使用Qiskit的NumPy Eigensolver计算H₂自旋态时标识反转问题

我使用Qiskit的NumPy Eigensolver计算H₂分子的精确本征值与本征态时，发现叠加态的自旋标识和预期相反：正号叠加的态被判定为单态（Singlet，S²=0），负号叠加的态被判定为三重态（Triplet，S²=2），这和氢自旋异构体的常规描述不一致。

以下是我的实现代码：

分子定义与求解代码 #

# 定义H₂分子
driver = PySCFDriver(atom="H .0 .0 .0; H .0 .0 0.735", basis="sto3g")
problem = driver.run()

# 将哈密顿量映射到量子比特
mapper = JordanWignerMapper()
qubit_op = mapper.map(problem.hamiltonian.second_q_op())

expected_num_electrons = 2

def filter_criterion_custom(eigenstate, eigenvalue, aux_values):
    num_particles_aux = aux_values["ParticleNumber"][0]
    total_angular_momentum_aux = np.round(aux_values["AngularMomentum"][0],10)
    
    return (
        np.isclose(num_particles_aux, expected_num_electrons)
    )

# 经典能量计算
algo = NumPyEigensolver(k=10)
algo.filter_criterion = filter_criterion_custom
solver = ExcitedStatesEigensolver(mapper, algo)
result = solver.solve(problem_reduced)
S2_matrix = problem_reduced.properties.angular_momentum.second_q_ops()['AngularMomentum']
s2_matrix_qubit = mapper.map(S2_matrix)

检查S²的代码 #

for vec in result.eigenstates:
    psi = Statevector(vec[0]).data  # numpy数组

    print('S2:', np.round(np.vdot(psi, s2_matrix_qubit.to_matrix() @ psi),0))
    print(get_nonzero_elements(psi))
    print('---')

输出结果 #

S2: 0j
(array([ 5, 10]), array([-0.99364675+0.j,  0.11254389+0.j]))
---
S2: (2+0j)
(array([3]), array([1.+0.j]))
---
S2: (2+0j)
(array([12]), array([1.+0.j]))
---
S2: (2+0j)
(array([9, 6]), array([ 0.70710678+0.j, -0.70710678+0.j]))
---
S2: 0j
(array([6, 9]), array([0.70710678+0.j, 0.70710678+0.j]))
---
S2: 0j
(array([10,  5]), array([0.99364675+0.j, 0.11254389+0.j]))
---

内容的提问来源于stack exchange，提问作者Maria Gabriela Oliveira