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How does the time reversal symmetry work in the second quantization frame of non-relativistic quantum mechanics? In particular what is the time-reversed of a given Fock-state?

As an example let's consider a system of 2 bosons that can be in two different positions, tagged by L and R respectively. The generic state of such a system can be written as $$ |\psi\rangle = a|2_{L},0_{R}\rangle+b|1_{L},1_{R}\rangle+c|0_{L},2_{R}\rangle $$ where $|n_{L},m_{R}\rangle$ is the Fock state with n bosons in the site L and m in the site R.

If we call the time-reversal transformation $T$, what is the explicit form of $T|\psi\rangle$?

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  • $\begingroup$ References to textbooks that face this topic are very welcome. $\endgroup$
    – Giovanni Fiore
    Commented Jan 23, 2017 at 16:31
  • $\begingroup$ Fock state is a stationary state, which means it doesn't depend on time, so time-reversal will do nothing to it, maybe you wanna write the time evolution of a Fock state then you can think about time-reversal :). $\endgroup$
    – Ismasou
    Commented Jan 23, 2017 at 17:28
  • $\begingroup$ I know that time reversal operation is usually written as a complex conjugation times a unitary transformation depending on the basis you chose for you representation, so that you may apply it to whatever state, stationary or time-evolving. For example it should be applicable to the state I wrote in the question. Am I wrong? $\endgroup$
    – Giovanni Fiore
    Commented Jan 24, 2017 at 10:04
  • $\begingroup$ Time reversal is defined as taking t into -t, you don't have a time parameter here to take to -t. $\endgroup$
    – Ismasou
    Commented Jan 24, 2017 at 17:16

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In case the answer in the comments feels unsatisfactory ("you don't have a time-parameter here to take to $-t$"), there is a related question here that gives another starting point.

Let $T$ be the anti-unitary time-reversal operator, and look at the definition of a creation operators in terms of the canonical position and momentum of the respective particle (assuming unit mass): $$ a^\dagger = \sqrt{\frac{\omega}{2\hbar}}\left( q - \frac{i}{\omega} p \right) $$ The canonical position is unaffected by time-reversal, but the momentum changes sign, $$ T q T^{-1} = q \;, \;\;\; T p T^{-1} = -p $$ So we have right away $$ T a^\dagger T^{-1} = T \sqrt{\frac{\omega}{2\hbar}}\left( q - \frac{i}{\omega} p \right) T^{-1} = \sqrt{\frac{\omega}{2\hbar}}\left( T q T^{-1} + \frac{i}{\omega} T p T^{-1} \right) = \sqrt{\frac{\omega}{2\hbar}}\left( q - \frac{i}{\omega} p \right) = a^\dagger $$ Since the Fock states are built by repeated applications of $a^\dagger$ onto a vacuum state invariant under time-reversal, it follows that they are also invariant under time-reversal.

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  • $\begingroup$ let's suppose I want to know how the creation operator in the eigenstate of the momentum $\vec{k}$ transforms by $T$. Following this line of reasoning that uses fictious momenta and positions, it should be $$ T a^{\dagger}_{\vec{k}} T^{-1}=a^{\dagger}_{\vec{k}} $$ while it should be $$ T a^{\dagger}_{\vec{k}} T^{-1}=a^{\dagger}_{-\vec{k}} $$ because the momentum of a particle is mirrored by time-inversion. $\endgroup$
    – Giovanni Fiore
    Commented Jan 25, 2017 at 20:24
  • $\begingroup$ For normal modes defined under periodic boundary conditions and a spectrum symmetric in $\vec{k}$, $\omega(\vec{k}) = \omega(-\vec{k}) = \omega_k$, the canonical coordinates transform as $$Tq_{\vec{k}}T^{-1} \rightarrow q_{-\vec{k}}\;, \;\;\; Tp_{-\vec{k}}T^{-1} \rightarrow -p_{\vec{k}}$$ while $a_{\vec{k}}^\dagger$ is defined as $$a_{\vec{k}}^\dagger = \sqrt{\frac{\omega_k}{2\hbar}}\left( q_{\vec{k}} - \frac{i}{\omega_k} p_{-\vec{k}} \right)$$ $\endgroup$
    – udrv
    Commented Jan 25, 2017 at 23:23
  • $\begingroup$ This gives indeed $$Ta_{\vec{k}}^\dagger T^{-1} = \sqrt{\frac{\omega_k}{2\hbar}}\left( Tq_{\vec{k}}T^{-1} + \frac{i}{\omega_k} Tp_{-\vec{k}}T^{-1} \right) = \sqrt{\frac{\omega_k}{2\hbar}}\left( q_{-\vec{k}} - \frac{i}{\omega_k} p_{\vec{k}} \right) = a_{-\vec{k}}^\dagger $$ $\endgroup$
    – udrv
    Commented Jan 25, 2017 at 23:23
  • $\begingroup$ As a general rule, whenever possible start with canonical coordinates defined in terms of particle coordinates, $\vec{x}_i$ and $\vec{p}_i$, which always transform as $T \vec{x}_i T^{-1} \rightarrow \vec{x}_i$ and $T \vec{p}_i T^{-1} \rightarrow -\vec{p}_i$. The transformations for the $q_{\vec{k}}$, $p_{\vec{k}}$, and $a_{\vec{k}}^\dagger$ follow immediately. $\endgroup$
    – udrv
    Commented Jan 25, 2017 at 23:32
  • $\begingroup$ Could you please give me a reference (book, articles or whatever) for this method? I find it difficult to interpret time-reversal of the fictious positions and momenta, what you call canonical coordinates. $\endgroup$
    – Giovanni Fiore
    Commented Jan 26, 2017 at 9:22

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