# What is the constraint on the Gauge Potential in the Covariant Gauges?

One of the most common gauges in QED computations are the $R_{\xi}$ gauges obtained by adding a term $$-\frac{(\partial_\mu A^{\mu})^2}{2\xi}$$ to the Lagrangian. Different choices of $\xi$ correspond to different gauges ($\xi=0$ is Landau, $\xi=1$ is Feynman etc.) The propagator for the gauge field is different depending on the choice of gauge. The choice of Landau gauge forces $\partial_\mu A^\mu=0$, but I have never seen a similiar statement for the other gauges. I would like to know what constraint on the gauge field is produced by the other covariant gauges. For instance, what is the constraint on $A_\mu$ when $\xi=1,2,3,...$ etc. Is it still $\partial_\mu A^\mu=0$ or something different (it seems like it should be different)?

• In Euclidean signature, you can regard the $R_{\xi}$ gauge fix term as a Gaussian distribution of $\partial_\mu A^\mu$, with zero mean and $\xi$ variance. Landau gauge $\xi=0$ corresponds to zero variance, i.e. $\partial_\mu A^\mu = 0$ with $100\%$ possibility. Commented Feb 18, 2020 at 15:55

I) The un-gauge-fixed QED Lagrangian density reads

$${\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \bar{\psi}(i\gamma^{\mu}D_{\mu} -m)\psi.\tag{1}$$ The gauge-fixed QED Lagrangian density in the $$R_{\xi}$$-gauge reads

$${\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 , \tag{2}$$

where the Faddeev-Popov term is

$${\cal L}_{FP}~=~ -d_{\mu}\bar{c}~d^{\mu}c, \tag{3}$$

and

$$\chi~:=~d_{\mu}A^{\mu}~\approx~0 \tag{4}$$

II) In the path integral with $$R_{\xi}$$-gauge-fixing, the Lorenz gauge-fixing condition (4) is only imposed in a quantum average sense. In general the Lorenz gauge-fixing condition may be violated by quantum fluctuations, except in the Landau gauge $$\xi=0^+$$, where such quantum fluctuations are exponentially suppressed (in the Wick-rotated Euclidean path integral).

III) If we introduce a Lautrup-Nakanishi auxiliary field $$B$$, the QED Lagrangian density in the $$R_{\xi}$$-gauge reads

$${\cal L}~=~ {\cal L}_0 +{\cal \cal L}_{FP} +\frac{\xi}{2}B^2+B\chi \quad\stackrel{\text{int. out } B}{\longrightarrow}\quad {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 ,\tag{5}$$

cf. this related Phys.SE post. The Euler-Lagrange equation for the $$B$$-field reads

$$-\xi B~\approx~\chi.\tag{6}$$

Since there are no in- and out-going external $$B$$-particles, one may argue that the $$B$$-field is classically zero, and therefore that the Lorenz condition $$\chi\approx 0$$ is classically imposed, cf. eq. (6), independently of the value of the gauge parameter $$\xi$$. Quantum mechanically for $$\xi>0$$, the eq. (4) does only hold in average, as explained above.

• Note added: The Lorenz function $\chi$ and the $B$-field are invariant under Wick rotation. To make the Gaussian integration over $B$ convergent, we should choose $B$ to be imaginary. But then the Euler-Lagrange eq. (6) equates something real to something imaginary, which is rubbish, except if they are both zero. In other words, solutions to the eq. (6) should be taken with a grain of salt. Nevertheless, the Gaussian integral representation remains valid even if the stationary point is complex. Commented Mar 1, 2014 at 20:34
• So what you are saying is that $\chi\approx 0$ no matter which gauge paramater $\xi$ is chosen, as long as it is positive. I still don't get what would be the Euler-Lagrange equation if $\xi\to0$ though..
– PPR
Commented Oct 5, 2014 at 20:40
• Note added: In the Minkowski signature, assuming that complex conjugation reverse the factors of supernumbers, we see that $c$ ($\bar{c}$, $B$) should be imaginary (real), respectively. The variables $\bar{c}$ and $B$ are Wick rotated. Commented May 30, 2015 at 20:57
• Note added: From quartic to cubic interaction: 1. Complex scalar: $\quad {\cal L}={\cal L}_2-\frac{1}{2}\lambda |\phi|^4$; Auxiliary real scalar: $\quad \widetilde{\cal L}={\cal L}_2 +\frac{1}{2}\varphi^2 \pm\sqrt{\lambda}\varphi |\phi|^2$; The variables $\varphi$ is Wick rotated. 2. YM idea: $\quad \varphi^{a}_{\mu\nu}:=f^{abc}A^b_{\mu}A^c_{\nu}$; 3. YM idea: $\quad \varphi^{ab}:=A^a_{\mu}A^{b\mu}$; Commented Jul 29, 2020 at 15:41