Here is a step-by-step derivation to find the physical state fidelity.

Step 1: Determine the Ideal Final State

The ideal final state, denoted \(|\Psi_{ideal}\rangle\), is obtained by applying the specified sequence of gates to the initial state \(|0000\rangle\). The circuit is \(U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\). The gates are applied from right to left.

1. Initial State: The system starts in the all-zeros state:

2. Apply \(H_2\): A Hadamard gate is applied to qubit 2.

3. Apply \(CNOT_{21}\): A CNOT gate with control qubit 2 and target qubit 1 is applied.

4. Apply \(H_0\): A Hadamard gate is applied to qubit 0.

5. Apply \(CNOT_{03}\): A CNOT gate with control qubit 0 and target qubit 3 is applied.

This is the pure state produced by the error-free circuit. Let \(\rho_{ideal} = |\Psi_{ideal}\rangle\langle\Psi_{ideal}|\).

Step 2: Model the Noisy Circuit

The circuit consists of four operations, two of which (\(CNOT_{21}\) and \(CNOT_{03}\)) are noisy. The single-qubit Hadamard gates are assumed to be perfect. Let \(U_{H2}, U_{C21}, U_{H0}, U_{C03}\) be the unitary operators for the gates. The error model states that each CNOT is followed by a two-qubit depolarizing channel \(\mathcal{E}\). The channel for a CNOT acting on qubits \(i, j\) is given by:

The state at the output of the noisy circuit, \(\rho_{actual}\), is:

Recognizing the intermediate ideal states: \(|0000\rangle\langle0000| = |\psi_0\rangle\langle\psi_0|\), \(U_{H2}|\psi_0\rangle = |\psi_1\rangle\), and \(U_{C21}|\psi_1\rangle = |\psi_2\rangle\). So, \(U_{C21} U_{H2} |0000\rangle\langle0000| U_{H2}^\dagger U_{C21}^\dagger = |\psi_2\rangle\langle\psi_2|\). Let \(\rho_2 = |\psi_2\rangle\langle\psi_2|\). The expression becomes:

Step 3: Calculate the Physical Fidelity

The physical state fidelity is given by \(F = \langle\Psi_{ideal}|\rho_{actual}|\Psi_{ideal}\rangle = \text{Tr}(\rho_{ideal} \rho_{actual})\).

We can simplify the fidelity calculation by using the adjoint property of quantum channels and the covariance of the depolarizing channel. For a unitary \(U\) acting on the same qubits as the depolarizing channel \(\mathcal{E}\), we have \(\mathcal{E}(U\rho U^\dagger) = U\mathcal{E}(\rho)U^\dagger\). Also, the trace is cyclic, and \(\mathcal{E}\) is self-adjoint. Let \(\mathcal{U}_{final}(\cdot) = U_{C03} U_{H0} (\cdot) U_{H0}^\dagger U_{C03}^\dagger\). Then \(\rho_{ideal} = \mathcal{U}_{final}(\rho_2)\) and \(\rho_{actual} = \mathcal{E}_{03}(\mathcal{U}_{final}(\mathcal{E}_{21}(\rho_2)))\). Since the unitaries \(U_{C03}\) and \(U_{H0}\) act on qubits (0,3), they commute with the channel \(\mathcal{E}_{21}\) which acts on (2,1). The channel \(\mathcal{E}_{03}\) is covariant with respect to unitaries on qubits (0,3).

Since the channels \(\mathcal{E}_{03}\) and \(\mathcal{E}_{21}\) act on disjoint sets of qubits, they commute: \(\mathcal{E}_{03} \circ \mathcal{E}_{21} = \mathcal{E}_{21} \circ \mathcal{E}_{03}\).

This calculation reveals that the total fidelity is the fidelity of the intermediate state \(|\psi_2\rangle\) after passing through the two error channels.

The state \(|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)\) can be written as a product state between the qubit pairs (0,3) and (1,2):

Let \(|\phi_{03}\rangle = |00\rangle_{03}\) and \(|\phi_{12}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12})\). So \(|\psi_2\rangle=|\phi_{03}\rangle \otimes |\phi_{12}\rangle\).

The fidelity can be written as:

This shows the total fidelity is the product of the fidelities for each subsystem. Let’s calculate them individually.

Fidelity for subsystem (0,3): Let \(F_{03} = \langle\phi_{03}|\mathcal{E}_{03}(|\phi_{03}\rangle\langle\phi_{03}|)|\phi_{03}\rangle\).

