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On page 519 of Peskin and Schroeder, the authors have the following discussion on the nilpotent BRST operator $Q$ that commutes with the Hamiltonian $H$.

Many eigenstates of $H$ must be annihilated by $Q$ so that $Q^2=0$ can be satisfied. Let $H_1$ be the subspace of states that are not annihilated by $Q$. Let $H_2$ be the subspace of states of the form $$\tag{16.51}|\psi_2\rangle=Q|\psi_1\rangle$$ where $|\psi_1\rangle$ is in $H_1$.

and later

The subspace $H_2$ is quite peculiar, because any two states in this subspace have zero inner product: $$\tag{16.52} \langle\psi_{2a}|\psi_{2b}\rangle=\langle\psi_{1a}|Q|\psi_{2b}\rangle=0$$

I found the above discussion to be very unconvincing. First if any two states in $H_2$ have zero inner product, then if any state in $H_2$ inner products itself should also give 0, which implies that any state in $H_2$ is zero if we assume positive definite inner products. Furthermore, in (16.52) we seem to be assuming $Q$ is Hermitian, but it is a linear algebra fact that the only nilpotent Hermitian operator is 0.

If we are working over a scalar product space, then this doesn't make sense either, as the Hermitian conjugate in a scalar product space isn't well defined.

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The main point is that in the BRST formulation the sesquilinear inner product $\langle \cdot,\cdot\rangle$ of the extended (graded, complex) Hilbert space ${\cal H}$ is assumed to be of indefinite signature rather than positive definite; Only the physical Hilbert space ${\cal H}_{\rm phys}\cong {\rm Ker}Q/{\rm Im Q}$ is positive definite.

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    $\begingroup$ So is $Q$ defined on the entire space or just the physical subspace? Hermitian conjugate shouldn't really make sense on a not positive definite space. $\endgroup$ Commented Nov 6, 2022 at 19:01
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Nov 6, 2022 at 19:23

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