Does the following limit exist in the BRST formalism?

Question

Consider the BRST operator $\Omega$ (which has ghost number $+1$) and the gauge fermion operator $\rho$ which has ghost number $-1$. Given an exact state $|\Phi\rangle$ (i.e. $|\Phi\rangle=\Omega|\Psi\rangle$) is it reasonable to suppose that $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]_+}|\Phi\rangle=0\tag{1} $$ I feel this should be true since

1. $e^{-t[\Omega,\rho]_+}$ is damping for large $t$.
2. $|\Phi\rangle$ is in the universality class of $0$. (two states are equivalent if related by an exact state.)
3. $e^{-t[\Omega,\rho]_+}|\Phi\rangle=|\Phi\rangle+\text{an exact state}$ (this follows from expanding the exponential and using $\Omega^2=0$ and $\Omega|\Phi\rangle=0$), which means $e^{-t[\Omega,\rho]_+}$ moves $|\Phi\rangle$ around within its universality class, cf. this related Phys.SE post.

Answer 1

1. First of all, note that $H:=[\Omega,\rho]_{+}=H^{\dagger}$ is self-adjoint, because $\Omega=\Omega^{\dagger}$ and $\rho=\rho^{\dagger}$ are.

2. For $e^{-tH}\to 0$ for $t\to\infty$ to hold we need furthermore that $H$ is a semi-positive definite operator. It is not clear that that is true in general. It would imply $$\lim_{t\to\infty}e^{-tH}|\Psi\rangle=|\Psi\rangle.$$

3. It is not clear why OP's eq. (1) should hold in general for BRST-exact states.

Answer 2

If we choose $\rho = {}^*\Omega$ (the co-BRST operator), then the answer is yes! This is because $[\Omega,{}^*\Omega]_+$ is the BRST Laplacian which is a semi-positive definite (self-adjoint) operator.