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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?

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  • $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Commented Feb 20, 2016 at 16:21
  • $\begingroup$ 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 $M_k$.) $\endgroup$
    – Nik Weaver
    Commented Feb 20, 2016 at 16:45
  • $\begingroup$ 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. $\endgroup$
    – Janus
    Commented Feb 20, 2016 at 17:01
  • $\begingroup$ "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. $\endgroup$
    – Nik Weaver
    Commented Feb 20, 2016 at 17:15
  • $\begingroup$ That's what it says on page 9. Postulates II.4 and II.5. $\endgroup$
    – Janus
    Commented Feb 20, 2016 at 17:20

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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.

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