Chapter 02 — The many-body problem

The exact electronic Schrödinger equation is solvable in closed form for $\mathrm{H}_2^+$ and tractable to extreme precision for $\mathrm{H}_2$. For every other molecule in the periodic table, we are forced to approximate. This chapter is about what the approximations are approximating — and about why, even with unlimited computer time, we could not write down the exact wavefunction for a thirty-atom molecule.

By the end of chapter 01 we had written down the time-independent electronic Schrödinger equation in atomic units, $\hat H_{\text{el}}\,\Psi = E\,\Psi$, with the Hamiltonian

\[\hat H_{\text{el}} \;=\; -\frac{1}{2}\sum_{i=1}^{N} \nabla_i^2 -\sum_{i=1}^{N}\sum_{A=1}^{M} \frac{Z_A}{|\mathbf r_i - \mathbf R_A|} +\sum_{i<j}^{N} \frac{1}{|\mathbf r_i - \mathbf r_j|}\]

operating on a wavefunction $\Psi(\mathbf x_1, \dots, \mathbf x_N)$ that depends on $3N$ continuous electronic coordinates (and, in principle, $2^N$ spin components). We also wrote down postulate P6 — identical fermions must be represented by a totally antisymmetric wavefunction — and used it to forbid two electrons from occupying the same one-electron state. Everything in the present chapter is a consequence of those two pieces of machinery. The central claim is that the exact ground-state wavefunction of an interacting electronic system is a linear combination of Slater determinants, and that the number of such determinants grows exponentially with system size, so that the exact wavefunction is not a thing we can write down, store, or even sample in any but the smallest molecules. The rest of quantum chemistry and density-functional theory is, broadly, a collection of strategies for not doing the exact calculation while still getting a useful answer. The Hartree–Fock method of chapter 03 is the mean-field starting point; the rest of these notes is, in various ways, an attempt to add electron correlation on top of it without paying the full exponential cost.

Reading note. This chapter assumes the postulates and the electronic Hamiltonian of chapter 01. It introduces two new pieces of notation — spin-orbitals and Slater determinants — and one new piece of vocabulary — the correlation energy. All three are defined the first time they appear, and the section-level cross-references are repeated where they help.

2.1 The claim

The exact ground-state energy of an $N$-electron molecule in the clamped-nuclei (Born–Oppenheimer) picture is the lowest eigenvalue of the electronic Hamiltonian in atomic units,

\[\label{eq:ch-02-hamiltonian} \boxed{\hat H_{\text{el}} = -\frac{1}{2}\sum_{i=1}^{N} \nabla_i^2 -\sum_{i=1}^{N}\sum_{A=1}^{M} \frac{Z_A}{|\mathbf r_i - \mathbf R_A|} +\sum_{i<j}^{N} \frac{1}{|\mathbf r_i - \mathbf r_j|},}\]

with the additional nuclear–nuclear repulsion added as a classical constant at the end,

\[\label{eq:ch-02-vnn} V_{NN} \;=\; \sum_{A<B}^{M} \frac{Z_A Z_B}{|\mathbf R_A - \mathbf R_B|}.\]

The corresponding exact wavefunction is a linear combination of all $N$-electron Slater determinants ${\Phi_I}$ that can be built from a chosen orthonormal one-electron basis of size $K$:

\[\label{eq:ch-02-fci} \boxed{\Psi_0 \;=\; \sum_{I=1}^{N_{\text{det}}} c_I\, \Phi_I, \qquad N_{\text{det}} \;=\; \binom{K}{N_\alpha}\binom{K}{N_\beta},}\]

with real or complex coefficients $c_I$ and $N_\alpha + N_\beta = N$. The expansion \eqref{eq:ch-02-fci} is full configuration interaction (full CI), and it is exact in the basis-set limit. Its cost is, however, the source of the exponential wall:

\[\label{eq:ch-02-exp-wall} N_{\text{det}}(K, N) \;=\; \binom{K}{N_\alpha}\binom{K}{N_\beta} \;\sim\; \frac{K^N}{N!} \qquad (K \gg N).\]

Equation \eqref{eq:ch-02-exp-wall} is the headline of the chapter. Everything we will do in the rest of these notes — Hartree chapter 03, Kohn–Sham chapter 04, chapter 05, basis sets chapter 06, solids chapter 07, pseudopotentials chapter 08 — is a strategy for not doing the full sum \eqref{eq:ch-02-fci} while still computing something useful.

