Physics Asked by Fibonacci M on February 22, 2021
Thanks for looking into this question and for your patience since I am just learning all of this stuff. I have a bi-partite state (density matrices with trace 1 and positive):
$$
rho_{AB} = rho_A otimes rho_B,
$$
in the presence of an interacting Hamiltonian:
$$ mathcal{H}= hat{A} otimes hat{B},$$
where $hat{A}, rho_{A} in mathcal{B(H_A)}$ as well as $hat{B}, rho_{B} in mathcal{B(H_B)}$ where $mathcal{B(H)}$ denotes bounded operators on the Hilbert space $mathcal{H}$ . $rho_B$ may or may not be a mixed state of the form $rho_B = sum_i a_i |i_B rangle langle i_B| $. The question is if the time evolution of $rho_{AB}$ is of the following form (and how can I prove this?) :
$$rho_{AB}(t) = e^{it hat{A}} rho_A e^{-it hat{A}} , , otimes e^{it hat{B}} rho_B , e^{-it hat{B}} $$
Is this in general true? Do things change if the density matrix is now tripartite ie. $rho_{ABC} = rho_A otimes rho_B otimes rho_C$.
No, this is wrong.
$e^{Aotimes B}$ generally cannot be expressed in terms of $e^{A}$ and $e^B$, such as as $e^Aotimes e^B$. Just try a random example.
One way to make it plausible why is to note that the Taylor series of $e^{Aotimes B}$ only contains terms $A^notimes B^n$, which you cannot easily obtain from $e^A$ and $e^B$.
Answered by Norbert Schuch on February 22, 2021
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