Within the path integral framework, the reason you have to use Fadeev-Popov quantization for spin 1 is because the matrix of the coefficients of the quadratic part of the free-Lagrangian is non-invertible. But doesn't an [itex]i\epsilon [/itex] prescription take care of that? The same thing happens with the Klein-Gordan field: the matrix of coefficients of the quadratic part is not invertible because it has a zero eigenvalue: k^2-m^2=0 when the particle has a k such that it is on shell. Adding [itex]i\epsilon [/itex] ensures k^2-m^2+[itex]i\epsilon [/itex] can never be zero, so the matrix is invertible. Doesn't the [itex]i\epsilon [/itex] prescription also prevent [itex]g^{\mu \nu}k^2-k^\mu k^\nu +i\epsilon=0[/itex] for spin 1?(adsbygoogle = window.adsbygoogle || []).push({});

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# Why does i-epsilon prescription not cure spin 1 propagator

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