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

I Propagator operator in Heinsenberg picture

  1. May 1, 2017 #1
    Hello! I read that in Heisenberg picture the propagator from x to y is given by ##<0|\phi(x)\phi(y)|0>##, where ##\phi## is the Klein-Gordon field. I am not sure I understand why. I tried to prove it like this:
    ##|x>=\phi(x,0)|0>## and after applying the time evolution operator we have ##U(t)|x>=e^{-iHt}\phi(x,0)|0>##. And the propagator should show the overlapping between ##|x>## and ##|y>## at time t. This would be
    ##<y|U(t)|x>=<0|\phi(y,0)e^{-iHt}\phi(x,0)|0> = <0|e^{-iHt}\phi(y,0)\phi(x,0)|0>## which is not what I was supposed to obtain. I am also not sure about the time dependence of ##\phi## in the propagator. Can someone explain this to me?
     
  2. jcsd
  3. May 1, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Where have you read this? What you've written down is the Wightman function and one Green function in the Schwinger-Keldysh real-time-contour formalism. What's the purpose of this function in your context?

    Usually in vacuum QFT you work with the time-ordered propagator (which in the vacuum is the same as the Feynman propagator), i.e.,
    $$\mathrm{i} \Delta_F(x)=\langle 0|T_c \hat{\phi}(x) \hat{\phi}(0) \rangle,$$
    where I assume the uncharged KG field (where ##\hat{\phi}=\hat{\phi}^{\dagger}##).
     
    Last edited: May 1, 2017
  4. May 1, 2017 #3
    It is from Peskin book on QFT. On page 27 (at least this is what I understood, please let me know if that means something different)
     
  5. May 1, 2017 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Peskin and Schroeder are not always careful enough. However, here they give the resolution in the discussion on the following pages (see p. 29).
     
  6. May 1, 2017 #5
    Wait I am confused. So what is wrong about this. Like what is this object ##<0|\phi(x) \phi(y)|0>##, if it is not a propagator? And how do they come up with it (as it seems to be the main tool for what follows)?
     
  7. May 2, 2017 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    It is not enough to call something propagator but you have to specify which propagator it is. In the Feynman rules, leading to causal S-matrix elements it's not this propagator (a socalled Wightman function) that occurs but the time-ordered propagator. Usually you define the Wightman function by
    $$\mathrm{i} \Delta_{12}(x)=\langle 0|\hat{\phi}(x) \hat{\phi}(0)| 0\rangle.$$
    Then the time-ordered propagator (which for vacuum QFT is the Feynman propagator) is defined as
    $$\mathrm{i} \Delta_{11}(x)=\mathrm{i} \Delta_{\text{F}}(x)=\langle 0|T_c \hat{\phi}(x) \hat{\phi}(0)|0 \rangle=\mathrm{i}[\Theta(x^0) \Delta_1(x)+\Theta(-x^0) \Delta_1(-x)].$$
    As is worked out in Peskin&Schroeder, this leads to a causal description of scattering events in terms of the S-matrix. The trick is that you introduce antiparticles (in the case of the hermitean Klein-Gordon field you describe neutral scalar bosons, for which the particle and antiparticle are the same) in addition to particles to make everything causal, where causal means that local experiments that are seperated by far distances lead to uncorrelated results for transition probabilities, the socalled "Linked-Cluster Principle". This is closely connected with the microcausality condition, which is realized by the vanishing of field-operator commutators (or anti-commutators for fermion fields) for space-like separated arguments, i.e., for the scalar field
    $$[\hat{\phi}(x),\hat{\phi}(y)]=0 \quad \text{for} \quad (x-y)^2<0,$$
    where my Minkowski product is defined by the west-coast convention, i.e., ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## as in Peskin&Schroeder. That's why Peskin&Schroeder proves that the vacuum expectation value of this field commutator vanishes for space-like separated arguments.

    For a much more careful (but also rather more advanced) line of arguments concerning all these issues, see

    S. Weinberg, Quantum Theory of Fields, vol. 1, Cambridge University Press
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Propagator operator in Heinsenberg picture
  1. Propagator operator (Replies: 1)

Loading...