# Why does the bit flip channel produce a uniform contraction of $1-2p$?

Studying the bit flip channel using the Nielsen & Chuang's.

And ran into the picture with the caption stating $$yz$$ plane is uniformly contracted by a factor of $$1-2p$$. I don't quite understand how the factor, $$1-2p$$, comes about and is derived. I'd appreciate any pointers here. Thank you!

• Parametrize state $\rho$ as $\frac{1}{2}\left( \begin{matrix} 1+z& x-iy\\ x+iy& 1-z\\ \end{matrix} \right)$ and then try to let it pass through the channel with Kraus operator description. Commented May 19, 2022 at 6:05
• @narip Thank you! Commented May 19, 2022 at 18:57

The operation elements corresponding to the bit flip channel are, $$E_0=\sqrt{p}I=\sqrt{p}\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ and $$E_1=\sqrt{1-p}X=\sqrt{1-p}\begin{bmatrix}0&1\\1&0\end{bmatrix}$$
And the density matrix of the initial state is, $$\rho=\dfrac{I+\vec{r}.\vec{\sigma}}{2}=\dfrac{1}{2}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}$$
\begin{align} \mathcal{E}(\rho)&=E_0\rho E_0^\dagger+E_1\rho E_1^\dagger=\rho +X\rho X\\ &=\frac{p}{2}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}+\frac{1-p}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}\\ &=\frac{p}{2}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}+\frac{1-p}{2}\begin{bmatrix}1-z&x+iy\\x-iy&1+z\end{bmatrix}\\ &=\begin{bmatrix}1+z(2p-1)&x-iy(2p-1)\\x+iy(2p-1)&1-z(2p-1)\end{bmatrix}\\ \sigma_x&=\begin{bmatrix}0&1\\1&0\end{bmatrix},\sigma_y=\begin{bmatrix}0&-i\\i&0\end{bmatrix},\sigma_z=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\\ \mathcal{E}(\rho)&=\frac{I+x\sigma_x+y(2p-1)\sigma_y+z(2p-1)\sigma_z}{2}=\frac{I+\vec{r}'.\vec{\sigma}}{2}\\ \end{align} $$\vec{r}=(x,y,z)\xrightarrow{\mathcal{E}}\vec{r}'=(x,(2p-1)y,(2p-1)z)$$