# How to calculate Noether current for Yang Mills theory

The Lagrangian of the Yang-Mills fields is given by \begin{align} \mathcal{L}=-\frac{1}{4} F^a_{\mu\nu}~ F_a^{\mu\nu} +\bar{\psi}(i\gamma^{\mu} D_{\mu})\psi- m~ \psi \bar{\psi} \end{align}

where:

\begin{align} D_\mu \psi = \partial_\mu \psi - i g~ t^a_{ij}~ A_\mu^a~ \psi\\ F_{\mu\nu}^a = \partial_\mu A^a_\nu - \partial_\nu A_\mu^a + i g f^{abc} A^b_\mu A_\nu^c \end{align}

I try to get the Noether current here: so starting by the equations of motions:

$\frac{\partial \mathcal{L}}{\partial\psi} - \partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu\psi)} = 0 ~~~~~~~~~~(1)\\ \frac{\partial \mathcal{L}}{\partial\bar{\psi}} - \partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu\bar{\psi})} = 0~~~~~~~~~(2) \\ \frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial \mathcal{L}}{\partial(\partial_\nu A_\mu)} = 0 ~~~~~~~(3)\\$

These yield:

$g \bar{\psi} \gamma^\mu A_\mu^a - m \bar{\psi} -i \partial_\mu \bar{\psi} \gamma^\mu = 0~~~~~~~~(4)\\ g \gamma^\mu A_\mu^a \psi - m \psi = 0~~~~~~~~~~~(5)\\ g\bar{\psi} \gamma^\mu t^a_{ij} \psi_j + g f^{abc} A^b_\nu F^{c~\mu\nu}= \partial_\nu F^{a~ \nu \mu}~~~~~~~~(6)$

Clearly (6) is the right equation of motion of the Yang Mills theory with a conserved current : $J^{a~ \mu} = g\bar{\psi} \gamma^\mu t^a_{ij} \psi_j$ , see for instance Peskin's book Equation(15.51) . Now I have extra terms in (4) and (5) what's wrong I made?

The current is just the LHS of your eq. $(6)$. In fact, the antisimmetry of $F^{\mu \nu}$ implies $\partial _\mu \partial _\nu F^{\mu \nu}=0$.

Observe that the gauge fields carry a non-zero charge, since they transform non-trivially under global transformations. In fact, they transform according to the adjoint representation of the group of simmetry. The part of the current which involves only $\psi$ is the covariant current $\mathscr J ^\mu$, which enters in:$$D_\mu F^{\mu \nu} =\mathscr J ^\nu.$$ This current (differently from the Noether current $J^\mu$) transforms like the Yang-Mills tensor under gauge transformations.

You have forgotten to add the term

$i\partial_\mu\psi \gamma^\mu$

in (5).

Equation (4) is correct. Note that $D_\mu = \partial_\mu + iA_\mu$.

• so $\bar{\psi}(i\gamma^{\mu} \partial_{\mu})\psi = \bar{\psi}(i\gamma^{\mu} \partial_{\mu})\psi + \psi(i\gamma^{\mu} \partial_{\mu})\bar{\psi}$ ? I just thought $\partial_\mu$ act only on $\psi$. Even so what will happen for Equ (4) and (5) ? shouldn't they added to get the sum on all the equations of motions of the three field $\psi,\bar{\psi}, A_\mu$ ?
– S.S.
Jan 6 '17 at 13:23
• The Version for the Fermion kinetic action $\bar{\psi}i \gamma^\mu \partial_\mu \psi + \psi i \gamma^\mu \partial_\mu \bar{\psi}$ can also be used; it is naturally a hermitean quantity. The Version given in your question is valid if and only if $\int d^4x \partial_\mu(\bar{\psi} \gamma^\mu \psi)=0$ (fermion number conservation) is satisfied. Jan 6 '17 at 14:39
• Ok, why then Equs. (4), (5) and (6) has been added to get the final equation of motion ? it seems now (4) and (5) have been subtracted .
– S.S.
Jan 6 '17 at 14:45
• This kinetic action I considered in my comment is only an alternative Lagrangian density; not an equation of Motion. Jan 6 '17 at 15:07
• Sorry, i don't get it. what I mean is when we got three equations of motions due to the three fields in the Lagrangian $\psi,\bar{\psi}, A_\mu$ , why only the last one Equ. (6) kept as the right final equation of motion of the Lagrangian while (4) and (5) in my solution have disapered ?
– S.S.
Jan 6 '17 at 19:14

In the Lagrangian of the question, there is a typo, it should be $\bar{\psi}\psi$ for the last term.

The equation (4) is basically correct except the sign of $i \gamma^{\mu} \partial_{\mu} \psi$ should be positive. By taking complex conjugate of it , you get the equation of motion as $i \gamma^{\mu} D_{\mu} \psi - m \psi =0$.

The equation (5) should be the same when you do a partial integral for the term $\bar{\psi} i \gamma^{\mu} \partial_{\mu} \psi=i\gamma^{\mu} \partial{\left(\bar{\psi}\psi\right)}- \partial{\bar{\psi} i \gamma^{\mu} \psi}$ and then compute the Euler-Lagrange equation.

The equation (6) is basically correct. It should be $\partial_{\nu} F^{a \nu \mu} + g \bar{\psi} \gamma^{\mu} t^a \psi + i g F^{c \mu \nu} f^{abc} A^b_{\nu}=0$.

There are two equations of motion in total for two fields, which makes sense.