FockMap

Chapter 3: From Spatial to Spin-Orbital Integrals

In this chapter, you’ll expand spatial integrals into the spin-orbital form used for encoding.

In This Chapter

The molecular integrals computed by quantum chemistry codes are in the spatial orbital basis (2 orbitals for H₂). But the fermionic operators act on spin-orbitals (4 for H₂), because each spatial orbital can hold one electron of each spin.

Spin-Orbital Indexing

Each spatial orbital $p$ gives rise to two spin-orbitals: $p\alpha$ (spin up) and $p\beta$ (spin down).

We use interleaved indexing:

Spin-orbital index Spatial orbital Spin
0 0 ($\sigma_g$) $\alpha$
1 0 ($\sigma_g$) $\beta$
2 1 ($\sigma_u$) $\alpha$
3 1 ($\sigma_u$) $\beta$

The conversion rules:

One-Body Expansion

The spin-orbital one-body integral is:

\[h^\text{spin}_{pq} = h^\text{spatial}_{p/2,\, q/2} \times \delta(\sigma_p, \sigma_q)\]

In words: the integral equals the spatial integral if the spins match, and zero otherwise. An electron cannot change its spin through one-body interactions (in the non-relativistic limit).

For H₂, this gives 4 non-zero entries — all diagonal:

$p$ $q$ $h^\text{spin}_{pq}$ (Ha) Origin
$0\alpha$ $0\alpha$ $-1.2563$ $h^\text{spatial}_{00}$, same spin
$0\beta$ $0\beta$ $-1.2563$ $h^\text{spatial}_{00}$, same spin
$1\alpha$ $1\alpha$ $-0.4719$ $h^\text{spatial}_{11}$, same spin
$1\beta$ $1\beta$ $-0.4719$ $h^\text{spatial}_{11}$, same spin

Two-Body Expansion

The spin-orbital two-body integral in physicist’s notation is:

\[\langle pq \mid rs\rangle_\text{spin} = \left[\frac{p}{2}\frac{r}{2}\bigg\mid\frac{q}{2}\frac{s}{2}\right]_\text{spatial} \times \delta(\sigma_p, \sigma_r) \times \delta(\sigma_q, \sigma_s)\]

Both electrons must independently conserve spin: electron 1 (indices $p, r$) keeps its spin, and electron 2 (indices $q, s$) keeps its spin.

This generates more non-zero integrals than one might expect, because cross-spin terms are allowed. For example, $\langle 0\alpha\, 1\beta \mid 0\alpha\, 1\beta\rangle$ is non-zero: electron 1 stays spin-$\alpha$ and electron 2 stays spin-$\beta$.

Common error: If you include only same-spin blocks ($\alpha\alpha$ and $\beta\beta$) and omit the cross-spin blocks ($\alpha\beta$ and $\beta\alpha$), your Hamiltonian will contain only Z-type (diagonal) Pauli terms and no XX/YY excitation terms. The eigenvalues will be wrong. This was our actual first-implementation bug.

For H₂, there are 32 non-zero spin-orbital two-body integrals (before symmetry reduction).

The Complete Spin-Orbital Hamiltonian

Combining one-body (4 terms) and two-body (32 terms, with $\frac{1}{2}$ prefactor), plus the nuclear repulsion constant:

\[\hat{H} = V_{nn}\cdot\hat{I} + \sum_{pq} h^\text{spin}_{pq}\, a^\dagger_p a_q + \frac{1}{2}\sum_{pqrs} \langle pq \mid rs\rangle_\text{spin}\, a^\dagger_p a^\dagger_q a_s a_r\]

with $V_{nn} = 0.7151$ Ha.

Integral Tables

Molecular Parameters

Parameter Value
Bond length $R$ 0.7414 Å = 1.401 Bohr
Nuclear repulsion $V_{nn}$ 0.7151043391 Ha
Spatial orbitals 2 ($\sigma_g$, $\sigma_u$)
Spin-orbitals 4
Electrons 2

Spatial One-Body Integrals $h_{pq}$ (Ha)

  $q = 0$ ($\sigma_g$) $q = 1$ ($\sigma_u$)
$p = 0$ $-1.2563390730$ $0$
$p = 1$ $0$ $-0.4718960244$

Spatial Two-Body Integrals $[pq \mid rs]$ (Ha)

Integral Value
$[00|00]$ $0.6744887663$
$[11|11]$ $0.6973979495$
$[00|11] = [11|00]$ $0.6636340479$
$[01|10] = [10|01] = [01|01] = [10|10]$ $0.6975782469$

All other elements are zero by symmetry. These integrals are reproduced by the companion code — see the H₂ Molecule lab.

With the spin-orbital tables ready, we can now do the core step: encode each fermionic term into Pauli strings and assemble the full qubit Hamiltonian.


Previous: Chapter 2 — The Notation Minefield Next: Chapter 4 — Building the H₂ Qubit Hamiltonian