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I understand that the time reversal operator, $\Theta$, acting on a state has the effect $t\rightarrow -t$ (and also takes complex conjugate). In the Heisenberg representation the states kets $|x,t_0\rangle_H$ are independent of time, while the base kets evolve backwards in time $|x,t\rangle_H=e^{\frac{iH(t-t_0)}{\hbar}}|x,t_0\rangle_H$.

I am wondering what the effect of the time reversal operator on the base ket in the Heisenberg representation is.

I suspect it is something like

\begin{equation} \begin{aligned} \Theta|x,t\rangle_H&=e^{\frac{(-i)H(-t)}{\hbar}}\Theta|x,0\rangle_H\\ &=e^{\frac{iHt}{\hbar}}|x,0\rangle_H\\ &=|x,t\rangle_H \end{aligned} \end{equation}

Hence the time reversal operator has no effect on the Heisenberg picture states because they are independent on time. Is this true?

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  • $\begingroup$ Something is unclear to me - in the Heisenberg picture, the states are completely time independent. Therefore, $|\psi_H(t)\rangle$ doesn't make any sense. Indeed, $|\psi_{H}(t)\rangle = |\psi_{H}(0)\rangle = |\psi_{S}(0)\rangle$. Where does the time-dependent phase comes from? $\endgroup$
    – user245141
    Nov 3, 2019 at 8:44
  • $\begingroup$ I had made a mistake in the previous comment, but the site was locking me out from editing the comment (then I deleted it which was probably bad). You are correct that for the state kets $|\phi_H(t)\rangle=|\phi_H(0)\rangle=|\phi_S(0)\rangle$ since they are time independent. I am not sure how to represent the base kets in this different notation anymore $\endgroup$
    – Adam
    Nov 3, 2019 at 13:44
  • $\begingroup$ The original comment that Yu-v is refering to is: The notation might be ambiguous and is in the likeness of this question. A different notation would have the Heisenberg state kets as $|\psi_H(t)⟩=|\psi_s(0)⟩$, and the base kets as $|\psi_H(t)⟩=e^{iHt/ℏ}|\psi_s(0)⟩$. $\endgroup$
    – Adam
    Nov 3, 2019 at 13:59

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