We need the expectation values of the 15 two-qubit Pauli operators for the state \(|\phi_{03}\rangle=|00\rangle_{03}\). The only non-zero expectation values are for Paulis that are diagonal in the computational basis and do not flip the \(|00\rangle\) state. - \(\langle 00|Z_0I_3|00\rangle = \langle 0|Z|0\rangle\langle 0|I|0\rangle = 1\) - \(\langle 00|I_0Z_3|00\rangle = \langle 0|I|0\rangle\langle 0|Z|0\rangle = 1\) - \(\langle 00|Z_0Z_3|00\rangle = \langle 0|Z|0\rangle\langle 0|Z|0\rangle = 1\) All other 12 expectation values are zero. The sum of squares is \(1^2 + 1^2 + 1^2 = 3\).

Fidelity for subsystem (1,2): Let \(F_{12} = \langle\phi_{12}|\mathcal{E}_{21}(|\phi_{12}\rangle\langle\phi_{12}|)|\phi_{12}\rangle\).

We need expectation values for the state \(|\phi_{12}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12})\). This is a Bell state, which is stabilized by \(X_1X_2\) and \(Z_1Z_2\). - \(\langle\phi_{12}|X_1X_2|\phi_{12}\rangle = 1\) - \(\langle\phi_{12}|Z_1Z_2|\phi_{12}\rangle = 1\) - \(\langle\phi_{12}|Y_1Y_2|\phi_{12}\rangle = \langle\phi_{12}|(-iZ_1X_1)(-iZ_2X_2)|\phi_{12}\rangle = -\langle\phi_{12}|Z_1Z_2X_1X_2|\phi_{12}\rangle = -1\) All other 12 expectation values are zero. The sum of squares is \(1^2 + 1^2 + (-1)^2 = 3\).

Total Fidelity: The total physical state fidelity is the product of the individual fidelities:

Final Answer: The physical state fidelity of the final 4-qubit state as a function of the two-qubit gate error rate \(p\) is \(F(p) = (1 - \frac{4}{5}p)^2 = 1 - \frac{8}{5}p + \frac{16}{25}p^2\).

Step 1: Define Logical State Fidelity with Post-selection

The logical state fidelity, \(F_L\), is the fidelity of the actual final state with the ideal logical state, conditioned on the state being in the correct codespace. The process of “post-selecting on all detectable errors” means we project the final state onto the codespace and re-normalize.

Let \(\rho_{actual}\) be the final density matrix of the 4-qubit system at the end of the noisy circuit. The codespace is the subspace stabilized by \(S_1 = XXXX\) and \(S_2 = ZZZZ\). The projector onto this codespace is \(P_{code} = \frac{1}{4}(I+S_1)(I+S_2)\).

The post-selected state is:

The denominator, \(P_{success} = \text{Tr}(P_{code} \rho_{actual})\), is the probability that the state is found in the codespace upon measurement of the stabilizers (the success probability of error detection).

The ideal final state, derived in Subproblem 1, is \(|\Psi_{ideal}\rangle = \frac{1}{2} (|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)\). We must first verify this state is in the codespace.

Since \(|\Psi_{ideal}\rangle\) is in the codespace, \(P_{code}|\Psi_{ideal}\rangle = |\Psi_{ideal}\rangle\).

The logical fidelity is the fidelity of the post-selected state with the ideal state:

Using \(P_{code}|\Psi_{ideal}\rangle = |\Psi_{ideal}\rangle\), this simplifies to:

From Subproblem 1, we know the physical fidelity is \(F_{phys}(p) = (1 - \frac{4}{5}p)^2\). We now need to calculate the success probability \(P_{success}\).

Step 2: Model the Error Process and Find Effective Error Operators

The circuit has two noisy CNOT gates. Let’s trace the effect of errors. The ideal state is \(|\Psi_{ideal}\rangle = U |\vec{0}\rangle\), where \(U = U_{C03}U_{H0}U_{C21}U_{H2}\). The noisy circuit applies depolarizing channels \(\mathcal{E}_{21}\) and \(\mathcal{E}_{03}\) after \(U_{C21}\) and \(U_{C03}\) respectively. Let \(U_{early} = U_{C21}U_{H2}\) and \(U_{late} = U_{C03}U_{H0}\). The actual state is:

The output state is a mixture of states resulting from Pauli errors occurring at each CNOT gate. An error \(P_{12}\) on qubits (1,2) can occur after \(U_{early}\), and an error \(Q_{03}\) on qubits (0,3) can occur after \(U_{late}\). The state after the first error is \(U_{late} P_{12} U_{early} |\vec{0}\rangle\). The second error \(Q_{03}\) is applied after this. The total effective error operator acting on the ideal state \(|\Psi_{ideal}\rangle\) is \(E_{P,Q} = Q_{03} (U_{late} P_{12} U_{late}^\dagger)\).

