# Equal-time commutation relations, Feynman propagator for gauge parameter $\lambda = 1$, physical meaning

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 (physical, if possible) significance of the simplifying that happens here with the equal-time commutators when we take $\lambda = 1$?

• There cannot be a physical significance since a gauge choice has no physical meaning by definition. – ACuriousMind Oct 15 '15 at 21:21
• As pointed by @ACuriousMind there is no physical meaning attached to $\lambda$, you can think of it as a Lagrange multiplier, e.g. $\lambda \rightarrow \infty$ corresponds to the Lorentz gauge, $\lambda =1$ is Feynman gauge. In the end whatever the choice of $\lambda$ be they all correspond to the same physical theory (at most up to a unitary transformation), so $\lambda$ is not contained in any observable. – Jon Snow Oct 16 '15 at 3:16