Reasoning (model)
Decomposing and Solving the 4-Qubit Fidelity Problem
Alright, let’s break down this problem. My goal is to figure out how well a specific quantum state, prepared by a particular circuit, holds up under the influence of noise. This circuit prepares a 4-qubit state, and the “noise” comes in the form of a depolarizing channel, acting on pairs of qubits after each CNOT gate. The end goal is to express the fidelity, how similar the actual noisy state is to the perfect one, as a function of the error rate p.
The key to unlocking this problem is understanding the circuit’s building blocks and the error model. The circuit, in reverse order, goes: \(H_2\), \(CNOT_{21}\), \(H_0\), \(CNOT_{03}\). The errors pop up after those CNOT gates. I can treat the initial state as all \(|0\rangle\)s, or \(|0000\rangle\). After the noisy CNOT gates, the two-qubit depolarizing channel has a chance of leaving the quantum state alone or, otherwise, applying a random Pauli operator to it, as I know. It’s essentially a mixed bag of potential Pauli operators.
First, I need to figure out what the “ideal” state should look like without any errors. Tracing through the circuit, the first two gates, \(H_2\) and \(CNOT_{21}\), create a Bell pair on qubits 1 and 2. Then, the second set of gates, \(H_0\) and \(CNOT_{03}\), makes a Bell pair on qubits 0 and 3. Since these two sets of operations act on separate sets of qubits, the final “ideal” state turns out to be a tensor product: a Bell pair on qubits 0,3 tensor a Bell pair on qubits 1,2. I can write this as \(|\psi_{ideal}\rangle = |\Phi^+\rangle_{03} \otimes |\Phi^+\rangle_{12}\) where \(|\Phi^+\rangle\) means a standard Bell state.
With the ideal state in hand, I can focus on modelling how the circuit gets ruined by errors. The state starts as all \(|0\rangle\), so \(\rho_0 = |0000\rangle\langle0000|\). Then I can step by step apply the circuit and error channels. The steps are, first a block of operations and an error on qubits 1, 2, and then a second block of operations and an error on qubits 0, 3. The noise is like a filter that distorts the quantum state at each CNOT operation. The channel, written in my code, is \(\mathcal{E}(\rho) = (1-p) \rho + p \sum_{P \in \{I,X,Y,Z\}^{\otimes 2} \setminus \{II\}} \frac{P \rho P^\dagger}{15}\). The important part is that with probability \(p\), one of the 15 Pauli operators (besides the identity) will randomly flip or rotate the states.
Fidelity, \(F\), is the inner product of the ideal state with the noisy density matrix \(\rho_{final}\): \(F = \langle\psi_{ideal}|\rho_{final}|\psi_{ideal}\rangle\). The key insight is that since the operations and the errors act on different qubit pairs independently (0,3 and 1,2), the whole process factorizes. That means I can calculate the fidelity for one pair, and then square the answer. This huge simplification comes from the fact that the two error channels act on different qubit spaces: it’s a product of fidelities, and I just have to do it once!
To do this, I break the final state up into smaller steps and solve for the total fidelity. First, I compute what happens if I apply a single noisy CNOT to an initial state \(\rho_0 = |0000\rangle\langle0000|\), which can be written in a density matrix. After doing so, I then calculate the average over all 2-qubit Paulis (apart from the identity). Then, I compute the fidelity. From this, I compute that the fidelity on the bell state after the error is \(F_{gate} = 1 - \frac{4p}{5}\). Finally, since the errors are independent, the total fidelity of the 4-qubit state will be the square of that number! Therefore, the final answer is \(F = (1 - \frac{4p}{5})^2\).
Step-by-Step Derivation
The goal is to compute the physical state fidelity \(F = \langle\psi_{ideal}|\rho_{final}|\psi_{ideal}\rangle\), where \(|\psi_{ideal}\rangle\) is the state prepared by the noiseless circuit and \(\rho_{final}\) is the state prepared by the noisy circuit.
1. Determine the ideal final state \(|\psi_{ideal}\rangle\)
The circuit operations are applied from right to left to the initial state \(|\psi_{in}\rangle = |0000\rangle\).
The unitary for the circuit is \(U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\).
- Initial state:
\[
|\psi_0\rangle = |0000\rangle = |0\rangle_0 |0\rangle_1 |0\rangle_2 |0\rangle_3
\]
- Apply the Hadamard gate \(H_2\) to qubit 2:
\[
|\psi_1\rangle = H_2 |\psi_0\rangle = |0\rangle_0 |0\rangle_1 \frac{1}{\sqrt{2}}(|0\rangle_2+|1\rangle_2) |0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)
\]
- Apply the \(CNOT_{21}\) gate (control on qubit 2, target on qubit 1):
\[
|\psi_2\rangle = CNOT_{21} |\psi_1\rangle = \frac{1}{\sqrt{2}}(CNOT_{21}|0000\rangle + CNOT_{21}|0010\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)
\]
This state can be factored as a tensor product:
\[
|\psi_2\rangle = |0\rangle_0 |0\rangle_3 \otimes \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12}) = |00\rangle_{03} \otimes |\Phi^+\rangle_{12}
\]
where $|\Phi^+\rangle_{12} = \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12})$ is a Bell state on qubits 1 and 2.
