I'm wonderdering if the time reversal operator $T$ commutes or anticommutes with the time derivative operator. On one hand I think they commute, because $$ T \frac {\text d} {\text d t} f(t) = T \lim_{\Delta \rightarrow 0} \frac {f(t+\Delta) - f(t)}{\Delta} \overset{\text{hope so}}= \lim_{\Delta \rightarrow 0} T \frac {f(t+\Delta) - f(t)}{\Delta} = \lim_{\Delta \rightarrow 0} \frac {f^*(t+\Delta) - f^*(t)}{\Delta^*} \\ \overset{t \in \mathbb R} = \lim_{\Delta \rightarrow 0} \frac {f^*(t+\Delta) - f^*(t)}{\Delta} = \frac {\text d}{\text d t} T f(t) $$ for every function $f(t)$.

On the other hand for every valid wavefunction $f(x,t)$ the Schrodinger Equation $$ i \hbar \frac {\text d}{\text d t} f(x,t) = H f(x,t) $$ holds, which means that I can replace $H$ with $i \hbar \frac {\text d}{\text d t}$ when acting on a valid wavefunction. If we now have a system that has time reversal invariance, we know that the Hamiltonian $H$ commutes with the time reversal operator, which means that $$ i \hbar \frac {\text d}{\text d t} T = H T = TH = T i \hbar \frac {\text d}{\text d t} = -i \hbar T \frac {\text d}{\text d t} $$ which means that $T$ and $\frac {\text d} {\text d t}$ anticommute.

What am I doing wrong? What is the (anti)commutation rule for the time reversal operator and the time derivative?


Your first equation is wrong. Define $g = Tf$. Applying the basic differentiation rule $(u \circ v)'(t) = u'(v(t))v'(t)$ for composed functions to $g(t)=f^*(-t)$, we have $$(Tf)'(t)=g'(t)\overset{\text{rule}}=-(f^*)'(-t)=-(f')^*(-t)=-(T(f'))(t)$$ for all $t$, hence $(Tf)' = -T(f')$.

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  • $\begingroup$ Thanks for your suggestion. But if you define $g:=Tf$ then $g(t)=f^*(t)$ and not $f(-t)$. $\endgroup$ – psicolor Nov 3 '16 at 14:12
  • $\begingroup$ Isn't time inversion changing $t$ in $-t$? See for example T-symmetry. $\endgroup$ – user130529 Nov 3 '16 at 14:58
  • $\begingroup$ No. The operator which in the literature is called "time inversion operator" is just an antiunitary operator under which spatial eigenstates are invariant, i.e. $T | x \rangle = | x \rangle$. If $f(x,t)$ is a wavefunction, then $Tf(x,t) = f^*(x,t)$. Inversion of the temporal argument is neccessary to obtain a "physical" wavefunction, but it's not the job of the time inversion operator. $\endgroup$ – psicolor Nov 3 '16 at 17:07
  • $\begingroup$ Are you sure about the sign of $t$? This reference says $Tf(x,t)=f^{∗}(x,-t)$. I was considering real $f$ in my answer, but the result remains the same for non-real $f$. $\endgroup$ – user130529 Nov 3 '16 at 18:00
  • $\begingroup$ @psicolor I added the conjugate case to my answer to make it correct for all complex valued functions. As you can see, the conclusion remains the same: $(Tf)′=−T(f′)$. $\endgroup$ – user130529 Nov 3 '16 at 19:03

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