quantum field theory - Conservation of BRST current in QED

I am trying to understand the conservation of the BRST current in QED but am having some trouble. This is what I have so far, QED lagrangian density in Lorenz gauge is,

$$L = \frac{1}{4}F_{\mu\nu}F^{\mu\nu} +\frac{1}{2\xi}(\partial_{\mu}A^{\mu})^2 + \partial^{\mu}\overline{c}\partial_{\mu} c$$

I have been working in $\xi=1$ gauge with the following BRST transformations,

$$\delta A_{\mu} = \partial_{\mu}c$$ $$\delta c = 0$$ $$\delta \overline{c} = \partial_{\mu}A^{\mu}$$

Using Noether's Theorem, I think the BRST current should be

$$j^{\mu} = \partial^{\mu}A^{\nu}\partial_{\nu} c + \partial^{\mu}c\partial_{\nu}A^{\nu}$$

I can not show that this current is conserved, using the equations of motion $\Box_x A_{\mu} = 0$ and $\Box_x c = 0$. I am left with,

$$\partial_{\mu}j^{\mu} = \partial^{\mu}A^{\nu}\partial_{\mu}\partial_{\nu}c + \partial^{\mu}c\partial_{\mu}\partial_{\nu}A^{\nu}$$

which I don't think is equal to 0. I'm not sure what I have done wrong here so any help would be greatly appreciated.

Answer

I) The gauge-fixed pure Maxwell action is

$$\tag{1} S[A,c,\bar{c}]~=~\int \! d^4x~ {\cal L} $$

with Lagrangian density$^1$

$$\tag{2} {\cal L}~=~{\cal L}_0 -\frac{\chi^2}{2\xi}-d_{\mu}\bar{c}~d^{\mu}c, \qquad {\cal L}_0~:=~-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}, \qquad \chi~:=~d_{\mu} A^{\mu}, \qquad \xi~>~0,$$

consisting of (i) the Maxwell term, (ii) the gauge-fixing term, and (iii) the Faddeev-Popov determinant term. The Euler-Lagrange equations read$^2$

$$ 0~\approx~\frac{\delta S}{\delta A_{\mu}} ~=~ d_{\nu}F^{\nu\mu} +\frac{d^{\mu}\chi}{\xi}, $$ $$\tag{3} 0~\approx~\frac{\delta S}{\delta c}~=~-\Box \bar{c}, \qquad 0~\approx~\frac{\delta S}{\delta\bar{c}}~=~\Box c. $$

(Here the $\approx$ symbol means equality modulo equations of motion.)

II) The gauge-fixed Grassmann-odd BRST transformation ${\bf s}$ reads$^3$

$$\tag{4} {\bf s} A_{\mu}~=~d_{\mu}c,\qquad {\bf s} c~=~0,\qquad {\bf s}\bar{c}~=~\frac{\chi}{\xi}, \qquad {\bf s}\chi~=~\Box c~\approx~0.$$

The BRST variation of the Lagrangian density (2) is a total divergence

$$\tag{5} {\bf s}{\cal L}~=~d_{\mu}f^{\mu}, \qquad f^{\mu}~:=~-\frac{\chi}{\xi} d^{\mu}c,$$

i.e. the BRST transformation ${\bf s}$ is a quasi-symmetry of the gauge-fixed Maxwell action (1), cf. this Phys.SE answer.

III) The bare Noether current for the BRST quasi-symmetry reads

$$ j^{\mu}~:=~\frac{\partial {\cal L}}{\partial(d_{\mu}A_{\nu})} {\bf s} A_{\nu} +\frac{\partial {\cal L}}{\partial(d_{\mu}c)} {\bf s}c +\frac{\partial {\cal L}}{\partial(d_{\mu}\bar{c})} {\bf s} \bar{c}$$ $$\tag{6} ~=~ - (F^{\mu\nu}+\frac{\chi}{\xi}\eta^{\mu\nu})d_{\nu} c - \frac{\chi}{\xi} d^{\mu}c, $$

which is Grassmann-odd. The full BRST Noether current reads:

$$\tag{7} J^{\mu}~:=~j^{\mu}-f^{\mu}~=~ -F^{\mu\nu}d_{\nu} c - \frac{\chi}{\xi} d^{\mu}c. $$

It is conserved on-shell

$$\tag{8} d_{\mu}J^{\mu} ~=~ -\frac{\delta S}{\delta A_{\mu}}{\bf s} A_{\mu} -\frac{\delta S}{\delta c}{\bf s}c -\frac{\delta S}{\delta\bar{c}}{\bf s}\bar{c} ~\approx~0,$$

cf. Noether's first theorem.

--

$^1$ A comment about signs: Classically, the overall sign of the action does not matter, although relative signs between terms are important. Quantum mechanically, the signs of the Maxwell term and the gauge fixing term are important in order to achieve unitarity, i.e. the sign in front of the kinetic term $\sum_{i=1}^3\dot{A}_i^2$ should be positive, while the sign in front of the potential term $\chi^2$ should be negative. See also e.g. this Phys.SE post. The (possibly complex) coefficient in front of the Faddeev-Popov determinant term should be correlated with the reality/Hermiticity conditions imposed on the Faddeev-Popov ghost and antighost.

$^2$ We use for simplicity here the convention that derivatives and BRST transformation ${\bf s}$ are left derivations, i.e.

$$\tag{9} {\bf s}(fg)~=~{\bf s}(f)~g + (-1)^{|f|}f ~{\bf s}(g). $$

$^3$ Note that the gauge-fixed BRST transformation ${\bf s}$ is only nilpotent on-shell in the antighost sector $$\tag{10} {\bf s}^2 \bar{c}~=~\frac{\Box c}{\xi}~\approx~0.$$ It is possible to obtain a BRST formulation that is off-shell nilpotent by including a Lautrup-Nakanishi (LN) auxiliary field $B$. For completeness, let us mention that the Batalin-Vilkovisky (BV) Lagrangian density reads

$$\tag{10} {\cal L}_{BV}~=~{\cal L}_0 + A^{\mu}_{\ast} d_{\mu} c + B\bar{c}^{\ast}, $$

with corresponding nilpotent Grassmann-odd BRST transformations $$\tag{11} {\bf s} A_{\mu}~=~d_{\mu}c,\qquad {\bf s} c~=~0,\qquad {\bf s}\bar{c}~=~-B, \qquad {\bf s}B~=~0.$$

The gauge-fixing fermion

$$\tag{12} \psi ~=~ \int \! d^4x~\bar{c}(\frac{\xi}{2}B+\chi) $$

yields the corresponding gauge-fixed Lagrangian density

$$\tag{13} {\cal L}_{\rm gf} ~=~ \left. {\cal L}_{BV} \right|_{\phi^{\ast}~=~\frac{\delta \psi}{\delta \phi}} ~=~ {\cal L}_0 - d_{\mu}\bar{c}~d^{\mu}c +\frac{\xi}{2}B^2+B\chi\quad\stackrel{\text{int. out } B}{\longrightarrow}\quad {\cal L}, $$

which becomes the Lagrangian density (2) after we integrate out the LN auxiliary $B$-field, which has eom $$\tag{14} B~\approx~ -\frac{\chi}{\xi}, $$ cf. e.g. this Phys.SE post.