= Solution
The first two <CNOT gates> encode $u|0\rangle+v|1\rangle$ as
$$
|\psi_L\rangle=u|000\rangle+v|111\rangle,
$$
the three-qubit <bit-flip repetition code>. In the printed detection circuit the first <quantum ancilla> records $q_3\oplus q_2$ and the second records $q_2\oplus q_1$. Both start in $|0\rangle$. For an incoming bit-error pattern $e=(e_1,e_2,e_3)$, the two logical basis branches give the same <error syndrome>
$$
s(e)=(e_2\oplus e_3,\ e_1\oplus e_2).
$$
Adding $111$ to a computational string leaves both parities unchanged, so the ancillas acquire no information about $u$ or $v$. The joint state after extraction is
$$
X_1^{e_1}X_2^{e_2}X_3^{e_3}|\psi_L\rangle\otimes|s(e)\rangle_{a_1a_2}.
$$
The syndrome measurement therefore preserves the unknown logical amplitudes. For zero or one bit flip the corrections are
$$
\boxed{\begin{array}{c|c|c}
(a_1,a_2)&\text{error}&\text{correction}\\\hline
00&I&I\\
01&X_1&X_1\\
10&X_3&X_3\\
11&X_2&X_2
\end{array}}
$$
Each indicated correction squares to the identity and restores the encoded state. The order of the two ancilla labels matters: the printed first ancilla is the $q_2,q_3$ parity, not the $q_1,q_2$ parity. This is <coherent syndrome extraction for the three-qubit repetition code> with the circuit's specific parity choices.
For later use, an error pattern and its complement have the same syndrome. Minimum-weight recovery corrects patterns of weight zero or one. Patterns of weight two or three instead leave $X_1X_2X_3$, the logical bit flip that exchanges the two codewords. The code protects against the stated bit-flip channel, not arbitrary single-qubit phase errors.
Back to article page
