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Virtual particle propagators in QFT

  1. Jul 15, 2011 #1
    I am reading a nice book (Quarks and Leptons, by Halzen and Martin) about particle physics. It states that the general form of the propagator of a virtual particle is:
    \dfrac{i\sum_{\text{spins}}}{p^2 - m^2}

    I see that this is the case for the Dirac propagator:

    \dfrac{i(\displaystyle{\not}{p} + m)}{p^2 - m^2} = \dfrac{i\sum_{s}u_s(p)\bar{u}_s(p)}{p^2 - m^2}

    but how can I prove that always holds?
  2. jcsd
  3. Jul 15, 2011 #2


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    Matthaeus , I agree, Halzen and Martin is a very well written book, and highly useful. Although it does tend to be a bit informal, and you are not going to catch them "proving" anything. So in the same spirit, here is not a proof but an argument.

    Let's say you have an operator L, which is generally a differential operator, and in particular will be the Klein-Gordon operator. You want to find the field |φ> at one point produced by a source |S> at another. The equation is L|φ> = |S>. The solution can be written formally as |φ> = L-1|S>. Expand L in terms of its eigenstates, namely L = ∑|n>Ln<n| where |n> are the eigenstates and Ln the eigenvalues. Then L-1 = ∑|n>Ln-1<n|. Thus |φ> = G|S> where G is the Green's function or propagator, G = ∑|n>Ln-1<n|.

    For the Klein-Gordon operator the eigenstates are the plane wave solutions, Ln = k2 + m2, and the numerator ∑|n><n| is the sum over spins. Halzen and Martin go on to write this out in more detail for the only cases that are physically realistic: spin 0, 1/2 and 1.
  4. Jul 16, 2011 #3
    Yes, I was also thinking about something along these lines.
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