Chapter 1: The Electronic Structure Problem
In this chapter, you’ll turn the molecular problem into a finite spin-orbital model.
In This Chapter
- What you’ll learn: How H₂ goes from a continuous molecular Schrödinger equation to a finite spin-orbital model.
- Why this matters: This is the starting point for every encoding and Hamiltonian step that follows.
- Try this next: Continue to Chapter 2 — The Notation Minefield to avoid common integral-convention mistakes.
The Schrödinger Equation for Molecules
The full molecular Hamiltonian for a molecule with $M$ nuclei (charges $Z_A$, masses $M_A$, positions $\mathbf{R}_A$) and $N$ electrons (mass $m_e$, positions $\mathbf{r}_i$) is:
\[\hat{H} = -\sum_{A=1}^{M} \frac{\hbar^2}{2M_A} \nabla_A^2 -\sum_{i=1}^{N} \frac{\hbar^2}{2m_e} \nabla_i^2 + \sum_{A<B} \frac{Z_A Z_B e^2}{|\mathbf{R}_A - \mathbf{R}_B|} - \sum_{i,A} \frac{Z_A e^2}{|\mathbf{r}_i - \mathbf{R}_A|} + \sum_{i<j} \frac{e^2}{|\mathbf{r}_i - \mathbf{r}_j|}\]For H₂, this means two protons ($A$ and $B$, separated by distance $R$) and two electrons (1 and 2). The Hamiltonian is a function of six electronic coordinates plus the internuclear distance $R$.
For the hydrogen atom (one electron, one proton), this equation can be solved analytically — giving the familiar $1s$, $2s$, $2p$, … orbitals. For two electrons, exact analytical solution is already impossible. The electron–electron repulsion term $e^2/\lvert \mathbf{r}_1 - \mathbf{r}_2 \rvert$ couples the two electrons, making the equation non-separable.
The Born–Oppenheimer Approximation
Protons are roughly 1836 times heavier than electrons. On the timescale of electronic motion, the nuclei are nearly stationary. The Born–Oppenheimer approximation treats the nuclear positions ${\mathbf{R}_A}$ as fixed parameters rather than dynamical variables.
The result is the electronic Hamiltonian:
\[\hat{H}_\text{el} = -\sum_{i=1}^{N} \frac{\hbar^2}{2m_e} \nabla_i^2 - \sum_{i,A} \frac{Z_A e^2}{|\mathbf{r}_i - \mathbf{R}_A|} + \sum_{i<j} \frac{e^2}{|\mathbf{r}_i - \mathbf{r}_j|}\]The nuclear–nuclear repulsion $V_{nn} = Z_A Z_B e^2 / R$ is just a constant for fixed $R$.
For H₂ at the equilibrium bond length $R = 0.7414$ Å (= 1.401 Bohr):
\[V_{nn} = \frac{e^2}{R} = 0.7151 \text{ Ha}\]Note: The Born–Oppenheimer approximation is the standard starting point for essentially all electronic structure theory, classical and quantum alike. It is not limiting for our purposes.
Basis Sets: Turning Continuous into Discrete
The electronic Hamiltonian acts on wavefunctions of $3N$ continuous variables. To make the problem finite-dimensional, we expand the molecular orbitals in a finite set of known functions — the basis set.
Atomic Orbitals
The hydrogen atom eigenstates ($1s$, $2s$, $2p$, …) have exponential (Slater-type) radial dependence $e^{-\zeta r}$, but integrals involving products of exponentials on different centres are analytically intractable. The practical solution: approximate each Slater-type orbital by a sum of Gaussians $e^{-\alpha r^2}$. Gaussians have the wonderful property that the product of two Gaussians is another Gaussian.
STO-3G
The “Slater-Type Orbital, 3 Gaussians” basis set approximates each atomic orbital by 3 Gaussian functions. It is the smallest meaningful basis set. For hydrogen, STO-3G provides one basis function per atom: the $1s$ orbital.
Molecular Orbitals for H₂
With one $1s$ orbital on each hydrogen atom, the Linear Combination of Atomic Orbitals (LCAO) procedure gives two molecular orbitals:
\[\sigma_g = \frac{1s_A + 1s_B}{\sqrt{2(1+S)}} \qquad \text{(bonding)}\] \[\sigma_u = \frac{1s_A - 1s_B}{\sqrt{2(1-S)}} \qquad \text{(antibonding)}\]where $S = \langle 1s_A \mid 1s_B \rangle$ is the overlap integral. The bonding orbital $\sigma_g$ has lower energy (electron density concentrated between the nuclei), while the antibonding orbital $\sigma_u$ has a node at the midpoint.
With 2 molecular orbitals and 2 spin states ($\alpha$ = spin-up, $\beta$ = spin-down), we have $2 \times 2 = 4$ spin-orbitals.
The Configuration Space
Two electrons distributed among 4 spin-orbitals can be arranged in $\binom{4}{2} = 6$ ways:
| Configuration | Notation | Description |
|---|---|---|
| $|1100\rangle$ | $\sigma_{g\alpha}\, \sigma_{g\beta}$ | Both in bonding orbital (ground state) |
| $|1010\rangle$ | $\sigma_{g\alpha}\, \sigma_{u\alpha}$ | One in each, same spin |
| $|1001\rangle$ | $\sigma_{g\alpha}\, \sigma_{u\beta}$ | One in each, opposite spin |
| $|0110\rangle$ | $\sigma_{g\beta}\, \sigma_{u\alpha}$ | One in each, opposite spin |
| $|0101\rangle$ | $\sigma_{g\beta}\, \sigma_{u\beta}$ | One in each, same spin |
| $|0011\rangle$ | $\sigma_{u\alpha}\, \sigma_{u\beta}$ | Both in antibonding orbital |
The exact ground state of H₂ is a superposition of these six configurations. The Hartree–Fock approximation uses only the first ($\lvert1100\rangle$), capturing about 99% of the energy. The remaining 1% — the correlation energy — is what makes quantum simulation valuable.
Key observation: These occupation vectors $\lvert n_0 n_1 n_2 n_3\rangle$ look exactly like qubit computational basis states $\lvert q_0 q_1 q_2 q_3\rangle$. This is not a coincidence — it is why quantum simulation of chemistry works. But the correspondence is not as simple as setting qubit $j$ = occupation of orbital $j$, because fermions and qubits obey different algebraic rules (see Why Encodings?).
Second Quantization
Rather than tracking individual electrons, second quantization tracks which orbitals are occupied. The antisymmetry of the wavefunction (which would require $N!$ terms in a Slater determinant) is absorbed into the operators.
For a detailed treatment, see Theory: Second Quantization.
The key result: the electronic Hamiltonian becomes
\[\hat{H} = \sum_{pq} h_{pq}\, a^\dagger_p a_q + \frac{1}{2}\sum_{pqrs} \langle pq \mid rs\rangle\, a^\dagger_p a^\dagger_q a_s a_r\]where $h_{pq}$ are one-body integrals (kinetic energy + electron–nucleus attraction) and $\langle pq \mid rs\rangle$ are two-body integrals (electron–electron repulsion) in physicist’s notation.
For H₂ in STO-3G, the non-zero one-body integrals are:
| Integral | Value (Ha) | Physical meaning |
|---|---|---|
| $h_{00}$ | $-1.2563$ | $\sigma_g$ orbital energy |
| $h_{11}$ | $-0.4719$ | $\sigma_u$ orbital energy |
The off-diagonal elements $h_{01} = h_{10} = 0$ by symmetry.
With the physical model now in place, the next step is to make sure our integral notation is consistent before building any encoded Hamiltonian terms.