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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.

mathvc_
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1 Answers1

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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.

Norbert Schuch
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