The problem asks for the physical state fidelity of the final 4-qubit state, given a state preparation circuit and an error model. The initial state is implicitly assumed to be \(|0000\rangle\).

1. Determine the ideal target physical state:

The circuit is \(U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\). The operations are applied right-to-left. Let the initial state be \(|\psi_0\rangle = |0000\rangle\).

• Apply \(H_2\) to qubit 2:

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

• Apply \(H_0\) to qubit 0:

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

This is the target ideal physical state. We verify it's a valid codeword (stabilized by $S_1=XXXX$ and $S_2=ZZZZ$) and has the properties of a logical GHZ state (e.g. $X_A X_B |\psi_{ideal}\rangle = |\psi_{ideal}\rangle$).

2. Analyze the error model and fidelity calculation:

Each \(CNOT_{ij}\) gate is followed by a two-qubit depolarizing error channel \(\mathcal{E}_{ij}(\rho) = (1-p)\rho + \frac{p}{15} \sum_{P_{ij} \neq I_i I_j} P_{ij} \rho P_{ij}\). Hadamard gates are assumed perfect. The circuit sequence of operations is: \(H_2 \to CNOT_{21} \to \mathcal{E}_{21} \to H_0 \to CNOT_{03} \to \mathcal{E}_{03}\). Let \(\rho_0 = |0000\rangle\langle0000|\). The state evolution is: \(\rho_1 = H_2 \rho_0 H_2^\dagger\) \(\rho_2^{pre-error} = CNOT_{21} \rho_1 CNOT_{21}^\dagger\) \(\rho_2 = \mathcal{E}_{21}(\rho_2^{pre-error}) = (1-p)\rho_2^{pre-error} + \frac{p}{15} \sum_{P_{21}} P_{21} \rho_2^{pre-error} P_{21}\) \(\rho_3 = H_0 \rho_2 H_0^\dagger\) \(\rho_4^{pre-error} = CNOT_{03} \rho_3 CNOT_{03}^\dagger\) \(\rho_f = \mathcal{E}_{03}(\rho_4^{pre-error}) = (1-p)\rho_4^{pre-error} + \frac{p}{15} \sum_{P_{03}} P_{03} \rho_4^{pre-error} P_{03}\)

The physical state fidelity is \(F = \langle\psi_{ideal}|\rho_f|\psi_{ideal}\rangle\). Substituting the expression for \(\rho_f\):

Let \(U_{circuit} = CNOT_{03} H_0 CNOT_{21} H_2\). Then \(|\psi_{ideal}\rangle = U_{circuit}|0000\rangle\). Let \(|\psi_C\rangle = H_0 CNOT_{21} H_2 |0000\rangle = |\psi_3\rangle\) from above. Then \(\rho_4^{pre-error} = CNOT_{03} \rho_3 CNOT_{03}^\dagger\). And \(|\psi_{ideal}\rangle = CNOT_{03} |\psi_C\rangle\). So \(\langle\psi_{ideal}|\rho_4^{pre-error}|\psi_{ideal}\rangle = \langle CNOT_{03}\psi_C| CNOT_{03} \rho_3 CNOT_{03}^\dagger |CNOT_{03}\psi_C\rangle = \langle\psi_C|\rho_3|\psi_C\rangle\). Also, \(\langle\psi_{ideal}|P_{03} \rho_4^{pre-error} P_{03}|\psi_{ideal}\rangle = \langle\psi_C| (CNOT_{03}^\dagger P_{03} CNOT_{03}) \rho_3 (CNOT_{03}^\dagger P_{03} CNOT_{03}) |\psi_C\rangle\). Let \(P'_{03} = CNOT_{03}^\dagger P_{03} CNOT_{03}\). This is just a different Pauli on qubits (0,3).

