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.

  • $\begingroup$ Is there nothing on that in Michael Wolf's lecture notes? One thought is that if you can show that if the operator norm of $T$ is one, you are done, as the operator norm is submultiplicative and it bounds the largest eigenvalue. $\endgroup$ Commented Sep 23, 2022 at 11:15
  • $\begingroup$ @NorbertSchuch I haven't seen anything in the Wolf's notes that could give me some hint... Respect to the bound of the largest eigenvalue, it's already true that $\lambda_{max}=1$ because the composition of two CPTP maps is a CPTP map (so the spectral radius is 1). The key of the question is to find $|\lambda|<1$ in the traceless hyperplane (for eigenvalues different from $\lambda=1$, trace-preservation of $T$ imply that the rest of eigenvectors live in the traceless hyperplane). But the operator norm could only yield $\leq$. $\endgroup$
    – RMPsp
    Commented Sep 26, 2022 at 2:50
  • $\begingroup$ Won't the proof of the operator norm inequality also yield sth. about the degeneracy of the leading eigenvalue? I'd think it would. -- What is not clear to me on the spot is why $\lambda_{\max}=1$ would imply that the operator norm is also $1$ (as it is the largest singular value), though it might of course be sth. which holds for CP maps. If this is true, I'm confident that the fact that the composed map has only one largest eigenvalue can be proven. $\endgroup$ Commented Sep 26, 2022 at 7:31

1 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 $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.