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Nov 21, 2019 at 19:04 comment added Kphysics 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.
Nov 21, 2019 at 15:25 comment added stanislav-iablokov 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.
Nov 21, 2019 at 10:29 comment added Kphysics 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 $\alpha$.
Nov 21, 2019 at 9:13 history edited Qmechanic
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Nov 21, 2019 at 9:09 history asked stanislav-iablokov CC BY-SA 4.0