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I am studying entanglement and its measurements in the context of a lattice model of the Dirac theory. The idea is that one has two bands, symmetric with respect to $E=0$, and the groundstate is obtained by filling the lower band, $$|GS\rangle=\prod_k\hat{\gamma}^\dagger_{k,-}|0\rangle,$$ where $\hat\gamma^\dagger_{k,\pm}$ annihilates a particle with momentum $k$ in the lower band. Later, a particle-hole transformation can be applied to interpret the filled lower band as a vacuum of antiparticles, $\hat{\gamma}_{k,+}\rightarrow\hat{b}_k$ (particles), $\hat{\gamma}_{k,-}\rightarrow\hat{d}_{-k}^\dagger$ (antiparticles), recovering thus the typical picture from QED.

If one chooses a partition and calculates the Entanglement Entropy of it, one obtains different results depending on the mass of the field (e.g. near the critical point, for $m\approx0$, one recovers predictions of logarithmic scaling from the underlying CFT, while for $m\gg 1$ a strict area law is satisfied).

However, I am now interested in excited states, i.e., states of the form $$|\psi\rangle\propto\prod_kf(k)\hat\gamma_{k,+}^\dagger\hat{\gamma}_{k,-}|GS\rangle,$$ with $f(k)$ a certain distribution of excitations in momentum space. After the particle-hole transformation, this state is understood as containing pairs of particles-antiparticles with opposite momentum. For example, if $f(k)=\delta_{k,q}$, one obtains a state with only one pair of particle-antiparticle with momentum $q$. This state is still an eigenstate of the Hamiltonian, $$\hat{H}=\sum_k\omega_k\left(\hat{\gamma}_{k,+}^\dagger\hat{\gamma}_{k,+}-\hat{\gamma}_{k,-}^\dagger\hat{\gamma}_{k,-}\right),$$ with energy $2\omega_q$, and therefore is an eigenstate of the evolution operator.

Here is where my question appears: the time-evolution of this state seems trivial, since the state is an eigenstate. Therefore, the associated covariance matrix (which is used in lattice models to calculate the Entanglement Entropy) has a trivial evolution in time, and therefore the Entanglement Entropy does not evolve in time. This goes against my intuition, which tells me that, since this state contains pairs of particles-antiparticles with opposite momentum, the entanglement between a partition and its complementary partition should grow as time evolves, in a way somehow related to the group velocity of these particles. Why isn't this the case?

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  • $\begingroup$ I mean you have kind of given the answer in your question: it's an eigenstate, nothing happens. So to help us answer we need to know why you expect a different answer. With that in mind, how exactly are you partitioning this space? In real space? In momentum? $\endgroup$ Commented Jul 1, 2023 at 10:36
  • $\begingroup$ I understand that, and see that mathematically things such as the correlators remain the same because of that a reason. However, thinking about it physically, and thinking within the particle-antiparticle picture, an excited state should correspond to a situation of a pair with oposite momentum, and therefore would expect the entanglement entropy between two spatial partitions (e.g. right and left halves in a lattice discretization) to evolve somehow, in a way related to the well-defined group velocity of the particles… $\endgroup$
    – TopoLynch
    Commented Jul 1, 2023 at 15:33
  • $\begingroup$ @BySymmetry May it be because the particle content, well defined in momentum-space, is delocalized in position space, so rather than inducing a time evolution of the entanglement entropy due to propagation of pairs of particle-antiparticle, these pairs are defined along all the chain and its neat effect is simply to increment the Entanglement Entropy in a time-independent way? If it is so, I have still some trouble understanding in this case what the associated group velocity would mean here... $\endgroup$
    – TopoLynch
    Commented Jul 3, 2023 at 15:24
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    $\begingroup$ These excited states are standing waves: they have no particle motion, no dynamics, no entanglement entropy growth. If you instead considered an initial state with a pair of localized wavepackets, then the excitations would indeed propagate and you would see the entanglement grow in time. $\endgroup$
    – mbintz
    Commented Jul 4, 2023 at 16:40

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