1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.