Step 1: Understand the given problem and the circuit for state preparation
The problem involves preparing a logical two-qubit \(|00\rangle_{AB}\) state in the [[4,2,2]] code using a specific circuit and assessing its logical fidelity as a function of the physical two-qubit gate error rate \(p\). The circuit given is \(M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1)\), where the operations are applied from right to left.
Step 2: Analyze the circuit and its operation
First, let’s break down the circuit:
- \(H_1\) applies a Hadamard gate to qubit 1.
- \(CNOT_{12}\) applies a CNOT with control on qubit 1 and target on qubit 2.
- \(CNOT_{10}\) applies a CNOT with control on qubit 1 and target on qubit 0.
- \(CNOT_{23}\) applies a CNOT with control on qubit 2 and target on qubit 3.
- \(CNOT_{34}\) applies a CNOT with control on qubit 3 and target on qubit 4 (ancilla).
- \(CNOT_{04}\) applies a CNOT with control on qubit 0 and target on qubit 4.
- \(M_4\) measures qubit 4.
Step 3: Determine the effect of the circuit on the initial state
Initially, all qubits are in \(|0\rangle\). After \(H_1\), qubit 1 is in \(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\).
The sequence of CNOTs and \(H_1\) prepares the state. Let’s analyze the circuit’s effect step by step.
Step 4: Calculate the state after each operation
- After \(H_1\): \(|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_1 \otimes |0\rangle_2 \otimes |0\rangle_3 \otimes |0\rangle_4\)
- After \(CNOT_{12}\): \(|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{12} \otimes |0\rangle_3 \otimes |0\rangle_4\)
- After \(CNOT_{10}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle_2 \otimes |0\rangle_3 \otimes |0\rangle_4\)
- After \(CNOT_{23}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle|0\rangle)_{23} \otimes |0\rangle_4\)
- After \(CNOT_{34}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle|0\rangle)_{23} \otimes |0\rangle|0\rangle)_{34}\)
- After \(CNOT_{04}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle|0\rangle + |1\rangle|1\rangle|1\rangle)_{013} \otimes |0\rangle)_{2} \otimes |0\rangle)_{4}\), simplifying the effect on qubits 0, 1, 3, and 4.
Step 5: Understand the effect of measurement \(M_4\)
The measurement \(M_4\) projects qubit 4 onto \(|0\rangle\) or \(|1\rangle\). For the state preparation to succeed, \(M_4\) must yield \(|0\rangle\). This implies that the overall state before measurement must be such that the probability of measuring \(|1\rangle\) on qubit 4 is zero or negligible for a successful preparation.
Step 6: Analyze the error model and its impact on the circuit
Each \(CNOT\) gate is followed by a two-qubit depolarizing error channel with probability \(p\) of producing one of the 15 non-identity two-qubit Paulis. We need to consider how these errors affect the state preparation and the measurement outcome.
Step 7: Calculate the probability of successful state preparation and its fidelity
To calculate the logical state fidelity, we must consider the probability of errors that are detectable by the code and the ancilla measurement. The [[4,2,2]] code detects single-qubit errors but cannot correct them. The ancilla helps in detecting errors during state preparation.
Step 8: Assess the impact of errors on the logical fidelity
Given that the circuit involves 5 CNOT gates, each with a probability \(p\) of error, we need to consider how these errors propagate and affect the logical state.
Step 9: Derive the expression for logical fidelity as a function of \(p\)
The logical fidelity is related to the probability that no undetectable error occurs. For the [[4,2,2]] code, errors are detectable if they are not in the stabilizer or logical operator set. The probability of an error in a single CNOT is \(p\), and there are 5 CNOTs.
Step 10: Calculate the probability of success and logical fidelity
The probability that a CNOT gate has no error is \(1-p\). Thus, the probability that all 5 CNOT gates have no error is \((1-p)^5\). However, we need to consider the effect of errors and the post-selection on \(|0\rangle\) for qubit 4.
Step 11: Simplify the expression for logical fidelity
Since the state is post-selected on the ancilla being \(|0\rangle\), and given that detectable errors will cause the preparation to fail or be rejected, we focus on the cases where the state is accepted.
The final answer is: \(\boxed{1-p}\)