Exercise 10.6#
Consider an arbitrary encoded logical qubit \(|\psi_L\rangle\) of Shor’s code as
(1)#\[
|\psi_L\rangle = \alpha |0_L\rangle + \beta |1_L\rangle
\]
where
(2)#\[\begin{split}
\begin{align}
|0_L\rangle &= \frac{(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)}{2\sqrt{2}}\\
|1_L\rangle &= \frac{(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)}{2\sqrt{2}}
\end{align}
\end{split}\]
Consider a phase flip error on any of first three qubits, since \(Z|0\rangle = |0\rangle\) and \(Z|1\rangle = -|1\rangle\), we will have noisy \(|0_L\rangle\) and \(|1_L\rangle\) state as
(3)#\[\begin{split}
\begin{align}
|0_L\rangle &\to \frac{(|000\rangle - |111\rangle)(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)}{2\sqrt{2}}\\
|1_L\rangle &\to \frac{(|000\rangle + |111\rangle)(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)}{2\sqrt{2}}
\end{align}
\end{split}\]
Then apply \(Z_{1}Z_{2}Z_{3}\) to noisy states, we have
(4)#\[\begin{split}
\begin{align}
&\frac{Z_{1}Z_{2}Z_{3}(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)}{2\sqrt{2}} \\
&= \frac{(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)(|000\rangle + |111\rangle)}{2\sqrt{2}}=|0_L\rangle \\
&\frac{Z_{1}Z_{2}Z_{3}(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)}{2\sqrt{2}} \\
&= \frac{(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)(|000\rangle - |111\rangle)}{2\sqrt{2}}=|1_L\rangle
\end{align}
\end{split}\]
Therefore, applying \(Z_{1}Z_{2}Z_{3}\) to noisy state recover noisy \(|0_L\rangle\) and \(|1_L\rangle\) state from a phase flip error on any of the first three qubits, and it can naturally recover a phase flip error on any of the first three qubits of arbitrary state \(|\psi\rangle = \alpha |0_L\rangle + \beta |1_L\rangle\).