Step-by-Step Derivation
We label the four physical qubits by 0,1,2,3, and start in the product state
\[
\rho_0 = |0000\rangle\langle0000|\,.
\]
The ideal preparation circuit is
\[
U_{\rm ideal} \;=\; \Bigl(\mathrm{CNOT}_{0\to3}\Bigr)\,(H_0)\,\Bigl(\mathrm{CNOT}_{2\to1}\Bigr)\,(H_2)\,,
\]
where the gates act (right‐to‐left) on \(\rho_0\). We assume only the two CNOT gates are noisy: immediately after each CNOT\(_{i\to j}\) we apply a two‐qubit depolarizing channel of error‐rate \(p\),
\[
\mathcal E^{(ij)}(\sigma)
\;=\;(1-p)\,\sigma\;+\;\frac p{15}\sum_{P\in\{I,X,Y,Z\}^{\otimes2}\setminus\{I\otimes I\}}
P\,\sigma\,P\,.
\]
Thus the actual final state is
\[
\rho_f
=\;\mathcal E^{(0,3)}\!\bigl(
\mathrm{CNOT}_{0\to3}\,
H_0\,
\mathcal E^{(2,1)}\!\bigl(\mathrm{CNOT}_{2\to1}\,H_2\,\rho_0\,H_2^\dagger\,\mathrm{CNOT}_{2\to1}^\dagger\bigr)\,
H_0^\dagger\,
\mathrm{CNOT}_{0\to3}^\dagger
\bigr)\,.
\]
The \emph{physical‐state fidelity} is
\[
F(p)\;=\;\bigl\langle\psi_{\rm ideal}\bigr|\,
\rho_f\,
\bigl|\psi_{\rm ideal}\bigr\rangle
\,,\qquad
|\psi_{\rm ideal}\rangle
=\;U_{\rm ideal}\,|0000\rangle\,.
\]
1. Unitary‐covariance of the Pauli‐depolarizing channel
Because each \(\mathcal E^{(ij)}\) is a uniform average over the non‐identity two‐qubit Pauli group, it is covariant under any Clifford unitary on qubits \(i,j\). In particular for a Clifford \(V\) on \((i,j)\),
\[
V^\dagger\,\mathcal E^{(ij)}(\sigma)\,V
=\mathcal E^{(ij)}\!\bigl(V^\dagger\,\sigma\,V\bigr)\,.
\]
Both CNOT and \(H\) are Cliffords. Hence in the fidelity
\[
F
=\Tr\Bigl[\,|\psi_{\rm ideal}\rangle\langle\psi_{\rm ideal}|\;\rho_f\Bigr]
\]
we may cyclically commute the ideal unitaries \(U_{\rm ideal}\) past the \(\mathcal E\)‐channels, reducing the problem to computing the fidelity of the \emph{intermediate} pure state immediately before each noise channel.
2. State after the first two gates (before \(\mathrm{CNOT}_{0\to3}\))
Compute
\[
|\phi\rangle
\;=\;\mathrm{CNOT}_{2\to1}\,H_2\,|0000\rangle.
\]
- \(H_2\,|0000\rangle = \tfrac1{\sqrt2}(|0000\rangle + |0010\rangle)\).
- \(\mathrm{CNOT}_{2\to1}\) flips qubit 1 iff qubit 2=1, so
\[
|\phi\rangle
=\frac1{\sqrt2}\bigl(|0\,0\,0\,0\rangle+|0\,1\,1\,0\rangle\bigr)
=\;|0\!\rangle_0\;\otimes\;\frac{|00\rangle_{21}+|11\rangle_{21}}{\sqrt2}\;\otimes\;|0\!\rangle_3.
\]
Thus
\[
|\phi\rangle
=\;|\phi_{21}\rangle\;\otimes\;|\phi_{03}\rangle,
\]
with a Bell‐state \(|\phi_{21}\rangle=|\Phi^+\rangle\) on qubits \((2,1)\) and \(|\phi_{03}\rangle=|00\rangle\) on \((0,3)\).
3. Fidelity factorizes on disjoint subsystems
Because the two noise channels act on disjoint qubit‐pairs \((2,1)\) and \((0,3)\), and the pre‐noise state \(|\phi\rangle\) is a product across that split, the total fidelity factorizes:
\[
F(p)
=\bigl\langle\phi_{21}\bigr|\,
\mathcal E^{(2,1)}\bigl(|\phi_{21}\rangle\langle\phi_{21}|\bigr)\,
\bigl|\phi_{21}\bigr\rangle
\;\times\;
\bigl\langle\phi_{03}\bigr|\,
\mathcal E^{(0,3)}\bigl(|\phi_{03}\rangle\langle\phi_{03}|\bigr)\,
\bigl|\phi_{03}\bigr\rangle
\;\equiv\;F_{21}(p)\,F_{03}(p)\,.
\]
4. Single‐channel fidelities
A two‐qubit depolarizing channel \(\mathcal E\) with error‐rate \(p\) has Kraus operators
\[
K_{I}= \sqrt{1-p}\,I\,,\quad
K_{P}=\sqrt{\tfrac p{15}}\,P\quad(P\neq I\text{ a 2‐qubit Pauli}),
\]
and for any pure stabilizer state \(|\psi\rangle\) on 2 qubits the fidelity is
\[
\langle\psi|\mathcal E(|\psi\rangle\langle\psi|)\,|\psi\rangle
=(1-p)\;+\;\frac p{15}\sum_{P\neq I}
\bigl|\langle\psi|P|\psi\rangle\bigr|^2.
\]
Since a 2‐qubit stabilizer state has exactly three non‐identity Pauli stabilizers,
\(\langle\psi|P|\psi\rangle=\pm1\) for those 3 and \(0\) for the other 12, one finds
\[
F_{\rm stab}(p)
=1-p\;+\;\frac p{15}\times3
=1-\frac{12}{15}\,p
=1-\frac{4}{5}\,p.
\]
In our case both subsystems are 2‐qubit stabilizer states:
– \(|\phi_{21}\rangle=|\Phi^+\rangle\)
– \(|\phi_{03}\rangle=|00\rangle\)
hence
\[
F_{21}(p)=1-\tfrac45p,
\qquad
F_{03}(p)=1-\tfrac45p.
\]
5. Final expression
Therefore the physical state fidelity of the full 4‐qubit output is
\[
\boxed{%
F(p)\;=\;\Bigl(1-\tfrac45\,p\Bigr)^2
\;=\;1-\frac{8}{5}\,p+\frac{16}{25}\,p^2\,.}
\]
Final Answer:
\[
\boxed{F_{\rm phys}(p)=\bigl(1-\tfrac45\,p\bigr)^{2}\,.}
\]