Now substitute \(\rho_3\): \(\rho_3 = H_0 \rho_2 H_0^\dagger = H_0 \left( (1-p)\rho_2^{pre-error} + \frac{p}{15} \sum_{P_{21}} P_{21} \rho_2^{pre-error} P_{21} \right) H_0^\dagger\). Since \(H_0\) acts on qubit 0 and \(P_{21}\) acts on qubits 1 and 2, they commute: \(H_0 P_{21} H_0^\dagger = P_{21}\). So \(\rho_3 = (1-p)H_0 \rho_2^{pre-error} H_0^\dagger + \frac{p}{15} \sum_{P_{21}} P_{21} H_0 \rho_2^{pre-error} H_0^\dagger P_{21}\). Let \(|\psi_C\rangle = H_0 \rho_2^{pre-error} H_0^\dagger = |\psi_3\rangle\langle\psi_3|\). \(\rho_3 = (1-p)|\psi_C\rangle\langle\psi_C| + \frac{p}{15} \sum_{P_{21}} P_{21} |\psi_C\rangle\langle\psi_C| P_{21}\).

Plugging this into the fidelity expression:

This simplifies to:

Let’s call the sums \(A_1, A_2, A_3\):

3. Calculate the expectation values for \(|\psi_C\rangle\):

The state is \(|\psi_C\rangle = \frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)\). This state can be factorized:

Let \(|\phi_{03}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{03} + |10\rangle_{03})\) (on qubits 0 and 3) and \(|\phi_{12}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12})\) (on qubits 1 and 2). So \(|\psi_C\rangle = |\phi_{03}\rangle \otimes |\phi_{12}\rangle\).

• Calculate \(A_1 = \sum_{P_{21}} |\langle\psi_C|P_{21}|\psi_C\rangle|^2\): \(P_{21}\) acts on qubits (1,2). So \(\langle\psi_C|P_{21}|\psi_C\rangle = \langle\phi_{03}|I|\phi_{03}\rangle \langle\phi_{12}|P_{21}|\phi_{12}\rangle = 1 \cdot \langle\phi_{12}|P_{21}|\phi_{12}\rangle\). We need to calculate \(\langle\phi_{12}|P_{21}|\phi_{12}\rangle\) for all 15 non-identity Paulis \(P_{21}\). For \(|\phi_{12}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\):

  • \(P_{21} = X_1 X_2\): \(\langle\phi_{12}|X_1 X_2|\phi_{12}\rangle = 1\).
  • \(P_{21} = Y_1 Y_2\): \(\langle\phi_{12}|Y_1 Y_2|\phi_{12}\rangle = -1\).
  • \(P_{21} = Z_1 Z_2\): \(\langle\phi_{12}|Z_1 Z_2|\phi_{12}\rangle = 1\). For all other 12 two-qubit Paulis (\(I_1 X_2, I_1 Y_2, I_1 Z_2, X_1 I_2, X_1 Y_2, X_1 Z_2, Y_1 I_2, Y_1 X_2, Y_1 Z_2, Z_1 I_2, Z_1 X_2, Z_1 Y_2\)), the expectation value \(\langle\phi_{12}|P_{21}|\phi_{12}\rangle = 0\). Thus, \(A_1 = 1^2 + (-1)^2 + 1^2 + 12 \cdot 0^2 = 3\).

• Calculate \(A_2 = \sum_{P'_{03}} |\langle\psi_C|P'_{03}|\psi_C\rangle|^2\): \(P'_{03} = CNOT_{03}^\dagger P_{03} CNOT_{03}\) acts on qubits (0,3). \(P'_{03}\) is one of the 15 non-identity Pauli operators on qubits 0 and 3. So \(\langle\psi_C|P'_{03}|\psi_C\rangle = \langle\phi_{03}|P'_{03}|\phi_{03}\rangle \langle\phi_{12}|I|\phi_{12}\rangle = \langle\phi_{03}|P'_{03}|\phi_{03}\rangle\). We need to calculate \(\langle\phi_{03}|P'_{03}|\phi_{03}\rangle\) for all 15 non-identity Paulis \(P'_{03}\). For \(|\phi_{03}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)\):

