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 qc.SE


1 Answer 1


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.


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