# 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.

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The BRST current can't contain any derivatives of $c$ in the simple Abelian case, especially not second derivatives. Try to recheck how you derive Noether's currents in other situations - why your incorrect way to do so produces more derivatives than the right way. ;-) Just to be sure, when you do it right, the BRST current is just $c$ times the ordinary $U(1)$ current if the group is Abelian. Moreover, in the non-Abelian case where the extra $f_{ijk}$ term is present, the term is proportional to $c\bar c$... – Luboš Motl Sep 2 '13 at 15:26
@Lubos Motl: Why can the BRST current not contain any derivatives of c? Also to calculate the current I just used $j^{\mu} = \frac{\partial L}{\partial(\partial_{\mu}\phi_a)}\delta \phi_a$. The derivatives of c appear because of the variation of the vector field, $\delta A^{\mu}$. – user29131 Sep 2 '13 at 21:42
Dear user, there can't be derivatives because the right rule is $Q=\sum c^i L_i$ in the Abelian case, see en.wikipedia.org/wiki/BRST_quantization#Example - Unfortunately, you didn't manage to solve the problem I recommended you otherwise you would learn what's wrong about your procedure to compute the BRST current. The spatial deriatives have to be explicitly removed from the formula - one must study the coefficient in front of $\partial_\mu \epsilon$ in the symmetry variation of the action. Indeed the terms with $\epsilon$ without derivatives cancel in $\delta S$ since it's a symmetry – Luboš Motl Sep 3 '13 at 11:10
Again, could you please develop some work of yourself? I am telling you that you don't know how to do the Noether procedure right - even in simpler cases - and you should start with that. BRST is just another level that you can't built properly if you are confused about simpler things. Incidentally, your formula for the current in the comment is downright nonsensical because the LHS is finite but RHS is infinitesimal. The right formula removes the infinitesimal transformation parameter $\epsilon$ but only after an integration by parts takes place. – Luboš Motl Sep 3 '13 at 11:11

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

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

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$^1$ 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.

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+1...I had forgotten that we have to include the current $f_\mu$. I have a question : I see in some references, like this one, formula (2.5) page 4, something like $j^\mu =\partial_\mu F^{\nu\mu}c - (\partial.A)\partial^\mu c$ or $j^\mu =\partial^\mu(\partial.A)c - (\partial.A)\partial^\mu c$. How can I justify the term proportionnal to $c$ and second derivative of $A_\mu$? – Trimok Sep 4 '13 at 5:13
Thanks, I also forgot about the current $f_{\mu}$. I shall try this again being more careful this time! – user29131 Sep 4 '13 at 20:31