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We can build propagators (in this case, for massive vector fields) with at least two approaches:

(1) from canonical quantization (or exact solutions, if you like)

$$\left[[(i\partial)^2-m^2]\delta_{\space\space\nu}^\mu + (1-\frac{1}{\alpha})(i\partial^\mu)(i\partial_\nu)\right]W_{\nu}(x)= 0$$

$$D_{\mu\nu}(x-x') = \sum_{\vec{p},\lambda}W_{\mu,\lambda}^{(+)}(x) W_{\nu,\lambda}^{(+)*}(x')\theta(t-t') + W_{\mu,\lambda}^{(-)}(x) W_{\nu,\lambda}^{(-)*}(x')\theta(t'-t)$$ (2) or from path integral's propagator equation: $$\left[[(i\partial)^2-m^2]\delta_{\space\space\nu}^\mu + (1-\frac{1}{\alpha})(i\partial^\mu)(i\partial_\nu)\right]D_{\space\space\rho}^\nu(x-x')=\delta_{\space\space\rho}^\nu \delta(x-x')$$

When $\alpha\neq\infty$ there is 4th polarization's contribution to (2), i.e. terms involving $\alpha$. It is possible to solve equation in (1) for 4th polarization but it wouldn't have positive normalization (as long as this is not a physical solution that's OK). But I want to use this 4th polarization to build the propagator which will be identical to the one obtained by method (2).

Any ideas on how to incorporate (normalize, add/substract) the solution for 4th polarization into sum over the momenta and polarizations in (1)?

P.S. Is it somehow related to ghosts?

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  • $\begingroup$ Whether you do canonical or path integral formalism, the same issues arise. If $\alpha=\infty$ then the operator is singular, and as soon as $\alpha \neq \infty$ then the 4th (and 3rd) polarization propagates. Thats just the way it is, there is no magic cure. t'Hooft (and before him Feynman) got the Nobel prise for proving you get the same results for any $\alpha$. $\endgroup$
    – Kphysics
    Commented Nov 21, 2019 at 10:29
  • $\begingroup$ The question was whether it was possible to add some term (constructed with 4th polarization) to the propagator built with canonical quantiaztion so it would coincide with the one found via solving the propagator equation. $\endgroup$ Commented Nov 21, 2019 at 15:25
  • $\begingroup$ For $\alpha=1$, which is Feynman gauge, you must get $$D = \frac{g_{\mu\nu}{p^2+i\epsilon}$$, and it is the same in both formalisms, if you are getting something different then you have a misunderstanding somewhere. $\endgroup$
    – Kphysics
    Commented Nov 21, 2019 at 19:04

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