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 infidelity \(p\) of two-qubit gates. 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 understand the circuit’s operation step by step. The initial state is \(|0000\rangle\) (assuming all qubits are initialized to \(|0\rangle\)), and an ancilla qubit \(|0\rangle\) is introduced, making the initial state \(|00000\rangle\). The circuit applies \(H_1\), creating a superposition on qubit 1.
Step 3: Apply the Hadamard gate \(H_1\) to the initial state
Applying \(H_1\) to \(|00000\rangle\) gives \(\frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)\).
Step 4: Understand the effect of subsequent \(CNOT\) gates
The \(CNOT_{12}\) gate applies to \(\frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)\), resulting in \(\frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)\). Then, \(CNOT_{10}\) is applied, giving \(\frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)\) because \(CNOT_{10}\) does not change the state since the control qubit (1) is \(|1\rangle\) only in the second term, but it targets qubit 0 which is \(|0\rangle\), so it remains unchanged. \(CNOT_{23}\) then gives \(\frac{1}{\sqrt{2}}(|00000\rangle + |01110\rangle)\), and \(CNOT_{34}\) results in \(\frac{1}{\sqrt{2}}(|00000\rangle + |01111\rangle)\). Finally, \(CNOT_{04}\) gives \(\frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)\), which is a GHZ state across qubits 0, 1, 2, 3, and 4.
Step 5: Consider the effect of measurement \(M_4\) on the ancilla qubit
Measuring qubit 4 in \(|0\rangle\) projects the state onto \(|0000\rangle\), which is the encoded \(|00\rangle_{AB}\) in the [[4,2,2]] code because \(|0000\rangle\) is a valid codeword.
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.
Step 7: Calculate the probability of error-free operation for the circuit
There are 5 \(CNOT\) gates. The probability that none of them experience an error is \((1-p)^5\).
Step 8: Understand the effect of detectable errors
The circuit is designed to be fault-tolerant with the ancilla detecting errors. Errors that are detectable by the code or the ancilla measurement will cause the state preparation to fail or be rejected.
Step 9: Determine the logical fidelity as a function of \(p\)
To calculate the logical fidelity, we need to consider the probability of successful state preparation (both the ancilla measuring \(|0\rangle\) and the state being correctly prepared within the code’s logical space) and the effect of errors.
Step 10: Calculate the probability of measuring \(|0\rangle\) on the ancilla
For the state \(\frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)\), measuring \(|0\rangle\) on qubit 4 projects the remaining qubits onto \(|0000\rangle\), which is a valid codeword for \(|00\rangle_{AB}\). Errors that cause the ancilla to measure \(|1\rangle\) will result in rejection.
Step 11: Assess the impact of two-qubit depolarizing errors on the logical state fidelity
Given that the state preparation circuit involves 5 \(CNOT\) gates, and each has a probability \(p\) of error, we need to consider how these errors affect the logical state. The probability of no error is \((1-p)^5\). For small \(p\), the dominant term is \(1-5p\).
Step 12: Evaluate the logical fidelity considering post-selection
The logical fidelity after post-selection on \(|0\rangle\) for the ancilla and detectable errors within the code is related to the probability of error-free or correctable operations. The [[4,2,2]] code detects errors, and with the ancilla, the circuit is designed to reject states with detectable errors.
Step 13: Derive the expression for logical fidelity
The logical fidelity \(F\) can be approximated by considering the probability of successful preparation and the effect of errors. For small \(p\), \(F \approx 1 - O(p)\), where \(O(p)\) represents terms linear in \(p\) or higher.
Step 14: Calculate the specific logical fidelity as a function of \(p\)
To first order in \(p\), the fidelity is affected by the errors in the \(CNOT\) gates. The probability of an error in any \(CNOT\) gate is \(p\), and there are 5 such gates. However, not all errors lead to a failure in state preparation due to post-selection.
The final answer is:
Final Answer: \(1 - \frac{15}{15}p \times 5 + O(p^2) = 1 - 5p + O(p^2)\)