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Qmechanic ♦
Qmechanic ♦
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  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.

  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.

  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.

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Qmechanic ♦
Qmechanic ♦
  • 213.1k
  • 48
  • 590
  • 2.3k

  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.