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Why and how can a quantum system (like the quantum vacuum of our universe) governed by the Schroedinger equation break the time-reversal $T$ symmetry (or $CP$ symmetry) microscopically at all?

p.s. the $T$ symmetry (or $CP$ symmetry) breaking is known and verified in neutral Kaons. See also the Wikipedia texts on $CP$ violation and strong $CP$ problems.

The Schroedinger equation governs the quantum system (regardless of relativistic or not) looks: $$ i \frac{\partial}{\partial t} |\Psi{}(t) \rangle = H(t) |\Psi{}(t) \rangle. $$ It is easy to check the Schroedinger equation is time-reversal $T$ invariant if

  1. $H(t)$ has no explicit time-dependent, or time-reversal symmetric, then $ H(-t)=H(t)=H$.
  2. Since we do not want to make $t=0$ becomes a special fixed point, we also require the time translational symmetry $H(t)=H(t+\Delta)$ for some arbitrary time interval $\Delta$.

We can do a time-reversal transformation on the whole equation, $$ T i T^{-1} T\frac{\partial}{\partial t} |\Psi{}(t) \rangle = T H(t) T^{-1} T|\Psi{}(t) \rangle $$ $$ =T i T^{-1} T\lim_{\Delta \to 0}\frac{ |\Psi{}(t+\Delta) \rangle- |\Psi{}(t) \rangle}{\Delta} = H(-t) |\Psi{}(-t) \rangle $$ $$=Ti T^{-1} \lim_{\Delta \to 0}\frac{ |\Psi{}(-t-\Delta) \rangle- |\Psi{}(-t) \rangle}{\Delta} = H(-t) |\Psi{}(-t) \rangle $$ $$ =T i T^{-1} (-)\frac{\partial}{\partial \tilde t} |\Psi{}(\tilde t) \rangle \Big \vert_{\tilde t = -t} = H(-t) |\Psi{}(-t) \rangle. $$

As we already knew, the Schroedinger equation is time-reversal $T$ invariant requiring that $$T i T^{-1} =-i$$

My question is that there seems to have no room to ask any quantum system governed by the Schroedinger equation to break the $T$ symmetry (or $CP$ symmetry) microscopically if we already demand a microscopically time-reversal symmetric Hamiltonian $ H(-t)=H(t)=H$.

Now my puzzle is,

  • our quantum universe does break $T$ symmetry (or $CP$ symmetry) microscopically.
  • our quantum universe is governed by the Schroedinger equation (regardless of relativistic or not). How can such a microscopically time-reversal symmetric Hamiltonian $ H(-t)=H(t)=H$ break the $T$ symmetry?
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  • $\begingroup$ Could you point me to a reference stating what you say in the last question? Namely, "our quantum universe is governed by the Schrödinger equation (regardless of relativistic or not).". I would like to learn how one arrives to such statement in general. I understand its validity in QM because it can be tested, what about the relativistic case and more over the Universe case? Any reference would be very welcome $\endgroup$
    – ohneVal
    Commented May 13, 2021 at 9:14
  • $\begingroup$ I meant to say Schrödinger equation works for both relativistic and nonrelativistic case. BUT not including general relativity! $\endgroup$ Commented May 13, 2021 at 13:50

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For a good treatment of time-reversal symmetry check Sakurai's "Modern Quantum Mechanics" chapter 4.4.

The name "Time reversal symmetry" is a bit misleading. The symmetry is not about "if you reverse the time, your equation of motion will reverse as well". The idea of the symmetry is rather "if you reverse all your spins and all momentums of the particles, the system will move in reverse". It is not difficult to violate this property, just consider a charged particle moving in magnetic field. If you flip the momentum of the particle, the particle will not follow it's initial trajectory backwards.

One constructs the time reversal operator by requiring a certain action onto the momentum, position and spin operator, as well as Hamiltonian. Because the operator is essentially "motion reversal", one defines the operator by the following requirements: $$ \begin{array}{c} T x T ^{-1} = x\\ T p T ^{-1} = -p\\ T S T ^{-1} = -S \end{array} $$ Together with antiunitarity these relationship fix the form of the "time-reversal operator". A system is considered time-reversal if $iHT=-TiH$ (which is the same as $TH=HT$). Once again, particles in magnetic field do not have this property.

Edit. When applied to a particle in magnetic field, the transformation $THT^{-1}$ changes $H=(p-iA)^2$ to $THT^{-1}=(p+iA)^2$, which is different from $H$, which means $TH\neq HT$, which implies $T|\psi(t)\rangle\neq|\psi(-t)\rangle$. The last equation was implicitly assumed in the derivation in the OP.

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  • $\begingroup$ Thanks, I am asking something different. I am familiar with Sakurai's "Modern Quantum Mechanics" chapter 4.4.. I meant that how can a quantum system like our standard model breaks $T$ if we demand a quantum Hamiltonian description such that this Hamiltonian is microscopically time reversal invariant $H(t)=H(-t)$. I suspect that the answer has something to do with the complex part of Hamiltonian. But it seems a bit weird because Hamiltonian $H$ is hermitian. $\endgroup$ Commented May 12, 2021 at 13:40
  • $\begingroup$ Are you asking which term in the Standard model specifically breaks T symmetry, or are you asking a general question? If the general question, in your derivation you assumed $T|\psi(t)\rangle=|\psi(-t)\rangle$, but it is not true in general case. It is only a property of systems with time-reversal symmetry. The time-reversal operator does not reverse the time, it reverses momentums and spins. $\endgroup$
    – Pavlo. B.
    Commented May 12, 2021 at 14:23
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    $\begingroup$ If your question was specifically about the Standard Model, I am not a specialist, but you can check CP symmetry violation: en.wikipedia.org/wiki/CP_violation . Since all QFT models are CPT symmetric, violation of CP symmetry implies also violation of T symmetry. $\endgroup$
    – Pavlo. B.
    Commented May 12, 2021 at 14:25
  • $\begingroup$ Not all QFT models, only QFT models with Lorentz symmetry, just in case, en.wikipedia.org/wiki/CPT_symmetry. Moreover with the requisite of being local and with a Hermitian hamiltonian, so the argument becomes a bit circular. $\endgroup$
    – ohneVal
    Commented May 13, 2021 at 9:08

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