# Non-trace-preserving, process tomography

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

Cross-posted on qc.SE

You have used no-where that $$c_{jk}$$ comes from a trace-preserving map. Thus, if you can obtain the correct $$c_{jk}$$ for a not trace-preserving map, you are done.
On the other hand, that is no problem: When you apply your map $$\mathcal E(\rho_j)$$, it will return $$\sum\hat c_{jk} \rho_k$$ only with a certain success probability $$p_j$$. Thus, if you run the map often enough, you can estimate both $$\hat c_{jk}$$ and $$p_j$$. It is then easy to see that $$\mathcal E(\rho_j) = p_j \sum \hat c_{jk} \rho_k$$, and thus, $$c_{jk} = p_j \hat c_{jk}$$. Then, you can just apply the procedure you outlined to infer $$\chi$$ from $$c$$, by inverting $$B$$.
On the other hand, if you live in a magic postselected world where it you only see successful applications of $$\mathcal E$$ -- note that we don't live in such a world, and living in such a world would have all kind of strange consequences -- there would be no way to infer $$\mathcal E$$: To see why, you can simply rescale all Kraus operators by a factor (i.e., rescale the success probability), and there will be no way to tell this factor.