# Is there a general way to find the fixed points of a quantum channel in terms of its Kraus operators?

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?

• Regarding your second question, the identity is always a fixed point of a trace preserving channel. Jul 21 '21 at 11:19
• Ah, of course, that makes sense. Thanks!
– asph
Jul 21 '21 at 11:24
• @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|. Jul 21 '21 at 13:30
• 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. Jul 21 '21 at 13:32
• @anon1802 In fact, that's an if and only if. Jul 21 '21 at 14:40

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 $$a$$.