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Let us write the possible states of a qubit in the Bloch representation as $$\newcommand{\bs}[1]{{\boldsymbol{#1}}}\rho_{\bs r}\equiv \frac{I+\bs r\cdot\bs \sigma}{2},$$ where $\bs\sigma=(\sigma_1,\sigma_2,\sigma_3)$ are the Pauli matrices. I can then represent $\rho_{\bs r}$ as the vector $$\rho_{\bs r}\doteq \frac12\begin{pmatrix}1 \\ \bs r\end{pmatrix}.$$ In this representation, I can represent a linear operator acting on $2\times 2$ matrices, $\mathcal E\in\mathrm{Lin}(\mathrm{Lin}( \mathbb C^2))$, as a matrix $$\mathcal E\doteq \begin{pmatrix}1 & \bs 0^T \\ \bs t & \Delta\end{pmatrix},$$ where $\Delta\in\mathrm{Lin}(\mathbb C^2)$ and $\bs t\in\mathbb R^2$. This form of $\mathcal E$ automatically preserves the normalisation. The action of $\mathcal E$ on $\rho_{\bs r}$ can then be written as $$\mathcal E(\rho_{\bs r}) = \rho_{\Delta\bs r + \bs t}.$$ Is there a good way to see how the CPTP constraint on $\mathcal E$ translates into constraints on $\bs t$ and $\Delta$ in this representation?

Clearly, for $\mathcal E(\rho)$ to still be a state we need $\|\Delta \bs r+\bs t\|\le 1$. Is this the only requirement?

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This representation of $\mathcal{E}$ is also known as the Pauli transfer matrix or PTM; see this document by Greenbaum for a brief introduction.

As stated in section $2.3.2$ of that document, the TP constraint is automatically met by having the top row be $(1,0\ldots 0)$. This is because trace preservation dictates that for $\rho_{\boldsymbol{r}}$ we have $\mathrm{tr}(\mathcal{E}(\rho_{\boldsymbol{r}})) = \mathrm{tr}(\rho_{\boldsymbol{r}})$. By linearity of the trace it must thus also hold for $\boldsymbol{e}_{i}$ for $i \in \{0,1,2,3\}$, which are just the Paulis $\boldsymbol{\sigma}$ in the space where the PTM acts upon. Since they are traceless, TP gives $\mathrm{tr}(\mathcal{E}(\boldsymbol{e}_{i})) = \mathrm{tr}(\boldsymbol{e}_{i}) = d\delta_{0i}$, which is evidently only true if $\mathcal{E}^{PTM}_{0,i} = \delta_{0i}$.

As also stated in the section $2.3.2$, the CP constraint doesn't work as nicely. The constraint $||\Delta\boldsymbol{r} + \boldsymbol{t}||\leq 1$ actually also permits positive (but not-completely positive) maps. The classical example of the transpose gives: $$ \boldsymbol{\Delta}=\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 &0 & 1 \end{bmatrix}, \boldsymbol{t}= \boldsymbol{0} $$ for which $||\Delta\boldsymbol{r} + \boldsymbol{t}|| \leq 1$ obviously holds.

I would rephrase the channel into some other representation for which the CP constraint does work nicely. However, matters are made worse because, for instance, conversion between the PTM and $\chi$-representation is not always trivial (See section $2.1.3$ of the document).

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  • $\begingroup$ But if we assume $\Delta$ is positive, does $\|\Delta r + t\| \leq 1$ become a sufficient condition? Is there a counter example? $\endgroup$
    – Doris
    Jan 15 at 3:22
  • $\begingroup$ @Doris A counterexample would be $t=0$ and $\Delta={\rm diag}(0.6,0.6,0.1)$. Then $\sup_{\|r\|=1}\|\Delta r+t\|=\sup_{\|r\|=1}\|\Delta r\|=\|\Delta\|_\infty=0.6\leq 1$ so the corresponding map is positive trace preserving, but its Choi matrix has an eigenvalue $-0.05$ meaning the map is not completely positive. The key here are the Fujiwara-Algoet conditions $|s_1\pm s_2|\leq |1\pm s_3|$ on the singular values $s_j$ of $\Delta$ which---in the case $t=0$---are necessary and sufficient for complete positivity $\endgroup$ Apr 15 at 8:26

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