# How does the CPTP constraint reflect on the matrix representation of a qubit channel in the Pauli basis?

Let us write the possible states of a qubit in the Bloch representation as $$\newcommand{\bs}{{\boldsymbol{#1}}}\rho_{\bs r}\equiv \frac{I+\bs r\cdot\bs \sigma}{2},$$ 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?

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).