Step-by-Step Derivation
1. Ideal State Preparation
First, let’s determine the ideal final state of the four physical qubits, assuming no errors. The initial state is \(|\psi_{init}\rangle = |0000\rangle\). The circuit applies a sequence of gates: \(U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\). The state evolves as follows:
- Apply \(H_2\): The Hadamard gate on qubit 2 acts on the initial state.
\[
|\psi_1\rangle = H_2 |\psi_{init}\rangle = |00\rangle \otimes \left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)_2 \otimes |0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)
\]
- Apply \(CNOT_{21}\): The controlled-NOT gate with control qubit 2 and target qubit 1 acts on \(|\psi_1\rangle\).
\[
|\psi_2\rangle = CNOT_{21} |\psi_1\rangle = \frac{1}{\sqrt{2}}(CNOT_{21}|0000\rangle + CNOT_{21}|0010\rangle)
\]
The control qubit (2) is $|0\rangle$ for the first term and $|1\rangle$ for the second.
\[
|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)
\]
- Apply \(H_0\): The Hadamard gate on qubit 0 acts on \(|\psi_2\rangle\).
\[
|\psi_3\rangle = H_0 |\psi_2\rangle = \frac{1}{\sqrt{2}}(H_0|0000\rangle + H_0|0110\rangle) = \frac{1}{\sqrt{2}}\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}|000\rangle + \frac{|0\rangle+|1\rangle}{\sqrt{2}}|110\rangle\right)
\]
\[
|\psi_3\rangle = \frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)
\]
- Apply \(CNOT_{03}\): The final CNOT gate with control qubit 0 and target qubit 3 acts on \(|\psi_3\rangle\).
\[
|\Psi_{ideal}\rangle = CNOT_{03} |\psi_3\rangle = \frac{1}{2}(CNOT_{03}|0000\rangle + CNOT_{03}|1000\rangle + CNOT_{03}|0110\rangle + CNOT_{03}|1110\rangle)
\]
The control qubit (0) is $|0\rangle$ for the first and third terms, and $|1\rangle$ for the second and fourth terms.
\[
|\Psi_{ideal}\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)
\]
This is the ideal final state of the circuit.
2. Error Model and Fidelity Calculation Strategy
The circuit contains two CNOT gates, each followed by a two-qubit depolarizing error channel. Let’s denote the gates as \(H_2\), \(C_1 = CNOT_{21}\), \(H_0\), \(C_2 = CNOT_{03}\). The error channels are \(\mathcal{E}_1\) on qubits (2,1) and \(\mathcal{E}_2\) on qubits (0,3).
The evolution of the density matrix \(\rho\) is:
- \(\rho_0 = |0000\rangle\langle 0000|\).
- \(\rho_1 = H_2 \rho_0 H_2^\dagger = |\psi_1\rangle\langle\psi_1|\).
- \(\rho_{2,ideal} = C_1 \rho_1 C_1^\dagger = |\psi_2\rangle\langle\psi_2|\).
- \(\rho_2 = \mathcal{E}_1(\rho_{2,ideal})\).
- \(\rho_3 = H_0 \rho_2 H_0^\dagger\).
- \(\rho_{4,ideal} = C_2 \rho_3 C_2^\dagger\).
- \(\rho_{final} = \mathcal{E}_2(\rho_{4,ideal})\).
The depolarizing channel on a two-qubit state \(\rho\) is given by:
\[
\mathcal{E}(\rho) = (1-p)\rho + p \sum_{P \in \mathcal{P}_2 \setminus \{II\}} \frac{1}{15} P \rho P^\dagger
\]
where \(\mathcal{P}_2\) is the set of 16 two-qubit Pauli operators.
We want to calculate the physical state fidelity \(F = \langle \Psi_{ideal} | \rho_{final} | \Psi_{ideal} \rangle\).
3. Fidelity After the First Noisy Gate
Let’s first calculate the fidelity of the state after the first CNOT and its error channel, followed by the remaining ideal gates. Let this intermediate state be \(\rho' = C_2 H_0 \mathcal{E}_1(C_1 H_2 \rho_0 H_2^\dagger C_1^\dagger) H_0^\dagger C_2^\dagger\). The fidelity with the ideal final state is \(F_1 = \langle \Psi_{ideal} | \rho' | \Psi_{ideal} \rangle\).
