Step-by-Step Derivation
1. Noiseless circuit and ideal output state
We start from the physical input
\[
|\psi_{\rm in}\rangle = |0_0\,0_1\,0_2\,0_3\rangle
\]
and apply, in order (right–to–left),
\[
U = \bigl(\mathrm{CNOT}_{0\to3}\bigr)\,(H_0)\,\bigl(\mathrm{CNOT}_{2\to1}\bigr)\,(H_2)\,.
\]
Labeling qubits as \((0,1,2,3)\):
- After \(H_2\) on qubit 2:
\[
|\psi_1\rangle
= \frac1{\sqrt2}\bigl|0\,0\,0\,0\bigr\rangle
+ \frac1{\sqrt2}\bigl|0\,0\,1\,0\bigr\rangle.
\]
- After \(\mathrm{CNOT}_{2\to1}\) (control\,=\,2, target\,=\,1):
\[
|\psi_2\rangle
= \frac1{\sqrt2}\bigl|0\,0\,0\,0\bigr\rangle
+ \frac1{\sqrt2}\bigl|0\,1\,1\,0\bigr\rangle.
\]
- After \(H_0\) on qubit 0:
\[
|\psi_3\rangle
= \frac12\Bigl(
|0\,0\,0\,0\rangle + |1\,0\,0\,0\rangle
+|0\,1\,1\,0\rangle + |1\,1\,1\,0\rangle
\Bigr).
\]
- After \(\mathrm{CNOT}_{0\to3}\) (control = 0, target = 3):
\[
|\psi\rangle
= \frac12\Bigl(
|0\,0\,0\,0\rangle + |1\,0\,0\,1\rangle
+|0\,1\,1\,0\rangle + |1\,1\,1\,1\rangle
\Bigr).
\]
One checks immediately that
\[
|\psi\rangle
= \frac{|0_0\,0_3\rangle + |1_0\,1_3\rangle}{\sqrt2}\;\otimes\;
\frac{|0_1\,0_2\rangle + |1_1\,1_2\rangle}{\sqrt2}
= |\Phi^+\rangle_{0,3}\;\otimes\;|\Phi^+\rangle_{1,2}\,,
\]
i.e.\ the final state is a product of two Bell pairs on qubit–pairs \((0,3)\) and \((1,2)\).
2. Noise model
Each CNOT gate is followed by an independent two‐qubit depolarizing channel on its control–target pair. Concretely, if \(G\) is a noiseless CNOT on qubits \((i,j)\) then the noisy gate acts as
\[
\mathcal{E}_{ij}(\rho)
= (1-p)\,G\,\rho\,G^\dagger
+ \frac{p}{15}\sum_{(a,b)\neq(0,0)}
\bigl(\sigma_a^{(i)}\otimes\sigma_b^{(j)}\bigr)\,
G\,\rho\,G^\dagger\,
\bigl(\sigma_a^{(i)}\otimes\sigma_b^{(j)}\bigr)\!,
\]
where \(\{\sigma_0,\sigma_1,\sigma_2,\sigma_3\}=\{I,X,Y,Z\}\) and \(p\) is the two‐qubit error rate.
Since the two CNOTs act on disjoint qubit‐pairs, and the Hadamards are assumed perfect, all errors on \((2,1)\) commute through the remaining gates to the end as Paulis on qubits \((1,2)\), and likewise errors on \((0,3)\) remain on that pair. Thus the final density matrix is
\[
\rho_{\rm out}
=\sum_{P\in\{I,X,Y,Z\}^{\otimes2}}\sum_{Q\in\{I,X,Y,Z\}^{\otimes2}}
\;w_{P}\,w_{Q}\;
\bigl(Q_{0,3}\otimes P_{1,2}\bigr)\,
|\psi\rangle\langle\psi|\,
\bigl(Q_{0,3}\otimes P_{1,2}\bigr),
\]
where
\[
w_{I\,I} = 1-p,\quad
w_{P\neq I\,I} = \frac p{15}
\]
for each two‐qubit Pauli \(P\) (resp.\ \(Q\)).
3. Fidelity as a sum over Pauli‐error patterns
The physical fidelity is
\[
F(p)
= \langle\psi|\rho_{\rm out}|\psi\rangle
= \sum_{P,Q} w_{P}\,w_{Q}\;
\bigl|\langle\psi|(Q_{0,3}\otimes P_{1,2})|\psi\rangle\bigr|^2.
\]
But \(|\psi\rangle = |\Phi^+\rangle_{0,3}\otimes|\Phi^+\rangle_{1,2}\), and on each Bell pair the only Pauli operators that satisfy
\[
\sigma\otimes\sigma\;|\Phi^+\rangle = \pm|\Phi^+\rangle
\]
are precisely \(\{\;I\otimes I,\;X\otimes X,\;Y\otimes Y,\;Z\otimes Z\}\). Hence:
– No errors (\(P=I\otimes I\), \(Q=I\otimes I\)):
weight \((1-p)^2\), contributes \((1-p)^2\).
– Single error on \((1,2)\) only (\(P\neq I\!I\), \(Q=I\!I\)):
there are 15 non‐identity \(P\), of which 3 (\(XX,YY,ZZ\)) lie in the Bell–pair stabilizer.
total contribution
\[
(1-p)\;\frac p{15}\;\times 3
= \frac p5\,(1-p).
\]
– Single error on \((0,3)\) only (\(P=I\!I\), \(Q\neq I\!I\)):
likewise contributes
\[
\frac p5\,(1-p).
\]
– Errors on both pairs (\(P\neq I\!I\), \(Q\neq I\!I\)):
there are \(3\times3=9\) favorable choices among the \(15\times15=225\) total, each with weight \((p/15)^2\).
total contribution
\[
9\;\bigl(\tfrac p{15}\bigr)^2 = \frac{p^2}{25}.
\]
Summing all contributions:
\[
F(p)
= (1-p)^2
+ 2\,\frac p5\,(1-p)
+ \frac{p^2}{25}
= 1 - \frac{8}{5}\,p + \frac{16}{25}\,p^2.
\]
Final Answer:
\[
\boxed{%
F(p)\;=\;(1-p)^2 \;+\;\frac{2p}{5}\,(1-p)\;+\;\frac{p^2}{25}
\;=\;1 \;-\;\tfrac{8}{5}\,p \;+\;\tfrac{16}{25}\,p^2
}\,.
\]