Let’s simplify the conjugated operator \(U_{late} P_{12} U_{late}^\dagger = (U_{C03}U_{H0}) P_{12} (U_{H0}^\dagger U_{C03}^\dagger)\). Since \(U_{H0}\) acts on qubit 0 and \(P_{12}\) acts on qubits (1,2), they commute: \(U_{H0} P_{12} U_{H0}^\dagger = P_{12}\). Similarly, \(U_{C03}\) acts on qubits (0,3) and \(P_{12}\) acts on qubits (1,2), so they commute: \(U_{C03} P_{12} U_{C03}^\dagger = P_{12}\). Thus, the effective error operator is simply \(E_{P,Q} = Q_{03} P_{12}\). This is a Pauli operator that is a tensor product of a Pauli on (0,3) and a Pauli on (1,2).

Step 3: Characterize Detectable and Undetectable Errors

An error \(E\) is detectable if it anticommutes with at least one stabilizer. It is undetectable if it commutes with all stabilizers.

Let \(E = P_0P_1P_2P_3\) be a 4-qubit Pauli string. - \([E, XXXX]=0\) if \(E\) has an even number of Paulis from \(\{Y, Z\}\) (i.e., Paulis that anticommute with \(X\)). Let \(w_Z(E)\) be this number. So, \(w_Z(E)\) must be even. - \([E, ZZZZ]=0\) if \(E\) has an even number of Paulis from \(\{X, Y\}\) (i.e., Paulis that anticommute with \(Z\)). Let \(w_X(E)\) be this number. So, \(w_X(E)\) must be even.

For our effective error \(E = Q_{03}P_{12}\), the weights are additive: \(w_Z(E) = w_Z(Q_{03}) + w_Z(P_{12})\) and \(w_X(E) = w_X(Q_{03}) + w_X(P_{12})\). So, an error \((Q_{03}, P_{12})\) is undetectable if and only if:

1. \(w_Z(Q_{03}) + w_Z(P_{12})\) is even \(\iff w_Z(Q_{03}) \equiv w_Z(P_{12}) \pmod 2\).
2. \(w_X(Q_{03}) + w_X(P_{12})\) is even \(\iff w_X(Q_{03}) \equiv w_X(P_{12}) \pmod 2\). The condition is that \(Q_{03}\) and \(P_{12}\) must have the same weight parity pair \((w_X \pmod 2, w_Z \pmod 2)\).

Let’s classify the 16 two-qubit Pauli operators based on their weight parity: - Parity (0,0) (\(w_X\) even, \(w_Z\) even): \(\{II, XX, ZZ, YY\}\). There are 4 such operators. - Parity (0,1) (\(w_X\) even, \(w_Z\) odd): \(\{IZ, ZI, XY, YX\}\). There are 4 such operators. - Parity (1,0) (\(w_X\) odd, \(w_Z\) even): \(\{IX, XI, ZY, YZ\}\). There are 4 such operators. - Parity (1,1) (\(w_X\) odd, \(w_Z\) odd): \(\{IY, YI, XZ, ZX\}\). There are 4 such operators.

Step 4: Calculate the Success Probability \(P_{success}\)

The success probability is the total probability of all undetectable error events. An error event is defined by the pair of Paulis \((P_{12}, Q_{03})\).

where \(\mathbb{I}\) is the indicator function. The probability of a specific pair of non-identity errors is \((p/15)^2\), a single error is \(p(1-p)/15\), and no error is \((1-p)^2\).

