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My question is basically this:

Does a maximally entangled state stay maximally entangled under the time evolution?

Assume our Hilbert space is $\mathcal{H}_A \otimes \mathcal{H}_B$. At the time $t=0$, the states is $\rho_{AB}$ such that $\text{Tr}_A \rho_{AB}=\frac{1}{n_A}id_B$ and $\text{Tr}_B \rho_{AB}=\frac{1}{n_B}id_A$. Now assume at the time $t$, the density matrix is $\rho_{AB}(t)$. Is it right that $\rho_{AB}(t)$ is a maximally entangled state? (If not, under what assumptions one can guarantee this?)

TEMP: Is it right?

$\text{Tr}_A \rho_{AB}(t)=\sum_A \left<e^A_i(t),\rho_{AB}(t)e^A_i(t)\right>=\left<U(t)e^A_i(0),U(t)\rho_{AB}(0)U^{-1}(t)U(t)e^A_i(0)\right>=\left<e^A_i(0),U^{-1}(t)U(t)\rho_{AB}(0)e^A_i(0)\right>=\left<e^A_i(0),\rho_{AB}(0)e^A_i(0)\right>=\text{Tr}_A \rho_{AB}(0)=\frac{1}{n_A}id_B$

So Maximally entangled states remains maximally entangled.


marked as duplicate by Norbert Schuch, Kyle Kanos, ZeroTheHero, John Rennie, Jon Custer May 13 '18 at 16:07

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  • $\begingroup$ Surely there are evolutions that map entangled states in non-entangled states (the easy example is the following: take a non-entangled state, and evolve it with an interacting hamiltonian; the resulting state is entangled for almost all times $t$, this one evolved back with $-t$ gives you a non-entangled state). Therefore my guess is that also maximal entanglement is not in general preserved by time evolution $\endgroup$ – yuggib May 10 '18 at 16:37
  • $\begingroup$ @yuggib I got your intuition and it makes sense but what about "maximally" entangled state? $\endgroup$ – mathvc_ May 10 '18 at 17:13

General time evolution can generate a general, i.e., arbitrary, unitary transformation. This can map any quantum state to any other state. Thus, no state is special, including maximally entangled ones.

The situation is different if the Hamiltonian acts on the two parts separately. Then it creates a unitary evolution $U_A\otimes U_B$, which cannot change the entanglement.


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