- Apply the Hadamard gate \(H_0\) to qubit 0:
\[
|\psi_3\rangle = H_0 |\psi_2\rangle = \left(H_0|00\rangle_{03}\right) \otimes |\Phi^+\rangle_{12} = \frac{1}{\sqrt{2}}(|00\rangle_{03} + |10\rangle_{03}) \otimes |\Phi^+\rangle_{12}
\]
- Apply the \(CNOT_{03}\) gate (control on qubit 0, target on qubit 3):
\[
|\psi_{ideal}\rangle = CNOT_{03} |\psi_3\rangle = \left(CNOT_{03} \frac{1}{\sqrt{2}}(|00\rangle_{03} + |10\rangle_{03})\right) \otimes |\Phi^+\rangle_{12}
\]
\[
|\psi_{ideal}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{03} + |11\rangle_{03}) \otimes |\Phi^+\rangle_{12} = |\Phi^+\rangle_{03} \otimes |\Phi^+\rangle_{12}
\]
The ideal final state is a tensor product of two Bell states, one on qubits (0,3) and one on qubits (1,2).
2. Model the evolution of the density matrix with errors
The circuit consists of two blocks of operations separated by an error channel.
* Block 1: \(U_{12} = CNOT_{21} H_2\) acts on qubits 1 and 2.
* Error 1: A two-qubit depolarizing channel \(\mathcal{E}_1\) on qubits 1 and 2 follows \(CNOT_{21}\).
* Block 2: \(U_{03} = CNOT_{03} H_0\) acts on qubits 0 and 3.
* Error 2: A two-qubit depolarizing channel \(\mathcal{E}_2\) on qubits 0 and 3 follows \(CNOT_{03}\).
The error channel for a two-qubit gate is given by:
\[
\mathcal{E}(\rho) = (1-p) \rho + \frac{p}{15} \sum_{P \in \mathcal{P}_2 \setminus \{I_4\}} P \rho P^\dagger
\]
where \(\mathcal{P}_2 = \{I,X,Y,Z\}^{\otimes 2}\) is the set of 16 two-qubit Pauli operators.
Let’s trace the state \(\rho(t)\) through the circuit:
-
Initial state: \(\rho_{in} = |0000\rangle\langle0000| = |00\rangle\langle00|_{03} \otimes |00\rangle\langle00|_{12}\).
-
After \(U_{12}\): The state becomes \(\rho_1^{ideal} = U_{12}\rho_{in}U_{12}^\dagger\). Since \(U_{12}\) only acts on qubits 1,2, this is:
\[
\rho_1^{ideal} = |00\rangle\langle00|_{03} \otimes \left(U_{12}|00\rangle\langle00|_{12}U_{12}^\dagger\right) = |00\rangle\langle00|_{03} \otimes |\Phi^+\rangle\langle\Phi^+|_{12}
\]
- After the first error channel \(\mathcal{E}_1\) on qubits 1,2:
\[
\rho_1 = \mathcal{E}_1(\rho_1^{ideal}) = |00\rangle\langle00|_{03} \otimes \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12})
\]
- After \(U_{03}\): The unitary \(U_{03}\) acts only on qubits 0,3.
\[
\rho_2^{pre-error} = U_{03} \rho_1 U_{03}^\dagger = \left(U_{03}|00\rangle\langle00|_{03}U_{03}^\dagger\right) \otimes \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12})
\]
\[
\rho_2^{pre-error} = |\Phi^+\rangle\langle\Phi^+|_{03} \otimes \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12})
\]
- After the second error channel \(\mathcal{E}_2\) on qubits 0,3:
\[
\rho_{final} = \mathcal{E}_2(\rho_2^{pre-error}) = \mathcal{E}_2(|\Phi^+\rangle\langle\Phi^+|_{03}) \otimes \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12})
\]
3. Calculate the final fidelity
The fidelity is \(F = \langle\psi_{ideal}|\rho_{final}|\psi_{ideal}\rangle\). Substituting the expressions for \(|\psi_{ideal}\rangle\) and \(\rho_{final}\):
\[
F = \left( \langle\Phi^+|_{03} \otimes \langle\Phi^+|_{12} \right) \left( \mathcal{E}_2(|\Phi^+\rangle\langle\Phi^+|_{03}) \otimes \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12}) \right) \left( |\Phi^+\rangle_{03} \otimes |\Phi^+\rangle_{12} \right)
\]
Due to the tensor product structure, the fidelity factors into two identical terms:
\[
F = \left( \langle\Phi^+|_{03}| \mathcal{E}_2(|\Phi^+\rangle\langle\Phi^+|_{03}) |\Phi^+\rangle_{03} \right) \times \left( \langle\Phi^+|_{12}| \mathcal{E}_1(|\Phi^+\rangle\langle\Phi^+|_{12}) |\Phi^+\rangle_{12} \right)
\]
Let’s compute one of these factors, which represents the fidelity of a Bell state after passing through the error channel. Let \(F_{gate} = \langle\Phi^+| \mathcal{E}(|\Phi^+\rangle\langle\Phi^+|) |\Phi^+\rangle\).
