In many texts in quantum field theory or string theory, it is stated that the BRST charge $Q$ must annihilate physical states because the states are required to be BRST invariant. Since $Q$ generates BRST transformations, then $Q|\psi\rangle = 0$ for any candidate for a physical state. This is before discussion of the BRST cohomology.

However, from ordinary QM, states that are invariant under symmetries are not annihilated by the generators, but are eigenstates of the generators. For example, the state $\psi(x)= e^{ipx}$ is invariant under spatial translations, which are generated by momentum operator $p$. It can be seen from this that $\psi(x)$ is an eigenstate of momentum operator $p$.

So, why do we not demand that physical states in BRST quantisation are instead eigenstates of $Q$, instead of being annihilated by it? I am aware of the derivation involving the commutator of $Q$ with the variation in gauge-fixing which allows you to prove it, but I am looking for a more intuitive interpretation of the result. I would think that since $Q$ generates BRST transformations, a BRST-invariant state would not be affected when $Q$ acts on it.

  • $\begingroup$ May be related to an old question of mine: physics.stackexchange.com/q/719972 (I say may be because I have just skimmed the OPs post, I didn't read it well enough) $\endgroup$
    – schris38
    Commented Nov 20, 2023 at 8:29

1 Answer 1


For starters, the BRST charge operator is Grassmann-odd, so an eigenvalue would be Grassmann-odd as well, which is unphysical.

  • $\begingroup$ Is there any intuition on why we demand that $Q|\psi\rangle$ must be zero compared to the QM case? Is the reason only because $Q$ is Grassman-odd? Or perhaps, if there were a Grassmann-odd generator in QM, would there be a corresponding symmetry-invariant state that gets annihilated by this generator? $\endgroup$ Commented Nov 20, 2023 at 8:38
  • $\begingroup$ The Gauss law (say, in lattice QCD, or any Hamiltonian formulation of gauge theories) is written as $G|\psi\rangle=0$, and here $G$ is bosonic, so this isn't really the reason, right? $\endgroup$ Commented Nov 20, 2023 at 19:43

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