What is the point of decomposing the Kraus operators of a channel in terms of a fixed basis?

Question

Say we have a linear map $\mathcal{E}$ describing the dynamics of a quantum system,

$$\rho \rightarrow \mathcal{E}(\rho)$$

As expressed in the operator-sum representation,

$$\mathcal{E}(\rho) = \sum_i A_i \rho A^\dagger_i.$$

This paper considers an equivalent description of $\mathcal{E}$ using a fixed set of operators $\tilde{A}_i$, which form a basis for the set of operators on the state space, so that

$$ A_i = \sum_m a_{im}\tilde{A}_m $$

for some set of complex numbers $a_{im}$. My question is, what aspect of $\tilde{A}_i$ is fixed, in comparison to $A_i$? This alternate representation seems to just be factoring out the imaginary part of $A_i$... is something else going on?

Answer 1

The point is that the operators $A_i$ are unknown. We are writing the $A_i$ in terms of a basis of known operators $\tilde{A}_m$ that we have chosen. The problem of determining the $A_i$ reduces to the problem of determining the coefficients $a_{im}$ in this basis. This is not any different from determining an unknown vector by writing the vector as a linear combination of a fixed set of basis vectors, then finding the coefficients in this basis.