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Massive vector (Proca) propagator

  1. Sep 19, 2008 #1

    CompuChip

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    Hi all,
    I'm stuck with this following problem:

    1. The problem statement, all variables and given/known data
    Consider the Proca action,
    [tex] S[A_\mu] = \int \, \mathrm d^4x \left[ - \frac14 F_{\mu\nu} F^{\mu\nu} + \frac12 m^2 A_\mu A^\mu \right] [/tex]
    where [itex]F_{\mu\nu} = 2 \partial_{[\mu} A_{\nu]}[/itex] is the anti-symmetric electromagnetic
    field tensor.

    Derive the propagator for the vector field [itex]A_\mu[/itex].

    2. Relevant equations

    I did a Fourier transform to get
    [tex] \left[ (- k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu \right] \tilde D_{\nu\lambda}(k) = \delta^\mu_\lambda. [/tex] (*)

    Zee's book on QFT gives the result on page 13, as if it were trivial, but I can't do the calculation (satisfactorily).

    3. The attempt at a solution

    I tried to follow the hint in the question: "the calculation involves deriving an identity for [itex]k^\nu \tilde D_{\nu\mu}[/itex]".
    I contracted (*) with [itex]k_\mu[/itex] which got me
    [tex]k^\nu \tilde D_{\nu\lambda} = k_\lambda[/tex]
    or (contracting with [itex]k^\lambda[/itex])
    [tex]k^\lambda k^\nu D_{\nu\lambda} = k^2[/tex]
    but I still didn't really see how to solve for [itex]\tilde D_{\nu\lambda}[/itex].
     
  2. jcsd
  3. Sep 19, 2008 #2

    George Jones

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    I get

    [tex]k^\nu \tilde D_{\nu\lambda} = \frac{k_\lambda}{m^2},[/tex]

    and then I think everything works out okay.
     
  4. Sep 19, 2008 #3

    CompuChip

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    Thanks, I'll check that calculation.
    My problem was how to extract the propagator from that contraction, though.

    Anyway, let me get some sleep now, as it's 1:30 :smile:
     
  5. Sep 19, 2008 #4

    George Jones

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    Substitute the identity and then contract with the metric.
     
  6. Sep 20, 2008 #5

    CompuChip

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    I checked my earlier calculation and the 1/m^2 missing was just a typo.
    Also, I see what you mean now and it turns out to be quite easy indeed.

    Thank you very much George!
     
  7. Feb 14, 2011 #6
    I am having the same problem.

    Could you elaborate on what you mean by substituting the identity?

    Edit: scratch that. I figured it out.
     
    Last edited: Feb 14, 2011
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