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The question is how many free real parameters a general CPTP map can maximally have.

Let's assume the CPTP map $\Phi:L(\mathcal{H}_A) \rightarrow L(\mathcal{H}_B)$ is given in the Kraus representation \begin{equation} \Phi(\rho) = \sum_{n=1}^r M_n\, \rho\, M_n^\dagger \;, \end{equation} with the completeness relation \begin{equation} \sum_{n=1}^r M_n^\dagger\, M_n = \mathbb{1}_A \;, \end{equation} and where $r=\mathrm{rank}(J(\Phi))$ is the rank of the Choi matrix of $\Phi$. Here, $L(\mathcal{H})$ denotes the set of linear operators mapping $\mathcal{H}$ onto itself. Following Watrous' book, I know that the maximal number of Kraus operators possibly needed to describe any CPTP map is upper bounded by the rank of the Choi matrix (that is why I chose $r$ as above).

Now my reasoning is this:

  • The Choi matrix $J(\Phi)$ is a linear operator mapping $\mathcal{H}_B \otimes \mathcal{H}_A$ onto itself. Thus the maximal rank of $J(\Phi)$ is $d_A d_B$, where $d_A = \mathrm{dim}(\mathcal{H}_A)$, and likewise for $B$. This is the maximal number of different Kraus operators $M_n$ one might need.
  • Each $M_n$ is a linear operator from $\mathcal{H}_A$ to $\mathcal{H}_B$. Thus it has in general $d_A d_B$ complex entries, i.e. $2 d_A d_B$ free real parameters.
  • The completeness relation given above fixes $d_A^2$ real parameters.
  • This leaves us with $d_A d_B \cdot 2 d_A d_B - d_A^2 = d_A^2 (2d_B^2 - 1)$ free parameters.

Is my reasoning correct or did I overlook something? I get the same result when considering the Stinespring representation and counting the free real parameters in the isometry.

Any help much appreciated.

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    $\begingroup$ Isn't it easier to count the parameters in the Choi state? (Which immediately tells you that you overcounted: Even a general hermitian matrix has only $d_A^2d_B^2$ parameters.) $\endgroup$ Feb 26 at 16:51
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    $\begingroup$ ... and the condition that the reduced state of the input is the identity should take out $d_A^2$ real parameters (as the reduced state is already hermitian). Thus, $d_A^2(d_B^2-1)$, i.e. up to the factor of $2$, this seems fine. --- Note that the Kraus representation is highly overparametrized, as there are many equivalent Kraus representations. This is likely where you get your factor 2, and why it is better to look at unique (non-overparametrized) representations of the channel. $\endgroup$ Feb 26 at 16:58
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    $\begingroup$ To give you an intuition why this overcounts: One way to obtain $M_n$s is from an eigenvalue decomposition of the Choi state. Let $d=d_Ad_B$ the dimension of the space. If you assign to each eigenvector $2d$ real parameters, you will overcount. The reason is that the eigenvectors have successively less independent parameters, as they have to be orthogonal to all of the previous ones. This takes out roughly one complex parameter per eigenvector, so you will -- as expected -- obtain roughly half the parameters. $\endgroup$ Feb 26 at 17:08

1 Answer 1

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An easier way of counting is to consider the Choi state: It is a hermitian $d_Ad_B\times d_Ad_B$ matrix, and thus has $(d_Ad_B)^2$ real parameters.

The condition that the reduced state of $A$ is the identity gives $d_A^2$ real constraints (real since the reduced state is already hermitian). You thus remain with $$ d_A^2(d_B^2-1) $$ real parameters.

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  • $\begingroup$ Thanks that sounds way simpler than my method. But this means that I must have made a mistake in my argument. I cannot figure out where I collected the additional factor of 2. $\endgroup$
    – Tobias95
    Feb 26 at 17:07
  • $\begingroup$ I just added a comment above explaining it. (Or rather giving a reasoning.) $\endgroup$ Feb 26 at 17:08
  • $\begingroup$ Yes, I just saw it. Thanks again for the clarification! $\endgroup$
    – Tobias95
    Feb 26 at 17:11

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