Step-by-Step Derivation
- Ideal (error-free) circuit
• All five data qubits (0-3) and the ancilla (4) are initialised in
\(\lvert 0\rangle^{\otimes 5}\).
• Running the gates from right to left,
\[
H_1\;\;CNOT_{12}\;\;CNOT_{10}\;\;CNOT_{23}\;\;CNOT_{34}\;\;CNOT_{04}
\]
\[
\]
while the ancilla returns to $\lvert 0\rangle_4$.
(The four–qubit GHZ state above is the code word for the logical state
$\lvert 00\rangle_{AB}$, since it is a simultaneous $+1$ eigenstate of
the stabilisers $XXXX$ and $ZZZZ$ and of the logical operators
$Z_A=ZZII$ and $Z_B=ZIZI$.)
- Error model
• After each of the five CNOT gates an independent two-qubit
depolarising channel acts:
with probability \(p\) one of the \(15\) non-identity two-qubit Pauli
operators is applied, each with probability \(p/15\).
• Single-qubit gates and measurements are taken to be perfect.
Up to first order in \(p\) it suffices to keep only a single error
happening somewhere in the circuit (terms of order \(p^{2}\) and higher
are neglected).
- What gets discarded?
At the end of the circuit we
a) measure the ancilla in the \(Z\) basis and keep the run only when the
outcome is \(0\), and
b) measure both stabilisers and keep the state only when the syndrome is
\(+1,+1\).
• Any Pauli on the ancilla that contains \(X\) or \(Y\) flips the ancilla to
\(\lvert 1\rangle\) and is therefore discarded.
• Any Pauli acting on the data block that has an odd
number of \(X\)’s (or \(Y\)’s) anticommutes with the stabiliser \(ZZZZ\) and
is discarded.
• Any Pauli on the data block that has an odd number of \(Z\)’s
(or \(Y\)’s) anticommutes with the stabiliser \(XXXX\) and is discarded.
Consequently every single-qubit error on the data qubits is detected,
i.e. no weight-1 error survives the post-selection.
Only weight-2 errors that commute with both stabilisers and do not put
an \(X\) or \(Y\) on the ancilla can slip through.
- Propagating the 15 Pauli errors of every CNOT gate
The five CNOTs appear in the time order
\[
(2)\;CNOT_{10}\;,\qquad
(3)\;CNOT_{23}\;,\qquad
(4)\;CNOT_{34}\;,\qquad
\]
Conjugating a Pauli \(P\) through a CNOT with control \(c\) and target \(t\)
\[
X_c &\longrightarrow X_cX_t,&\quad Z_c&\longrightarrow Z_c,\\
X_t &\longrightarrow X_t, &\quad Z_t&\longrightarrow Z_cZ_t
\]
allows us to see how an error that occurs after gate \((k)\) is carried
forward by the remaining gates.
• Gate \((1)\) - \(CNOT_{12}\) (error acts on qubits 1 & 2)
Among the \(15\) Paulis only
\[
Z_1Z_2
\]
remains a weight-2 operator on the data qubits, commutes with both
stabilisers and avoids placing $X/Y$ on the ancilla.
It acts as the logical operator $Z_A$ and therefore flips the logical
state although it is not detected.
• Gate \((2)\) - \(CNOT_{10}\) (error acts on qubits 1 & 0)
Only
\[
Z_1Z_0
\]
survives all tests. It implements the undetected logical $Z_A$.
• Gate \((3)\) - \(CNOT_{23}\) (error acts on qubits 2 & 3)
Only
\[
Z_2Z_3
\]
survives. It equals the logical $Z_B$ and is likewise undetected.
• Gate \((4)\) - \(CNOT_{34}\) (error on qubits 3 & 4)
The only Pauli that is never flagged is
\[
Z_4,
\]
which acts trivially on the code because it is removed when the ancilla
is projected onto $\lvert 0\rangle$. Hence it is *harmless*.
• Gate \((5)\) - \(CNOT_{04}\) (error on qubits 0 & 4)
Again only the pure phase error
\[
Z_4
\]
survives, which is harmless for the same reason.
Summary (first order in \(p\))
────────────────────────────────────────────
accepted & harmful : 3 patterns … \(Z_1Z_2,\;Z_1Z_0,\;Z_2Z_3\)
accepted & harmless: 2 patterns … \(Z_4\) (from gate 4 or 5)
discarded : the remaining \(15\times5-5=70\) patterns
────────────────────────────────────────────
- Probabilities to first order in \(p\)
• Each gate fails with probability \(p\); each particular Pauli appears
with probability \(p/15\).
a) Probability that an undetected logical error occurred:
\[
\]
b) Probability that the run is accepted (ancilla \(0\) and
trivial syndrome):
\[
(1-p)^{5}+5\!\left(\frac{p}{15}\right)
\;=\;1-5p+\frac{p}{3}
\;=\;1-\frac{14}{3}p
\]
(The five accepted single-Pauli patterns are the three harmful ones
plus the two $Z_4$ patterns.)
c) Probability that the accepted state is the correct logical state:
\[
(1-p)^{5}+2\!\left(\frac{p}{15}\right)
\;=\;1-5p+\frac{2}{15}p
\]
- Logical state fidelity
The logical fidelity is the conditional probability that the accepted
state equals \(\lvert 00\rangle_{AB}\):
\[
\frac{P_{\text{corr}}}{P_{\text{acc}}}
=\frac{1-\dfrac{73}{15}p}{1-\dfrac{14}{3}p}
\]
Hence, to leading order in the physical two-qubit gate error rate \(p\),
the post-selected preparation protocol suppresses the error probability
by a factor of \(5\).
Final Answer:
The logical two-qubit state fidelity after post-selection is
\[
F_{\text{logical}}(p)=1-\frac{p}{5}+O(p^{2}),
\]
i.e. the first-order logical infidelity is \(p/5\).