To solve this problem, we need to analyze the given state preparation circuit and understand how quantum errors impact the fidelity of the final prepared state. The circuit is designed to prepare a logical two-qubit GHZ state \((|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}\) in a [[4,2,2]] quantum code. Let’s go through the derivation step by step.

Step-by-Step Derivation:

1. Circuit Description:

The circuit provided is:

This indicates the quantum operations are applied in this order: - Apply Hadamard \(H\) on qubit 2. - Apply \(CNOT_{21}\) with qubit 2 as the control and qubit 1 as the target. - Apply Hadamard \(H\) on qubit 0. - Apply \(CNOT_{03}\) with qubit 0 as the control and qubit 3 as the target.

2. Error Model:

After each \(CNOT\) gate, the circuit is followed by a two-qubit depolarizing channel, which introduces one of the 15 non-identity two-qubit Pauli errors with probability \(p/15\). The depolarizing channel for two qubits is given by:

where the sum runs over all combinations of two-qubit Paulis, excluding the identity.

3. State Preparation without Errors:

Consider the initial state \(|0000\rangle\). The operations transform this as follows: - \(H_2\): Converts \(|0\rangle_2 \to \frac{|0\rangle_2 + |1\rangle_2}{\sqrt{2}}\). - \(CNOT_{21}\): Entangles qubits 2 and 1:

• \(H_0\): Converts \(|0\rangle_0 \to \frac{|0\rangle_0 + |1\rangle_0}{\sqrt{2}}\).
• \(CNOT_{03}\): Further entangles qubits 0 and 3:

4. State with Errors:

Each \(CNOT\) introduces an error \(\mathcal{E}\), so the sequence \(CNOT_{21}\) and \(CNOT_{03}\) each may add a depolarizing channel. At this point, there are two main \(CNOT\) operations contributing to errors: - (1-p) represents the probability of no error after both gates. - \(2p/15\) for a single error occurring in the first gate sequence. - \(p^2/225\) for errors in both \(CNOT\) operations.

5. Fidelity Calculation:

Fidelity between expected \(|\psi_{ideal}\rangle\) and error-induced \(\rho\) is calculated by:

For small \(p\), assuming independence of error sources: - Fidelity contributions for each \(CNOT\) give:

Final Answer:

The physical state fidelity \( F \) of the final physical 4-qubit state with respect to the target logical two-qubit GHZ state is approximately:

This formula hints at the degradation of fidelity due to errors introduced by the \(CNOT\) gates in the presence of a depolarizing channel following each such operation.

To determine the logical state fidelity of the final 2-qubit logical state after running the circuit while post-selecting on all detectable errors, we analyze how the quantum error detection properties of the [[4,2,2]] code handle errors introduced by the circuit.

Step-by-Step Derivation:

1. State Preparation Circuit:

The circuit sequence is:

The operations execute in reverse order, affecting qubits as follows: - Initial state: \(|0000\rangle\). - \(H_2\): Transforms qubit 2:

• \(CNOT_{21}\): Entangles qubits 2 (control) and 1 (target):

• \(H_0\): Acts on qubit 0:

• \(CNOT_{03}\): Entangles qubits 0 (control) and 3 (target):

2. Error Model with Post-Selection:

Each \(CNOT\) is followed by a depolarizing channel. However, since we are post-selecting on no errors being detected by the code, the stabilizers \(XXXX\) and \(ZZZZ\) must return \(+1\) when employed.

Post-Selection: - Only errors not detectable by the stabilizers are post-selected. - Detectable errors causing syndrome changes are discarded. - For one error in either \(CNOT\), an error might be detected by messing up a stabilizer condition.

3. Logical Operators and Error Detection Abilities:

• Logical State with Ideal Code: \((|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}\).
• Logical Errors and Undetected Errors: The crucial undetectable error combinations which don’t trigger stabilizers are either no error or very rare uncorrelated double-qubit errors considering the stabilizer constraints of the code.

4. Fidelity Calculation:

• Undetected Errors: Error modalities that don’t trigger a detectable change in the syndrome are logical errors that occur less often than detectable errors.

• Probability of Detectable Error: Given the error model for a single \(CNOT\), detectable errors precipitate a logical error with less than a 2-fold probability reduction for each 2-qubit gate.

• The probability of not detecting an error, which scapegoats contribution from undetected errors outweighs full fidelity degradation to logical state:

where additional negligible contributions originate from unmonitored stabilizer-breach events after single round undetected stabilization, reflecting the coherence preservation.

Final Answer:

The logical state fidelity \(F_{\text{logical}}\) of the final 2-qubit logical state post selection is approximately:

This reflects the code’s ability to post-select the final state while discarding detectable errors, retaining a fidelity higher than simple physical fidelity compromised by errors.

Here is the finalized code to reflect the derived logical state fidelity:

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of the two-qubit gate error rate $p$ in Sympy format.

    Inputs
    ----------
    p: sympy.Symbol, the two-qubit gate error rate, $p$

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = 1 - (2/15) * p  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical

This code provides the logical state fidelity as a function of the gate error rate \( p \) using the SymPy library.