The Lagrangian of the Yang-Mills fields is given by
$$
\mathcal{L}=-\frac{1}{4}(F^a_{\mu\nu})^2+\bar{\psi}(i\gamma^{\mu}
D_{\mu}-m)\psi-\frac{1}{2\xi}(\partial\cdot A^a)^2+
\bar{c}^a(\partial\cdot D^{ab})c^b
$$
where the metric is $(-,+,+,+)$, and the conventions are the following:
$$
[D_{\mu},D_{\nu}]=-igF_{\mu\nu},\quad
D_{\mu}=\partial_{\mu}-igA^a_{\mu}t^a, \quad
D^{ab}_{\mu}=\delta^{ab}\partial_{\mu}-gf^{abc}A^c_{\mu}
$$

Let $\epsilon$ be an infinitesimal anticummuting parameter, and consider the BRST transformation:
$$
\delta\psi=ig\epsilon c^at^a\psi,\quad
\delta A^a_{\mu}=\epsilon D^{ab}_{\mu}c^b,\quad
\delta c^a=-\frac{1}{2}g\epsilon f^{abc}c^bc^c,\quad
\delta\bar{c}^a=\frac{1}{\xi}\epsilon\partial^{\mu}A^a_{\mu}
$$

I have calculated the corresponding Noether current as
$$
j_{BRST}^{\mu}=-g\bar{\psi}\gamma^{\mu}c^at^a\psi-F^{a\mu\nu}D^{ab}_{\nu}c^b-
\frac{1}{\xi}(\partial\cdot A^a)D^{ab\mu}c^b+
\frac{1}{2}gf^{abc}(\partial^{\mu}\bar{c}^a)c^bc^c
$$

I am not sure whether the result is correct or not, so I would like to check that $\partial_{\mu}j^{\mu}_{BRST}=0$. Even though I have used the equation of motion
$$
\partial_{\mu}F^{a\mu\nu}=-g\bar{\psi}\gamma^{\nu}t^a\psi-
gf^{abc}A^b_{\mu}F^{c\mu\nu}-\frac{1}{\xi}\partial^{\nu}
(\partial\cdot A^a)-gf^{abc}(\partial^{\nu}\bar{c}^b)c^c
$$
$$
(i\gamma^{\mu}D_{\mu}-m)\psi=0,\quad \partial^{\mu}D^{ab}_{\mu}c^b=0
$$
and spent about four hours, I still cannot get it right. Could someone help me check this? Thanks a lot.

This post imported from StackExchange Physics at 2014-07-03 17:59 (UCT), posted by SE-user soliton