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For the dynamics of open quantum systems, the Kraus operators $K_\kappa$ can be derived from the unitary orbit $U(t)\rho U(t)^\dagger$ for $\rho=\rho_S\otimes\rho_E$ of the composite system given by the time-propagators $U(t)=\exp(-\mathrm{i} tH)$ via tracing over the environmental states $|\mu\rangle$: $$K_\kappa(t)=K_{\mu\nu}(t):=\langle\mu|U(t)|\nu\rangle$$

I wanted to derive an expression for $K_\kappa$ directly in terms of the interaction Hamiltonian $H_I=\sum_\alpha S_\alpha\otimes E_\alpha$ for the whole Hamiltonian $H=H_S\otimes I_E+I_S\otimes H_E+H_I$ (assuming time-independent $H$). So I wrote $U(t)$ as the exponential $\sum_k (-\mathrm{i} tH)^k/k!$ and employed binomial expansions for the powers of sums of Hamiltonians to obtain

$$K_{\mu\nu}(t)=\sum_{\alpha,k,j\leq k,i\leq j} (-\mathrm{i} t)^k/k!{k\choose j}{j\choose i}\langle\mu|H_S^iI_S^{j-i}S_\alpha^{k-j}\otimes I_E^iH_E^{j-i} E_\alpha^{k-j}|\nu\rangle$$

How can this be simplified to get a clean formula only depending on $H_S$ and $S_\alpha$? It should be possible, as the Kraus operators only act on the reduced Hilbert space alone. In addition, how would one proceed for time-dependent $H(t)$ and their time-ordered products?

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  • $\begingroup$ Perhaps one can break the unitary operator into separate unitaries using Baker-Campbell-Hausdorff and then see if it simplifies the trace calculation. Note that there be nontrivial commutation relations to compute between the interacting parts and the free parts of the Hamiltonian. Did you consider the effect of such commutation relations in the final expression you got? $\endgroup$ – flippiefanus Jun 30 at 4:09
  • $\begingroup$ Ouch, very bad mistake on my side. The binomial expansion is not even possible due to the non-vanishing commutators between free and interaction hamiltonians... $\endgroup$ – quantumorsch Jun 30 at 23:54

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