Is there a canonical way to describe an open, non-relativistic quantum system with density matrix $\rho(t)$ entirely in terms of the light that it emits and absorbs (and vice versa?) Or is it possible in general for a density matrix trajectory $\rho(t)$ to be induced by several (e.g. possibly contrived and time dependent) photon baths?


1 Answer 1


For Markovian systems, this is possible in a certain sense. If the quantum system is linearly coupled to the bath (which is usually the case, e.g. in light-matter interaction, in cavity-bath interaction etc.), one can write an input-output relation

$$\hat{b}_\mathrm{out}(t) = \hat{b}_\mathrm{in}(t) + i\kappa\hat{a}(t)$$

where $b$ are the bath operators, $a$ is the system operator and $\kappa$ the coupling between them.

This equation implies that if you are able to measure/specify the correlation function of the input and output operators, you are able to reconstruct the corresponding correlation function of the system operator.

  • $\begingroup$ In this context, would the subscripts 'in' and 'out' refer to the bath operator before and after the system is introduced (e.g. as a perturbation), or should they be interpreted differently? In the terms you have introduced, my question might be rephrased as, "if $\hat b$ represents the time evolution of a photon field, then is the operator $\hat a$ well defined over a wide range of additional perturbations $ik_\gamma \hat a_\gamma(t)$, or is the operator $\hat a$ always dependent on other perturbations?" $\endgroup$
    – TLDR
    Nov 16, 2020 at 1:05
  • $\begingroup$ @TLDR For the physical meaning of the in- and out-operator check out my answer here: physics.stackexchange.com/a/534741/101770 They are essentially asymptotic degrees of freedom in the sense of scattering. So you can think of it as the radiation you would see "far away" from the system. $\endgroup$ Nov 16, 2020 at 10:31
  • $\begingroup$ With regards to your second question: I am not 100% sure what $ik_\gamma \hat{a}_\gamma(t)$ signifies, but will attempt to answer anyway. My statement in the answer was that you can in principle reconstruct correlation functions of $\hat{a}(t)$ from the in- and out-operator correlation functions. If instead you want to calculate $\hat{a}(t)$ by solving the systems dynamics, you have to solve its equations of motions. Indeed, other perturbations become relevant then. However, the statement I made is general, at least for Markovian systems. $\endgroup$ Nov 16, 2020 at 10:34
  • $\begingroup$ So should $\hat b(t)$ and $\hat a(t)$ be understood as Heisenberg picture operators, and $t$ as approaching infinity? My second question was whether the operator $\hat a$ is dependent on other perturbations that might act on the system ($ik_\gamma \hat a_\gamma(t)$ represents a sum in Einstein notation.) For instance, with (asymptotic?) system operators $\hat a(t)$ and $\hat c(t)$, would the 'output' operator in which both interactions are present be expressed as $\hat b_{\text{in}}(t) + i\kappa \hat a(t) + i\sigma \hat c(t)$, or something else that accounts for interactions? $\endgroup$
    – TLDR
    Nov 16, 2020 at 22:23
  • $\begingroup$ @TLDR The precise equations of motion are fixed by the Hamiltonian. I would suggest that you include a few example Hamiltonians in your questions to clarify which type of couplings/perturbations exactly you are thinking about. In the above comment, it is not clear to me what you mean. $\endgroup$ Nov 17, 2020 at 10:20

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