The problem asks for the physical state fidelity of a 4-qubit state prepared using a specific circuit, in the presence of two-qubit depolarizing errors after each CNOT gate.

First, let’s identify the ideal final state. The circuit operations occur in reverse order of how they are written:

The initial state is \(|0000\rangle\).

1. Apply \(H_2\) (Hadamard on qubit 2):

2. Apply \(CNOT_{21}\) (control 2, target 1):

3. Apply \(H_0\) (Hadamard on qubit 0):

4. Apply \(CNOT_{03}\) (control 0, target 3):

This is the ideal physical 4-qubit state \(|\psi_{ideal}\rangle\). The problem states that this circuit prepares the logical GHZ state \((|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}\). Let’s verify this. Using the standard mapping for this code: \(|00\rangle_{AB} = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)\) \(|11\rangle_{AB} = \frac{1}{\sqrt{2}}(|0110\rangle + |1001\rangle)\) Then the target logical GHZ state is:

This matches \(|\psi_{ideal}\rangle\).

Now, let’s incorporate the errors. Each \(CNOT_{ij}\) gate is followed by a two-qubit depolarizing error channel \(\mathcal{E}_{ij}\). For a two-qubit state \(\rho\), the channel is:

where \(\mathcal{P}_{ij}^*\) is the set of 15 non-identity two-qubit Pauli operators acting on qubits \(i,j\).

Let \(U_A = H_2\), \(U_B = CNOT_{21}\), \(U_C = H_0\), \(U_D = CNOT_{03}\). The full ideal unitary is \(U_{ideal} = U_D U_C U_B U_A\). The input state is \(\rho_0 = |0000\rangle\langle0000|\). The noisy state evolution:

1. After \(U_A\): \(\rho_A = U_A \rho_0 U_A^\dagger = |\psi_1\rangle\langle\psi_1|\).
2. After \(U_B\): \(\rho_B = U_B \rho_A U_B^\dagger = |\psi_2\rangle\langle\psi_2|\).
3. After error \(\mathcal{E}_{21}\): \(\rho_B' = \mathcal{E}_{21}(\rho_B)\).
4. After \(U_C\): \(\rho_C' = U_C \rho_B' U_C^\dagger\).
5. After \(U_D\): \(\rho_D' = U_D \rho_C' U_D^\dagger\).
6. After error \(\mathcal{E}_{03}\): \(\rho_{final} = \mathcal{E}_{03}(\rho_D')\).

The physical state fidelity is \(F = \langle \psi_{ideal} | \rho_{final} | \psi_{ideal} \rangle = \mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| \rho_{final})\).

Let’s expand \(\rho_{final}\):

Substitute \(\rho_D'\):

Note that \(U_D U_C \rho_B U_C^\dagger U_D^\dagger = |\psi_{ideal}\rangle\langle\psi_{ideal}|\). Also, \(P_{21}\) acts on qubits 1,2. \(H_0\) (\(U_C\)) acts on qubit 0. \(CNOT_{03}\) (\(U_D\)) acts on qubits 0,3. Since these operators act on disjoint sets of qubits, they commute: \(U_D U_C P_{21} U_C^\dagger U_D^\dagger = P_{21}\). So, \(\rho_D' = (1-p) |\psi_{ideal}\rangle\langle\psi_{ideal}| + p \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} P_{21} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21}\).

Now substitute \(\rho_D'\) into the fidelity expression:

Let’s analyze the terms:

1. \(\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| \rho_D') = (1-p)\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| |\psi_{ideal}\rangle\langle\psi_{ideal}|) + p \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} \mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21})\)

