I've given myself a severe headache jumping between East/West Coast sign conventions; I have picked up an extra minus sign and could do with a hand. I am currently using $\eta=\textrm{Diag}[-,+,+,+]$

So I have the symmetrised Dirac Lagrangian: $$ \mathcal{L} = -i ~\bar{\psi}\left(\frac{1}{2}\gamma^\mu \overrightarrow{\partial_\mu} -\frac{1}{2}\gamma^\mu \overleftarrow{\partial_\mu} + m\right)\psi $$

I am considering the infinitesimal transformation: $$ \delta \psi = i \theta \psi ; ~\delta\bar{\psi}=-i\theta\bar{\psi} $$

Clearly we have: $$ \frac{\partial\mathcal{L}}{\partial\bar{\psi}} = -i\left(\frac{1}{2}\gamma^\mu \overrightarrow{\partial_\mu} + m \right) \psi;~ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})} = \frac{i}{2}\gamma^\mu\psi $$

My confusion begins when I consider the derivatives w.r.t. $\psi$; I was under the impression that these Fermonic fields were Grassmann odd and therefore I pick up a minus sign as I move the derivative "through" $\bar{\psi}$. In doing this I find: $$ \frac{\partial\mathcal{L}}{\partial\psi} = +i~\bar{\psi}\left(-\frac{1}{2}\gamma^\mu \overleftarrow{\partial_\mu} + m \right);~ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)} = \frac{i}{2}\bar{\psi}\gamma^\mu $$

I then find the Noether current to be:

$$ \begin{split} J^\mu =& \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\delta\psi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})}\delta\bar{\psi} \\ =& \frac{\theta}{2}\left(-\bar{\psi}\gamma^\mu\psi + \gamma^\mu \psi \bar{\psi}\right) \end{split} $$ Then: $$ \begin{split} \gamma^\mu \psi\bar{\psi} &= \gamma^\mu \psi \psi^\dagger \gamma^0 \\ &= - \psi^\dagger \gamma^\mu \gamma^0 \psi \\ &= + \psi^\dagger \gamma^0 \gamma^\mu \psi + 2 \eta^{\mu 0}\psi^\dagger \psi\\ &= +\bar{\psi} \gamma^\mu \psi \end{split} $$

Clearly something is wrong here. Either I made a mistake in taking the derivatives or I have messed up the rearrangement of the fields.

I ought to get the result: $$ J^\mu \propto \bar{\psi} \gamma^\mu \psi $$ Some advice would be really appreciated, I've been banging my head against this for too long trying to find the erroneous minus one.

  • $\begingroup$ You need a relative sign between the two terms in the Noether current. You have to be careful when deriving the form of the current - you should act with $\delta\psi \frac{\partial}{\partial\left(\partial_\mu \psi\right)}$ on $\mathcal{L}$. $\endgroup$
    – Toffomat
    Commented Oct 25, 2017 at 14:31
  • $\begingroup$ Is this due to the ordering of the $\gamma^0$ and $\gamma^\mu$ matrices? Tong has the same form as me (although I think he's using the other metric convention). Whilst changing the order fixes my issue it still leaves me with little understanding of whether I have been doing the rest correctly. damtp.cam.ac.uk/user/pz229/Teaching_files/QFT3.pdf $\endgroup$
    – Vielbein
    Commented Oct 25, 2017 at 15:43
  • $\begingroup$ essentially the same question behind physics.stackexchange.com/questions/301860 $\endgroup$ Commented Oct 25, 2017 at 20:06
  • $\begingroup$ Ah I need to sit and rederive the form of the Noether current then? I saw that thread before but it only affirmed my notion that I was taking the derivatives correctly. $\endgroup$
    – Vielbein
    Commented Oct 25, 2017 at 20:32


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