# Derivative with respect to a spinor of the free Dirac lagrangian

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?

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.

User AccidentalFourierTransform has already given a good answer. Here we will stress a few points:

1. The main issue is that the superdifferential reads $$\mathrm{d}\theta\frac{\partial^Lf(\theta,\bar{\theta})}{\partial\theta} +\mathrm{d}\bar{\theta}\frac{\partial^Lf(\theta,\bar{\theta})}{\partial\bar{\theta}} ~=~ \mathrm{d}f(\theta,\bar{\theta}) ~=~ \frac{\partial^Rf(\theta,\bar{\theta})}{\partial\theta}\mathrm{d}\theta+\frac{\partial^Rf(\theta,\bar{\theta})}{\partial\bar{\theta}}\mathrm{d}\bar{\theta},\tag{1}$$ where the order of factors depends crucially on whether one uses left or right derivatives. See also e.g. this Phys.SE post.

E.g. the order (1) becomes important when constructing the canonical stress-energy-momentum (SEM) tensor, which is the Noether current for spacetime translations, cf. OP's question. Another example is when constructing the Hamiltonian formalism for fermions. See e.g. this Phys.SE post and links therein.

2. The other issue is that $\theta$ and $\bar{\theta}$ are strictly speaking not independent Grassmann-odd variables; however one may treat them as independent. This issue is similar to this Phys.SE post for Grassmann-even variables.