- #1

TriTertButoxy

- 194

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Hi. I am stuck.

By inverting the spin-1 differential operator I was able to derive (quite easily) the propagator for the spin-1 field (in a spontaneously broken gauge theory) in the [itex]R_\xi[/itex] gauges for the arbitrary gauge parameter [itex]\xi[/itex]. The result is

[tex]

\tilde{D}^{\mu\nu}(p)=\frac{-i}{p^2-m^2+i\epsilon}\left(g^{\mu\nu}-(1-\xi)\frac{p^\mu p^\nu}{p^2-\xi m^2+i\epsilon}\right).

[/tex]

But now, when I try to calculate the two-point correlator using the mode expansion for the spin-1 field, I can't quite get the same answer.

[tex]

\langle 0|T(\hat{A}_\mu(x)\hat{A}_\nu(y))|0\rangle=\int \frac{d^4p}{(2\pi)^4}\frac{ie^{-i p.(x-y)}}{p^2-m^2+i\epsilon}\left(\epsilon_\mu^{[0]}(\mathbf{p})\epsilon_\nu^{*[0]}(\mathbf{p})-\sum_{\lambda=1,2,3}\epsilon_\mu^{[\lambda]}(\mathbf{p})\epsilon_\nu^{*[\lambda]}(\mathbf{p})\right)

[/tex]

I need to do a polarization sum, but can't quite figure out how to get the gauge-dependence in there. What are the polarization vectors? and how do I derive them?

By inverting the spin-1 differential operator I was able to derive (quite easily) the propagator for the spin-1 field (in a spontaneously broken gauge theory) in the [itex]R_\xi[/itex] gauges for the arbitrary gauge parameter [itex]\xi[/itex]. The result is

[tex]

\tilde{D}^{\mu\nu}(p)=\frac{-i}{p^2-m^2+i\epsilon}\left(g^{\mu\nu}-(1-\xi)\frac{p^\mu p^\nu}{p^2-\xi m^2+i\epsilon}\right).

[/tex]

But now, when I try to calculate the two-point correlator using the mode expansion for the spin-1 field, I can't quite get the same answer.

[tex]

\langle 0|T(\hat{A}_\mu(x)\hat{A}_\nu(y))|0\rangle=\int \frac{d^4p}{(2\pi)^4}\frac{ie^{-i p.(x-y)}}{p^2-m^2+i\epsilon}\left(\epsilon_\mu^{[0]}(\mathbf{p})\epsilon_\nu^{*[0]}(\mathbf{p})-\sum_{\lambda=1,2,3}\epsilon_\mu^{[\lambda]}(\mathbf{p})\epsilon_\nu^{*[\lambda]}(\mathbf{p})\right)

[/tex]

I need to do a polarization sum, but can't quite figure out how to get the gauge-dependence in there. What are the polarization vectors? and how do I derive them?

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