One of the most common gauges in QED computations are the $R_{\xi}$ gauges obtained by adding a term \begin{equation} -\frac{(\partial_\mu A^{\mu})^2}{2\xi} \end{equation} 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)?


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

$$\tag{1} {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}^2 + \bar{\psi}(iD\!\!\!\!/ \ \ -m)\psi.$$ The gauge-fixed QED Lagrangian density in the $R_{\xi}$-gauge reads

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

where the Faddeev-Popov term is

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


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

is the Lorenz gauge-fixing condition.

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

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

The Euler-Lagrange equation for the $B$-field reads

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

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.

  • $\begingroup$ Are you saying that I cannot impose the covariant gauges classically? I am asking what would be the classical constraint on A in such gauges. I don't really need to know about the quantum theory unless you are saying the only way to impose this gauge is through the path integral. $\endgroup$ – Dan Feb 28 '14 at 7:02
  • $\begingroup$ @Qmechanic A comment on the answer (v2). I'm rusty on this stuff, but I imagine your last statement about the classical situation somehow came from the fact that for any $\xi\neq \infty$, the classical equations of motion are (possibly up to a sign error) $\Box A^\mu + (\frac{1}{\xi}-1)\partial^\mu(\partial\cdot A) = j^\mu$, so that taking $\partial_\mu$ of both sides gives $\Box(\partial\cdot A) = 0$ and finally one can then somehow argue that this gives $\partial\cdot A = 0$ given suitable initial/boundary data? $\endgroup$ – joshphysics Feb 28 '14 at 7:34
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Mar 1 '14 at 20:34
  • $\begingroup$ 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.. $\endgroup$ – PPR Oct 5 '14 at 20:40
  • $\begingroup$ 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. $\endgroup$ – Qmechanic May 30 '15 at 20:57

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