# Tag Info

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The idea is that $$\frac{d}{dt}\rho=\mathcal{L}\rho\qquad \Leftrightarrow\qquad \rho(t)=\exp(t\mathcal{L})\rho(0).$$ In that sense, the Lindbladian $\mathcal{L}$ generates evolution through $$\rho(dt)\approx \rho(0)+dt \mathcal{L}\rho(0),$$ which is the same sense as an infinitesmal generator of a Lie group. You might be familiar with something like a ...

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I think this question is generally difficult because there is no standard metric for non-Markovianity, for example this paper would suggest you try to express the evolution in a time-local canonical (Lindblad) form and then look at the negativity of the rates, but other metrics may not agree for certain channels. Perhaps it is easier to answer the simpler ...

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The given form of the Kraus operators, though not unique, tells us what is physically happening. In the case of pure-state outputs, each Kraus operator takes one of the possible basis states and swaps it with the desired output state. All basis states thus become $|0\rangle$ through some sort of swapping mechanism. In the case of mixed-state outputs, each ...

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Go back to the Lindblad master equation: $$\frac{d\rho}{dt}=i[H,\rho]+\sum_nL_n^\dagger\rho L_n-\frac12\sum_nL_nL_n^\dagger \rho-\frac12\sum_n\rho L_nL_n^\dagger.$$ The statement that the maximally mixed state is a fixed point is equivalent to saying that if $\rho=I/d$ then $\frac{d\rho}{dt}=0$. So, that's check that. \begin{align*} d\frac{d\rho}{dt}&=...

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I'll show you how to do it by brute force, since this will demonstrate a lot of techniques that will be useful for you if you have to derive something more complicated. The Lindblad evolution: $$\frac{\textrm{d}\rho}{\textrm{d}t} = -\frac{\textrm{i}}{\hbar}[H,\rho] + \sum_{\mu,\nu} h_{\mu,\nu} \left( L_\mu \rho L_\nu^\dagger -\frac{1}{2}\left\{ L^\dagger_\... 3 I guess from the operator sum representation \rho(t) = \sum_k K_k \rho(t_0) K_k^\dagger alone you won't be able to make any statement about non-Markovianity, since you are missing interesting features: (Domain) Is this dynamical map valid for all initial reduced density matrices? If so, the initial state might be a product state \rho(t_0) \otimes \rho_E ... 3 To be clear: You have a Hilbert space \mathcal{H}_A\otimes\mathcal{H}_B\otimes\mathcal{H}_C. The initial state is \rho_\text{tot}(0)=\rho_A\otimes\rho_B\otimes\rho_C. There is a Hamiltonian acting on the system of the form H_{\text{tot}}=H_{AB}\otimes\mathbb{I}_C+\mathbb{I}_A\otimes H_{BC} Instead of directly calculating the effect of the Hamiltonian ... 2 It's fundamentally similar to/the same as Baker–Campbell–Hausdorff (BCH). Generally, in quantum physics, this is most often used (or at least taught) with commuting Hamiltonians:$$e^{-i\left(H_1+H_2\right)t} = \sum_{n=1}^\infty\frac{\left(-it\right)^n}{n!}\left(H_1+H_2\right)^n = e^{-iH_1t}e^{-iH_2t}e^{\frac{1}{2}\left[H_1, H_2\right]t^2}\cdots where the ...

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This is a very generic description that captures essentially all possibilities of describing the Hamiltonian with a decay process (I supposed one should allow for a general $H_{1,2}$ rather than a simple tensor product of terms)). The basic idea is that if you've got a system ("1"), then if you just describe using a Hamiltonian on that system, the evolution ...

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I know this isn't really what you're thinking of with your question, but measurement-based quantum computing is pretty well studied. Under the many-worlds interpretation, the system counts as open for that protocol, because every time you perform a measurement you're becoming entangled with the system.

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That is a very old paper, corresponding to version one of the software. It is now on version 4.x. It is best to see the current documentation for how to use the Bloch sphere: http://qutip.org/docs/latest/guide/guide-bloch.html

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There are two possible answers. Let's say the universe evolves from $t=0$ to $t_f$ then the unitary evolution $U$ from $0$ to $t_f$ induces a CP evolution on the subsystem. To see this, note that the composition of CP maps is CP. Now, the reduced (system) evolution is $Tr_E U\rho_s\otimes\rho_E U^\dagger$ which is a composition of the map \$\rho_s\...

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