Reasoning (model)
My Calculated Fidelity for a Noisy GHZ State
Alright, let’s break this down. I’m tasked with figuring out the fidelity of a specific 4-qubit state, born from a circuit with Hadamards and CNOTs, and then peppered with two-qubit depolarizing noise after each CNOT. The goal is to see how well the final state matches our ideal GHZ state.
First, I need to get the “ideal” state. Since the initial state isn’t given, I’m assuming, as usual, all qubits start at \(|0\rangle\). My brain immediately jumps to calculating the successive states after each gate. I have the circuit: \(H_2\), \(CNOT_{21}\), \(H_0\), \(CNOT_{03}\). The ideal state should satisfy the definition of a GHZ in this particular [[4,2,2]] code. I went through the steps and derived the final ideal state to be \(\frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)\). This has to be the target state because the gates are correct and applying them as described produces this state. It’s the physical representation of \((|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}\) within the [[4,2,2]] code, as required by the problem.
Now, the noise. Each CNOT is followed by a two-qubit depolarizing channel acting on the control and target qubits. This is where things get messy, but not insurmountable. I need to figure out what happens when these error channels mess up the perfect state. My first instinct is to build a density matrix model. The depolarizing channel introduces all possible Pauli errors (\(X\), \(Y\), \(Z\), and combinations) with probability \(p\). The formula is: \(\mathcal{E}_{ij}(\rho) = (1-p)\rho + \frac{p}{15} \sum_{k=1}^{15} P_k \rho P_k\). The key is to keep track of where these errors occur relative to the circuit operations.
I traced the full process with density matrices, applying each gate and then the noise channel after the CNOT. This makes the math complicated because each gate transformation affects the next. The thing to remember is the order, \(H_2 \rightarrow CNOT \rightarrow\) Error Channel on (Q1,Q2) \(\rightarrow H_0 \rightarrow CNOT \rightarrow\) Error Channel on (Q0,Q3). The ideal final state is unaffected by gates that do not involve either qubit of the 2-qubit noise channel. My math shows that each noise channel affects only the gate prior to its operation. Because of this, the order of operations helps simplify things.
Now for fidelity calculation. I need to calculate \(F = \langle \psi_{ideal} | \rho_{actual} | \psi_{ideal} \rangle\). where \(|\psi_{ideal}\rangle\) is what I got earlier, and \(\rho_{actual}\) is my messy density matrix after everything. I know \(\rho_{actual}\) will be a mix of the ideal state and error states. This is a mix of the ideal state combined with the sum of all the Pauli matrices. The simplification stems from noting that the stabilizer state is an eigenstate of most of these operations. This is a stabilizer code! With a stabilizer code, I can apply the depolarizing channels in sequence, noting that the error channel on qubits (1,2) is separate from that on (0,3).
I went through the math and the process yielded, using properties of stabilizer states, \(F = (1-p)^2 + (\frac{p}{15})^2 3\). I did need to count the stabilizer operators. The stabilizers for a GHZ state are \(XXXX\) and \(ZZZZ\). After calculation, I know my state \(|\psi_{ideal}\rangle\) is stabilized by \(I\), \(XXXX\), \(ZZZZ\), and \(YYYY\). The final fidelity turns out to be \(1 - 2p + \frac{76}{75}p^2\). That’s my final answer. It takes a little elbow grease, but I’m confident in the math and the result!
The problem asks for the physical state fidelity of a prepared logical two-qubit GHZ state, accounting for depolarizing errors after each CNOT gate.
First, let’s determine the ideal final state.
The initial state is typically \(|0000\rangle\).
The circuit operations are applied in reverse order of the given equation: \(H_2\), then \(CNOT_{21}\), then \(H_0\), then \(CNOT_{03}\).
-
Initial State: \(|\psi_0\rangle = |0000\rangle\).
-
Apply \(H_2\) (on qubit 2):
\(|\psi_1\rangle = H_2 |\psi_0\rangle = I_0 \otimes I_1 \otimes H_2 \otimes I_3 |0000\rangle = |00\rangle \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) |0\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|0010\rangle)\).
-
Apply \(CNOT_{21}\) (control 2, target 1):
\(|\psi_2\rangle = CNOT_{21} |\psi_1\rangle = CNOT_{21} \frac{1}{\sqrt{2}}(|0000\rangle+|0010\rangle)\).
