of the photon absorption - however, the propagation of the change in the wave function must not exceed the speed of light to agree with special relativity? I realize that that wave function lives in Hilbert space, so I guess what I'm asking is: the speed at which 'information' about the absorption of the photon propagates through the relevant field in spacetime should be c or less - correct? Otherwise, e.g., nearby particles (other field excitations) would feel in influence of the change faster light.

  2. jcsd
  3. Oct 18, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
  4. Oct 18, 2016 #3
    Yes, the information propagates at less than or equal to c. But there's a caveat because of the wave function, and this may be the reason no one answered this simple question. If the electron were entangled with another one, far away, some people say that influence propagates instantaneously (non-locally). It gets into one of those endless QM interpretation debates. If you don't care about entanglement there's no problem. But if you're thinking of entanglement, my advice would be: don't.
  5. Oct 18, 2016 #4


    Staff: Mentor

    Where do you get this understanding from? It is not really correct. A correct statement would be that QFT does not give a meaningful "speed" to such a transition.

    If you are thinking about QFT (which you have to if you're thinking about relativistic QM), then there isn't a "wave function" in the sense of ordinary QM. An ordinary QM wave function is a function defined on the configuration space of the system; but in QFT, there is no unique "configuration space" because there is no preferred frame. You have to think of things in terms of spacetime. (Actually, QFT is more often done in momentum space, which is easier to work with.) In spacetime terms, QFT is consistent with relativistic causality, which means information doesn't propagate faster than ##c##. But "information" is not the same as "wave function" (or the things that play similar roles to wave functions in QFT); those things don't have a meaningful "speed of propagation".

    Even in non-relativistic QM, where you can talk about wave functions, probability distributions in space, orbitals, etc., "speed of propagation" still isn't a meaningful concept for those things. "Speed" (more precisely velocity) can be an observable, but an observable is not the same as a wave function (or those other things).
  6. Oct 18, 2016 #5


    Staff: Mentor

    In QFT, yes. But as I noted in my other post just now, "information" in QFT is not the same thing as the QFT counterpart of "wave function". Wave functions, or their QFT counterparts, don't have a meaningful speed of propagation, and you can't really assign a meaningful "time" to transitions of the sort the OP is describing.
  7. Oct 19, 2016 #6
    That's the point I was trying to make re. entanglement. But perhaps that doesn't cover all non-negligible cases? Does "speed of wave function propagation" ever become an issue in QFT, apart from entanglement?
  8. Oct 19, 2016 #7


    Staff: Mentor

    I would say no, whether entanglement is involved or not. In QFT, the property that ensures that relativistic causality holds is that spacelike separated measurements commute, i.e., their results are independent of the order in which they are performed. This holds even though there are nonzero amplitudes for correlation between spacelike separated measurements (which is the property of entanglement that often makes people think that there is some sort of FTL "propagation" going on). So in QFT the problem of "propagation speed of the wave function" doesn't really exist; the thing to look at is what measurement operators commute.
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