For a stabilizer state $|\psi\rangle$, for any Pauli operator $P$, either $P|\psi\rangle = \pm |\psi\rangle$ (if $P$ is in the stabilizer group of $|\psi\rangle$) or $P|\psi\rangle$ is orthogonal to $|\psi\rangle$. In the first case, $|\langle\psi|P|\psi\rangle|^2 = 1$; in the second, it is 0.
The state $|\psi_{ideal}\rangle = \frac{1}{2} (|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$ is a stabilizer state. We need to find which $P_{21}$ (Pauli on qubits 1 and 2) belong to the stabilizer group of $|\psi_{ideal}\rangle$.
Let's check $P_{21}=X_1 X_2$:
$X_1 X_2 |\psi_{ideal}\rangle = \frac{1}{2} (X_1 X_2|0000\rangle + X_1 X_2|1001\rangle + X_1 X_2|0110\rangle + X_1 X_2|1111\rangle)$
$= \frac{1}{2} (|0110\rangle + |1111\rangle + |0000\rangle + |1001\rangle) = |\psi_{ideal}\rangle$.
Thus, for $P_{21}=X_1 X_2$, $|\langle\psi_{ideal}|X_1 X_2|\psi_{ideal}\rangle|^2 = 1$.
For all other 14 non-identity $P_{21}$ operators (e.g., $X_1$, $Z_2$, $Y_1 Z_2$, etc.), it can be verified that they map $|\psi_{ideal}\rangle$ to an orthogonal state, so their contribution to the sum is 0.
Therefore, $\sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} |\langle\psi_{ideal}|P_{21}|\psi_{ideal}\rangle|^2 = \frac{1}{15} \cdot 1 = \frac{1}{15}$.
So, $\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| \rho_D') = (1-p) + \frac{p}{15} = 1 - \frac{14p}{15}$.

2. \(\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| P_{03} \rho_D' P_{03}) = (1-p)\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| P_{03} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{03}) + p \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} \mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| P_{03} P_{21} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21} P_{03})\).

Similarly, for $P_{03}$ (Pauli on qubits 0 and 3), we check which ones belong to the stabilizer group of $|\psi_{ideal}\rangle$.
Let's check $P_{03}=X_0 X_3$:
$X_0 X_3 |\psi_{ideal}\rangle = \frac{1}{2} (X_0 X_3|0000\rangle + X_0 X_3|1001\rangle + X_0 X_3|0110\rangle + X_0 X_3|1111\rangle)$
$= \frac{1}{2} (|1001\rangle + |0000\rangle + |1111\rangle + |0110\rangle) = |\psi_{ideal}\rangle$.
Thus, for $P_{03}=X_0 X_3$, $|\langle\psi_{ideal}|X_0 X_3|\psi_{ideal}\rangle|^2 = 1$. All other 14 non-identity $P_{03}$ operators lead to 0.
So, for the first part of term 2, we have $(1-p) \frac{1}{15}$.

Now consider the second part: $p \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} |\langle\psi_{ideal}|P_{03} P_{21}|\psi_{ideal}\rangle|^2$.
This sum will only have a non-zero term if both $P_{03}$ and $P_{21}$ combine to form an operator that is in the stabilizer group of $|\psi_{ideal}\rangle$.
We know $P_{03}=X_0 X_3$ is the only relevant $P_{03}$, and $P_{21}=X_1 X_2$ is the only relevant $P_{21}$.
If $P_{03}=X_0 X_3$ and $P_{21}=X_1 X_2$, then $P_{03} P_{21} = X_0 X_3 X_1 X_2 = X_0 X_1 X_2 X_3$, which is a stabilizer of the [[4,2,2]] code and also of $|\psi_{ideal}\rangle$.
$X_0 X_1 X_2 X_3 |\psi_{ideal}\rangle = |\psi_{ideal}\rangle$. So $|\langle\psi_{ideal}|X_0 X_1 X_2 X_3|\psi_{ideal}\rangle|^2 = 1$.
For any other combination of $P_{03}$ and $P_{21}$ chosen from $\mathcal{P}_{03}^*$ and $\mathcal{P}_{21}^*$, the product $P_{03}P_{21}$ will map $|\psi_{ideal}\rangle$ to an orthogonal state, leading to a zero contribution.
So, this sum is non-zero only for one specific combination of $P_{03}$ and $P_{21}$: $\frac{1}{15} \cdot \frac{1}{15} \cdot 1 = \frac{1}{225}$.
Thus, $p \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} |\langle\psi_{ideal}|P_{03} P_{21}|\psi_{ideal}\rangle|^2 = p \frac{1}{225}$.

