# Fixed point of quantum operations

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?

• I think you should find someone to talk to in person about this. The $M_k$ aren't assumed to be positive; the identity matrix is a fixed point (Schauder's theorem doesn't come into it), and your understanding of measurement is not correct. The terms of the sum do not represent different measurement outcomes. Feb 20 '16 at 16:21
• As to the technical question of whether $\frac{M_k\rho M_k^\dagger}{{\rm tr}(M_k\rho M_k^\dagger)}$ must have a nonzero fixed point, the answer is no. Take $M_k = \left[\matrix{0&1\cr 0&0}\right]$ and let $\rho$ be an arbitrary $2\times 2$ matrix. (If you add the nonstandard requirement that $M_k$ be positive, then the answer is yes; let $\rho = |\lambda\rangle\langle\lambda|$ where $|\lambda\rangle$ is any nonzero eigenvector for $M_k$.) Feb 20 '16 at 16:45
• Right. It's $M_k^{\dagger}M_k$ that is positive. What's wrong with my understanding of measurement? It's what this paper said. arxiv.org/abs/1110.6815 If the measurement was not recorded, it will be a sum but if it is recorded, there will be no sum. Just a single expression. The recording part is what I actually find mysterious. Feb 20 '16 at 17:01
• "If the measurement was not recorded, it will be a sum but if it is recorded, there will be no sum." Yes, but not a term of the sum you have written. The paper you referenced looks like it has a good, thorough explanation of all this. Feb 20 '16 at 17:15
• That's what it says on page 9. Postulates II.4 and II.5. Feb 20 '16 at 17:20

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.