When we derived Dirac Equation starting form the lagrangian, our QFT professor said:
"let's take the free lagrangian $$\mathscr L = i\bar\Psi\gamma^\mu\partial_\mu\Psi – m\bar\Psi\Psi$$ and perform
$$ \frac{\partial\mathscr L}{\partial (\partial_\mu\Psi)} = \frac{\partial (i\bar\Psi\gamma^\mu\partial_\mu\Psi)}{\partial (\partial_\mu\Psi)} = – i\bar\Psi\gamma^\mu ,$$
where the extra minus sign come from the fact that when we perform the derivative with respect to $\partial_\mu\Psi$ we 'pass through' $\bar\Psi$ and the exchange of two spinors give raise to a minus sign".
This doesn't change anything in computing Dirac equation, but when I tried to compute the stress energy tensor $T^{\mu\nu}$ I obtained (I'm using $\eta^{\mu\nu} = \mathrm{diag}(+1, -1, -1, -1$))
$$T^{\mu\nu} = \frac{\partial\mathscr L}{\partial (\partial_\mu\Psi)}\partial^\nu\Psi + \frac{\partial\mathscr L}{\partial (\partial_\mu\bar\Psi)}\partial^\nu\bar\Psi – \eta^{\mu\nu}\mathscr L = -i\bar\Psi\gamma^\mu\partial^\nu\Psi $$
since the lagrangian is zero on-shell.
Now I take the zero-zero component which is nothing but the energy density
$$\mathscr H = T^{00} = -i\bar\Psi\gamma^0\partial^0\Psi $$
but this energy not only is different from the one I found in every book, it is also negative which means it is certainly wrong. My question now is where did I go wrong?
Best Answer
If $\theta_1,\theta_2$ are a pair of Grassmann variables, then $$ \frac{\partial}{\partial\theta_2}(\theta_1\theta_2)=-\theta_1\qquad\tag{left derivative} $$ where the negative sign is due to the fact that partial derivatives anti-commute with odd variables.
Moreover, Taylor's theorem reads $$ f(\theta_1,\theta_2+\delta)=f(\theta_1,\theta_2)+\delta\frac{\partial f}{\partial\theta_2}+\cdots $$ where here the ordering of factors is important ($f$ is an even function).
Therefore, the correct expression of the energy-momentum tensor is $$ T^{\mu\nu} = \color{red}{-}\frac{\partial\mathscr L}{\partial (\partial_\mu\Psi)}\partial^\nu\Psi + \partial^\nu\bar\Psi\frac{\partial\mathscr L}{\partial (\partial_\mu\bar\Psi)} - \eta^{\mu\nu}\mathscr L = \color{red}{+}i\bar\Psi\gamma^\mu\partial^\nu\Psi $$
Note that most books use right derivatives, which makes this analysis simpler.