# Peskin and Schroeder's discussion of the BRST operator

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.

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.
• 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. Commented Nov 6, 2022 at 19:01