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


1 Answer 1


No, this is wrong.

$e^{A\otimes B}$ 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 $e^B$.

  • $\begingroup$ Then what does $\rho_{AB}(t)$ look like? How do I simplify $e^{it \mathcal{H} }\rho_{A} \otimes \rho_{B} \, e^{-it \mathcal{H}}$? $\endgroup$ Feb 12, 2021 at 8:24
  • $\begingroup$ Why do you want to simplify it? Why should this be possible? $\endgroup$ Feb 12, 2021 at 9:44
  • $\begingroup$ Hi! Thanks for all your help, I was hoping to simplify this for the case of two coupled quantum harmonic oscillators with an interaction term. Will solving the von-Neumann equation be possible? Also is there a nice reference where this kind of stuff including tensor products are discussed? $\endgroup$ Feb 13, 2021 at 16:01

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