1. Ideal (error–free) output state
We start in the computational vacuum
\[
|0000\rangle_{0\,1\,2\,3}\equiv |0_0 0_1 0_2 0_3 \rangle .
\]
Carrying the gates through from right to left,
- \(H_2\)
\[
|0000\rangle\xrightarrow{H_2}
\tfrac{1}{\sqrt 2}\!\left(|0000\rangle+|0010\rangle\right).
\]
- \(CNOT_{21}\) (control-2, target-1)
\[
\tfrac{1}{\sqrt 2}\!\left(|0000\rangle+|0110\rangle\right).
\]
- \(H_0\)
\[
\frac{1}{2}\Bigl(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle\Bigr).
\]
- \(CNOT_{03}\) (control-0, target-3)
\[
|\psi\rangle\;=\;\frac{1}{2}\Bigl(|0000\rangle+|1001\rangle+|0110\rangle+
|1111\rangle\Bigr).
\]
This \(|\psi\rangle\) is the [[4,2,2]]-encoded Bell (GHZ) state
\((|00\rangle_{AB}+|11\rangle_{AB})/\sqrt 2\).
2. Stabilizer of the target state
The [[4,2,2]] code stabilizers together with the logical-GHZ stabilizers give
four independent generators
\[
\begin{aligned}
S_1 &= XXXX,\\[4pt]
S_2 &= ZZZZ,\\[4pt]
S_3 &= IXXI,\\[4pt]
S_4 &= IZZI,
\end{aligned}
\]
so that the full stabilizer group
\(\mathcal S=\langle S_1,S_2,S_3,S_4\rangle\) has \(|\mathcal S|=16\) elements.
Those supported only on qubits \(\{1,2\}\) are
\[
\mathcal S_{12}=\{\,IIII,\;IXXI,\;IYYI,\;IZZI\},
\]
and those supported only on qubits \(\{0,3\}\) are
\[
\mathcal S_{03}=\{\,IIII,\;XIIX,\;YIIY,\;ZIIZ\}.
\]
3. Error model and locality of the faults
• Two faulty gates:
– after \(CNOT_{21}\) an error \(P_1\) acts on qubits \(\{2,1\}\),
– after \(CNOT_{03}\) an error \(P_2\) acts on qubits \(\{0,3\}\).
• Each error is the two-qubit depolarising channel
\[
\mathcal E(\rho)=(1-p)\rho+\frac{p}{15}\sum_{\substack{P\in\mathcal P_{2}\\P\neq I}}P\rho P,
\]
so
\[
\Pr[P_i=I]=(1-p), \qquad
\Pr[P_i=P\neq I]=\frac{p}{15}.
\]
• Because the two subsequent single-qubit gates act on qubit 0 only,
\(P_1\) is not propagated: at the end of the circuit
\[
P_1\; \text{acts on qubits } 1,2,\qquad
P_2\; \text{acts on qubits } 0,3,
\]
and the total error operator is
\[
P_{\text{tot}}=P_1P_2.
\]
4. Which errors leave the state unchanged?
For a stabilizer state \(|\psi\rangle\),
\[
\langle\psi|Q|\psi\rangle=
\begin{cases}
\pm1,&Q\in\mathcal S,\\
0,&Q\notin\mathcal S.
\end{cases}
\]
Hence the state fidelity is the probability that the realised error
operator lies in \(\pm\mathcal S\).
Because \(P_1\) and \(P_2\) act on disjoint qubit sets,
\[
P_{\text{tot}}\;\in\;\mathcal S
\iff
P_1\in\mathcal S_{12}\;\text{ and }\;P_2\in\mathcal S_{03}.
\]
5. Probability that an error is in the local
stabilizer subset
On two qubits there are 15 non-identity Paulis.
Exactly three of them belong to each local stabilizer subset:
\[
\#\bigl(\mathcal S_{12}\setminus\{I\}\bigr)=
\#\bigl(\mathcal S_{03}\setminus\{I\}\bigr)=3.
\]
Therefore
\[
\begin{aligned}
\Pr\bigl[P_1\in\mathcal S_{12}\bigr] &= (1-p)+3\frac{p}{15}
=1-\frac{4p}{5},\\[6pt]
\Pr\bigl[P_2\in\mathcal S_{03}\bigr] &= (1-p)+3\frac{p}{15}
=1-\frac{4p}{5}.
\end{aligned}
\]
6. Physical state fidelity
Because the two error events are independent,
\[
F(p)=\Pr\!\bigl[P_{\text{tot}}\in\mathcal S\bigr]
=\Pr\!\bigl[P_1\in\mathcal S_{12}\bigr]\;
\Pr\!\bigl[P_2\in\mathcal S_{03}\bigr]
=\left(1-\frac{4p}{5}\right)^2.
\]
Expanding,
\[
\boxed{\,F(p)=1-\frac{8}{5}\,p+\frac{16}{25}\,p^{2}\,}.
\]
Since at most one fault can occur per gate, this expression is exact to all
orders in \(p\) for this circuit.
Final Answer:
The physical four-qubit output state has fidelity
\[
F(p)=\left(1-\frac{4}{5}p\right)^{2}
=1-\frac{8}{5}\,p+\frac{16}{25}\,p^{2}.
\]