1
$\begingroup$

In the compressive QPT method, the trace-preserving constraint of the process matrix is $\sum_{n, m} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=1$. In this paper:https://arxiv.org/abs/1404.2877, as $\|\chi\|_{\operatorname{Tr}}=\operatorname{Tr}(\chi)$ being the object function, we must drop any equation that constrains the trace of the process matrix. They take an operator basis of traceless Hermitian matrices to ensure that there is only one equation relevant to the trace of the process matrix, which is dropped as a constraint. Thus define the estimator as follows: $$ \begin{aligned} \min . & \operatorname{Tr}(\chi) \\ \text { Subject to: } & \sum_{j, l}\left|f_{j l}-\operatorname{Tr}\left(D_{j l}^{\dagger} \chi\right)\right|^2 \leq \varepsilon \\ & \sum_{n, m \neq 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=0 \\ & \chi=\chi^{\dagger}, \chi \geq 0 \end{aligned} $$ where now $\chi$ and $D_{j l}$ are represented in a basis with $\Upsilon_1=1$ and the elements $\Upsilon_{m \neq 1}$ are orthogonal traceless Hermitian matrices. The sum in the second constraint include all the terms except $n=m=1$.

I don't know the reason for using $\sum_{n, m \neq 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=0$, can we just omit the constraint $\sum_{n, m} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=1$, and then renormalized $\chi$ after we get the estimator? By the way, why $\operatorname{Tr}(\hat{\chi})=d$?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.