6
$\begingroup$

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)?

$\endgroup$
6
$\begingroup$

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

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

and

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

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.