Molecular Hamiltonians as Paulis

We can finally assemble the object a quantum computer actually consumes: a qubit Hamiltonian, a weighted sum of Pauli strings. We build it for the simplest interesting molecule, molecular hydrogen $\text{H}_2$ , in the minimal STO-3G basis.

From integrals to Pauli strings

The recipe chains together everything from this module:

Pick a geometry and basis set; a classical package computes the molecular integrals $h_{pq}$ and $h_{pqrs}$ .
Assemble the second-quantized Hamiltonian $\sum_{pq} h_{pq} a_p^\dagger a_q + \tfrac12 \sum_{pqrs} h_{pqrs} a_p^\dagger a_q^\dagger a_r a_s$ .
Apply Jordan–Wigner (or Bravyi–Kitaev): each $a_p^\dagger$ , $a_p$ becomes a Pauli operator, and products of them expand into Pauli strings.
Collect like terms. The result is

\hat{H} = \sum_k c_k\, P_k, \qquad P_k \in \{I, X, Y, Z\}^{\otimes n}.

For $\text{H}_2$ in STO-3G there are 4 spin-orbitals, so the raw Jordan–Wigner Hamiltonian acts on 4 qubits and has 15 Pauli terms.

Reducing to two qubits

The 4-qubit Hamiltonian has symmetries — conservation of total electron number and of spin — that let us taper away two qubits. After this reduction the equilibrium $\text{H}_2$ Hamiltonian collapses to a compact two-qubit operator (coefficients in Hartree, after O'Malley et al. 2016):

\hat{H} = g_0\, I + g_1\, Z_0 + g_2\, Z_1 + g_3\, Z_0 Z_1 + g_4\, Y_0 Y_1 + g_5\, X_0 X_1.

The diagonal $Z$ terms set the energies of the basis states; the off-diagonal $X_0 X_1$ and $Y_0 Y_1$ terms mix $|01\rangle$ and $|10\rangle$ , which is what lets the true ground state be a superposition rather than a single Hartree–Fock configuration.

Reading off basis-state energies

For any computational basis state the off-diagonal terms vanish, and each $Z$ contributes its eigenvalue: $+1$ when the qubit is $|0\rangle$ , $-1$ when it is $|1\rangle$ . So the diagonal energy of a basis state $|q_0 q_1\rangle$ is

E_{\text{diag}} = g_0 + g_1 (-1)^{q_0} + g_2 (-1)^{q_1} + g_3 (-1)^{q_0 + q_1}.

The Hartree–Fock state for $\text{H}_2$ in this encoding is $|01\rangle$ (one electron pair in the bonding orbital). Its diagonal energy is the Hartree–Fock energy — a good starting estimate, but above the true ground state because it ignores the mixing from $X_0 X_1$ and $Y_0 Y_1$ .

Try it

Using the two-qubit $\text{H}_2$ Hamiltonian above, compute the energy of the Hartree–Fock state $|01\rangle$ . Only the diagonal $I$ , $Z_0$ , $Z_1$ , $Z_0 Z_1$ terms contribute; the $X_0 X_1$ and $Y_0 Y_1$ terms have zero diagonal matrix element. Return the energy as a number.

Run your code to see the quantum state.

This Hartree–Fock energy is the baseline that VQE will lower toward the true ground state.