\(CNOT_{21}|0000\rangle = |0000\rangle\) (qubit 2 is 0, qubit 1 unchanged).
\(CNOT_{21}|0010\rangle = |0110\rangle\) (qubit 2 is 1, qubit 1 flipped).
So, \(|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle)\).
-
Apply \(H_0\) (on qubit 0):
\(|\psi_3\rangle = H_0 |\psi_2\rangle = H_0 \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle)\).
\(H_0|0000\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|1000\rangle)\).
\(H_0|0110\rangle = \frac{1}{\sqrt{2}}(|0110\rangle+|1110\rangle)\).
So, \(|\psi_3\rangle = \frac{1}{2}(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle)\).
-
Apply \(CNOT_{03}\) (control 0, target 3):
\(|\psi_{ideal}\rangle = CNOT_{03} |\psi_3\rangle = CNOT_{03} \frac{1}{2}(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle)\).
\(CNOT_{03}|0000\rangle = |0000\rangle\).
\(CNOT_{03}|1000\rangle = |1001\rangle\).
\(CNOT_{03}|0110\rangle = |0110\rangle\).
\(CNOT_{03}|1110\rangle = |1111\rangle\).
So, \(|\psi_{ideal}\rangle = \frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)\).
This is the ideal target physical state. Let’s verify it corresponds to the logical GHZ state \((|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}\).
For the given logical operators, the typical encoded states in the [[4,2,2]] code are:
\(|00\rangle_{AB} = \frac{1}{\sqrt{2}}(|0000\rangle+|1111\rangle)\)
\(|11\rangle_{AB} = \frac{1}{\sqrt{2}}(|0110\rangle+|1001\rangle)\)
Indeed, for these states \(Z_A|00\rangle_{AB}=|00\rangle_{AB}\), \(Z_B|00\rangle_{AB}=|00\rangle_{AB}\), \(Z_A|11\rangle_{AB}=-|11\rangle_{AB}\), \(Z_B|11\rangle_{AB}=-|11\rangle_{AB}\).
The target logical GHZ state is \((|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}\). Substituting the physical representations:
\(|\psi_{target}\rangle = \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}}(|0000\rangle+|1111\rangle) + \frac{1}{\sqrt{2}}(|0110\rangle+|1001\rangle) \right)\)
\(|\psi_{target}\rangle = \frac{1}{2}(|0000\rangle+|1111\rangle+|0110\rangle+|1001\rangle)\).
This matches \(|\psi_{ideal}\rangle\).
Next, we incorporate the error model. Each CNOT gate is followed by a two-qubit depolarizing error channel.
Let \(\rho_0 = |0000\rangle\langle0000|\).
The circuit has two CNOT gates: \(CNOT_{21}\) and \(CNOT_{03}\).
The operations are \(U_1=H_2\), \(U_2=CNOT_{21}\), \(U_3=H_0\), \(U_4=CNOT_{03}\).
The density matrix evolves as:
\(\rho_0 \xrightarrow{U_1} \rho_1 = U_1 \rho_0 U_1^\dagger = |\psi_1\rangle\langle\psi_1|\)
\(\rho_1 \xrightarrow{U_2} \rho_2^{pre-err} = U_2 \rho_1 U_2^\dagger = |\psi_2\rangle\langle\psi_2|\)
\(\rho_2^{pre-err} \xrightarrow{\mathcal{E}_{21}} \rho_2 = \mathcal{E}_{21}(\rho_2^{pre-err})\), where \(\mathcal{E}_{21}(\sigma) = (1-p)\sigma + \frac{p}{15} \sum_{P_{12}} P_{12} \sigma P_{12}\) (sum over 15 non-identity Pauli operators on qubits 1 and 2).
\(\rho_2 \xrightarrow{U_3} \rho_3 = U_3 \rho_2 U_3^\dagger\)
\(\rho_3 \xrightarrow{U_4} \rho_4^{pre-err} = U_4 \rho_3 U_4^\dagger\)
\(\rho_4^{pre-err} \xrightarrow{\mathcal{E}_{03}} \rho_{actual} = \mathcal{E}_{03}(\rho_4^{pre-err})\), where \(\mathcal{E}_{03}(\sigma) = (1-p)\sigma + \frac{p}{15} \sum_{Q_{03}} Q_{03} \sigma Q_{03}\) (sum over 15 non-identity Pauli operators on qubits 0 and 3).
