Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 58 / 3

The first two CNOT gates encode $u|0⟩+v|1⟩$ asthe 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 $|0⟩$. For an incoming bit-error pattern $e=(e_1,e_2,e_3)$, the two logical basis branches give the same error syndromeAdding $111$ to a computational string leaves both parities unchanged, so the ancillas acquire no information about $u$ or $v$. The joint state after extraction isThe syndrome measurement therefore preserves the unknown logical amplitudes. For zero or one bit flip the corrections areEach 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 $X_1{X}_2{X}_3$, the logical bit flip that exchanges the two codewords. The code protects against the stated bit-flip channel, not arbitrary single-qubit phase errors.

The PDF places the extra $X$ on $q_2$ after that qubit has interacted with $a_1$ but before it interacts with $a_2$. Its timing is essential. With no incoming error, the first extracted parity is zero. The fault then changes the data toThe remaining parity measurement gives $a_2=1$ on both branches. Thus the measured error syndrome is $01$, which the rule in the preceding part interprets as an error on $q_1$. Applying $X_1$ leavesThis is outside the original encoded subspace: neither of its supporting basis strings is $000$ or $111$. The incorrect syndrome causes a second data error rather than restoring the input. This timed bit flip during repetition-code syndrome extraction demonstrates why a simple parity circuit need not tolerate its own faults.

The same failure can be checked for every incoming channel error. Let $f\in \{0,1\}$ indicate the internal fault and let $c(s)$ be the correction bit pattern from the table. The observed syndrome and the final pre-correction data pattern areBecause $s$ is linear over binary addition and $s(c(s_{\mathrm{obs}}))=s_{\mathrm{obs}}$, the residual pattern $r=d\oplus c(s_{\mathrm{obs}})$ satisfiesWhenever the fault occurs, the residual syndrome is $10$, so recovery never returns the state to the encoded subspace, even if a channel error also occurred. The diagram therefore must not be treated as an extra independent pre-extraction data flip that would always be diagnosed normally.

Use the probability of failing to restore an arbitrary unknown input in the original encoded subspace as the error criterion. With no extraction fault, the bit-flip repetition code succeeds exactly on channel-error patterns of weight zero or one. Independent physical flips giveThe probability $B$ is the logical bit-flip probability for the ideal recovery. The preceding calculation shows that every occurrence of the specified internal fault leaves the code space. Independence of that fault therefore gives exact encoded-state failure with a faulty parity check:A single unencoded qubit has bit-flip error probability $p$. Thus the encoded scheme improves this error criterion precisely whenFor $0<p<1/2$, $p-B=p(1-p)(1-2p)>0$ and $A=(1-p{)}^2(1+2p)>0$. Rearranging givesEquality gives no improvement. At $p=0$ the unencoded channel is already perfect; at $p=1/2$ ideal recovery only ties it. For $1/2<p\le 1$, the stipulated majority correction is no better even without the internal fault, and this fault cannot improve it. Hence there is no nonnegative $q$ giving a strict improvement in those ranges.

The criterion concerns restoration of every possible logical state: a logical bit flip can accidentally preserve a particular logical $X$ eigenstate, but that is not protection of an unknown input. The threshold assumes the displayed single fault location and ideal encoding, measurement and conditional correction. Adding another recovery circuit or comparing a different average-fidelity metric would define a different noise model or performance criterion.