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Poles of the propagator

  1. Jul 7, 2012 #1
    From "Le Bellac, Quantum and statistical field theory, 10.5.2-Massive vector field":
    "The longitudinal part of the propagator [itex]k_{\mu}D^{\mu\nu}[/itex] has no pole at
    [itex]k^2=m^2[/itex], so the longitudinal part doesn't constitute a dynamical degree of freedom."

    I have two questions:
    1) Why the propagator doesn't represent a dynamical degree of freedom if it hasn't any pole?
    How do you demonstrate that physical particles correspond to the pole of the propagator?

    2) The propagator [itex]D^{\mu\nu}[/itex] is a rank-2 tensor. The longitudinal part is [itex]k_{\mu}D^{\mu\nu}[/itex] and it is a vector, so, how can it be a propagator?
  2. jcsd
  3. Jul 7, 2012 #2


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    The components of the vector field satisfy the Klein-Gordan equation

    $$(-\partial_\nu \partial^\nu + m^2) A_\mu =0.$$

    By Lorentz-invariance, the mass appearing there must be the same for all components. The momentum-space propagator is the Fourier transform of the 2-point function ##\langle A_\mu(x)A_\nu(y)\rangle##. Because of the KG equation above, all physical components (and linear combinations of them) of the propagator must either have a pole at ##m^2## or vanish.

    The longitudinal part is a linear combination of propagators, or equivalently, the propagator for a linear combination of components of ##A_\mu##. Since there is no pole, the corresponding combination of vector fields, ## k^\mu A_\mu## cannot satisfy a non-trivial KG equation. So it cannot be a physical degree of freedom.
  4. Jul 8, 2012 #3
    Also you can study the spectral representation by Kallen-Lehmann (for example in Bjorken-Drell book). It is an exact result (not a perturbative one) and it shows that any Green function have always a pole at the physical mass of the particle.
  5. Jul 11, 2012 #4
    I've read the Kallen-Lehmann representation and I've understood why the propagator has poles. However I don't fully understand fzero's answer. I don't understand this passage

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