Combining all terms:

This expression can be recognized as a perfect square:

The final answer is \(\boxed{\left(1 - \frac{14p}{15}\right)^2}\)

This problem asks for the logical state fidelity of the prepared state, assuming post-selection on all detectable errors. This means we only consider events where the final physical state remains within the code space. The logical fidelity in this scenario is defined as:

where \(|\psi_{ideal}\rangle\) is the target ideal physical state and \(\mathcal{P}_{\mathcal{C}}\) is the projector onto the code space.

The numerator, \(\langle \psi_{ideal} | \rho_{final} | \psi_{ideal} \rangle\), is the physical state fidelity, which was calculated in Subproblem 1. We will re-evaluate it carefully here. The denominator, \(\mathrm{Tr}(\mathcal{P}_{\mathcal{C}} \rho_{final})\), is the probability that the final state \(\rho_{final}\) is in the code space, i.e., it is not detected as an error.

The ideal physical state, as derived in Subproblem 1, is:

This state is in the code space, meaning it is a \(+1\) eigenstate of both code stabilizers \(S_1 = X_0 X_1 X_2 X_3\) and \(S_2 = Z_0 Z_1 Z_2 Z_3\).

The noisy final state is:

where \(\mathcal{P}_{ij}^*\) are the 15 non-identity Pauli operators on qubits \(i,j\).

1. Re-evaluation of Physical State Fidelity (Numerator)

The physical state fidelity is \(F_{phys} = \langle \psi_{ideal} | \rho_{final} | \psi_{ideal} \rangle = \mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| \rho_{final})\). The term \(\mathrm{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| P |\psi_{ideal}\rangle\langle\psi_{ideal}| P) = |\langle\psi_{ideal}| P |\psi_{ideal}\rangle|^2\) is non-zero (equal to 1) if and only if \(P|\psi_{ideal}\rangle = \pm |\psi_{ideal}\rangle\), i.e., \(P\) is a stabilizer (or anti-stabilizer) of \(|\psi_{ideal}\rangle\).

We identify the Pauli operators that act on qubits \(\{0,3\}\) or \(\{1,2\}\) and stabilize (or anti-stabilize) \(|\psi_{ideal}\rangle\): * For \(P_{21}\) (acting on qubits 1,2): * \(X_1X_2 |\psi_{ideal}\rangle = |\psi_{ideal}\rangle\). * \(Y_1Y_2 |\psi_{ideal}\rangle = -|\psi_{ideal}\rangle\). * \(Z_1Z_2 |\psi_{ideal}\rangle = |\psi_{ideal}\rangle\). These three operators contribute to the sum. All other 12 non-identity \(P_{21}\) operators map \(|\psi_{ideal}\rangle\) to an orthogonal state. So, \(\sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} |\langle\psi_{ideal}|P_{21}|\psi_{ideal}\rangle|^2 = \frac{3}{15}\). * For \(P_{03}\) (acting on qubits 0,3): By symmetry, these are \(X_0X_3\), \(Y_0Y_3\), \(Z_0Z_3\). So, \(\sum_{P_{03} \in \mathcal{P}_{03}^*} \frac{1}{15} |\langle\psi_{ideal}|P_{03}|\psi_{ideal}\rangle|^2 = \frac{3}{15}\). * For \(P_{03}P_{21}\): The product \(P_{03}P_{21}\) must stabilize \(|\psi_{ideal}\rangle\). This occurs if both \(P_{03}\) and \(P_{21}\) stabilize \(|\psi_{ideal}\rangle\). There are \(3 \times 3 = 9\) such products. So, \(\sum_{P_{03} \in \mathcal{P}_{03}^*} \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15^2} |\langle\psi_{ideal}|P_{03}P_{21}|\psi_{ideal}\rangle|^2 = \frac{9}{225}\).

