# Anticommutation relations for fermionic fields imply that Hamiltonian / Lagrangian can at most be linear?

Fermionic field operators do obey anticommutation relations, so for a chosen Field operator (and the field momentum), we have: $$\{\Psi_a, \Psi_b\} = \{\pi_a, \pi_b\}= 0$$ with the $$\Psi_a$$ being different field components, and the $$\pi_a$$ being the respective field momentum. $$\{\cdot ,\cdot\}$$ denotes the anticommutator.

Since these anticommutation relations imply that $$\Psi_a \Psi_a$$ is zero, does this mean that any hamiltonian or lagrangian I write down can at most be linear in fields / momenta?

EDIT: To be more clear about what I mean with "Linearity". Let's say the field has two independent components, $$\Psi_1$$ and $$\Psi_2$$. I can imagine that there will be terms like $$\Psi_1 \Psi_2$$, but any term like $$\Psi_1 \Psi_2 \Psi_1$$ would also be forbidden, because I can swap it into a configuration like "$$(-1)^n \Psi_1 \Psi_1 \Psi_2$$". Is that right?

• No, if there are more than one Grassmann-variable (which is often the case), the Lagrangian/Hamiltonian can contain non-linear higher-order terms. With $$n$$ Grassmann-variables, one can build an $$n$$th-order polyonomial.
• $\uparrow$ Yes. – Qmechanic Sep 28 at 12:28
• $\uparrow$ Right. – Qmechanic Sep 29 at 10:41