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Whilst there is a lot of information on time-reversal in relation to the Schroedinger equation, I am a bit unsure about its effects on the Heisenberg equation of motion, mostly because I'm not sure how the time-reversal operator works exactly.

I think I am happy with $TpT^{-1}=-p$ and $TxT^{-1}=x$, which leads to $TaT^{-1}=a^\dagger$, if $a$ is the annihilation operator for a harmonic oscillators with coordinates $x,p$. Going to Heisenberg's equation of motion, I think it should be

$$T\dot{a}T^{-1}=\dot{a}^\dagger=Ti[H,a]T^{-1}=i\omega a^\dagger. (?)$$

Is the last equality correct? If so, how would I derive it? I'm aware that presumably

$$ Ti[H,a]T^{-1}=T(-i\omega a)T^{-1}=i\omega TaT^{-1}=i\omega a^\dagger,$$

but on the other hand

$$ Ti[H,a]T^{-1}=-iT(Ha-aH)T^{-1}=-i(Ha^\dagger-a^\dagger H)=-i[H,a^\dagger],$$

which does not agree with the previous result. Does $T$ not commute with $H$, even if $H$ is time-reversal invariant?

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  • $\begingroup$ You have $a = \sqrt{\omega/2\hbar}\left(q +(i/\omega) p \right) $ and $TaT^{-1} = \sqrt{\omega/2\hbar}\left( TqT^{-1} - (i/\omega) TpT^{-1} \right] = \sqrt{\omega/2\hbar}\left(q - (i/\omega) (-p) \right] = a$, not $TaT^{-1} = a^\dagger$. Similarly for $a^\dagger$. $\endgroup$
    – udrv
    Commented Jan 17, 2017 at 17:09
  • $\begingroup$ In case anyone else runs into this problem, there are more mistakes above, although I haven't carefully checked it. Probably $T\dot a T^{-1}=-\dot a$ as well. $\endgroup$
    – Daniel
    Commented Feb 3, 2017 at 10:43

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