Is every quantum channel covariant with respect to some non-trivial Hamiltonian?

Question

When asking whether every channel is covariant with respect to some non-trivial unitary channel I mean the following:

Does there for every CPTP map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ exist a unitary $U\in\mathbb C^{n\times n}$ such that $U\neq e^{i\phi}{\bf1}$ for any $\phi\in\mathbb R$ and $\Phi(U\rho U^\dagger)=U\Phi(\rho)U^\dagger$ for all $\rho$?

As the exponential map, of course, provides a one-to-one relation between non-trivial unitaries and non-trivial Hamiltonians ($H\neq E\cdot{\bf1}$ for any $E\in\mathbb R$) it is neither surprising nor difficult to see that the above question is equivalent to:

Does there for every CPTP map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ exist $H\in\mathbb C^{n\times n}$ Hermitian such that $H\neq E\cdot{\bf1}$ for any $E\in\mathbb R$ and $\Phi([H,\rho])=[H,\Phi(\rho)]$ for all $\rho$?

To give some context: covariance is central, e.g., to quantum thermodynamics as it encodes time-translation symmetry $\Phi(e^{-itH}\rho e^{itH})=e^{-itH}\Phi(\rho)e^{itH}$. From this perspective the above question asks about whether every quantum channel is symmetric with respect to some non-trivial closed-system evolution.

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(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

This quesiton passes all of Norbert Schuch's canonical channel examples except for the last one: consider the entanglement breaking channel $$ \Phi(\rho):={\rm tr}(F\rho)\sigma_1+{\rm tr}(({\bf1}-F)\rho)\sigma_2 $$ with $F=|0\rangle\langle 0|=\sigma_1$ and $\sigma_2=|+\rangle\langle +|$. Its Pauli transfer matrix reads $$ \mathcal P(\Phi)=\begin{pmatrix} 1 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{pmatrix}\,. $$ To see that this is a counterexample to our problem we have to show that if any $H\in\mathbb C^{2\times 2}$ Hermitian satisfies $\Phi([H,\cdot])=[H,\Phi(\cdot)]$, then $H$ has to be a multiple of the identity. Starting from $$ H=\begin{pmatrix}h_{11}&h_{12}\\h_{12}^*&h_{22}\end{pmatrix} $$ one readily computes $$ \mathcal P([H,\cdot])=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i (h_{11}-h_{22}) & -2 i\, {\rm Im}\, h_{12} \\ 0 & i (h_{11}-h_{22}) & 0 & -2 i\, {\rm Re}\,h_{12} \\ 0 & 2 i\, {\rm Im}\,h_{12} & 2 i\, {\rm Re}\,h_{12} & 0 \end{pmatrix} $$ and thus \begin{align*} 0&=\mathcal P(0)\\ &=\mathcal P\big( \Phi\circ[H,\cdot]-[H,\cdot]\circ \Phi \big)\\ &=\mathcal P(\Phi)\mathcal P([H,\cdot])-\mathcal P([H,\cdot])\mathcal P(\Phi)\\ &=\begin{pmatrix} 0 & 0 & 0 & 0 \\ i \,{\rm Im}\,h_{12} & -i \,{\rm Im}\,h_{12} & -i \,{\rm Re}\,h_{12} & i \,{\rm Im}\,h_{12} \\ i \,{\rm Re}\,h_{12}-\frac{i}{2} (h_{11}-h_{22}) & 0 & 0 & \frac{i}{2} (h_{11}-h_{22})+i \,{\rm Re}\,h_{12} \\ -i\, {\rm Im}\,h_{12} & i\, {\rm Im}\,h_{12} & i\, {\rm Re}\,h_{12} & i\, {\rm Im}\,h_{12} \end{pmatrix}. \end{align*} But this is only possible ${\rm Re}\,h_{12}=0={\rm Im}\,h_{12}$ and, subsequently, $\frac{i}{2} (h_{11}-h_{22})=0$, i.e. $h_{11}=h_{22}=:E\in\mathbb R$ meaning $H=E\cdot{\bf1}$, as desired.

Alternatively, one may consider a Pauli diagonal channel $\Phi$—that is, there exist $a,b,c\in\mathbb R$ such that $ \mathcal P(\Phi)={\rm diag}(1,a,b,c) $—in which case \begin{align*} \mathcal P\big( &\Phi\circ[H,\cdot]-[H,\cdot]\circ \Phi \big)\\ &=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i (a-b) (h_{11}-h_{22}) & -2 i (a-c) \,{\rm Im}\,h_{12} \\ 0 & -i (a-b) (h_{11}-h_{22}) & 0 & -2 i (b-c) \,{\rm Re}\,h_{12} \\ 0 & -2 i (a-c) \,{\rm Im}\,h_{12} & -2 i (b-c) \,{\rm Re}\,h_{12} & 0 \end{pmatrix}. \end{align*} Hence if $a,b,c$ are pairwise distinct, then (and only then) is $\Phi$ not covariant with respect to any non-trivial Hamiltonian. E.g., $\mathcal P(\Phi)={\rm diag}(1,2/3,1/3,0)$ would be another counterexample to our conjecture, and even a unital one at that.