# How does canonical quantization work with Grassmann variables?

Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.

First, the books introduce the Dirac Lagrangian, $$\mathcal{L} = \bar{\psi}(i \not\partial - m) \psi.$$ To compute the canonical momentum, we note that $$\mathcal{L} \supset \psi^\dagger \gamma^0 (i \partial_0 \gamma^0 \psi) = i \psi^\dagger \dot{\psi}$$ in mostly negative signature. Therefore, the canonical momentum is $$\frac{\partial \mathcal{L}}{\partial \dot{\psi}} = i \psi^\dagger.$$ One then goes on to perform canonical quantization.

Later, the books introduce the Majorana Lagrangian, which in Peskin and Schroeder (problem 3.4) has the form $$\mathcal{L} = \chi^\dagger i \bar{\sigma} \cdot \partial \chi + \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*).$$ The Majorana mass term vanishes at the classical level because $$\sigma^2$$ is an antisymmetric matrix. The only way out is to postulate that the two-component spinor $$\chi$$ is really a Grassmann variable, so that the two terms in the mass term have the same sign after anticommutation. It is usually stated that, in general, every spinor in a classical Lagrangian has to be a Grassmann number.

However, this contradicts the earlier treatment of the Dirac Lagrangian. If we treat $$\psi$$ as a Grassmann number, then we pick up a sign upon anticommuting the Grassmann derivative, so $$\frac{\partial \mathcal{L}}{\partial \dot{\psi}} = \frac{\partial}{\partial \dot{\psi}} (i \psi^\dagger \dot{\psi}) = - i \psi^\dagger \frac{\partial}{\partial \dot{\psi}} \dot{\psi} = - i \psi^\dagger.$$ This extra negative sign completely changes the result of canonical quantization, e.g. it leads to a disastrous negative definite energy. The same problem seems to occur in problem 3.4 of Peskin. If one correctly accounts for the Grassmann sign flip when performing canonical quantization, then one arrives at anticommutation relations that are opposite those given in the problem.

I've searched through a stack of quantum field theory textbooks, and frustratingly, not one of them even mentions this apparent inconsistency, because they all cover the Majorana Lagrangian (and Grassmann numbers) after they've finished the Dirac Lagrangian, so there's no opportunity for this issue to come up. One could avoid this issue by saying that Grassmann numbers only appear in the path integral, but then it becomes impossible to canonically quantize the Majorana theory because the mass term vanishes, which seems even worse. What's going on here?

• I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $\psi$ are Grassmann numbers, then for the momentum $\Pi$ you may get either $\Pi=i\bar{\psi}$ or $\Pi=-i\bar{\psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one) Jan 9, 2019 at 17:25
• @Luthien Yes, naively it seems to me that we get the latter, but need the former. Jan 9, 2019 at 17:28
• Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein. Jan 9, 2019 at 17:54

Let's say we have the function: $$$$f(\theta_1, \theta_2)=f_0+f_1\theta_1+f_2\theta_2+f_3\theta_1\theta_2$$$$ Then the left derivative with respect to $$\theta_1$$ is $$$$\frac{\partial_L f}{\partial\theta_1}=f_1+f_3\theta_2$$$$ while the right derivative with respect to $$\theta_1$$ is $$$$\frac{\partial_R f}{\partial\theta_1}=f_1-f_3\theta_2$$$$
When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as $$$$\Pi=\frac{\partial_L L}{\partial\dot{\theta}}$$$$ then the canonical Hamiltonian has to be $$$$H=\dot{\theta}\Pi-L$$$$ (you can easily see it's consistent with the definition of momentum with left derivatives) and NOT $$$$H=\Pi\dot{\theta}-L$$$$ (which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.