Why does i-epsilon prescription not cure spin 1 propagator

1. Nov 25, 2014

geoduck

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 $i\epsilon$ 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 $i\epsilon$ ensures k^2-m^2+$i\epsilon$ can never be zero, so the matrix is invertible. Doesn't the $i\epsilon$ prescription also prevent $g^{\mu \nu}k^2-k^\mu k^\nu +i\epsilon=0$ for spin 1?

2. Nov 25, 2014

Einj

The Fadeev-Popov method is introduced for another reason. It is not used for general spin-1 field but for gauge fields (not all spin-1 fields are gauge fields). The main idea is that, since you have gauge invariance, the action insider the functional integral is constant over an infinite set of configurations for the gauge fields. On this set the action cannot provide the usual convergence and hence the functional integral is infinite.
Using the Fadeev-Popov method you fix the gauge and manage to factorize the contribution from this infinte set, that then cancels with the normalization of the functional integral.