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The pure gauge QED Lagrangian density, including the ghost fields, is

\begin{align*} \mathcal{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. \end{align*}

The action associated with this Lagrangian is invariant under the BRST transformation

\begin{align*} \delta A_\mu=\epsilon\partial_\mu c,\qquad \delta c=0,\qquad \delta\overline{c}=-\frac{1}{\xi}\epsilon\partial^\mu A_\mu. \end{align*}

I calculate the conserved current associated with this symmetry, which is

\begin{align*} J^\mu=-F^{\mu\nu}\partial_\nu c-\frac{1}{\xi}\partial^\nu A_\nu\partial^\mu c, \end{align*}

and checked that it's conserved. Now I want to calculate the conserved charge

\begin{align*} Q=\int d^3\vec{x} J_0=-\int d^3\vec{x}(\partial_0 A_\nu-\partial_\nu A_0)\partial^\nu c+\partial^\nu A_\nu \partial_0 c \end{align*}

(I choose $\xi=1$,) in terms of the modes $a_\mu(p),a_\mu^\dagger(p),c(k)$ and $c^\dagger(k)$, which appear in the expansions

\begin{align*} A_\mu(x) & =\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_\mu(p)e^{ipx}+a_\mu^\dagger(p)e^{-ipx}),\\ c(x) & =\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}(c(k)e^{ikx}+c^\dagger(k)e^{-ikx}). \end{align*}

Finally I want to show that these charge Q satisfy the commutation and anticommutation relations associated with the BRST symmetry, namely

\begin{align*} & [Q,a_\mu(p)] = -p_\mu c(p),\\ & [Q,a_\mu^\dagger(p)] = p_\mu c^\dagger(p),\\ & \{Q,\overline{c}(p)\} = p^\mu a_\mu(p),\\ & \{Q,\overline{c}^\dagger(p)\} = p^\mu a_\mu^\dagger(p),\\ & \{Q,c(p)\} =\{Q,c^\dagger(p)\}=0. \end{align*}

I think the right answer is

\begin{align*} Q=\int d^3\vec{p}(a_\mu^\dagger c+a_\mu c^\dagger)p^\mu \end{align*}

because with this I was able to show the commutation and anticommutation relations, however I was not able to get to this expression.

I did the following, first I split the charge as

\begin{align*} Q=Q_{i0i}-Q_{0ii}+Q_{000}-Q_{ii0} \end{align*}

where, for example

\begin{align} \begin{split} Q_{ii0} & =\int d^3\vec{x}\partial_iA_i\partial_0c\\ & = \int\frac{d^3\vec{q}}{(2\pi)^3}\frac{d^3\vec{p}}{(2\pi)^3} d^3\vec{x}\frac{1}{\sqrt{2\omega_p}}\frac{1}{\sqrt{2\omega_q}}\partial_i\left(a_i e^{ipx}+a_i^\dagger e^{-ipx}\right)\partial_0(ce^{iqx}+c^\dagger e^{-iqx})\\ & =\int\frac{d^3\vec{q}}{(2\pi)^3}\frac{d^3\vec{p}}{(2\pi)^3} d^3\vec{x}\frac{1}{\sqrt{2\omega_p}}\frac{1}{\sqrt{2\omega_q}}\Big(p_i\omega_qa_i c^\dagger e^{i(p-q)x}+p_i\omega_qa_i^\dagger ce^{i(q-p)x}-p_i\omega_q a_ice^{i(p+q)x}-p_i\omega_qa_i^\dagger c^\dagger e^{-i(p+q)x}\Big)\\ & =\int d^3\vec{q}\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\frac{1}{\sqrt{2\omega_q}}\Big(p_i\omega_qa_i c^\dagger e^{i(\omega_p-\omega_q)t}+p_i\omega_qa_i^\dagger ce^{i(\omega_q-\omega_p)t}\big)\delta(\vec{p}-\vec{q})-\big(p_i\omega_qa_ice^{i(\omega_p+\omega_q)t}+p_i\omega_qa_i^\dagger c^\dagger e^{-i(\omega_p+\omega_q)t}\big)\delta(\vec{p}+\vec{q})\Big)\\ & =\int d^3\vec{p}\frac{1}{2}\big(p_ia_i(\vec{p}) c^\dagger(\vec{p})+p_ia_i^\dagger(\vec{p}) c(\vec{p})-p_ia_i(\vec{p})c(-\vec{p})e^{i2\omega_pt}-p_ia_i^\dagger (\vec{p})c^\dagger(-\vec{p}) e^{-i2\omega_pt}\big) \end{split}. \label{calculation} \end{align}

But now I have these time dependent exponentials and I don't know how to get ride of then! Is there any error in the above calculations? Also, you have any good reference on the subject where this calculations are made with detail? Thanks!

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  • $\begingroup$ They should cancel with the corresponding terms in $Q_{0ii}$. There the spatial derivative acts on $c$, so the momentum will have the opposite sign. $\endgroup$ – Subhaneil Lahiri Jun 9 '19 at 18:58
  • $\begingroup$ Thanks Subhaneil Lahiri! It works! :) $\endgroup$ – Davi Bastos Jun 10 '19 at 14:57

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