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)?
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1In 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. – MadMax Feb 18 '20 at 15:55
1 Answers
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
$$ \chi~:=~d_{\mu}A^{\mu}~\approx~0 \tag{4}$$
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
$$ {\cal L}~=~ {\cal L}_0 +{\cal \cal L}_{FP} +\frac{\xi}{2}B^2+B\chi \quad\stackrel{\text{int. out } B}{\longrightarrow}\quad {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 ,\tag{5} $$
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.
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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. – Qmechanic Mar 01 '14 at 20:34
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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 Oct 05 '14 at 20:40
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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. – Qmechanic May 30 '15 at 20:57
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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: $\quad \varphi^{ab}:=A^a_{\mu}A^{b\mu}$; – Qmechanic Jul 29 '20 at 15:41