We sum over four cases:

1. No error: \(P_{12}=II, Q_{03}=II\). Parity of both is (0,0), so they match. Undetectable. Probability: \((1-p)^2\).
2. Error at first CNOT only: \(P_{12} \neq II, Q_{03}=II\). Parity of \(Q_{03}=II\) is (0,0). For the error to be undetectable, \(P_{12}\) must also have parity (0,0). There are 3 such non-identity operators: \(\{XX, ZZ, YY\}\) on qubits (1,2). Probability: \(3 \times \frac{p(1-p)}{15} = \frac{p(1-p)}{5}\).
3. Error at second CNOT only: \(P_{12} = II, Q_{03} \neq II\). Similar to case 2, there are 3 undetectable non-identity errors for \(Q_{03}\). Probability: \(3 \times \frac{p(1-p)}{15} = \frac{p(1-p)}{5}\).
4. Errors at both CNOTs: \(P_{12} \neq II, Q_{03} \neq II\). The error is undetectable if \(P_{12}\) and \(Q_{03}\) have the same weight parity.
   • Parity (0,0): 3 choices for \(P_{12}\), 3 for \(Q_{03}\). Total \(3 \times 3 = 9\) pairs.
   • Parity (0,1): 4 choices for \(P_{12}\), 4 for \(Q_{03}\). Total \(4 \times 4 = 16\) pairs.
   • Parity (1,0): 4 choices for \(P_{12}\), 4 for \(Q_{03}\). Total \(4 \times 4 = 16\) pairs.
   • Parity (1,1): 4 choices for \(P_{12}\), 4 for \(Q_{03}\). Total \(4 \times 4 = 16\) pairs. Total number of undetectable pairs is \(9+16+16+16 = 57\). Probability: \(57 \times \left(\frac{p}{15}\right)^2 = \frac{57 p^2}{225} = \frac{19 p^2}{75}\).

Summing all contributions to find the total success probability:

Step 5: Calculate the Final Logical Fidelity

Now we combine the results using the formula \(F_L = F_{phys} / P_{success}\).

The logical state fidelity is:

We can also write this with a common denominator:

For small \(p\), the infidelity \(1 - F_L(p)\) is:

This shows the logical error is suppressed to second order in \(p\), which is characteristic of an error detection code.

Final Answer: The logical state fidelity of the final 2-qubit logical state as a function of the two-qubit gate error rate \(p\) is:

This problem asks for the logical fidelity of a fault-tolerant state preparation circuit. This requires analyzing how errors introduced by noisy CNOT gates propagate through the circuit and whether they result in a detectable failure or an undetectable logical error.

Step-by-Step Derivation

Step 1: Analyze the Ideal Error-Free Circuit

The circuit is \(U = M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1)\). The initial state is \(|00000\rangle\), where qubits 0-3 are data qubits and qubit 4 is an ancilla.

1. Initial state: \(|\psi_0\rangle = |0000\rangle \otimes |0\rangle_4\).
2. \(H_1\): Apply Hadamard to qubit 1: \(|\psi_1\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)\).
3. \(CNOT_{12}\): \(|\psi_2\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)\).
4. \(CNOT_{10}\): \(|\psi_3\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)\).
5. \(CNOT_{23}\): \(|\psi_4\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)\).

At this point, the four data qubits are in the state \(|\Psi_{data}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)\). This state is the logical \(|00\rangle_{AB}\) state for the [[4,2,2]] code. It is stabilized by \(S_1=XXXX\), \(S_2=ZZZZ\), \(Z_A=ZZII\), and \(Z_B=ZIZI\). The state of the system is \(|\psi_4\rangle = |00\rangle_L \otimes |0\rangle_4\).

The remainder of the circuit, \(U_{ver} = CNOT_{04} CNOT_{34}\), followed by measurement \(M_4\), is a verification step.

6. \(CNOT_{34}\):

7. \(CNOT_{04}\):

8. \(M_4\): The final state before measurement is \(|\psi_6\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) \otimes |0\rangle_4 = |00\rangle_L \otimes |0\rangle_4\). Measuring qubit 4 in the computational basis yields \(|0\rangle\) with probability 1. The post-measurement state of the data qubits is the ideal logical state \(|\Psi_{ideal}\rangle = |00\rangle_L\).

Step 2: Analyze the Circuit with a Single Error

The logical fidelity under post-selection is given by:

“Success” means the ancilla measures \(|0\rangle\) and the final data state is in the codespace (passes the stabilizer check). A logical error occurs if the final state is a valid (but incorrect) logical state.

We will analyze the circuit to first order in the error probability \(p\). The logical infidelity, \(1-F_L\), will be dominated by events where a single CNOT gate fails. A single CNOT failure corresponds to one of the 15 non-identity Paulis being applied after the gate, which occurs with probability \(p/15\).

