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

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?

• This is a question about language rather than physics. The adjective "fixed" does not apply to any of the operators themselves, It applies to the set of operators. The collection of operators is chosen (i.e. "fixed") and is not altered during whatever the authors are doing with them.. May 5 '21 at 18:04

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.
• ... or just writing the matrix $A_i$ in terms of its entries. May 5 '21 at 18:49