  • \(P'_{03} = I_0 Z_3\): \(\langle\phi_{03}|I_0 Z_3|\phi_{03}\rangle = 1\). (Since \(q_3\) is always 0)
  • \(P'_{03} = X_0 I_3\): \(\langle\phi_{03}|X_0 I_3|\phi_{03}\rangle = 1\).
  • \(P'_{03} = X_0 Z_3\): \(\langle\phi_{03}|X_0 Z_3|\phi_{03}\rangle = 1\). (Since \(q_3\) is always 0) For all other 12 two-qubit Paulis, the expectation value \(\langle\phi_{03}|P'_{03}|\phi_{03}\rangle = 0\). Thus, \(A_2 = 1^2 + 1^2 + 1^2 + 12 \cdot 0^2 = 3\).

• Calculate \(A_3 = \sum_{P'_{03}, P_{21}} |\langle\psi_C|P'_{03}P_{21}|\psi_C\rangle|^2\): Since \(P'_{03}\) acts on (0,3) and \(P_{21}\) acts on (1,2), they commute. \(\langle\psi_C|P'_{03}P_{21}|\psi_C\rangle = \langle\phi_{03}|P'_{03}|\phi_{03}\rangle \langle\phi_{12}|P_{21}|\phi_{12}\rangle\).

This sum is non-zero only when both factors are non-zero.
The first factor is non-zero for 3 Paulis (value 1).
The second factor is non-zero for 3 Paulis (value 1).
So $A_3 = (3 \cdot 1^2) \cdot (3 \cdot 1^2) = 9$.

4. Combine results for the final fidelity:

Substitute \(A_1=3, A_2=3, A_3=9\) into the fidelity formula:

Expand the terms:

Collect terms by powers of \(p\):

As a decimal:

The final answer is \(\boxed{1 - 1.6p + 0.64p^2}\)

To determine the logical state fidelity with post-selection, we need to calculate:

where \(|\psi_{ideal}\rangle\) is the ideal target physical state, \(\rho_f\) is the final noisy physical state, and \(P_S\) is the projector onto the codespace.

1. Numerator: Physical State Fidelity (\(F\))

The numerator \(\langle\psi_{ideal}|\rho_f|\psi_{ideal}\rangle\) is precisely the physical state fidelity calculated in Subproblem 1. From Subproblem 1, the physical fidelity is:

2. Denominator: Probability of No Detectable Error (\(P_{success}\))

The denominator \(\text{Tr}(P_S \rho_f P_S)\) is the probability that the final state \(\rho_f\) lies within the codespace (i.e., no error is detected). An error \(E\) is undetectable (i.e., it maps a codeword to another codeword, or \(E|\psi_{ideal}\rangle\) remains in the codespace) if and only if \(E\) commutes with all stabilizers of the code. The set of all Pauli operators that commute with the stabilizer group \(S\) is called the normalizer \(N(S)\).

As established in the thought process for Subproblem 1, the state evolution can be considered as the ideal state \(|\psi_{ideal}\rangle\) acted upon by an effective Pauli error \(E_{eff} = P_2 P_1\), where \(P_1\) is a Pauli error on qubits (1,2) and \(P_2\) is a Pauli error on qubits (0,3). The final noisy state \(\rho_f\) is a mixture of states \(E_{eff}|\psi_{ideal}\rangle\langle\psi_{ideal}|E_{eff}^\dagger\), with coefficients \(c(P_1,P_2)\):

where \(c(I,I) = (1-p)^2\), \(c(P_1,I) = \frac{p}{15}(1-p)\) for \(P_1 \ne I\), \(c(I,P_2) = \frac{p}{15}(1-p)\) for \(P_2 \ne I\), and \(c(P_1,P_2) = \frac{p^2}{15^2}\) for \(P_1 \ne I, P_2 \ne I\).