The Hadamard gate \(H_0\) acts on qubit 0. The CNOT gate \(CNOT_{03}\) acts on qubits 0 and 3. The error operator \(P_{12}\) acts on qubits 1 and 2. Since these operations act on disjoint sets of qubits, they commute.
\(U_3 P_{12} U_3^\dagger = H_0 P_{12} H_0^\dagger = P_{12}\).
\(U_4 P_{12} U_4^\dagger = CNOT_{03} P_{12} CNOT_{03}^\dagger = P_{12}\).
So, \(\rho_3 = (U_3 U_2 U_1) \rho_0 (U_3 U_2 U_1)^\dagger = U_3 \mathcal{E}_{21}(U_2 \rho_1 U_2^\dagger) U_3^\dagger = (1-p) |\psi_3\rangle\langle\psi_3| + \frac{p}{15} \sum_{P_{12}} P_{12} |\psi_3\rangle\langle\psi_3| P_{12}\).
Similarly, \(\rho_4^{pre-err} = (1-p) |\psi_{ideal}\rangle\langle\psi_{ideal}| + \frac{p}{15} \sum_{P_{12}} P_{12} |\psi_{ideal}\rangle\langle\psi_{ideal}| P_{12}\).
Let \(|\Psi\rangle = |\psi_{ideal}\rangle\).
\(\rho_{actual} = \mathcal{E}_{03}(\rho_4^{pre-err}) = (1-p)\rho_4^{pre-err} + \frac{p}{15} \sum_{Q_{03}} Q_{03} \rho_4^{pre-err} Q_{03}\).
Substitute \(\rho_4^{pre-err}\):
\[
\rho_{actual} = (1-p) \left[ (1-p)|\Psi\rangle\langle\Psi| + \frac{p}{15} \sum_{P_{12}} P_{12} |\Psi\rangle\langle\Psi| P_{12} \right] \\ + \frac{p}{15} \sum_{Q_{03}} Q_{03} \left[ (1-p)|\Psi\rangle\langle\Psi| + \frac{p}{15} \sum_{P_{12}} P_{12} |\Psi\rangle\langle\Psi| P_{12} \right] Q_{03}.
\]
Since \(P_{12}\) and \(Q_{03}\) act on disjoint sets of qubits, they commute.
\[
\rho_{actual} = (1-p)^2 |\Psi\rangle\langle\Psi| + (1-p)\frac{p}{15} \sum_{P_{12}} P_{12} |\Psi\rangle\langle\Psi| P_{12} \\ + (1-p)\frac{p}{15} \sum_{Q_{03}} Q_{03} |\Psi\rangle\langle\Psi| Q_{03} + \left(\frac{p}{15}\right)^2 \sum_{P_{12}} \sum_{Q_{03}} Q_{03} P_{12} |\Psi\rangle\langle\Psi| P_{12} Q_{03}.
\]
The physical state fidelity is \(F = \langle \Psi | \rho_{actual} | \Psi \rangle\).
Since \(|\Psi\rangle\) is a stabilizer state (it’s a code word in a stabilizer code), for any Pauli operator \(O\), \(\langle \Psi | O |\Psi\rangle\) is either \(\pm 1\) if \(O|\Psi\rangle=\pm|\Psi\rangle\), or \(0\) if \(O|\Psi\rangle\) is orthogonal to \(|\Psi\rangle\). Therefore, \(|\langle \Psi | O |\Psi\rangle|^2\) is either 1 or 0.
\[
F = (1-p)^2 + (1-p)\frac{p}{15} \sum_{P_{12}} |\langle \Psi | P_{12} |\Psi\rangle|^2 \\ + (1-p)\frac{p}{15} \sum_{Q_{03}} |\langle \Psi | Q_{03} |\Psi\rangle|^2 + \left(\frac{p}{15}\right)^2 \sum_{P_{12}} \sum_{Q_{03}} |\langle \Psi | Q_{03} P_{12} |\Psi\rangle|^2.
\]
We need to count how many operators \(P_{12}\), \(Q_{03}\), and \(Q_{03}P_{12}\) stabilize \(|\Psi\rangle\) (i.e., cause \(|\langle \Psi | O |\Psi\rangle|^2 = 1\)).
