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

Question

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?

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 $a$.