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It is said that when an unknown process is unitary, its $\chi$ matrix is rank-$1$ and possesses only one positive eigenvalue. See eg https://arxiv.org/abs/2306.07867. So when the process matrix has rank $1$, is the matrix unique? If the process matrix is not unique, what is the relationship between these rank-$1$ process matrices?

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TL;DR: The elements of the process matrix with respect to an operator basis $E_i$ are just the coefficients in the expansion of the channel, viewed as an operator on $\mathcal{H}\otimes\mathcal{H}$, in the basis $E_i\otimes\overline{E_j}$. Therefore, the process matrix depends on the chosen operator basis, but is otherwise unique. This is true irrespective of its rank.

Definition

The process matrix of a channel $\mathcal{E}(\rho)=\sum_kK_k\rho K_k^\dagger$ with respect to an operator basis $E_i$ can be obtained by expressing the Kraus operators as a linear combination $K_k=\sum_ia_{ki}E_i$ of the basis elements $E_i$ and collecting the terms $E_i\rho E_j^\dagger$ from all Kraus operators $$ \mathcal{E}(\rho)=\sum_{ijk}a_{ki}\overline{a_{kj}}E_i\rho E_j^\dagger=\sum_{ij}\chi_{ij}E_i\rho E_j^\dagger\tag1 $$ where $\chi_{ij}:=\sum_ka_{ki}\overline{a_{kj}}$ is called the process matrix.

Uniqueness

Now, $E_i$ is a basis in the Hilbert space $L(\mathcal{H})$, so $E_i\otimes\overline{E_j}$ is a basis in the Hilbert space $L(\mathcal{H}\otimes\mathcal{H})$. By vectorizing and suppressing $\rho$ in $(1)$, we obtain an equality between a representation$^1$ of $\mathcal{E}$ as an operator on $\mathcal{H}\otimes\mathcal{H}$ on the LHS and a linear combination of the basis elements $E_i\otimes\overline{E_j}$ with $\chi_{ij}$ as the coefficients on the RHS. The representation of $\mathcal{E}$ on the LHS is unique$^2$. The coefficients in the linear combination on the RHS depend on the basis, but are otherwise unique. Thus, the process matrix of a channel depends on the chosen operator basis $E_i$, but is otherwise$^3$ unique.

Rank

The definition $\chi_{ij}:=\sum_ka_{ki}\overline{a_{kj}}$ implies that the rank of $\chi_{ij}$ is the Choi rank of $\mathcal{E}$. In particular, the rank of the process matrix of a unitary channel is necessarily one.


$^1$ This is sometimes denoted with $K(\mathcal{E})$ and called the natural representation of the channel $\mathcal{E}$. Intuitively, it is what $\mathcal{E}$ becomes when we forget that it acts on operators and pretend that it is merely a linear operator on vectors (stripped of their own operatorial structure by vectorization).
$^2$ Since vectorization is a bijection.
$^3$ Sometimes folks implicitly assume that $E_i$ is the Pauli basis. Obviously, under this more restrictive definition, the process matrix is unique without any qualification.

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  • $\begingroup$ Thank you, I see. But there can be lots of unitary processes, right? Can you give some examples of unitary process? $\endgroup$
    – karry
    Nov 27, 2023 at 2:03
  • $\begingroup$ Yes, there are infinitely many unitary processes. Every quantum gate is an example. $\endgroup$ Nov 27, 2023 at 2:10
  • $\begingroup$ It is said that if the map is unitary, there is only one term of Kraus operators, namely, $\mathcal{E}(\rho)=U\rho U^{\dagger}$. So if there are infinitely many unitary processes, it means that it is common we don't need the sum of the Kraus representation. And when it comes to the noise channel, we can use the sum. Therefore, the Kraus representation can be used to represent every quantum channel, and when the channel is quantum gate, it is only one term. Right? $\endgroup$
    – karry
    Nov 27, 2023 at 2:38
  • $\begingroup$ Yes, that's right. $\endgroup$ Nov 27, 2023 at 2:43
  • $\begingroup$ Ok. Thanks. : ) $\endgroup$
    – karry
    Nov 27, 2023 at 2:47

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