1. Step–by–Step Derivation
1.1 Circuit and error locations
• The preparation circuit (written left–to–right in time order) is
H₂ → CNOT₂1 → H₀ → CNOT₀3.
• The hardware model inserts a two–qubit depolarising channel after each CNOT.
For the CNOT acting on qubits (i,j) the channel is
\[
\mathcal D_{ij}(\rho)=
(1-p)\rho+\frac{p}{15}\sum_{P\in\mathcal{P}_{ij}\setminus\{II\}} P\rho P ,
\]
where \(\mathcal{P}_{ij}\) is the 16-element two–qubit Pauli group on qubits \(i,j\).
Thus two independent errors appear:
• \(E_1\) (on qubits 2,1) after CNOT₂1,
• \(E_2\) (on qubits 0,3) after CNOT₀3.
Single–qubit Hadamards are assumed perfect.
1.2 Effective Pauli acting on the final state
Because the gates following \(E_1\) act only on qubits 0 and 3, \(E_1\) commutes with all later gates.
Hence the net error on the final state is simply the tensor product
\[
P = E_2\,E_1 = (E_2^{(0,3)})\otimes(E_1^{(1,2)}),
\qquad
E_1\in\mathcal{P}_{21},\;E_2\in\mathcal{P}_{03}.
\]
The two error channels are independent, so
\(\Pr\!\bigl(P=E_2E_1\bigr)=\Pr(E_1)\Pr(E_2).\)
1.3 Ideal target state \(|\psi\rangle\)
After the noiseless circuit the four physical qubits are in the logical Bell (GHZ) state.
It is the stabiliser state generated by
\[
\begin{aligned}
S_1 &= XXXX, \\
S_2 &= ZZZZ, \\
S_3 &= X_A X_B = IXXI, \\
S_4 &= Z_A Z_B = IZZI .
\end{aligned}
\]
The full stabiliser group \(\mathcal S\) has \(2^4=16\) elements.
For a stabiliser state \(|\psi\rangle\),
\[
\langle\psi|P|\psi\rangle =
\begin{cases}
\pm1 & \text{if }P\in\mathcal S,\\[2pt]
0 & \text{otherwise}.
\end{cases}
\]
Therefore
\[
F(p)=\langle\psi|\rho|\psi\rangle
=\sum_{E_1,E_2}\Pr(E_1)\Pr(E_2)\,
\bigl|\langle\psi|E_2E_1|\psi\rangle\bigr|^2
=\Pr\!\bigl(P\in\mathcal S\bigr).
\]
1.4 Which Pauli products lie in the stabiliser?
Every stabiliser element has the form
\[
P_{\sigma,\tau}
= \sigma\otimes\tau\otimes\tau\otimes\sigma,
\qquad
\sigma,\tau\in\{I,X,Y,Z\}.
\]
Hence \(P=E_2E_1\in\mathcal S\) iff
• the two-qubit Pauli on qubits (0,3) is \(\sigma\sigma\) for some \(\sigma\), and
• the two-qubit Pauli on qubits (1,2) is \(\tau\tau\) for some \(\tau\).
1.5 Probability that a two–qubit depolarising channel outputs \(\sigma\sigma\)
There are 16 two-qubit Paulis.
Exactly four of them have identical single-qubit factors:
\(II,\;XX,\;YY,\;ZZ.\)
\[
\Pr(\text{``same on both qubits''}) =
(1-p) \;+\; 3\bigl(p/15\bigr)
= 1-p+\tfrac{p}{5}
= 1-\tfrac{4p}{5}.
\]
1.6 Total fidelity
The two error channels are independent, so
\[
F(p)=\Bigl(1-\tfrac{4p}{5}\Bigr)^2
=1-\frac{8}{5}p+\frac{16}{25}p^{2}.
\]
2. Final Answer
Final Answer: The physical four-qubit fidelity after the noisy preparation circuit is
\[
\boxed{F(p)=\left(1-\frac{4p}{5}\right)^{2}
=1-\frac{8}{5}p+\frac{16}{25}p^{2}.}
\]