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  Noether currents for the BRST tranformation of Yang-Mills fields

+ 5 like - 0 dislike

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
asked Apr 6, 2013 in Theoretical Physics by soliton (110 points) [ no revision ]
You should first understand the (covariant-derivative-based) conservation law for the normal Yang-Mills current $j^\mu$ and the conservation of the BRST current should morally be a similar calculation except that the current is multiplied by $c$ and traced - and the coefficient of the $cc\bar c$ term is halved to make it work.

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

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