2
$\begingroup$

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

$\endgroup$

1 Answer 1

8
$\begingroup$

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 $e^B$.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.