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  A question about propagator of Maxwell field in different gauge

+ 2 like - 0 dislike

The propagator of Maxwell theory is different, depending on the gauge fixing procedure used. Then why will the S-matrix elements be the same for the same process in different gauges?

This post imported from StackExchange Physics at 2014-04-13 14:32 (UCT), posted by SE-user user34669
asked Apr 10, 2014 in Theoretical Physics by user34669 (205 points) [ no revision ]
Have you computed a scattering amplitude using the general propagator? You'll find that the $\xi$ parameter cancels in the final expression.

This post imported from StackExchange Physics at 2014-04-13 14:32 (UCT), posted by SE-user JamalS

1 Answer

+ 2 like - 0 dislike

I'll answer this question by example. Some standard gauge choices are the $R_\xi$ gauge and axial gauge with propagators $$ \Delta^\xi_{\mu\nu} (k) = - \frac{i }{ p^2 - i \varepsilon} \left[ g_{\mu\nu} - \left( 1 - \xi \right) \frac{ k_\mu k_\nu }{ k^2 } \right] $$ $$ \Delta^{\text{axial}}_{\mu\nu} (k) = - \frac{i }{ p^2 - i \varepsilon} \left[ g_{\mu\nu} - \frac{k_\mu k_\nu + (k \cdot n )( k_\mu n_\nu + k_\nu n_\mu ) - k^2 n_\mu n_\nu }{ k^2 + k \cdot n } \right] $$ The crucial thing is that the $\mu,\nu$ indices in the propagator are always contracted with "the rest of the amplitude" which satisfies $k_\mu {\cal M}^{\mu\nu} = 0$ (in both axial and $R_\xi$ gauges) and $n_\mu {\cal M}^{\mu\nu} = 0$ (in axial gauge). Thus, the second term always vanishes upon contraction and the only non-zero contribution is the first term, which are the same in both the propagators.

This post imported from StackExchange Physics at 2014-04-13 14:32 (UCT), posted by SE-user Prahar
answered Apr 10, 2014 by prahar21 (545 points) [ no revision ]

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