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

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 $$(\xi=0)$$.

Independence of gauge-fixing is easiest to see 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.