What is the significance of the simplification of the equal-time commutation relations when we take the Feynman propagator for gauge parameter $\lambda = 1$?

Question

Classical electromagnetism (with no sources) follows from the actions $$S = \int d^4x\left(-{1\over4}F_{\mu\nu}F^{\mu\nu}\right),\text{ where }F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$ The Lagrangian for $A_\mu$, including a gauge fixing term, is $$\mathcal{L} = -{1\over4}F^2 - {\lambda\over2}(\partial_\mu A^\mu)^2.$$ Computation of the equal time commutators $[\dot{A}_\mu(\textbf{x}), A_\nu(\textbf{y})]$ and $[\dot{A}_\mu(\textbf{x}), \dot{A}_\nu(\textbf{y})]$ for general $\lambda$ gets $$[\dot{A}_\mu (\textbf{x}), {A}_\nu (\textbf{y})] = i\left(g^{\mu\nu} - {{\lambda - 1}\over{\lambda}}g^{\mu0}g^{\nu0}\right)\delta^3(\textbf{x} - \textbf{y}),$$ $$[\dot{A}_\mu (\textbf{x}), \dot{A}_\nu (\textbf{y})] = i{{\lambda - 1}\over{\lambda}}(g^{\mu 0}g^{\nu k} + g^{\nu 0}g^{\mu k})\partial_k \delta^3(\textbf{x} - \textbf{y}).$$ Taking $\lambda = 1$, we have $$[\dot{A}_\mu (\textbf{x}), {A}_\nu (\textbf{y})] = ig^{\mu v}\delta^3(\textbf{x} - \textbf{y}), \text{ }[\dot{A}_\mu (\textbf{x}), \dot{A}_\nu (\textbf{y})] = 0.$$ My question is, what is the significance of the simplifying that happens here with the equal-time commutators when we take $\lambda = 1$?

Comments

What do you mean? The whole point gauge symmetry is that your choice of gauge doesn't affect the physics at all, so you are free to exploit gauge freedom to simplify calculations.

Answer 1

Comparing this to electromagnetism with source (in slightly different notation) with the Lagrangian

$$\mathcal L = -\frac{1}{4}F^2_{\mu\nu} - \frac{1}{2\xi}(\partial_{\mu}A_{\mu})^2 -J_{\mu}A_{\mu}$$
and the corresponding time-ordered Feynman propagator for a photon

$$i \Pi^{\mu\nu}((p) = \frac{-i}{p^2 + i\varepsilon}\left [ g^{\mu\nu} - (1-\xi)\frac{p^{\mu}p^{\nu}}{p^2}\right]$$

it can be seen that the parameter $\lambda$ and $\xi$ are related as $\xi = \frac{1}{\lambda}$.

So the cas $\lambda = 1$ corresponds to the Feynman-'t Hooft gauge with $\xi = 1$. The only advantage of this is that it makes the propagator consisting of only one term, but there is as said in the comments no physical meaning of this.