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Step-by-Step Derivation
- Ideal encoded GHZ state in the [[4,2,2]] code.
The circuit
\[
|0000\rangle \xrightarrow{H_2}
\xrightarrow{\rm CNOT_{2\to1}}
\xrightarrow{H_0}
\xrightarrow{\rm CNOT_{0\to3}}
|\,\psi\>\;=\;\frac{1}{2}\bigl(|0000\> + |0110\> + |1001\> + |1111\>\bigr)
\]
prepares the logical GHZ
\[
\bigl(|00\>_{AB}+|11\>_{AB}\bigr)/\sqrt2
\;=\;\frac1{2}\bigl(|0000\>+|0110\>+|1001\>+|1111\>\bigr)\,.
\]
- Error model.
Each CNOT gate is followed by an i.i.d.\ two-qubit depolarizing channel
\[
\mathcal D(\rho)=(1-p)\,\rho\;+\;\frac{p}{15}\sum_{E\neq I\otimes I}E\,\rho\,E\,.
\]
The Hadamards are assumed perfect.
- Final noisy state as a Pauli‐mixture.
Let the first faulty gate (CNOT\(_{2\to1}\)) insert error \(E_1\) on qubits \(\{1,2\}\), drawn from
\[
\Pr[E_1=I]=1-p,\quad
\Pr[E_1=P]=\tfrac p{15}\quad(P\in\{X,Y,Z\}^{\otimes2}\setminus\{II\}).
\]
Likewise the second faulty gate (CNOT\(_{0\to3}\)) inserts \(E_2\) on qubits \(\{0,3\}\) with the same law.
- Fidelity as sum over “harmless” errors.
The final density matrix is
\[
\rho=\sum_{E_1,E_2}\Pr[E_1]\Pr[E_2]\;\bigl(E_2\,U\,E_1\,V\,|0000\>\<0000|\,V^\dagger\,E_1\,U^\dagger\,E_2\bigr)\,,
\]
where \(V=H_2\), \(U={\rm CNOT}_{2\to1}\,H_0\,{\rm CNOT}_{0\to3}\). Since the second half of the circuit, \(H_0\) and CNOT\(_{0\to3}\), acts only on qubits \(\{0,3\}\), it commutes with any Pauli on \(\{1,2\}\). Hence
\[
E_1\;\xrightarrow{\;U\;} \;E_1,\qquad
E_2\;\xrightarrow{\;\cdot\;} \;E_2,
\]
and the net error before comparison with \(|\psi\>\) is \(E=E_2\otimes E_1\).
- Which errors leave \(|\psi\>\) invariant?
The stabilizer group of \(|\psi\>\) (the logical‐GHZ) is generated by the four independent Pauli stabilizers
\[
S_1=X_0X_1X_2X_3,\quad
S_2=Z_0Z_1Z_2Z_3,\quad
S_3=X_1X_2,\quad
S_4=Z_1Z_2.
\]
One checks that the entire stabilizer group is exactly
\[
G=\bigl\{A_{0,3}\otimes B_{1,2}\,\bigm|\,A,B\in\{II,\;XX,\;ZZ,\;YY\}\bigr\}.
\]
Any \(E\in G\) satisfies \(E\,|\psi\>=\pm|\psi\>\), so contributes unit fidelity overlap.
- Compute the fidelity.
Since \(E_2\) lives on \(\{0,3\}\) and \(E_1\) on \(\{1,2\}\), they are independent. The total fidelity is
\[
F(p)
=\sum_{\substack{E_2\in\{II,\,XX,\,YY,\,ZZ\}\\E_1\in\{II,\,XX,\,YY,\,ZZ\}}}
\Pr[E_2]\,\Pr[E_1]
\;=\;\Bigl(\Pr[II]+\!3\,\tfrac p{15}\Bigr)^2
=\Bigl((1-p)+3\frac p{15}\Bigr)^2
=\bigl(1-\tfrac{4p}{5}\bigr)^2.
\]
Final Answer:
\[
\boxed{F_{\rm phys}(p)\;=\;\bigl(1 - \tfrac{4}{5}\,p\bigr)^2\,.}
\]