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Massive vector (Proca) propagator 
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#1
Sep1908, 09:37 AM

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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 antisymmetric 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
Sep1908, 01:14 PM

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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. 


#3
Sep1908, 06:23 PM

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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 


#4
Sep1908, 06:39 PM

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Massive vector (Proca) propagator



#5
Sep2008, 03:41 AM

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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! 


#6
Feb1411, 08:59 AM

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I am having the same problem.
Could you elaborate on what you mean by substituting the identity? Edit: scratch that. I figured it out. 


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