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Operators related to physical transformations in quantum mechanics are usually unitary and linear except time-reversal which is both antiunitary and antilinear. What is the explanation for this difference?

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The statement is true if the Hamiltonian operator has spectrum bounded below ( due to thermodynamical reasons) but not above as it happens in many relevant physical systems. Here is the proof.

The time reversal operator $T$ satisfies, by definition, $$T e^{iHt} = e^{-iHt} T\tag{1}$$ for every real $t$ and must be either unitary or anti unitary due to Wigner's theorem. The identity above implies, according to the nature of $T$, $$ e^{\pm iTH T^{-1}t} = e^{-iHt}$$ for all reals $t$. Stone theorem yields $$\mp THT^{-1} = H.\tag{2}$$ If the operator were unitary, we would have $$THT^{-1} = - H.$$ Since unitary transformations preserve the spectrum, the found identity would also imply $$\sigma(H) = -\sigma(H)$$ that is not permitted since the spectrum we are considering is not symmetric under change of sign by hypothesis.

In general, unitary time reversal is possible for Hamiltonian operators with symmetric spectrum. However, since they must have spectrum bounded below, it is possible for Hamiltonian operators with symmetric and bounded spectrum.

ADDENDUM. As remarked by Elio Fabri, since (pure) states are unit vectors $\psi$ up to phases, (1) is a too restrictive condition and it has to be relaxed into $$T e^{iHt}\psi = e^{ic_\psi(t)}e^{-iHt} T\psi \tag{1'}\:.$$ It is not difficult to prove, taking advantage of Stone theorem, that $c_\psi(t)$ does not depend on $\psi$ and that $c(t) = ct$ is the only possibility for the phase $c(t)$. Therefore, taking the derivative of both sides we have $$\mp THT^{-1}= -H + cI\:.$$ If $T$ is unitary and $\sigma(H)$ is bounded below but not above, we have a contradiction $\sigma(H) = -\sigma(H) +c$. The only possibility is that $T$ is antiunitary so that $$\sigma(H) = \sigma(H) - c$$
Since both sides must have finite $\inf$, we conclude that $c=0$ and $$THT^{-1} = H$$ and the time reversal operation satisfies however $$T e^{iHt} = e^{-iHt} T\:.$$ If $T$ is unitary, $\sigma(T)$ must satisfy $$\sigma(H) = -\sigma(H) +c$$ for some $c\in \mathbb R$. So, if $T$ is unitary, $\sigma(H)$ is bounded below if and only if it is bounded above and it is symmetric with respect to some point which is not necessarily $0$.

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    $\begingroup$ Yes, you can relax that way the requirement. Actually a more correct way to state the theorem concerns evolution in the projective space which, when passing to the Hilbert space, becomes exactly the condition you wrote. $\endgroup$ Nov 5, 2018 at 15:28
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    $\begingroup$ However, with this apparently weaker requirement, using Stone's theorem you have $\pm THT^{-1} = -H + cI$. If $\sigma(H)$ is bounded below but not above and $T$ is unitary, we find a contradiction. So $T$ must be antununitary and the written identity is however possible only if $c=0$. $\endgroup$ Nov 5, 2018 at 15:35
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    $\begingroup$ Actually one should also prove that $c(t)$ does not depend form $\psi$. The true requirement is $Te^{iHt}\psi = \chi(t)^{(\psi)} e^{-iHt} T\psi$ for some unit complex number $\chi^{(\psi)}(t)$. The fact that it does not depend on $\psi$ has the same proof as the analogous property in the proof of Wigner theorem, where the phases of the (anti)unitary operators representing symmetries do not depend on the vectors... $\endgroup$ Nov 5, 2018 at 16:24
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    $\begingroup$ Now, $Te^{iHt}\psi = \chi(t)^{(\psi)} e^{-iHt} T\psi$ implies that $\langle (e^{-iHt} T)^{-1}\psi| Te^{iHt} \phi\rangle = \chi(t) \langle \psi|\phi \rangle$. From this identity one sees that $\chi(t)$ is differentiable. $\endgroup$ Nov 5, 2018 at 16:29
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    $\begingroup$ Next, $t \mapsto \chi(t) e^{-iHt} = e^{\pm i THT^{-1}t}$ is evidently a unitary strongly continuous group so that it admits generator by Stone theorem. $\chi(t) e^{-iHt} = e^{-iH't}$. A direct computation, using the fact that $\chi$ is differentiable, proves that $H' = H + cI$ for some constant $c \in \mathbb R$ so that $\chi(t) = e^{ict}$. I hope this help understand the procedure @Elio Fabri. The remark is important, because it is also related to the fact that, in non-relativistic theory, $H$ is however defined up to additive constants and $c$ may enter the picture that way... $\endgroup$ Nov 5, 2018 at 16:33

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