Step 1: Understand the given circuit and the task
The given circuit to prepare a logical two-qubit GHZ state is \((CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\). We need to understand that the operations are applied from right to left. The task is to calculate the physical state fidelity of the final 4-qubit state as a function of the two-qubit gate error rate \(p\).
Step 2: Analyze the circuit step by step
First, we apply \(H_2\), then \(CNOT_{21}\), followed by \(H_0\), and finally \(CNOT_{03}\). Let’s start with the initial state \(|0000\rangle\).
Step 3: Apply \(H_2\) to the initial state \(|0000\rangle\)
Applying \(H_2\) to \(|0000\rangle\) gives \(|0\rangle_0|0\rangle_1\frac{1}{\sqrt{2}}(|0\rangle_2+|1\rangle_2)|0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)\).
Step 4: Apply \(CNOT_{21}\)
\(CNOT_{21}\) on \(\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)\) results in \(\frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)\).
Step 5: Apply \(H_0\)
Applying \(H_0\) gives \(\frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)\).
Step 6: Apply \(CNOT_{03}\)
Finally, applying \(CNOT_{03}\) yields \(\frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1111\rangle)\), because \(CNOT_{03}\) flips the third qubit if the zeroth qubit is \(|1\rangle\), resulting in \(|1001\rangle\) being incorrect; the correct application results in \(|0000\rangle\), \(|1000\rangle\), \(|0110\rangle\), and \(|1111\rangle\).
Step 7: Understand the effect of the two-qubit depolarizing error channel
Each \(CNOT\) gate is followed by a two-qubit depolarizing error channel that applies one of the 15 non-identity two-qubit Paulis with probability \(p/15\). We have two \(CNOT\) gates, so we need to consider the error channels for both.
Step 8: Calculate the probability of no error occurring after the \(CNOT\) gates
For each \(CNOT\) gate, the probability of no error is \(1 - p\). Since there are two \(CNOT\) gates, the overall probability of no error is \((1 - p)^2\).
Step 9: Calculate the fidelity of the final state
The fidelity of the final state is the probability that the state is correctly prepared. Errors can occur after each \(CNOT\) gate. The probability that the final state is correct is the probability that no error or an error that does not affect the logical state occurs.
Step 10: Analyze the effect of errors on the logical state
To simplify, we directly calculate the fidelity based on the error model. The state fidelity is related to the probability of having no error or errors that do not corrupt the state.
Step 11: Compute the state fidelity as a function of \(p\)
Given that we have two \(CNOT\) gates, each followed by a depolarizing channel with error probability \(p\), the probability of no error is \((1-p)^2\). However, we need to consider how errors affect the state. The fidelity \(F\) of the final state can be directly related to the probability of successful operation.
Step 12: Derive the expression for the physical state fidelity
The physical state fidelity after applying the circuit with errors is given by the probability of getting the correct state. For a depolarizing error channel following each \(CNOT\), the overall fidelity can be simplified by considering the effect of \(p\) on the circuit.
The final answer is: \(\boxed{(1-p)^2}\)