Let's break down the problem step by step: ## 1. **Hamiltonian Analysis** Given: \[ H = \mu \sum_{i=1}^3 \sigma_x^{(i)} + \lambda \sigma_x^{(1)} \sigma_y^{(2)} \sigma_z^{(3)} \] - \(\sigma_k^{(i)}\) is the \(k\)th Pauli matrix on qubit \(i\). - The first term is a sum of single-qubit \(X\) rotations on each qubit. - The second term is a three-qubit interaction. We are to simulate the unitary evolution for a small timestep \(\delta t\) (using Trotterization if needed): \[ U(\delta t) = e^{-i H \delta t} \] --- ## 2. **Trotterization** Since the two terms don’t commute, we use first-order Trotter approximation: \[ U(\delta t) \approx e^{-i \mu \sum_{i=1}^3 \sigma_x^{(i)} \delta t} \cdot e^{-i \lambda \sigma_x^{(1)} \sigma_y^{(2)} \sigma_z^{(3)} \delta t} \] --- ## 3. **Quantum Circuit for \(e^{-i \mu \sum_{i=1}^3 \sigma_x^{(i)} \delta t}\)** This can be split into three single-qubit \(X\) rotations: \[ e^{-i \mu \sigma_x^{(i)} \delta t} = R_x^{(i)}(2 \mu \delta t) \] - Each qubit gets an \(R_x\) rotation by angle \(2\mu \delta t\). --- ## 4. **Quantum Circuit for \(e^{-i \lambda \sigma_x^{(1)} \sigma_y^{(2)} \sigma_z^{(3)} \delta t}\)** We need to implement a three-qubit multi-Pauli rotation. The general strategy is: - **Basis change** so that the observable becomes \(Z \otimes Z \otimes Z\). - **Apply controlled rotation**. - **Undo basis change**. **Step-by-step:** ### a. **Basis Change** To map \(\sigma_x \otimes \sigma_y \otimes \sigma_z\) to \(Z \otimes Z \otimes Z\), apply: - Qubit 1: \(H\) (Hadamard, maps \(X\) to \(Z\)) - Qubit 2: \(S^\dagger\) then \(H\) (maps \(Y\) to \(Z\)) - Qubit 3: Do nothing (already \(Z\)) ### b. **Apply \(e^{-i \lambda ZZZ \delta t}\)** Implement \(e^{-i \lambda ZZZ \delta t}\). This can be implemented using a Toffoli (CCNOT) and a single-qubit rotation: - Use two CNOTs to control on qubits 1 and 2 and target on qubit 3, then rotate qubit 3 by \(R_z(2\lambda \delta t)\), then undo the CNOTs. - Or use ancilla and Toffoli if available. **Standard decomposition:** - CNOT(2,3) - CNOT(1,3) - \(R_z(2\lambda \delta t)\) on qubit 3 - CNOT(1,3) - CNOT(2,3) ### c. **Undo Basis Change** - Qubit 1: \(H\) - Qubit 2: \(H\) then \(S\) - Qubit 3: nothing --- ## 5. **Full Circuit Steps** ### **Step 1:** Apply \(R_x(2\mu \delta t)\) to all three qubits. ### **Step 2:** Basis change for three-qubit term: - Qubit 1: \(H\) - Qubit 2: \(S^\dagger\), then \(H\) ### **Step 3:** Controlled \(R_z\) rotation - CNOT(2,3) - CNOT(1,3) - \(R_z(2\lambda \delta t)\) on qubit 3 - CNOT(1,3) - CNOT(2,3) ### **Step 4:** Undo basis change - Qubit 1: \(H\) - Qubit 2: \(H\), then \(S\) --- ## 6. **Final Quantum Circuit (Summary)** 1. **For each qubit \(i=1,2,3\):** - Apply \(R_x(2\mu \delta t)\) to qubit \(i\). 2. **Basis change:** - Qubit 1: \(H\) - Qubit 2: \(S^\dagger\), \(H\) 3. **Three-qubit interaction:** - CNOT(2,3) - CNOT(1,3) - \(R_z(2\lambda \delta t)\) on qubit 3 - CNOT(1,3) - CNOT(2,3) 4. **Undo basis change:** - Qubit 1: \(H\) - Qubit 2: \(H\), \(S\) --- ### **Final Answer (Circuit Diagram in Steps):** ``` |q1⟩ ---Rx(2μδt)---H--------------------■-----------■-----------------H---| | | | |q2⟩ ---Rx(2μδt)---S†---H--------------■-----------|-----------------H---S-| | | | |q3⟩ ---Rx(2μδt)---------------------X---Rz(2λδt)---X----------------------| ``` - S† is \(S^\dagger\), the adjoint (inverse) of the phase gate. --- ### **Explanation:** - **Single-qubit Rx gates** simulate the \(\mu \sum \sigma_x\) part. - **Basis change and controlled-Rz** simulate the \(\lambda \sigma_x^{(1)} \sigma_y^{(2)} \sigma_z^{(3)}\) part. - **Trotterization** allows us to split the evolution operator for one timestep. --- Let me know if you want the circuit in Qiskit or another format!