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.


$$E_{m}\rho _{j}E_{n}^{\dagger }=\sum _{{k}}B_{{m,n,j,k}}\rho _{k},$$ 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 physics.SE


1 Answer 1


Yes, I think you have the correct idea. I'll just add some additional details.

As you suggest, "non-trace-preserving" means that the map of interest is applied with some probability which (in general) depends on the input state. Furthermore, we presumably know if the correct map has been applied or not. Otherwise, we would have to describe our output density matrix in a way that expresses our lack of knowledge of the map applied, perhaps by summing different outcomes weighted by their likelihood.

Having clarified the problem, the solution is that for each basis state $\rho_j$, you want to estimate (a) the effect of the map assuming success and (b) the probability that the map is successfully applied. This can be done by repeating the quantum process, counting the number of successful runs, and only doing the analysis for those instances when the correct map is applied.

Doing things this way, you will learn an effective trace-preserving map $\tilde{\mathcal{E}}(\rho) = \rho_{out}$ and a "success probability function" $p(\rho) = \sum_j p_j \text{Tr}(\rho \rho_j)$, where $p_j$ is the estimated success probability for the $j$th basis state. The final estimate for your map then becomes $\mathcal{E}(\rho) = p(\rho) \tilde{\mathcal{E}}(\rho)$.


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