This is related to a previous question of mine.

We consider a density matrix $\sigma(t)$ operating on a Hilbert space $\mathscr{H}_{s}\otimes \mathscr{H}_b$ with Hamiltonian $H = H_s \otimes \mathbb{I}_B + \mathbb{I}_s \otimes H_b + g H_{int}$. We are interested in the reduced density matrix $\rho(t) := \mathrm{Tr}_{B}[ \sigma(t) ]$.

In the Born Approximation, the evolution of the reduced density matrix is described (in the interaction-picture) by $$ \frac{d\rho_I(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_b\bigg( \big[ V(t), [V(s), \rho_I(s) \otimes \varrho_{B}] \big] \bigg) $$ where $\rho_{I}(t) = e^{+ i H_s t} \rho(t) e^{- i H_s t}$, and $V(t)$ is just $H_{int}$ in the interaction-picture, and $\varrho_B$ is the initial state of the reservoir (where $\sigma(0) = \rho(0) \otimes \varrho_B$).

Suppose for concreteness that $\mathscr{H}_s$ is finite-dimensional, and the interaction is of the form $$ V(t) = \sum_{jk} A_{j} \otimes B_{k} $$ Then the above equation of motion takes the form $$ \frac{d\rho_I(t)}{dt} \simeq g^2 \sum_{jk\ell m} \int_0^t ds\ \bigg( \big[ A_{\ell}(s) \rho_I(s), A_{j}(t) \big] \langle B_{k}(t) B_{m}(s) \rangle + \big[ A_{j}(t), \rho_I(s) A_{\ell}(s) \big] \langle B_{m}(s) B_{k}(t) \rangle \bigg) $$ where we define the correlator $\langle B_{k}(t) B_{m}(s) \rangle := \mathrm{Tr}_B\left[ B_{k}(t) B_{m}(s) \varrho_B \right]$.

My Question: Suppose $\sigma(0)$ is trace 1, Hermitian and positive. This means that $\varrho(0)$ is also trace 1, Hermitian and positive. Are these properties preserved for $\varrho(t)$ with $t>0$?

I am not sure. Here are my thoughts:

  1. Preservation of Trace: taking the trace of the above evolution equation yields $\mathrm{Tr}[\dot{\varrho}_I(t)]=0$ because the equation contains commutators of finite-dimensional matrices. As far as I can tell this is only true because of my particular choice of $H_{int}$ which is a tensor product.

  2. Preservation of Hermicity: I would like to show explicitly that $\frac{d}{dt}\big[ \varrho(t)^\dagger - \varrho(t) \big]=0$ for the state evolved by the above approximate equation.

  3. Preservation of Positivity: I don't know if this is true at all, or how to show it.

In my previous post, it was pointed out that $\sigma(t)$ is trace 1, Hermitian and positive. Taking the partial trace of a positive operator yields a positive operator, which also implies that the reduced density matrix is Hermitian (since bounded positive operators are self-adjoint). I may be confused, but I think this is not an answer to my question: the evolved state $\rho(t)$ from the above approximate equation of motion does not contain the full information about $\sigma(t)$. I would like to know if these properties hold true for the $\rho(t)$ predicted by the above approximation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.