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?


1 Answer 1

  • 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.

  • In field theory (where the number of DOF is infinite), one can in principle construct higher-order terms using spacetime derivatives of the Grassmann-fields.

  • $\begingroup$ Do you mean for example different components of the fields? $\endgroup$ Sep 28, 2019 at 12:27
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Sep 28, 2019 at 12:28
  • $\begingroup$ I made a little edit, to get rid off any eventual misconceptions. If you could account for this in your answer, I will accept it. $\endgroup$ Sep 28, 2019 at 13:16
  • $\begingroup$ An n-order term in which every of the Grassmann variables appears just one time? $\endgroup$ Sep 29, 2019 at 10:38
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Sep 29, 2019 at 10:41

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