Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A What is the "real" Feynman propagator?

  1. Sep 11, 2018 #1
    The logic of the Feynman Propagator is confusing to me. Written in integral form as it is below

    $$\Delta _ { F } ( x - y ) = \int \frac { d ^ { 4 } p } { ( 2 \pi ) ^ { 4 } } \frac { i } { p ^ { 2 } - m ^ { 2 } } e ^ { - i p \cdot ( x - y ) },$$

    there are poles on the real axis. I have seen several different prescriptions for avoiding the poles using contour integration such as rotating the integral into the complex plane, traveling around the poles in tiny semicircular paths and adding an infinitesimal complex term in the denominator.

    $$\Delta _ { F } ( x - y ) = \int \frac { d ^ { 4 } p } { ( 2 \pi ) ^ { 4 } } \frac { i e ^ { - i p \cdot ( x - y ) } } { p ^ { 2 } - m ^ { 2 } + i \epsilon }$$


    How can we say that these modified integrals are equal to the Feynman Propagator? Didn't we fundamentally change it to avoid the poles? We got rid of an infinity after all.
     
  2. jcsd
  3. Sep 12, 2018 #2

    strangerep

    User Avatar
    Science Advisor

    It's misleading to say that both your expressions are "the Feynman propagator". Your first expression is the Klein-Gordon Green's function. One must impose causality conditions on top of this to get a physically-useful propagator. That's why you'll see "advanced propagator", "retarded propagator", and "Feynman propagator" in textbooks, distinguished by different ##\Theta## functions that express the causality. Which textbook(s) are you using?
     
  4. Sep 13, 2018 #3
    Thanks! Watching Tobias Osborne's QFT lectures on youtube and following along with David Tong's lecture notes. Supplementing with Sredniki and Zee.
     
  5. Sep 16, 2018 #4

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The first expression in #1 is undefined and should never be written down in any serious physics textbook (well, unfortunately there are some such books). It is indeed an important issue to understand, which Green's function has to be used in which physical situation.

    For the evaluation of Feynman diagrams in vacuum QFT (in many-body QFT it becomes more complicated!) you need the Feynman propagator, i.e., the expectation value of time-ordered products of two field operators, and the time ordering uniquely determines how to circumvent the poles, i.e., the positive frequency modes with the retarded and the negative-frequency modes with the advanced propagator. The latter is a version of the Feynman-Stueckelberg trick to redefine the negative-frequency modes as positive-energy modes of antiparticles, moving "forward in time", i.e., to make the demand of (micro-)causality consistent with the demand to have a stable ground state in local microcausal relativistic QFTs and in consequence make the transition probabilities given by the square of S-matrix elements satisfy the linked-cluster principle.

    For a very good explanation of all this, see the first chapters of Weinberg, Quantum Theory of Fields Vol 1, or Duncan, The conceptional framework of QFT.
     
  6. Sep 19, 2018 #5
    There's also a very nice very direct way to see this in Landau volume 4 sec 75 on the Electron Propagator if (as I find :frown:) Weinberg/Duncan are hard to make sense of.
     
  7. Sep 20, 2018 #6
    So am I correct in the following? There are several KG Green's functions (my first eqn, Advanced, Retarded, Feynman) and the one which we choose is my 2nd equation because it is equivalent to the amplitude of a particle traveling into the future and an antiparticle with negative energy traveling into the past? If so, my only remaining question is why is the transition amplitude equal to the Green's function of the EOM. These seem like different things to me.
     
  8. Sep 21, 2018 #7
    Peskin and Schroeder chapter 2 on KG give a good explanation of why you need to take the Feynman propagator because of Causality, and Landau says the same thing a few sections earlier in the 'Chronological product' section quicker. As to why it's a Green function, if you apply the KG operator to it, because it has the wave function in it, it's zero at all points, except when they are at equal times because the time-ordering operator messes things up, which is already analogous to a Green functions behaviour - see section 75 on the electron propagator referenced above to see how you get to the final result.
     
  9. Sep 21, 2018 #8

    king vitamin

    User Avatar
    Science Advisor
    Gold Member

    In general, the correlation functions of a quantum field theory satisfy a (generically infinite) set of differential equations called the Schwinger-Dyson equations. These are derived in Chapter 22 of Srednicki, since you mentioned you're using that textbook. For free field theory, the Schwinger-Dyson equation is simply the statement that this transition amplitude is the Green's function for the Klein-Gordon operator (which is also shown in Srednicki).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted