Is the composition of two mixing CPTP maps a mixing CPTP map?

Question

A CPTP map $T$ is called mixing if and only if it has a single fixed point $\rho^*\in S(\mathcal{H})$ (where $S(\mathcal{H})$ is the convex subset of density matrix operators) to which it converges after repeated applications: $$ \lim_{n\rightarrow \infty}||T^n(\rho)-\rho^*||_1=0,\ \forall\rho\in S(\mathcal{H}), $$ where $||\cdot||_1$ denotes the trace norm. My question is: is the composition of two mixing CPTP maps a mixing CPTP map? My main concern comes from the fact that a CPTP map is mixing if and only if $\lambda=1$ (the eigenvalue of $\rho^*$) is the unique peripheral eigenvalue of the map (peripheral eigenvalues of CPTP maps are those which fulfill $|\lambda|=1$). See for instance Theorem 7 of https://dx.doi.org/10.1088/1367-2630/9/5/150 for a proof. Then, the rest of eigenvalues fulfill $|\lambda|<1$. So I don't know if the composition of two mixing channels can preserve this later condition. What I know is that this is true for strictly contractive channels, but I would like to know if there is a more general statement, or find at least a counterexample.

Answer 1

The composition of two mixing channels need not be mixing anymore, which can be related to the fact that the spectral radius can fail to be sub-multiplicative for non-normal matrices (more on this at the end).

To give an explicit counterexample, consider the linear maps $$ \Phi_1\begin{pmatrix}\rho_{11}&\rho_{12}\\ \rho_{21}&\rho_{22}\end{pmatrix}:=\begin{pmatrix} \frac{1}{2} (\rho_{11}+i (\rho_{12}-\rho_{21})+\rho_{22}) & 0 \\ 0 & \frac{1}{2} (\rho_{11}-i \rho_{12}+i \rho_{21}+\rho_{22}) \end{pmatrix} $$ and $$ \Phi_2\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}:=\begin{pmatrix} \frac{\rho_{11}+\rho_{22}}{2} & -\frac{1}{2} i (\rho_{11}-\rho_{22}) \\ \frac{1}{2} i (\rho_{11}-\rho_{22}) & \frac{\rho_{11}+\rho_{22}}{2} \end{pmatrix}\,. $$ These maps are obviously trace-preserving, and their respective Choi matrices $$ \mathsf C(\Phi_1)= \begin{pmatrix} \frac{1}{2} & 0 & \frac{i}{2} & 0 \\ 0 & \frac{1}{2} & 0 & -\frac{i}{2} \\ -\frac{i}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{i}{2} & 0 & \frac{1}{2}\end{pmatrix}\simeq \omega\oplus\omega^\top \qquad \mathsf C(\Phi_2)= \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} & 0 & 0 \\ \frac{i}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{i}{2} \\ 0 & 0 & -\frac{i}{2} & \frac{1}{2} \end{pmatrix}=\omega^\top\oplus\omega $$ are positive semi-definite because $$ \omega:=\begin{pmatrix} \frac{1}{2} & \frac{i}{2}\\ - \frac{i}{2} & \frac{1}{2} \end{pmatrix}\geq 0\,. $$ There are multiple ways to see that these channels are mixing, the easiest one being to check that their squares $\Phi_1(\Phi_1(\rho))={\rm tr}(\rho)\frac{\bf1}2=\Phi_2(\Phi_2(\rho))$ are a reset-channel, hence they are mixing with respect to the maximally mixed state $\frac{\bf1}2$. Similarly, one can check the respective Pauli transfer matrix $\mathsf P(\Phi_j)$---which is the representation matrix of $\Phi_j$ w.r.t. the orthonormal basis $\{\frac1{\sqrt2}{\bf1},\frac1{\sqrt2}\sigma_x,\frac1{\sqrt2}\sigma_y,\frac1{\sqrt2}\sigma_z\}$, meaning it has the same spectrum as $\Phi_j$--- $$ \mathsf P(\Phi_1)=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}\qquad \mathsf P(\Phi_2)=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ to see that both channels have simple eigenvalue $1$ and threefold eigenvalue $0$, hence they satisfy the mixing criterion you cited.

However, the composition of these channels is the dephasing channel $$ \Phi_1\Big(\Phi_2\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix} \Big)=\begin{pmatrix}\rho_{11}&0\\0&\rho_{22}\end{pmatrix} $$ as is readily verified (e.g., via $$ \mathsf P(\Phi_1\Phi_2)=\mathsf P(\Phi_1)\mathsf P(\Phi_2)=\begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{pmatrix}\,). $$ Thus $\Phi_1\Phi_2$ has twofold eigenvalue $1$ as every diagonal state is a fixed point; in particular, $\Phi_1\Phi_2$ is obviously not mixing anymore.

To get back to the underlying mathematics, the problem here is that the sub-matrices $$ \begin{pmatrix} 0&0\\1&0 \end{pmatrix}\,,\,\begin{pmatrix} 0&1\\0&0 \end{pmatrix} $$ of $\mathsf P(\Phi_1),\mathsf P(\Phi_2)$ have spectral radius (largest mod of eigenvalue) $0$, but their product $=|1\rangle\langle 1|$ has spectral radius $1$.