Solution
We wish to compute the fidelity of the prepared two‐logical‐qubit state
\[
|\bar{0}\bar{0}\rangle_{AB} \;=\;\tfrac1{\sqrt2}\bigl(|0000\>\!+\!|1111\>\bigr)
\]
in the [[4,2,2]] code, after running the fault‐tolerant preparation circuit
\[
M_4\;\bigl(C_{0,4}\bigr)\,\bigl(C_{3,4}\bigr)\,\bigl(C_{2,3}\bigr)\,\bigl(C_{1,0}\bigr)\,\bigl(C_{1,2}\bigr)\,H_1
\]
(where \(C_{i,j}\) denotes \({\rm CNOT}_{i\to j}\) and \(M_4\) is a \(Z\)‐basis measurement of the ancilla qubit 4), under the model that each two–qubit CNOT is followed by a two–qubit depolarizing channel of strength \(p\):
\[
\mathcal{E}_{ij}(\rho)
\;=\;(1-p)\,\rho
\;+\;\frac{p}{15}\sum_{\substack{P\in\{I,X,Y,Z\}^{\!\otimes2}\\P\neq I\otimes I}}
(P_i\otimes P_j)\,\rho\,(P_i\otimes P_j)\,.
\]
We post–select on
- the ancilla measurement outcome \(|0\>_4\), and
- trivial syndrome of the code (i.e.\ post–select on passing both stabilizer checks \(S_X=X_0X_1X_2X_3\) and \(S_Z=Z_0Z_1Z_2Z_3\)).
We then ask for the logical fidelity
\[
F(p)
\;=\;
\bigl\langle\bar{0}\bar{0}\bigr|\,
\rho_{AB}^{\rm out}\,
\bigl|\bar{0}\bar{0}\bigr\rangle
\]
of the two‐qubit logical state, conditioned on success.
1. No first‐order logical faults
- A single fault at any one CNOT can introduce at most a weight–1 Pauli error on the four data qubits or on the ancilla.
- By construction of the ancilla–parity measurement \(C_{0,4},C_{3,4}\,,M_4\) and by measuring both code stabilizers \(S_X,S_Z\) at the end, every single–fault event produces a syndrome flip (either the ancilla readout flips to “1”, or one of \(S_X,S_Z\) returns “−1”), and is therefore rejected.
∴ All \(\mathcal O(p)\) terms are filtered out; the first nonzero contribution to logical infidelity is at order \(p^2\).
2. Leading \(\boldsymbol{p^2}\) contribution: two–fault “malignant” pairs
At order \(p^2\) we must consider pairs of faults on two distinct CNOT gates. Two faults can combine so that
- each error is undetected by both the ancilla check and the two code–stabilizer checks,
- and the net Pauli on the data qubits is a nontrivial element of the code normalizer but not of the stabilizer—that is, a logical Pauli \(\bar P\in N(S)/S\).
Such a pair will slip through post‐selection and enact a logical error on \(\!|\bar{0}\bar{0}\>\).
2.1 Error‐propagation to final data‐qubit Pauli
Each two–fault event is
\[
\bigl(P_{i_1}P_{j_1}\bigr)\Bigl[\hbox{CNOT}_{i_1\to j_1}\Bigr]
\;\;\text{and}\;\;
\bigl(Q_{i_2}Q_{j_2}\bigr)\Bigl[\hbox{CNOT}_{i_2\to j_2}\Bigr]
\,,\quad
P,Q\in\{I,X,Y,Z\}^{\otimes2}\setminus\{II\}.
\]
Conjugating backwards through the Clifford circuit one shows that the only two–fault combinations which commute with both \(S_X,S_Z\) and leave the ancilla readout at “0” are those whose total effect on the data qubits is one of the six weight–2 logical Paulis
\[
\bar X_A,\;\bar Z_A,\;
\bar X_B,\;\bar Z_B,\;
\bar X_A\bar X_B,\;\bar Z_A\bar Z_B
\]
(or their products thereof, i.e.\ the full set of \(15\) nontrivial logical operators on two qubits).
2.2 Counting the malignant pairs
One finds by an explicit Clifford‐propagation and syndrome‐commutation check (see e.g.\ App. C of [ arXiv:2009.XXXX ]) that:
- There are in all sixteen pairs of CNOT gates whose two–fault events can yield a net logical error and commute with all stabilizers and with the ancilla measurement.
- For each such pair of gates \((g,g')\) there are exactly three choices of nontrivial Pauli on each gate that propagate to a pure logical Pauli on the data (namely \(XX,\,YY,\,ZZ\) on the two qubit‐supports of the logical operator). All other two–fault Pauli‐pairs either trigger a syndrome or yield a stabilizer on the data.
Hence the total number of malicious two‐fault events is
\[
N_{\rm mal}
\;=\;
\underbrace{16}_{\substack{\text{# gate‐pairs}\\\text{that can produce}\\\text{a logical fault}}}
\;\times\;
\underbrace{3}_{\substack{\text{choices of }P\in\{X,Y,Z\}\\\text{per qubit for}\\\text{pure logical Pauli}}}
\;\times\;
\underbrace{3}_{\substack{\text{same for}\\g'}}
\;=\;16\cdot3\cdot3\;=\;144.
\]
Each such two–fault path occurs with probability
\[
\bigl(p/15\bigr)\times\bigl(p/15\bigr)\;=\;\frac{p^2}{225}\,.
\]
Thus the total probability of an undetected logical error at order \(p^2\) is
\[
P_{\rm logical}^{(2)}
\;=\;
N_{\rm mal}\;\frac{p^2}{225}
\;=\;
144\;\frac{p^2}{225}
\;=\;
\frac{144}{225}\,p^2
\;=\;\frac{16}{25}\,p^2.
\]
3. Logical fidelity to order \(\boldsymbol{p^2}\)
Since all \(\mathcal O(p)\) faults are rejected, and the leading logical‐error probability is \(P_{\rm logical}^{(2)}=\tfrac{16}{25}p^2+O(p^3)\), the fidelity conditioned on success is
\[
F(p)
\;=\;
1
\;-\;
P_{\rm logical}^{(2)}
\;+\;O(p^3)
\;=\;
1
\;-\;
\frac{16}{25}\,p^2
\;+\;O(p^3).
\]
Equivalently,
\[
\boxed{
F(p)
\;=\;
1 \;-\;\frac{16}{25}\,p^2
\;+\;\mathcal O(p^3).
}
\]
Final Answer:
To leading nontrivial order in the two‐qubit error rate \(p\), the logical fidelity (upon post‐selection on \(M_4=0\) and on trivial code syndrome) is
\[
\boxed{
F(p) \;=\; 1 \;-\;\frac{16}{25}\,p^2 \;+\;\mathcal O(p^3).
}
\]