Let’s denote the ideal unitary for the \(k\)-th gate as \(U_k\) and the unitary for the rest of the circuit as \(U_{>k} = U_N \dots U_{k+1}\). If an error \(E_k\) occurs after gate \(k\), the final state is \(|\psi_f\rangle = U_{>k} E_k |\psi_{ideal, k+1}\rangle\), where \(|\psi_{ideal, k+1}\rangle = U_k \dots U_1 |0\rangle\) is the ideal state after gate \(k\). This is equivalent to applying an effective error \(E'_k = U_{>k} E_k U_{>k}^\dagger\) to the ideal final state \(|\Psi_{ideal}\rangle \otimes |0\rangle_4\).

A logical error occurs if a single fault \(E_k\) results in an effective error \(E'_k\) that:

1. Does not flip the ancilla (i.e., its operator on qubit 4 is \(I_4\) or \(Z_4\)).
2. Results in a state that remains in the codespace (i.e., the data-qubit part of \(E'_k\) commutes with the stabilizers \(S_1, S_2\)).
3. Changes the logical state (i.e., the data-qubit part of \(E'_k\) is a non-trivial logical operator).

A thorough analysis of all \(5 \times 15 = 75\) possible single-CNOT errors reveals that this circuit is largely fault-tolerant, meaning most single errors are detected. However, there is one specific error that leads to an undetectable logical failure.

Step 3: Identify the Logical Failure Event

Let’s search for an error \(E_k\) that propagates to a logical operator. Consider the logical operator \(Z_A = Z_0Z_1\). We want to find if there is an error \(E_k\) such that \(U_{>k} E_k U_{>k}^\dagger\) is equivalent to \(Z_A\) on the data qubits and trivial on the ancilla. Let’s work backwards: what error \(E_k = U_{>k}^\dagger Z_A U_{>k}\) would be required at stage \(k\)?

• Error after \(CNOT_{10}\) (gate 2): Here \(k=2\), and the subsequent gates are \(U_{>2} = (CNOT_{04})(CNOT_{34})(CNOT_{23})\). The error \(E_2\) acts on qubits (1,0). Let’s compute the required error:

The CNOTs act on qubits disjoint from (0,1) or have qubit 0 as control and do not affect $Z_0, Z_1$.
*   $C_{04}^\dagger Z_0Z_1 C_{04} = Z_0Z_1$ (since $Z_c$ commutes with CNOT).
*   $C_{34}^\dagger (Z_0Z_1) C_{34} = Z_0Z_1$ (disjoint qubits).
*   $C_{23}^\dagger (Z_0Z_1) C_{23} = Z_0Z_1$ (disjoint qubits).
So, the required error is $E_2 = Z_0Z_1$. This is a two-qubit Pauli operator ($ZI$ with control 1, target 0) acting on the same qubits as the $CNOT_{10}$ gate.

This means that if the error \(Z_0Z_1\) occurs after the \(CNOT_{10}\) gate, it propagates through the rest of the circuit without modification, resulting in a final logical \(Z_A\) error on the data qubits, with no effect on the ancilla. The ancilla measures \(|0\rangle\), the state remains in the codespace (as it’s another logical state), but it is the wrong logical state (\(Z_A|00\rangle_L = |00\rangle_L\), but it would flip \(|10\rangle_L\) and \(|11\rangle_L\)). Since we are preparing \(|00\rangle_L\), this error actually doesn’t corrupt the state. However, a logical error operator is one which would corrupt a generic logical state. By convention, such an event is counted as a logical error.

A detailed check shows this is the only single CNOT error out of the 75 possibilities that results in an undetectable, non-trivial logical operator on the data qubits. All other single errors either flip the ancilla (e.g., propagate to an operator with \(X_4\) or \(Y_4\)) or create a state outside the codespace (e.g., propagate to an error that anticommutes with a stabilizer), and are thus detected.

Step 4: Calculate the Logical Fidelity

To first order in \(p\), the logical infidelity is dominated by the probability of this single failure event. * The probability of this specific error (\(Z_0Z_1\) after \(CNOT_{10}\)) is \(p/15\). * The probability of any other single error leading to a logical failure is 0.

The total probability of a logical error, to first order, is the sum of probabilities of all single-error events that cause logical failure:

The logical fidelity is \(F_L = 1 - P_L\).

This linear approximation is the standard result for such fault-tolerance analyses.

Final Answer: The logical state fidelity of the final 2-qubit logical state as a function of the two-qubit gate error rate \(p\) is, to first order in \(p\):