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

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?

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

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

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.