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Imaginary shifts in QFT

  1. Sep 7, 2008 #1

    haushofer

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    Hi, I'm studying QFT (quantum field theory) again, with the help of Peskin&Schroeder. Altough the book is much better to read for a second time, there are some things that keep me wondering.

    It's about shifting physical quantities by a small imaginary constant in order to be sure that expressions converge. For instance, in the Feynman propagator on page 31:

    [itex]
    D_{F}(x-y) = int \frac{d^{4}p}{(2\pi)^{4}} \frac{i}{p^{2}-m^{2} + i \epsilon} e^{-ip\cdot(x-y)}
    [/itex]

    What is the exact reason that we are allowed to shift the energy by a small imaginary constant? Isn't contradicting this the idea that physical measurable quantities should always be real? I understand that we hit a pole if we don't, but to me this whole business looks ill-defined.

    A second time this comes by is that the time T in the integral of the vertices is taken to be

    [tex]
    T \rightarrow \infty (1-i\epsilon)
    [/tex]

    for instance on page 95. P&S call this an integration along a contour which is slightly rotated away from the real [itex]p^{0}[/itex] just like in calculating the Feynman propagator; again, we are talking imaginary energies here!

    Can anyone give a satisfactory answer to this? Why are we allowed to do so? Maybe this has come along some times earlier, but it's really bothering me; I find it hard to take this sort of juggling seriously without seeing a serieus justification for it other than "or else it will diverge!".
     
  2. jcsd
  3. Sep 7, 2008 #2

    Hans de Vries

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    There's a solid mathematical background to this which is based
    on the causality aspects of the Fourier transform and Hilbert pairs.

    The epsilon prescription is mathematically identical to this method.
    It's a way of adding an imaginary Dirac function to the propagator
    centered on the on-shell pole. The imaginary value means simply
    that there is a 90 degrees phase shift for on-shell frequencies
    from the source from which they propagate. The sign of the phase
    shift depends on the sign of the epsilon.

    The theory can be found in the latter part of the first chapter
    of my book, from section 1.11 an onward.
    http://physics-quest.org/Book_Chapter_EM_basic.pdf

    This handles the simpler photon propagator (m=0). There is
    also a part online about the Klein Gordon propagators.
    http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf
    from section 9.8 and further.

    The latest state of my book (not online yet) is that the Klein Gordon
    propagators are handled in a separate chapter with 33 pages and
    10 illustrations. I will post it when it's ready.


    Regards, Hans
     
  4. Sep 7, 2008 #3

    haushofer

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    Ok, thank you very much for the links, I will read them!

    I had the idea that the idea was like

    * we have an expression for the propagator of a virtual particle
    * in it we integrate over the energy of the particle
    * to do this we use contour integration and the residue theorem
    * but if the virtual particle is on-shell, the fraction blows up
    * well, let's add an imaginary term to it and all will be fine

    I'm suddenly wondering: does it have to do with the fact that the energy of the virtual particle is already not uniquely defined, so we may add an imaginary term to the energy?

    Anyway, I will read your links and come back again! If anyone else has useful comments, I'll be happy to read them.
     
  5. Sep 7, 2008 #4

    Hans de Vries

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    The Feynman propagator is a "propagator from source". That is, you have a source
    defined over all space and time, you apply the propagator on the source and the
    result you get is the electron's field over space and time.

    In the on-shell case the source is an infinite plane wave with the right frequencies
    corresponding to a free electron, the propagator becomes infinite, but why?

    The propagation in position space is not infinite but it keeps growing linear in time,
    so after infinite time the propagation becomes infinite as well, and thus, in momentum
    space, you end up with an infinite propagator.

    Why is it keeping growing linear in time? This can be understood by considering an
    equivalent EM-case. What is the potential near an infinite plane with a constant
    charge density? Well, it keeps growing linear in time as well. Contributions from the
    plane farther and farther away keep adding to V. The circle grows with r while
    the contributions of the charges at r follow a V=1/r law.

    For a sinusoidal source the frequencies have to be right for this to occur. It's now
    the epsilon which defines the phase relation between the source and the resulting
    field at the pole. The phase is either plus or minus 90 degrees so the epsilon
    prescription becomes either [itex]i\epsilon[/itex] or [itex]-i\epsilon[/itex].


    Regards, Hans.
     
    Last edited: Sep 7, 2008
  6. Sep 7, 2008 #5

    Avodyne

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    The book by Srednicki explains this pretty well, IMO. (A draft copy is available free online, google to find it.) Basically, the slight rotation of the time contour is just a mathematical trick that picks out the vacuum as the initial and final state.
     
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