Combining these terms for the physical fidelity:

2. Probability of Remaining in Code Space (Denominator)

The probability of the state being in the code space is \(P_{code} = \mathrm{Tr}(\mathcal{P}_{\mathcal{C}} \rho_{final})\). The term \(\mathrm{Tr}(\mathcal{P}_{\mathcal{C}} P |\psi_{ideal}\rangle\langle\psi_{ideal}| P)\) is non-zero (equal to 1) if and only if \(P|\psi_{ideal}\rangle\) is in the code space. Since \(|\psi_{ideal}\rangle\) is already in the code space, this means \(P\) must commute with both code stabilizers \(S_1 = X_0X_1X_2X_3\) and \(S_2 = Z_0Z_1Z_2Z_3\).

We identify the Pauli operators on qubits \(\{0,3\}\) or \(\{1,2\}\) that commute with both \(S_1\) and \(S_2\). A Pauli \(P\) commutes with \(S\) if the number of qubits on which \(P\) and \(S\) anticommute is even.

• For \(P_{21}\) (acting on qubits 1,2):
  • \(X_1X_2\): Commutes with \(S_1\) (2 anticommutations on \(X_1, X_2\)). Commutes with \(S_2\) (2 anticommutations on \(Z_1, Z_2\)). (Contributes)
  • \(Y_1Y_2\): Commutes with \(S_1\) (2 anticommutations on \(X_1, X_2\)). Commutes with \(S_2\) (2 anticommutations on \(Z_1, Z_2\)). (Contributes)
  • \(Z_1Z_2\): Commutes with \(S_1\) (2 anticommutations on \(X_1, X_2\)). Commutes with \(S_2\) (2 anticommutations on \(Z_1, Z_2\)). (Contributes)
  • \(X_1Y_2\): Commutes with \(S_1\) (2 anticommutations on \(X_1, X_2\)). Commutes with \(S_2\) (2 anticommutations on \(Z_1, Z_2\)). (Contributes)
  • \(Y_1X_2\): Commutes with \(S_1\) (2 anticommutations on \(X_1, X_2\)). Commutes with \(S_2\) (2 anticommutations on \(Z_1, Z_2\)). (Contributes)
  • All other 10 non-identity \(P_{21}\) operators (e.g., \(X_1\), \(Z_1\), \(X_1Z_2\)) anticommute with at least one stabilizer (odd number of total anticommuting pairs), meaning they map \(|\psi_{ideal}\rangle\) out of the code space. So, \(\sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15} \mathrm{Tr}(\mathcal{P}_{\mathcal{C}} P_{21} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21}) = \frac{5}{15}\).
• For \(P_{03}\) (acting on qubits 0,3): By symmetry, there are 5 such operators: \(X_0X_3, Y_0Y_3, Z_0Z_3, X_0Y_3, Y_0X_3\). So, \(\sum_{P_{03} \in \mathcal{P}_{03}^*} \frac{1}{15} \mathrm{Tr}(\mathcal{P}_{\mathcal{C}} P_{03} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{03}) = \frac{5}{15}\).
• For \(P_{03}P_{21}\): The product \(P_{03}P_{21}\) must commute with both \(S_1\) and \(S_2\). This occurs if both \(P_{03}\) and \(P_{21}\) commute with \(S_1\) and \(S_2\). There are \(5 \times 5 = 25\) such products. So, \(\sum_{P_{03} \in \mathcal{P}_{03}^*} \sum_{P_{21} \in \mathcal{P}_{21}^*} \frac{1}{15^2} \mathrm{Tr}(\mathcal{P}_{\mathcal{C}} P_{03} P_{21} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{21} P_{03}) = \frac{25}{225}\).

Combining these terms for the probability of remaining in the code space:

3. Logical State Fidelity

Now we combine the numerator (\(F_{phys}\)) and the denominator (\(P_{code}\)):

This result can be simplified by recognizing that the numerator and denominator are not zero for \(p \in [0, 1]\).

The final answer is \(\boxed{\frac{\left(1 - \frac{4p}{5}\right)^2}{\left(1 - \frac{2p}{3}\right)^2}}\).