Consider an unknown quantum process, i.e., a black box, acting on a physical quantum system described by a density matrix $\rho$ associated with a d-dimensional Hilbert space $\mathcal{H}$. A complete characterization of the process may be obtained by the Kraus representation of quantum operations in an open system. A generic map $\mathcal{E}$ acting on a generic state $\rho$ can be expressed by the Kraus representation $${\mathcal {E}}(\rho )=\sum _{i}A_{i}\rho A_{i}^{\dagger },$$where $\rho \in {\mathcal {B(H)}}$, the bounded operators on Hilbert space;
Let ${\displaystyle \displaystyle \{E_{i}\}}$ be an orthogonal basis for ${\displaystyle {\mathcal {B(H)}}}$. Write the ${\displaystyle \displaystyle A_{i}}$ operators in this basis
${\displaystyle \displaystyle A_{i}=\sum _{m}a_{im}E_{m}}$.
This leads to $ {\displaystyle {\mathcal {E}}(\rho )=\sum _{m,n}\chi _{mn}E_{m}\rho E_{n}^{\dagger }},$ where ${\displaystyle \chi _{mn}=\sum _{i}a_{mi}a_{ni}^{*}}$.
The goal is then to solve for ${\displaystyle \displaystyle \chi }$, which is a positive superoperator and completely characterizes ${\displaystyle {\mathcal {E}}}$ with respect to the ${\displaystyle \displaystyle \{E_{i}\}}$ basis.
For each of these input states ${\displaystyle \rho _{j}}$, sending it through the process gives an output state ${\displaystyle {\mathcal {E}}(\rho )}$ which can be written as a linear combination of the ${\displaystyle \rho _{k}}$, i.e. $\textstyle {\mathcal {E}}(\rho _{j})=\sum _{k}c_{{jk}}\rho _{k}$. By sending each $\rho _{j}$ through many times, quantum state tomography can be used to determine the coefficients $c_{jk}$ experimentally.
Write
$$E_{m}\rho _{j}E_{n}^{\dagger }=\sum _{{k}}B_{{m,n,j,k}}\rho _{k},$$ where $B$ is a matrix of coefficients. Then
$${\displaystyle \sum _{k}c_{jk}\rho _{k}={\mathcal {E}}(\rho _{j})=\sum _{m,n}\chi _{m,n}E_{m}\rho _{j}E_{n}^{\dagger }=\sum _{m,n}\sum _{k}\chi _{m,n}B_{m,n,j,k}\rho _{k}}.$$ Since $\rho _{k}$ form a linearly independent basis, $\displaystyle c_{{jk}}=\sum _{{m,n}}\chi _{{m,n}}B_{{m,n,j,k}}$. Inverting $B$ gives $\chi$ :
$$\chi _{{m,n}}=\sum _{{j,k}}B_{{m,n,j,k}}^{{-1}}c_{{jk}}.$$
This is the tomography process in this link
In the case of non-trace preserving $\sum_iA_i^{\dagger}A_i\leq \mathbb{I}$, it is important to consider not only the transformation acting on a generic input state, but also the probability of success of the map, $\mathrm{Tr}(\mathcal{E}(\rho))= \mathrm{Tr}(\sum_{m,n}\chi_{mn}E_{n} ^\dagger E_{m}\rho)$. My question is how we can extend this process for a non-trace preserving? Taking this probability into account, my idea is using $\rho_{out}=\dfrac{\mathcal{E}(\rho_j)}{\mathrm{Tr}(\mathcal{E}(\rho_j))}$, then $\mathcal{E}(\rho_j)=\mathrm{Tr}(\mathcal{E}(\rho_j))\rho_{out}$ but I'm not sure if is sufficent to extend this process for a non-trace preserving
