To determine the physical state fidelity of the final 4-qubit state at the end of the circuit, we need to assess the impact of errors at each stage of the provided quantum circuit. The circuit is given by:
\[
(CNOT_{03}) (H_0) (CNOT_{21}) (H_2).
\]
We’ll analyze the circuit step-by-step, considering the effects of the two-qubit depolarizing error channel after each CNOT gate. Recall that each two-qubit depolarizing channel acts as the identity channel with probability \(1 - p\) and applies one of the 15 non-identity two-qubit Pauli operations (over two qubits) each with probability \(p/15\).
Step-by-Step Analysis:
-
Initial State
- Begin with the state \( |0000\rangle \).
- This state is not affected by errors initially since no operations have been performed yet.
-
Apply Hadamard Gate \( H_2 \)
- Transformation: \( |0000\rangle \xrightarrow{H_2} \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle) \).
-
Apply \( CNOT_{21} \) and Error Channel
- \( CNOT_{21} \) acts on the previously prepared superposition:
\[
\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)
\]
- After this, the two-qubit depolarizing channel could induce an error. Let the error operator be \( E_{21} \) which is one of the Pauli errors with probability \( p/15 \) or identity with probability \( 1 - p \).
- Apply Hadamard Gate \( H_0 \)
- Transformation due to Hadamard on qubit 0:
\[
\frac{1}{2}(|0000\rangle + |0011\rangle + |1000\rangle + |1011\rangle)
\]
- Apply \( CNOT_{03} \) and Error Channel
- \( CNOT_{03} \) transforms the state:
\[
\frac{1}{2}(|0000\rangle + |0011\rangle + |1000\rangle + |1011\rangle) \rightarrow \frac{1}{2}(|0000\rangle + |0011\rangle + |1100\rangle + |1111\rangle)
\]
- Apply another error channel here. Let the error operator be \( E_{03} \).
- Account for Errors in Final State
- Each error \( E_{21} \) and \( E_{03} \) has a probability \( p \) of causing deviation from the intended logical state.
- The total process can thus change the prepared state’s fidelity. An error-free logical GHZ state corresponds to \( \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}) \).
Calculating Physical State Fidelity
The fidelity \( F \) of the actual, error-affected state with respect to the target state is defined by:
\[
F = \langle \text{GHZ}_\text{target} | \rho_\text{actual} | \text{GHZ}_\text{target} \rangle
\]
where \(\text{GHZ}_\text{target}\) is the error-free final state and \(\rho_\text{actual}\) is the actual state density matrix influenced by errors.
Errors occur with probability \( p \), and the GHZ state can tolerate only one qubit-flip error. Here, intermediate errors between the actual logical states can lead to logical state transitions (logical error) or leakage outside the logical space.
When computing fidelity considering these, the major check is to think of which errors can be tolerated to remain within an acceptable state near logical GHZ:
\[
F = (1-p)^k \text{ (for all qubits remaining error-free through the circuit) }
\]
where \(k\) is the number of \(CNOT\) gates, which have associated errors.
Final Simplified Fidelity:
\[
F = (1-p)^2
\]
Final Answer:
The physical state fidelity of the final physical 4-qubit state is approximately \( (1-p)^2 \), assuming all potential errors occur independently and are uncorrelated over the circuit.