What is the root of the non-trace-preserving bit-flip map

Question

I have a quantum channel defined by the Kraus operators: $$ U_1 = \begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix},\quad U_2 = \begin{bmatrix} 0 & p \\ p & 0 \end{bmatrix} $$ i.e. $$ U_1\rho U_1^* + U_2\rho U_2^*. $$ Now I want to find the square root of this channel i.e. a channel that applied twice results in the given channel. Of course, this can be solved via a system of equations but is there an easier way to do this?

Answer 1

Assuming w.l.o.g. that $p\in\mathbb{R}$, the linear map in the question may be rewritten as

$$ \mathcal{E}(\rho) = p^2\rho+p^2X\rho X = 2p^2\left(\frac12\rho + \frac12 X\rho X\right) $$

where $X$ is the Pauli matrix. Thus, the action of $\mathcal{E}$ can be understood as the composition $\mathcal{E}=\mathcal{S}_{2p^2}\circ\mathcal{X}_{\frac12}$ of a scaling map $\mathcal{S}_{2p^2}$ and a bit-flip map $\mathcal{X}_{1/2}$ where

$$ \mathcal{S}_{a}(\rho)=a\rho \\ \mathcal{X}_{b}(\rho)=(1-b)\rho + bX\rho X. $$

Now, since the two maps commute we try to take the square roots independently. Moreover, it is easy to see that

$$ \mathcal{S}_a\circ\mathcal{S}_{a'}=\mathcal{S}_{aa'} \\ \mathcal{X}_{1/2}\circ\mathcal{X}_{1/2}=\mathcal{X}_{1/2}. $$

We infer that $\mathcal{F} := \mathcal{S}_{p\sqrt2}\circ\mathcal{X}_{1/2}$ is the square root of $\mathcal{E}$, i.e. that $\mathcal{E}=\mathcal{F}\circ\mathcal{F}$. The map $\mathcal{F}$ can be written as

$$ \begin{align} \mathcal{F}(\rho)&=p\sqrt2\left(\frac12\rho + \frac12 X\rho X\right)\\ &=\frac{p}{\sqrt2}\rho + \frac{p}{\sqrt2}X\rho X \end{align} $$

and has Kraus operators

$$ V_1 = \frac{\sqrt{p}}{\sqrt[4]{2}}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad V_2 = \frac{\sqrt{p}}{\sqrt[4]{2}}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$

which are proportional to the Kraus operators $U_1$ and $U_2$ of $\mathcal{E}$. In particular, if the original map $\mathcal{E}$ is trace-preserving, i.e. if $p=\frac{1}{\sqrt2}$, then the map is idempotent and hence its own square root.