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 $n=m=1$. Finally, the estimated processmatrix $\hat{\chi}$ should be renormalized such that $\operatorname{Tr}(\hat{\chi} ) = d$.

I think there is something wrong, if the elements $\Upsilon_{m \neq 1}$ are orthogonal traceless Hermitian matrices, why $\sum_{n, m \neq 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=0$? For example, Pauli bases are orthogonal traceless Hermitian matrices, but $\operatorname{Tr}(P_m^{\dagger}P_n)=d\delta_{mn}$, as long as $m=n$, $P_m^{\dagger}P_m=I\ne 0$.

And if they want to ensure $\operatorname{Tr}(\chi)=d$, we need orthogonal traceless Hermitian matrices that satisfies $\operatorname{Tr}(\Upsilon_m^{\dagger}\Upsilon_n)=\delta_{mn}$ . But Pauli bases satisfy $\operatorname{Tr}(\Upsilon_m^{\dagger}\Upsilon_n)=d\delta_{mn}$ , SVD basis, computational basis satisfy $\operatorname{Tr}(\Upsilon_m^{\dagger}\Upsilon_n)=\delta_{mn}$ , but they are not traceless Hermitian matrices, so I am confused.


1 Answer 1


In compressed sensing (CS), we want to find the sparsest solution vector in the $\ell_1$ norm. Hence, the objective function is $\min ||\cdot||_1$. In this particular case, the "sparsity" is enforced through the minimization of the trace of the vector $\chi$.

Now, to answer your question. It is true that for Pauli matrices $P_n$, we have $P_n^{\dagger} P_n = I$. However, this does not mean that the sum $\sum_{n, m \neq 1} \chi_{n m} P_m^{\dagger} P_n$ can not be zero. Remember, that $\chi_{mn}$ are variables. Therefore, the constraint $$\sum_{n, m \neq 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n=0 \tag{1}$$ enforces that $\chi_{nm}$ are chosen such that the total sum of $\chi_{nm}\Upsilon_m^{\dagger}\Upsilon_n$ adds up to zero. For example, consider a hypothetical constraint $\chi_{1} I + \chi_{2} I=0$. We have $I \neq 0$, so does it mean that there are no non-zero variables $\chi_{1}$ and $\chi_{2}$ so that the equation evaluates to zero? One choice that evaluates the equation to zero is $\chi_{1}=-\chi_{2}$ and $\chi_{2} = a$ where $a$ is some number.

A more interesting question is, why do we have this constraint at all? What does it do?

To understand the meaning behind this constraint, we observe that the sum in (1) excludes $\Upsilon_1 = I$. By using an operator basis where $\Upsilon_1 = I$ and all other $\Upsilon_n$ are orthogonal traceless Hermitian matrices, the constraint formulation makes sure that $\chi_{nm}$ for $n,m \neq 1$ do not contribute to the trace-preserving aspects of the quantum operation. This is done to allow the minimization of $Tr(\chi)$.

More formally, suppose that we found a feasible $\chi$ (which means $\chi$ satisfies the constraints). Then, the trace-preservation property requires: $$\sum_{n, m = 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n = I.$$ Which is equivalent to writing: \begin{align} \chi_{11}\Upsilon_{1}^{\dagger}\Upsilon_{1} + \sum_{n, m \neq 1} \chi_{n m} \Upsilon_m^{\dagger} \Upsilon_n = \chi_{11}\Upsilon_{1}^{\dagger}\Upsilon_{1} + 0 = I. \end{align} In the above, I used the fact that $\chi$ satisfies the constraint in (1). Note that the trace preservation now only depends on the choice of $\chi_{11}$. Therefore, the actual trace preservation constraint is simplified to: $$\chi_{11} = 1.$$

  • $\begingroup$ I'm not sure if I understand your statements and notations. You can't just choose any matrix $\chi$. The matrix $\chi$ must minimize the objective function and satisfy the constraints. I'm not sure how well you are familiar with combinatorial optimization or linear programming, but essentially what your original question asks is linear programming problem. Looking at some basic linear programming examples might help to add some clarity. $\endgroup$
    – MonteNero
    Commented Apr 5 at 2:54
  • $\begingroup$ And if you have more questions, you are welcome to create a new question. $\endgroup$
    – MonteNero
    Commented Apr 5 at 2:56
  • $\begingroup$ What is $\Gamma$? Sorry, I'm not familiar with the context so I don't know the definitions you are using cause they were not defined in the question. $\endgroup$
    – MonteNero
    Commented Apr 5 at 3:11
  • $\begingroup$ It seems you edited your original question and now asking something else. $\endgroup$
    – MonteNero
    Commented Apr 5 at 3:13
  • $\begingroup$ sorry, I mean we can choose any $\Upsilon_m=I$. $\endgroup$
    – karry
    Commented Apr 5 at 3:24

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