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!

enter image description here

Please refer to Page 376, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang

  • 2
    $\begingroup$ 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. $\endgroup$
    – narip
    Commented May 19, 2022 at 6:05
  • $\begingroup$ @narip Thank you! $\endgroup$ Commented May 19, 2022 at 18:57

1 Answer 1


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, $\rho=\dfrac{I+\vec{r}.\vec{\sigma}}{2}=\dfrac{1}{2}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}$

The action of the bit flip channel can be defined as,

\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} $$ \vec{r}=(x,y,z)\xrightarrow{\mathcal{E}}\vec{r}'=(x,(2p-1)y,(2p-1)z) $$


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