The action of a quantum channel $\mathcal{E}$ on state given by a density operator $\rho$ is described by a completely positive trace preserving (CPTP) map: $$ \rho \rightarrow \sum_k R_k \rho R_k^\dagger = \mathcal{E(\rho)} . $$ The Kraus operators $R_k$ must obey the condition $\sum_kR_k^\dagger R_k = \mathcal{I}$ where $\mathcal{I}$ is the identity operator.

A fixed point $\rho_f$ of a channel is a state which is left unchanged by the action of the channel so that: $$ \mathcal{E}(\rho_f) = \rho_f $$ My question is: Given that we know the Kraus operators $R_k$ for a quantum channel $\mathcal{E}$, how can we find the fixed point(s) of that channel?

Is there a general process for obtaining an expression for $\rho_f$ in terms of $R_k$?

A related, maybe easier, question is whether we can prove that there exists a fixed point of a particular channel. Do all channels have fixed points?

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    $\begingroup$ Regarding your second question, the identity is always a fixed point of a trace preserving channel. $\endgroup$
    – anon1802
    Commented Jul 21, 2021 at 11:19
  • $\begingroup$ Ah, of course, that makes sense. Thanks! $\endgroup$
    – asph
    Commented Jul 21, 2021 at 11:24
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    $\begingroup$ @anon1802 "the identity is always a fixed point of a trace preserving channel." -- This is incorrect, take e.g. the channel which maps every input to |0><0|. $\endgroup$ Commented Jul 21, 2021 at 13:30
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    $\begingroup$ General reading: www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/MichaelWolf/…. Regarding your last question, yes, every channel has at least one positive semi-definite fixed point. (This is the quantum version of the Perron-Frobenius Theorem; see the linked notes.) For (numerically) finding fixed points, you can start by solving the corresponding eigenvalue equation for the eigenvector $\rho$. If the fixed point is unique, you get it immediately; otherwise, you have to massage the eigenvectors to get a positive semi-definite fixed point. $\endgroup$ Commented Jul 21, 2021 at 13:32
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    $\begingroup$ @anon1802 In fact, that's an if and only if. $\endgroup$ Commented Jul 21, 2021 at 14:40

1 Answer 1


A relevant result is given by Watrous in section 4.2.2 of his book. Here are two theorems proved in the section:

  1. Let $\Phi\in\mathrm T(\mathcal X)$ be a positive and trace-preserving map, with $\mathcal X$ a finite-dimensional Hilbert space. Then there is some state $\rho$ such that $\Phi(\rho)=\rho$.

  2. Let $\Phi$ be a unital channel (thus CPTP with $\Phi(I)=I$), and suppose its Kraus decomposition reads $$\Phi(X)=\sum_a A_a X A_a^\dagger.$$ Then, for any linear operator $X$, we have $\Phi(X)=X$ iff $[X,A_a]=0$ for every $a$.


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