Step-by-Step Derivation

1. Ideal (error–free) state prepared by the circuit
   Apply the gates, right-to-left, on the computational basis state
   \(|0000\rangle_{0123}\).

• \(H_{2}: \; |0000\rangle\;\longrightarrow\; \dfrac{|0000\rangle+|0010\rangle}{\sqrt 2}\)

• \(CNOT_{21}\;(2\!\to\!1):\)

 $\dfrac{|0000\rangle+|0010\rangle}{\sqrt 2}
    \longrightarrow
    \dfrac{|0000\rangle+|0110\rangle}{\sqrt 2}$

• \(H_{0}:\)

 $\dfrac{|0000\rangle+|0110\rangle}{\sqrt 2}
    \longrightarrow
    \dfrac{|0000\rangle+|1000\rangle
            +|0110\rangle+|1110\rangle}{2}$

• \(CNOT_{03}\;(0\!\to\!3):\)

 $\dfrac{|0000\rangle+|1000\rangle
         +|0110\rangle+|1110\rangle}{2}
    \longrightarrow
    |\psi_{\text{ideal}}\rangle
    =\dfrac{|0000\rangle+|1001\rangle
             +|0110\rangle+|1111\rangle}{2}\;.$

2. Noise model after each two-qubit gate
   After every CNOT we apply an independent two-qubit depolarising channel on the same two qubits:

where the sum runs over the 15 non-identity Paulis on that pair.

• First channel \(\mathcal E_{21}\) acts on qubits \((2,1)\)
• Second channel \(\mathcal E_{03}\) acts on qubits \((0,3)\)

The single–qubit Hadamards are taken to be perfect.

3. Net Pauli error at the end of the circuit
   Because the two noisy CNOTs touch disjoint pairs, the overall error operator on the four physical qubits is a direct product

where
\(P_{21}\in\{I,X,Y,Z\}^{\otimes 2}\) acts on qubits \((2,1)\) and
\(P_{03}\in\{I,X,Y,Z\}^{\otimes 2}\) acts on qubits \((0,3)\).

Each factor equals \(I\!\otimes\! I\) with probability \((1-p)\) and equals any one fixed non-identity two-qubit Pauli with probability \(p/15\).

4. Stabiliser of the target state
   The four-qubit state \(|\psi_{\text{ideal}}\rangle\) is the unique +1 eigenstate of the stabiliser group generated by

Any Pauli in \(S\) leaves the state unchanged (up to an irrelevant global phase). For fidelity with the pure target state we therefore need the net error \(P\) to be one of these four elements.

5. Probability that \(P\) is a stabiliser element

• \(P = I\)
Requires \(P_{21}=II\) and \(P_{03}=II\).
Probability: \((1-p)^2\).

• \(P = XXXX\)
Requires \(P_{21}=XX\) and \(P_{03}=XX\).
Probability: \((p/15)^2\).

• \(P = ZZZZ\)
Requires \(P_{21}=ZZ\) and \(P_{03}=ZZ\).
Probability: \((p/15)^2\).

• \(P = YYYY\)
Requires \(P_{21}=YY\) and \(P_{03}=YY\).
Probability: \((p/15)^2\).

Summing these mutually exclusive cases,

6. Physical state fidelity
   The output density matrix is a classical mixture over all error realisations. Its fidelity with the target pure state is exactly the probability that the realised Pauli lies in \(S\):

This expression is exact – there are only two noisy gates, so no higher powers of \(p\) appear.

Final Answer: The physical four-qubit state fidelity after completing the preparation circuit is

Step-by-Step Derivation

1. Ideal logical target
   After the (perfect) circuit

is executed on the computational vacuum \(|0000\rangle_{0123}\) the code qubits are in the physical state

which is the encoded Bell state
\(\bigl(|00\rangle_{AB}+|11\rangle_{AB}\bigr)/\sqrt2\).

2. Error model (per two–qubit gate)
   A two–qubit depolarising channel acts after each CNOT:

where \(\mathcal P_{ij}\) is the 16-element Pauli set on qubits \(\{i,j\}\). The two error operators are independent:

• \(P_{21}\) acts on qubits \((2,1)\) (after \(CNOT_{21}\))
• \(P_{03}\) acts on qubits \((0,3)\) (after \(CNOT_{03}\))

The net physical error is the four–qubit Pauli
\(P=P_{21}P_{03}\).

3. Detectable versus undetectable errors
   The code stabilisers are

A Pauli commutes with both \(S_X\) and \(S_Z\) (i.e. produces the all-zero syndrome) iff on every qubit pair its two letters are identical:

Proof: write a Pauli letter as the binary vector \((f_1,f_2)\) with
\(f_1(l)=1\) for \(l\in\{Y,Z\}\), \(f_2(l)=1\) for \(l\in\{X,Y\}\). A two-qubit Pauli commutes with \(S_X\) (\(S_Z\)) when the parity of \(f_1\) (\(f_2\)) across the pair is even. Only the four elements of \(\mathcal C\) satisfy both parity constraints.

Hence

• “accept” (no detected error) ⇔ \(P_{21},P_{03}\in\mathcal C\) or they lie in the same other parity class;
• “good” (no logical damage) ⇔ \(P_{21},P_{03}\in\mathcal C\).

4. Probabilities for one noisy gate

Therefore

Define

5. Acceptance probability
   The syndrome is trivial when the two parity‐class vectors add to zero. Because the four classes are equally likely apart from the weight \(\alpha\) of \(\mathcal C\) we get

6. Logical–error-free (“good”) probability
   The logical Bell state is stabilised by
   \(X_{A}X_{B}=IXXI\) and \(Z_{A}Z_{B}=IZZI\). Multiplying these with the code stabilisers produces the 16-element set

Any \(P\in\mathcal G\) leaves \(|\psi_{\mathrm{ideal}}\rangle\) (and therefore the logical state) unchanged. All such \(P\) arise exactly when both two-qubit errors lie in \(\mathcal C\). Hence

7. Logical fidelity after post-selection
   By definition

Multiplying numerator and denominator by 75 gives a compact polynomial ratio,

Series expansion for small \(p\):

i.e. the leading logical infidelity is second order in the physical gate error, as expected for a distance-2 error-detection code with post-selection.

Final Answer: The logical two-qubit state fidelity after post-selecting on a trivial syndrome is