$R_\xi$ gauges and the EM-field

Question

$R_\xi$-gauges are said to be a generalization of the Lorenz gauge. I dont quite get why we add the term $$ \mathcal L_{GF} = - \frac{(\partial_\mu A ^\mu)^2}{2\xi}\tag{1} $$ to the Lagrangian. If i calculate the EOM for the electromagnetic field $A^\mu$ adding the "gauge-breaking"-term (1) so for the Lagrangian $$ \mathcal L = - \frac 1 4 F_{\mu\nu}F^{\mu\nu} + j_\mu A^\mu - \frac{(\partial_\mu A^\mu)^2}{2\xi}\tag{2} $$ i get EOM $$ \partial_{\nu}(\partial^\mu A^\nu + \left(\frac 1 \xi - 1\right)\partial^\nu A^\mu) = j^\mu\tag{3} $$ which for $\xi = 1$ just gives the Lorenz gauge. Why is the $\xi = 1$ gauge now called Feynman-'t Hooft gauge? And what exactly is the use of this extra term since there is no "real" use for it?

Answer 1

In the Lagrangian of the path integral, a popular class of gauge-fixing terms is of the form$^1$ $${\cal L}_{GF}~=~-\frac{\chi^2}{2\xi}.$$ The word "gauge" is here confusingly used in 2 ways:

1. The gauge-fixing function $\chi$, e.g. Lorenz gauge $\chi=\partial_{\mu}A^{\mu}$, Coulomb gauge $\chi=\vec{\nabla}\cdot \vec{A}$, etc.

2. The gauge parameter $\xi>0$, e.g. Feynman – 't Hooft gauge $\xi=1$, Landau gauge $\xi=0^+$, etc.

Both gauge-fixing choices $(\chi,\xi)$ should be made. So e.g. OP is properly speaking considering Lorenz gauge $\chi=\partial_{\mu}A^{\mu}$ in Feynman – 't Hooft gauge $\xi=1$.

Different gauges are useful in answering different questions. It should be stressed that the gauge-fixing condition is in generical only imposed in a quantum average sense in the path integral, cf. e.g. my Phys.SE answers here & here.

Why the path integral needs gauge-fixing is e.g. discussed in this, this, this & this Phys.SE posts.

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$^1$ Some gauges may require non-trivial Faddeev-Popov (FP) terms. For a BRST formulation, see e.g. this Phys.SE post. More general types of gauge-fixing is possible in the Batalin-Vilkovisky (BV) formulation.