The state \(|\Psi\rangle = \frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)\) is in the +1 eigenspace of \(XXXX\) and \(ZZZZ\). It is also stabilized by \(YYYY = XXXX \cdot ZZZZ\). The stabilizer group of \(|\Psi\rangle\) is \(G_{|\Psi\rangle} = \{I, XXXX, ZZZZ, YYYY\}\).
-
Count for \(P_{12}\): Operators \(P_{12}\) are non-identity Paulis on qubits 1 and 2 (i.e., \(I_0 \otimes P_{12} \otimes I_3\)). For \(P_{12}\) to stabilize \(|\Psi\rangle\), \(I_0 P_{12} I_3\) must be in \(G_{|\Psi\rangle}\).
- If \(I_0 P_{12} I_3 = I_0 I_1 I_2 I_3\), then \(P_{12} = I_1 I_2\). But \(P_{12}\) must be non-identity. So this doesn’t count.
- If \(I_0 P_{12} I_3 = XXXX\), then \(I_0=X_0\) and \(I_3=X_3\), which is false. Similarly for \(ZZZZ\) and \(YYYY\).
Thus, there are 0 non-identity \(P_{12}\) operators that stabilize \(|\Psi\rangle\).
-
Count for \(Q_{03}\): Operators \(Q_{03}\) are non-identity Paulis on qubits 0 and 3 (i.e., \(Q_{03} \otimes I_1 I_2\)). For \(Q_{03}\) to stabilize \(|\Psi\rangle\), \(Q_{03} I_1 I_2\) must be in \(G_{|\Psi\rangle}\).
- Similar to \(P_{12}\), no non-identity \(Q_{03}\) operator will be in \(G_{|\Psi\rangle}\).
Thus, there are 0 non-identity \(Q_{03}\) operators that stabilize \(|\Psi\rangle\).
-
Count for \(Q_{03} P_{12}\): Operators \(Q_{03} P_{12}\) are products of a non-identity \(Q_{03}\) and a non-identity \(P_{12}\). These are 4-qubit Pauli operators \(Q_0 P_1 P_2 Q_3\). We need to count how many of these stabilize \(|\Psi\rangle\).
- If \(Q_0 P_1 P_2 Q_3 = I_4\), this implies \(Q_0=I_0, P_1=I_1, P_2=I_2, Q_3=I_3\). But \(Q_{03}\) and \(P_{12}\) must be non-identity. So this doesn’t count.
- If \(Q_0 P_1 P_2 Q_3 = XXXX\): This requires \(Q_0=X_0, P_1=X_1, P_2=X_2, Q_3=X_3\). In this case, \(Q_{03}=X_0 X_3\) (non-identity) and \(P_{12}=X_1 X_2\) (non-identity). This counts as 1.
- If \(Q_0 P_1 P_2 Q_3 = ZZZZ\): This requires \(Q_0=Z_0, P_1=Z_1, P_2=Z_2, Q_3=Z_3\). In this case, \(Q_{03}=Z_0 Z_3\) (non-identity) and \(P_{12}=Z_1 Z_2\) (non-identity). This counts as 1.
- If \(Q_0 P_1 P_2 Q_3 = YYYY\): This requires \(Q_0=Y_0, P_1=Y_1, P_2=Y_2, Q_3=Y_3\). In this case, \(Q_{03}=Y_0 Y_3\) (non-identity) and \(P_{12}=Y_1 Y_2\) (non-identity). This counts as 1.
Thus, there are 3 combinations of non-identity \(Q_{03} P_{12}\) that stabilize \(|\Psi\rangle\).
Substituting these counts into the fidelity formula:
\[
F = (1-p)^2 + (1-p)\frac{p}{15}(0) + (1-p)\frac{p}{15}(0) + \left(\frac{p}{15}\right)^2 (3).
\]
\[
F = (1-p)^2 + \frac{3p^2}{15^2}
\]
\[
F = (1-2p+p^2) + \frac{3p^2}{225}
\]
\[
F = 1 - 2p + p^2 + \frac{p^2}{75}
\]
\[
F = 1 - 2p + \left(1 + \frac{1}{75}\right) p^2
\]
\[
F = 1 - 2p + \frac{76}{75} p^2.
\]
The final answer is \(\boxed{1 - 2p + \frac{76}{75} p^2}\).