Why the $R_{\xi}$-contribution to the Lagrangian disappears when computing physical observables?

Question

In QED for example, you add the term $$\mathcal{L}_{GF}=-\frac{1}{2\xi}(A_{\mu}A^{\mu})^{2}$$ so you can compute the photon propagator. The question is basically, why you can compute physical observables where at the beginning a $\xi$-dependence appears, and the physical observable ends up with no such dependence, and it doesn't matter whether you choose 't Hooft-Feynman gauge $(\xi=1)$ or Unitary gauge $(\xi=0)$.

Answer 1

1. Independence of gauge-fixing is perhaps easiest to understand in the BRST formulation. It turns out that all of the gauge-fixing (including the $\xi$-parameter) can be tucked away inside a BRST-exact part of the action, where it cannot affect BRST-invariant (=physical) observables, cf. e.g. my Phys.SE answer here.

2. For proofs that does not use the BRST formulation, see e.g. Refs 1-2.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; section 9.4.

2. M. Srednicki, QFT, 2007; chapter 71. A prepublication draft PDF file is available here.