Using the cyclic property of the trace and that \(|\Psi_{ideal}\rangle = C_2 H_0 |\psi_2\rangle\), this simplifies to:
\[
F_1 = \langle \psi_2 | \mathcal{E}_1(|\psi_2\rangle\langle\psi_2|) | \psi_2 \rangle
\]
where \(|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)\). The error channel \(\mathcal{E}_1\) acts on qubits 2 and 1.
\[
F_1 = (1-p)\langle\psi_2|\psi_2\rangle + \frac{p}{15} \sum_{P \in \mathcal{P}_{21}\setminus\{II\}} \langle\psi_2| P |\psi_2\rangle\langle\psi_2| P^\dagger |\psi_2\rangle
\]
\[
F_1 = (1-p) + \frac{p}{15} \sum_{P \in \mathcal{P}_{21}\setminus\{II\}} |\langle\psi_2| P |\psi_2\rangle|^2
\]
The expectation value \(\langle\psi_2| P |\psi_2\rangle\) is non-zero only if \(P\) commutes with all stabilizers of \(|\psi_2\rangle\). The state \(|\psi_2\rangle\) can be written as \(|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{12} \otimes |0\rangle_3\). The stabilizers of the Bell state on qubits 1 and 2 are \(X_1X_2\) and \(Z_1Z_2\). The stabilizers for the full state are \(S^{(2)} = \langle Z_0, Z_3, X_1X_2, Z_1Z_2 \rangle\).
The Pauli operator \(P\) acts on qubits 1 and 2, so it naturally commutes with \(Z_0\) and \(Z_3\). For the expectation value to be non-zero, \(P\) must also commute with \(X_1X_2\) and \(Z_1Z_2\). The two-qubit Pauli operators on (1,2) that commute with both \(X_1X_2\) and \(Z_1Z_2\) are \(\{II, X_1X_2, Y_1Y_2, Z_1Z_2\}\).
Let’s evaluate the expectation values for the non-identity operators:
* \(P = X_1X_2\): \(X_1X_2 |\psi_2\rangle = |0\rangle_0 \otimes X_1X_2 \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{12} \otimes |0\rangle_3 = |\psi_2\rangle\). So \(\langle\psi_2|X_1X_2|\psi_2\rangle = 1\).
* \(P = Z_1Z_2\): \(Z_1Z_2 |\psi_2\rangle = |0\rangle_0 \otimes Z_1Z_2 \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{12} \otimes |0\rangle_3 = |\psi_2\rangle\). So \(\langle\psi_2|Z_1Z_2|\psi_2\rangle = 1\).
* \(P = Y_1Y_2 = (iX_1Z_1)(iX_2Z_2) = -X_1X_2 Z_1Z_2\). Since \(X_1X_2\) and \(Z_1Z_2\) are stabilizers, \(Y_1Y_2|\psi_2\rangle = -|\psi_2\rangle\). So \(\langle\psi_2|Y_1Y_2|\psi_2\rangle = -1\).
The sum of squared expectation values is:
\[
\sum_{P \in \mathcal{P}_{21}\setminus\{II\}} |\langle\psi_2| P |\psi_2\rangle|^2 = |1|^2 + |1|^2 + |-1|^2 = 3
\]
Substituting this into the expression for \(F_1\):
\[
F_1 = (1-p) + \frac{p}{15}(3) = 1 - p + \frac{p}{5} = 1 - \frac{4p}{5}
\]
4. Fidelity After the Second Noisy Gate
The final state is \(\rho_{final} = \mathcal{E}_2(\rho')\), where \(\mathcal{E}_2\) is the error channel on qubits (0,3). The final fidelity is:
\[
F = \langle \Psi_{ideal} | \mathcal{E}_2(\rho') | \Psi_{ideal} \rangle = (1-p)\langle \Psi_{ideal} | \rho' | \Psi_{ideal} \rangle + \frac{p}{15}\sum_{Q \in \mathcal{P}_{03}\setminus\{II\}} \langle \Psi_{ideal} | Q \rho' Q | \Psi_{ideal} \rangle
\]
The first term is \((1-p)F_1\). For the second term, we substitute the expression for \(\rho'\).
\[
\rho' = (1-p)|\Psi_{ideal}\rangle\langle\Psi_{ideal}| + \frac{p}{15}\sum_{P \in \mathcal{P}_{21}\setminus\{II\}} P' |\Psi_{ideal}\rangle\langle\Psi_{ideal}| (P')^\dagger
\]
where \(P' = C_2 H_0 P H_0^\dagger C_2^\dagger\). Since \(P\) acts on qubits (1,2) and \(C_2 H_0\) acts on (0,3), they commute, so \(P'=P\).
