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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 $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]}|\Phi\rangle=0\tag{1} $$ (where $[,]$ is the graded commutator, which in this case is an anti-commutator)? 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.

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.

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 $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]}|\Phi\rangle=0 $$ (where $[,]$ is the graded commutator, which in this case is an anti-commutator)? 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.

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 $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]}|\Phi\rangle=0\tag{1} $$ (where $[,]$ is the graded commutator, which in this case is an anti-commutator)? 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.

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. $\Omega|\Phi\rangle=0$) is it reasonable to suppose that $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]}|\Phi\rangle=|\Phi\rangle $$ (where $[,]$ is the graded commutator, which in this case is an anti-commutator)? 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.

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 $$ \lim_{t\to\infty}e^{-t[\Omega,\rho]}|\Phi\rangle=0 $$ (where $[,]$ is the graded commutator, which in this case is an anti-commutator)? 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.