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On the classical action in Feynman approach

  1. Feb 28, 2014 #1
    Hi All,

    In the Feynman, 'sum over paths' approach to quantum field theory, we compute amplitudes, generating functionals etc by feeding in a "classical action".

    By calling the Lagrangian that we feed in "classical", this mean that the fields that feature in that action are regarded as classical fields? (Is there even such a thing as a classical field for a proton, etc?!)

    I'm quite perplexed by this, so any sage words at all would be much appreciated!
  2. jcsd
  3. Feb 28, 2014 #2


    User Avatar
    Science Advisor

    Yes, for bosons the Feynman path integral is over classical trajectories, and we rotate into imnaginary time to calculate so that it becomes a stochastic process. Whether this works in quantum mechanics - ie. is meaningful in terms of Hilbert spaces, operators etc - is governed by the Osterwalder-Schrader conditions. http://www.einstein-online.info/spotlights/path_integrals. There are more comments on these conditions on p17-18 of http://www.rivasseau.com/resources/book.pdf.

    For fermions, one has to use a Grassmann variables, which are not so classical.
    Last edited: Feb 28, 2014
  4. Mar 5, 2014 #3
    Atyy, thank you very much for your answer and for the reference to the OS conditions. I have one more question in case you happen to know the answer to this too.

    In the Feynman approach, we feed in a classical action in order to calculate outcomes of quantum field-theoretic processes. Is there an approach to QFT in which one has a *non*-classical action -- a Lagrangian that is itself explicitly quantum field-theoretic? (I appreciate this is a weird question, but if you have any input I'd appreciate it very much!)

    Thanks again.
  5. Mar 5, 2014 #4


    Staff: Mentor

    In QFT the actions, just like classical physics, are constrained by symmetry - eg QED is based on gauge symmetry. Because of that actions are often the same or similar.

    But often is not always, for example the actions found in QCD do not have a classical analogue, being based on Yang-Mills fields of which EM is just a simple example.

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