Why is $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = 0$, for the Dirac Lagrangian $\mathcal{L} = \bar{\psi}(i \gamma^\mu \partial_\mu - m)\psi$?

This comes up in deriving the Noether current for $\psi \rightarrow e^{i\alpha}\psi$ for example.

My confusion comes from the fact that we can write the following term in the Lagrangian $i\bar{\psi}\gamma^\mu\partial_\mu\psi = -i(\partial_\mu \bar{\psi})\gamma^\mu\psi$ by integrating by parts which makes it look like $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = -i \gamma^\mu \psi$. In fact, this is how we get the equations of motion for $\bar{\psi}$.

  • $\begingroup$ Are you asking why $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = -i \gamma^\mu \psi$? $\endgroup$
    – joseph h
    Commented Oct 30, 2020 at 0:25
  • $\begingroup$ How can you integrate by parts without an integral? $\endgroup$
    – Qmechanic
    Commented Oct 30, 2020 at 1:01

1 Answer 1

  1. $\psi$ and $\bar \psi$ are thought as two independent variables in the Lagrangian.
  2. If you write a Lagrangian as $\mathcal{L}_1 =\bar\psi(...)\psi$, you should use it to calculate the Noether current or equation of motion. If you have the other one, $\mathcal{L}_2 =\psi(...)\bar\psi$, you have to perform the derivatives based on this one.
  3. The two results will be equivalent, they are Dirac dual to each other.

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