\[
F_{gate} = \langle\Phi^+| \left( (1-p) |\Phi^+\rangle\langle\Phi^+| + \frac{p}{15} \sum_{P \neq I} P |\Phi^+\rangle\langle\Phi^+| P^\dagger \right) |\Phi^+\rangle
\]
\[
F_{gate} = (1-p) \langle\Phi^+|\Phi^+\rangle\langle\Phi^+|\Phi^+\rangle + \frac{p}{15} \sum_{P \neq I} \langle\Phi^+|P|\Phi^+\rangle\langle\Phi^+|P^\dagger|\Phi^+\rangle
\]
Since \(\langle\Phi^+|\Phi^+\rangle = 1\) and \(\langle\Phi^+|P^\dagger|\Phi^+\rangle = (\langle\Phi^+|P|\Phi^+\rangle)^*\):
\[
F_{gate} = (1-p) + \frac{p}{15} \sum_{P \neq I} |\langle\Phi^+|P|\Phi^+\rangle|^2
\]
We need to calculate the expectation value of the 15 non-identity two-qubit Pauli operators for the Bell state \(|\Phi^+\rangle\). The state \(|\Phi^+\rangle\) is a \(+1\) eigenstate of the Pauli operators \(X \otimes X\) and \(Z \otimes Z\).
* \(\langle\Phi^+|X \otimes X|\Phi^+\rangle = 1\)
* \(\langle\Phi^+|Z \otimes Z|\Phi^+\rangle = 1\)
* Also, \(Y \otimes Y = (iZX)(iZX) = - (Z \otimes Z)(X \otimes X)\). Thus, \(\langle\Phi^+|Y \otimes Y|\Phi^+\rangle = -1\).
For any other two-qubit Pauli operator \(P\) (not in the set \(\{I\otimes I, X\otimes X, Y\otimes Y, Z\otimes Z\}\)), it can be shown that \(\langle\Phi^+|P|\Phi^+\rangle = 0\). This is because any such \(P\) will anti-commute with at least one of the stabilizers (\(X \otimes X\) or \(Z \otimes Z\)) of \(|\Phi^+\rangle\). If \(\{P, S\}=0\) and \(S|\Phi^+\rangle = |\Phi^+\rangle\), then \(\langle\Phi^+|P|\Phi^+\rangle = \langle\Phi^+|PS|\Phi^+\rangle = -\langle\Phi^+|SP|\Phi^+\rangle = -\langle\Phi^+|P|\Phi^+\rangle\), which implies \(\langle\Phi^+|P|\Phi^+\rangle = 0\).
The sum is over the 15 non-identity Paulis. The only non-zero contributions come from \(P \in \{X \otimes X, Y \otimes Y, Z \otimes Z\}\).
\[
\sum_{P \neq I} |\langle\Phi^+|P|\Phi^+\rangle|^2 = |\langle\Phi^+|X \otimes X|\Phi^+\rangle|^2 + |\langle\Phi^+|Y \otimes Y|\Phi^+\rangle|^2 + |\langle\Phi^+|Z \otimes Z|\Phi^+\rangle|^2
\]
\[
\sum_{P \neq I} |\langle\Phi^+|P|\Phi^+\rangle|^2 = |1|^2 + |-1|^2 + |1|^2 = 1+1+1=3
\]
Now, substitute this back into the expression for \(F_{gate}\):
\[
F_{gate} = (1-p) + \frac{p}{15}(3) = 1 - p + \frac{p}{5} = 1 - \frac{4p}{5}
\]
The total fidelity is the product of the fidelities from the two independent error channels:
\[
F = F_{gate} \times F_{gate} = \left(1 - \frac{4p}{5}\right)^2
\]
Expanding this expression gives:
\[
F = 1 - \frac{8p}{5} + \frac{16p^2}{25}
\]
Final Answer:
The physical state fidelity of the final 4-qubit state is:
\[
F(p) = \left(1 - \frac{4p}{5}\right)^2 = 1 - \frac{8p}{5} + \frac{16p^2}{25}
\]