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.

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 6, 2013 at 15:08
  • $\begingroup$ Not sure if it's correct but I get $j^\mu=\left(-F^{\mu\sigma}\,^{a}+B^{a}\eta^{\mu\sigma}-g\eta^{\mu\sigma}f^{bac}\bar{c}^{b}c^{c}\right)\left(D_{\sigma}\,^{ac}c^{c}\right)+g\bar{\psi}\gamma^{\mu}c^{a}t^{a}\psi-\frac{1}{2}g^{2}f^{cba}f^{ade}A^{\mu}\,^{b}\bar{c}^{c}c^{d}c^{e}-\frac{1}{2}gf^{abc}\bar{c}^{a}\partial^{\mu}c^{b}c^{c}+\frac{1}{2}gf^{abc}\left( \partial^{\mu} \bar{c}^{a}\right)c^{b}c^{c}$ $\endgroup$
    – PPR
    Jul 4, 2014 at 14:06

1 Answer 1


The crucial point is the anticommutation of $\epsilon$ with fermion fields ($\psi,\bar\psi,\bar{c}^a,c^a$).

First, we will rewrite $\mathcal{L}$ as (for simplicity, we define $B^a \equiv \xi^{-1}\partial^\mu A_\mu^a$)

$$\tag{1} \mathcal{L}=-\frac{1}{4}(F^a_{\mu\nu})^2+\bar{\psi}(i\gamma^{\mu} D_{\mu}-m)\psi-\frac{\xi}{2}B^aB^a- \partial^{\mu}\bar{c}^a D_{\mu}^{ab}c^b $$

which differs from the original $\mathcal{L}$ by a total derivative,

$$\tag{2} \partial^{\mu}(\bar{c}^a D_{\mu}^{ab}c^b) $$

so $\delta\mathcal{L}$ no longer equals $0$, but

$$\tag{3} \delta\mathcal{L} = -\delta \partial^{\mu}(\bar{c}^a D_{\mu}^{ab}c^b) = \partial^{\mu}( -\delta\bar{c}^a D_{\mu}^{ab}c^b) = \partial^{\mu}\left(-\epsilon B^a D_{\mu}^{ab}c^b\right) \equiv \partial^{\mu} K_{\mu} $$

We will use Jacobi identity occasionally,

$$\tag{4} f^{abd}f^{dce} + f^{bcd}f^{dae} + f^{cad}f^{dbe} = 0 $$

We will use right derivative in the following calculations. So the Noether current is defined as $$\tag{5} \epsilon j^{\mu} \equiv \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar\psi)}\delta\bar\psi + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}c^a)}\delta c^a + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar{c}^a)}\delta\bar{c}^a + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}A_{\nu}^a)}\delta A_{\nu}^a - K^{\mu} $$

Now we will calculate the individual parts of the current, moving $\epsilon$ to the left of each expression. We'll get an extra minus sign if $\epsilon$ passes a fermion field,

$$\begin{aligned} \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi &= (\bar\psi i \gamma^{\mu}) (ig \epsilon c^a t^a \psi) = \epsilon\, g \bar\psi \gamma^{\mu} c^a t^a \psi \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar\psi)}\delta\bar\psi &= 0 \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}c^a)}\delta c^a &= (-\partial^{\mu}\bar{c}^a)\left(-\frac{1}{2}g\epsilon f^{abc}c^bc^c\right) = -\frac{1}{2} \epsilon\, g f^{abc} (\partial^{\mu}\bar{c}^a) c^bc^c \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar{c}^a)}\delta\bar{c}^a &= (g^{\mu\nu}D_\nu^{ab} c^b) \left(\epsilon B^a\right) = - \epsilon (g^{\mu\nu}D_\nu^{ab} c^b) B^a = K^{\mu}\\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}A_{\nu}^a)}\delta A_{\nu}^a &= \left(-F^{a\mu\nu} - g^{\mu\nu} B^a\right)(\epsilon D_\nu^{ab} c^b) = \epsilon \left(-F^{a\mu\nu} - g^{\mu\nu}B^a\right)(D_\nu^{ab} c^b)\\ \end{aligned} \tag{6}$$

Inserting results from $(6)$ and $(3)$ into $(5)$ gives

$$\tag{7} j^\mu = \left(-F^{a\mu\nu} - g^{\mu\nu}B^a\right)D_\nu^{ab} c^b -\frac{1}{2} g f^{abc} (\partial^{\mu}\bar{c}^a) c^bc^c + g \bar\psi \gamma^{\mu} c^a t^a \psi $$

It's easy to derive the equations of motion,

$$\begin{aligned} 0= \frac{\partial\mathcal{L}}{\partial A_\nu^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu^a)} &= D_\mu^{ab}F^{b\mu\nu} + \partial^\nu B^a + g \bar\psi \gamma^\nu t^a \psi + g f^{abc}(\partial^\nu \bar{c}^b) c^c \\ 0= \frac{\partial\mathcal{L}}{\partial\bar\psi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \bar\psi)} &= -i \gamma^\mu\partial_\mu\psi - g A_{\mu}^a \gamma^\mu t^a \psi + m \psi \\ 0= \frac{\partial\mathcal{L}}{\partial\psi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \psi)} &= -i \partial_\mu \bar\psi \gamma^\mu + g A_\mu^a \bar\psi \gamma^\mu t^a - m \bar\psi \\ 0= \frac{\partial\mathcal{L}}{\partial c^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu c^a)} &= D_\mu^{ab} \partial^\mu \bar{c}^b \\ 0= \frac{\partial\mathcal{L}}{\partial\bar{c}^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \bar{c}^a)} &= -\partial^\mu D^{ab}_\mu c^b \\ \end{aligned} \tag{8a-e}$$

Now we will check the validation of $\partial_\mu j^\mu = 0 $,

$$\begin{aligned} \partial_\mu \Bigl[ g \bar\psi \gamma^\mu c^a t^a \psi \Bigr] &\stackrel{(8b,8c)}{=} -\Bigl[ g \bar\psi \gamma^\nu t^a \psi \Bigr] D_\mu^{ad} c^d \\ \partial_\mu \left[ - \frac{1}{2}g f^{abc} (\partial^\mu \bar{c}^a) c^b c^c \right] &\stackrel{(4,8d)}{=} -\Bigl[ g f^{abc}(\partial^\nu \bar{c}^b) c^c \Bigr] D_\mu^{ad} c^d \\ \partial_\mu \Bigl[ (-F^{a\mu\nu} -B^a g^{\mu\nu}) D_\nu^{ab} c^b \Bigr] &\stackrel{(8e)}{=} -(\partial_\mu F^{a\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d - F^{a\mu\nu} \partial_\mu D_\nu^{ad}c^d \\ &= -(D_\mu^{ab} F^{b\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d \\ &\quad - (gf^{abc}A_\mu^c F^{b\mu\nu}) D_\nu^{ad} c^d - F^{a\mu\nu} \partial_\mu D_\nu^{ad}c^d \\ &\stackrel{(4)}{=} -(D_\mu^{ab} F^{b\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d \\ \end{aligned} \tag{9a-c}$$

We'll get

$$ \partial_\mu j^\mu \stackrel{(9a+9b+9c,8a)}{=} -\left( \frac{\partial\mathcal{L}}{\partial A_\nu^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu^a)} \right) D_\nu^{ad}c^d \tag{10}$$

This is indeed the off-shell Noether identity, and is zero when the equations of motion are fulfilled.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.