The probability of no detectable error is the sum of coefficients \(c(P_1,P_2)\) for all effective errors \(E_{eff} = P_2 P_1\) that belong to the normalizer \(N(S)\). The [[4,2,2]] code has \(n=4\) physical qubits and \(k=2\) logical qubits. The size of the normalizer group \(N(S)\) for a CSS code is \(2^{n+k}\). For this code, it is \(2^{4+2} = 2^6 = 64\). Since any 4-qubit Pauli operator \(E_{eff}\) can be uniquely expressed as \(P_2 P_1\) (where \(P_1\) acts on \(q_1,q_2\) and \(P_2\) acts on \(q_0,q_3\)), we simply need to count how many of the \(4^4=256\) total Pauli operators are in \(N(S)\) and sum their corresponding probabilities.

Let’s categorize the \(N_C=64\) operators in \(N(S)\): * Case 1: \(P_1=I, P_2=I\) (Identity error) This is \(E_{eff} = I\). \(I \in N(S)\). There is 1 such term. Contribution to \(P_{success}\): \(1 \cdot c(I,I) = (1-p)^2\).

• Case 2: \(P_1 \ne I, P_2=I\) (Error only on qubits 1,2) Here \(E_{eff} = P_1\). We need \(P_1 \in N(S)\). \(P_1\) acts on qubits (1,2). The stabilizers are \(S_1 = X_0X_1X_2X_3\) and \(S_2 = Z_0Z_1Z_2Z_3\). For \(P_1\) to commute with \(S_1\), the number of qubits where \(P_1\) has \(Y\) or \(Z\) and \(S_1\) has \(X\) must be even. For \(P_1\) (on \(q_1,q_2\)) \(P_{11}P_{12}\), this means \(\delta(P_{11},X_1) + \delta(P_{12},X_2)\) must be even. For \(P_1\) to commute with \(S_2\), \(\delta(P_{11},Z_1) + \delta(P_{12},Z_2)\) must be even. As derived in the scratchpad, there are 4 such Pauli operators on \((q_1,q_2)\): \(\{I_1I_2, X_1X_2, Y_1Y_2, Z_1Z_2\}\). Since \(P_1 \ne I\), there are \(4-1=3\) such terms. Contribution to \(P_{success}\): \(3 \cdot c(P_1,I) = 3 \cdot \frac{p(1-p)}{15}\).

• Case 3: \(P_1 = I, P_2 \ne I\) (Error only on qubits 0,3) Here \(E_{eff} = P_2\). We need \(P_2 \in N(S)\). This calculation is analogous to Case 2. There are 4 such Pauli operators on \((q_0,q_3)\): \(\{I_0I_3, X_0X_3, Y_0Y_3, Z_0Z_3\}\). Since \(P_2 \ne I\), there are \(4-1=3\) such terms. Contribution to \(P_{success}\): \(3 \cdot c(I,P_2) = 3 \cdot \frac{p(1-p)}{15}\).

• Case 4: \(P_1 \ne I, P_2 \ne I\) (Errors on both sets of qubits) The total number of operators in \(N(S)\) is 64. The number of operators in Case 1 is 1. The number of operators in Case 2 (excluding \(I\)) is 3. The number of operators in Case 3 (excluding \(I\)) is 3. So, the number of operators where both \(P_1 \ne I\) and \(P_2 \ne I\) that result in \(E_{eff} \in N(S)\) is \(64 - 1 - 3 - 3 = 57\). Contribution to \(P_{success}\): \(57 \cdot c(P_1,P_2) = 57 \cdot \frac{p^2}{15^2}\).

Summing these contributions for \(P_{success}\):

Simplifying the fraction \(\frac{192}{225}\): divide by 3: \(\frac{64}{75}\).

3. Logical State Fidelity (\(F_L\))

Now, divide the numerator \(F\) by the denominator \(P_{success}\):

The final answer is \(\boxed{\frac{1 - \frac{8}{5}p + \frac{16}{25}p^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2}}\)