I wonder what is the Hartree Fock state in openfermion. Is it 110000 or 000011, assume 6 qubits and 2 particles?


1 Answer 1


The Hartree-Fock state in Openfermion has $1$'s in the left positions, and $0$'s in the right positions. In other words, the spin orbitals (i.e: the qubits) are ordered from least energy to most energy. So for your 6 qubit 2 particle example, the HF state is $|110000\rangle$.

Here is an example code that shows this for a 4 qubit 2 particle system.

import numpy as np
from openfermion import *
from openfermionpyscf import run_pyscf

#create Hamiltonian and compute HF energy
geometry = [
    ['H', [0,0,0]],
    ['H', [0,0,1]]
basis = 'sto-3g'
multiplicity = 1
charge = 0

molecule = MolecularData(geometry, basis, multiplicity, charge)
molecule = run_pyscf(molecule, run_scf=True)
HF_energy = molecule.hf_energy

#convert Hamiltonian to matrix form; this is a 4 qubit system
H = get_sparse_operator(molecule.get_molecular_hamiltonian())
n_qubits = 4

#Now, compute energy manually with formula E_psi = <psi|H|psi>

#compute energy with |1100>. This is a 2**n_qubits sized vector with 1 in the int('1100', 2) index
v1100 = np.zeros(2**n_qubits)
v1100[int('1100', 2)] = 1
E1100 = v1100 @ H @ v1100

#compute energy with |0011>. This is a 2**n_qubits sized vector with 1 in the int('0011', 2) index
v0011 = np.zeros(2**n_qubits)
v0011[int('0011', 2)] = 1
E0011 = v0011 @ H @ v0011

#print results
print("Hartree Fock Energy: {}\n".format(HF_energy))
print("<1100|H|1100> = {}\n".format(E1100))
print("<0011|H|0011> = {}\n".format(E0011))

The output is

Hartree Fock Energy: -1.0661086493179366

<1100|H|1100> = (-1.0661086493179368+0j)

<0011|H|0011> = (0.00400595045040808+0j)

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