Massive vector (Proca) propagator


by CompuChip
Tags: massive vector boson, proca action, proca lagrangian, propagator
CompuChip
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#1
Sep19-08, 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 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].
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George Jones
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#2
Sep19-08, 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.
CompuChip
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#3
Sep19-08, 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

George Jones
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#4
Sep19-08, 06:39 PM
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Massive vector (Proca) propagator


Quote Quote by CompuChip View Post
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
Substitute the identity and then contract with the metric.
CompuChip
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#5
Sep20-08, 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!
Helgi
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#6
Feb14-11, 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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