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.

  • $\begingroup$ 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
    Commented May 10, 2018 at 16:37
  • $\begingroup$ @yuggib I got your intuition and it makes sense but what about "maximally" entangled state? $\endgroup$
    – mathvc_
    Commented May 10, 2018 at 17:13

1 Answer 1


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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