\[
\rho' = (1-p)|\Psi_{ideal}\rangle\langle\Psi_{ideal}| + \frac{p}{15}\sum_{P \in \mathcal{P}_{21}\setminus\{II\}} P |\Psi_{ideal}\rangle\langle\Psi_{ideal}| P^\dagger
\]
Now we evaluate the term \(\langle \Psi_{ideal} | Q \rho' Q | \Psi_{ideal} \rangle\):
\[
\langle \Psi_{ideal} | Q \rho' Q | \Psi_{ideal} \rangle = (1-p)|\langle \Psi_{ideal} | Q | \Psi_{ideal} \rangle|^2 + \frac{p}{15}\sum_{P \neq II} |\langle \Psi_{ideal} | Q P | \Psi_{ideal} \rangle|^2
\]
The expectation values are non-zero only if the operators commute with the stabilizers of \(|\Psi_{ideal}\rangle\). The stabilizers of \(|\Psi_{ideal}\rangle\) are \(\mathcal{S} = \langle X_0X_3, Z_0Z_3, X_1X_2, Z_1Z_2 \rangle\).
-
For \(\langle \Psi_{ideal} | Q | \Psi_{ideal} \rangle\) with \(Q \in \mathcal{P}_{03}\setminus\{II\}\): \(Q\) must commute with \(X_0X_3\) and \(Z_0Z_3\). The operators are \(\{X_0X_3, Y_0Y_3, Z_0Z_3\}\). The sum of squares of their expectation values is \(|\langle X_0X_3\rangle|^2 + |\langle Y_0Y_3\rangle|^2 + |\langle Z_0Z_3\rangle|^2 = 1^2+(-1)^2+1^2 = 3\).
-
For \(\langle \Psi_{ideal} | Q P | \Psi_{ideal} \rangle\): The operator \(QP\) must commute with \(\mathcal{S}\). Since \(Q\) acts on (0,3) and \(P\) acts on (1,2), they commute. For \(QP\) to commute with all stabilizers in \(\mathcal{S}\), \(Q\) must commute with the (0,3) stabilizers (\(X_0X_3, Z_0Z_3\)) and \(P\) must commute with the (1,2) stabilizers (\(X_1X_2, Z_1Z_2\)).
- This gives 3 choices for \(Q \in \mathcal{P}_{03}\setminus\{II\}\).
- This gives 3 choices for \(P \in \mathcal{P}_{21}\setminus\{II\}\).
- For each of the \(3 \times 3 = 9\) pairs, \(|\langle \Psi_{ideal} | Q P | \Psi_{ideal} \rangle|^2 = |\langle Q \rangle \langle P \rangle|^2 = (\pm 1)^2(\pm 1)^2 = 1\).
- So, \(\sum_{Q\neq II}\sum_{P\neq II} |\langle \Psi_{ideal} | Q P | \Psi_{ideal} \rangle|^2 = 9\).
Now, let’s assemble the full expression for the fidelity \(F\):
\[
F = (1-p)F_1 + \frac{p}{15}\sum_{Q \neq II} \left[ (1-p)|\langle Q \rangle|^2 + \frac{p}{15}\sum_{P\neq II} |\langle QP \rangle|^2 \right]
\]
\[
F = (1-p)F_1 + \frac{p}{15} \left[ (1-p)\sum_{Q \neq II}|\langle Q \rangle|^2 + \frac{p}{15}\sum_{Q \neq II, P\neq II} |\langle QP \rangle|^2 \right]
\]
\[
F = (1-p)F_1 + \frac{p}{15} \left[ (1-p)(3) + \frac{p}{15}(9) \right]
\]
Substituting \(F_1 = 1 - 4p/5\):
\[
F = (1-p)\left(1 - \frac{4p}{5}\right) + \frac{3p(1-p)}{15} + \frac{9p^2}{225}
\]
\[
F = 1 - p - \frac{4p}{5} + \frac{4p^2}{5} + \frac{p(1-p)}{5} + \frac{p^2}{25}
\]
\[
F = 1 - \frac{9p}{5} + \frac{4p^2}{5} + \frac{p}{5} - \frac{p^2}{5} + \frac{p^2}{25}
\]
\[
F = 1 - \frac{8p}{5} + \frac{3p^2}{5} + \frac{p^2}{25}
\]
\[
F = 1 - \frac{8p}{5} + \frac{15p^2+p^2}{25} = 1 - \frac{8p}{5} + \frac{16p^2}{25}
\]
This can be factored as a perfect square:
\[
F = \left(1 - \frac{4p}{5}\right)^2
\]
The physical state fidelity is the product of the fidelities from each noisy gate operation. The infidelity introduced by each CNOT gate is \(1 - (1-4p/5) = 4p/5\).
Final Answer:
The physical state fidelity \(F\) of the final 4-qubit state is given by the function:
\[
F(p) = \left(1 - \frac{4p}{5}\right)^2 = 1 - \frac{8p}{5} + \frac{16p^2}{25}
\]