- #1
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Hi all,
I'm stuck with this following problem:
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].
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).
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].
I'm stuck with this following problem:
Homework Statement
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].
Homework 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).
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].