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Set $\hbar =1$. Let $U(t) = e^{-itH}$ be evolution under a Hamiltonian $H$ (for convenience let's assume $H$ is not degenerate). A time-translation invariant quantum operation $\mathcal{E}$ is one that effectively commutes with evolution under $H$, i.e. for any state $\rho$,

$$\mathcal{E}(U(t) \rho U^\dagger(t)) = U(t)\mathcal{E}(\rho)U^\dagger(t).$$

In my lecture notes I came across the statement that “time-translation invariant operations cannot generate any coherence between energy eigenspaces”. I took this to mean that these operations do not increase the coherence of quantum states.

For a qubit state, I thought a good measure of its coherence would be the magnitude of either of its off-diagonal elements, e.g. $|\rho_{0,1}|$. So my question is: how can I prove that $|\rho_{0,1}|$ is never increased by $\mathcal{E}$ in the qubit case?

I can show that $\mathcal{E}$ cannot change a qubit state $\rho$ with $|\rho_{0,1}| = 0$ to a state $\mathcal{E}(\rho)$ with $|\mathcal{E}(\rho)_{0,1}| \neq 0$ (in this case $\mathcal{E}(\rho)$ commutes with $H$, and so must be a completely incoherent state). But I haven't been able to go further than this.

Do I have a bad definition of coherence? Or have I misinterpreted the statement from my lecture notes?

MarianD
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Si Chen
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1 Answers1

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For qubits it will indeed be the case that $|\rho_{0,1}|$ has to decrease. You can see it from the general form of a corresponding master equation (this paper, arXiv), where you see that different non-invariant operators decouple.

In higher dimension, the total amount of coherence in each mode is non-increasing, see "Quantum coherence, time-translation symmetry and thermodynamics" (arXiv) for details (so you would have possibly sum over absolute value of some coherences gets smaller).

Agnieszka
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