A quantum operation is defined as
\begin{equation}
\varepsilon(\rho)=\sum\_{k}M\_k\rho M\_k^{\dagger}
\end{equation}
where $\varepsilon(\rho)$ takes an initial state $\rho$ to some final state $\rho'$ and $M\_k^{\dagger}M\_k$'s are positive, contractive operators that satisfy
\begin{equation}
\sum\_kM\_k^{\dagger}M\_k=\mathbb{I}.
\end{equation}
It is known through Schauder's fixed point theorem that this admits a fixed point. Note that because of the summation in the first equation, it means that the measurement was not recorded.


QUESTION: If I have a single quantum measurement that was recorded, then I would only have $\varepsilon(\rho)=\frac{M\_k\rho M\_k^{\dagger}}{\text{tr}(M\_k\rho M\_k^{\dagger})}$. Would this still admit a fixed point? Is there a way to prove it?



Fixed point of quantum operations
As clarified in the comments, the question refers to positive operator valued measures, not quantum operations. Basically the issue is whether, for a given $A \in M\_n = M\_n(\mathbb{C})$ the map $B \mapsto ABA^\*$ on $M\_n$ has an eigenvector. A counterexample is given by the matrix $A = \left[\matrix{0&1\cr 0&0}\right]$; for any matrix $B = \left[\matrix{a&b\cr c&d}\right]$ we have $ABA^\* = \left[\matrix{d&0\cr 0&0}\right]$, so that $0$ is the only eigenvalue.


However, if $A$ is positive then taking $B$ to be any projection onto an eigenvector of $A$ belonging to a nonzero eigenvalue yields a solution. Slightly more generally, $B$ could be any operator which satisfies $PBP = B$ where $P$ is the orthogonal projection onto some eigenspace for $A$. Putting $A$ in diagonal form shows that these are the only solutions.


