# Time evolution of density matrix for an interacting bipartite Hamiltonian

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. $$\rho_{ABC} = \rho_A \otimes \rho_B \otimes \rho_C$$.

$$e^{A\otimes B}$$ generally cannot be expressed in terms of $$e^{A}$$ and $$e^B$$, such as as $$e^A\otimes e^B$$. Just try a random example.
One way to make it plausible why is to note that the Taylor series of $$e^{A\otimes B}$$ only contains terms $$A^n\otimes B^n$$, which you cannot easily obtain from $$e^A$$ and $$e^B$$.
• Then what does $\rho_{AB}(t)$ look like? How do I simplify $e^{it \mathcal{H} }\rho_{A} \otimes \rho_{B} \, e^{-it \mathcal{H}}$? Feb 12, 2021 at 8:24