Tip. The notation $\binom{K}{N_\alpha}\binom{K}{N_\beta}$ in \eqref{eq:ch-02-fci} is not a typo. The electrons come in two spin flavours, $\alpha$ (spin up) and $\beta$ (spin down), and the determinant basis factors into the product of an $\alpha$-string (choose $N_\alpha$ orbitals out of $K$) and a $\beta$-string (choose $N_\beta$ orbitals out of $K$). This is the Slater determinant basis; section 2.2.4 below writes it out explicitly.

2.2 The derivation

The headline \eqref{eq:ch-02-fci} is the result of a chain of arguments that we now derive step by step. Each subsection is self-contained and ends with a clear statement of what the preceding steps have given us.

2.2.1 Born–Oppenheimer separation

The full, non-relativistic Hamiltonian of a molecule with $M$ nuclei (charges $Z_A$, masses $M_A$, positions $\mathbf R_A$) and $N$ electrons (positions $\mathbf r_i$) is

\[\label{eq:ch-02-full-H} \hat H \;=\; \hat T_N + \hat T_e + \hat V_{NN} + \hat V_{en} + \hat V_{ee} .\]

The five terms are the nuclear kinetic energy

\[\label{eq:ch-02-TN} \hat T_N \;=\; -\sum_{A=1}^{M} \frac{1}{2 M_A} \nabla_A^2 ,\]

the electronic kinetic energy

\[\label{eq:ch-02-Te} \hat T_e \;=\; -\sum_{i=1}^{N} \frac{1}{2} \nabla_i^2 ,\]

the nuclear–nuclear repulsion

\[\label{eq:ch-02-VNN} \hat V_{NN} \;=\; \sum_{A<B}^{M} \frac{Z_A Z_B}{|\mathbf R_A - \mathbf R_B|} ,\]

the electron–nuclear attraction

\[\label{eq:ch-02-Ven} \hat V_{en} \;=\; -\sum_{i=1}^{N}\sum_{A=1}^{M} \frac{Z_A}{|\mathbf r_i - \mathbf R_A|} ,\]

and the electron–electron repulsion

\[\label{eq:ch-02-Vee} \hat V_{ee} \;=\; \sum_{i<j}^{N} \frac{1}{|\mathbf r_i - \mathbf r_j|} .\]

All masses are in units of the electron mass $m_e$ and all charges in units of $e$; the units are atomic units, defined in chapter 01.

The full Schrödinger equation is

\[\label{eq:ch-02-full-TISE} \hat H\, \Psi(\mathbf r, \mathbf R) \;=\; E\, \Psi(\mathbf r, \mathbf R) ,\]

where $\mathbf r = (\mathbf r_1, \dots, \mathbf r_N)$ and $\mathbf R = (\mathbf R_1, \dots, \mathbf R_M)$ are the full electronic and nuclear coordinate tuples. The wavefunction $\Psi(\mathbf r, \mathbf R)$ has $3(N+M)$ continuous arguments, and its exact solution is a hard problem even for the smallest molecule, $\mathrm H_2$. The Born–Oppenheimer approximation is the one move that makes the problem tractable.

Argument. The nuclear mass is $M_A \gtrsim 1836\,m_e$ (hydrogen) or larger (every other element). The ratio $m_e/M_A$ is therefore small: $m_e/M_{\text H} \approx 5.4 \times 10^{-4}$, $m_e/M_{\text C} \approx 4.6 \times 10^{-5}$, and so on. In the language of classical mechanics, the electrons move much faster than the nuclei; in the language of quantum mechanics, the de Broglie wavelength of a thermal nucleus is much shorter than that of a thermal electron, so the nuclei behave almost classically on the electronic timescale. The Born–Oppenheimer Ansatz is therefore to write the total wavefunction as a product

\[\label{eq:ch-02-BO-Ansatz} \Psi(\mathbf r, \mathbf R) \;\approx\; \psi_{\text{el}}(\mathbf r; \mathbf R)\, \chi_{\text{nuc}}(\mathbf R) ,\]

where the electronic wavefunction $\psi_{\text{el}}(\mathbf r; \mathbf R)$ depends on the nuclear coordinates parametrically (the semicolon indicates that the $\mathbf R$ are held fixed, not integrated over) and the nuclear wavefunction $\chi_{\text{nuc}}(\mathbf R)$ depends only on the nuclear coordinates. Substituting \eqref{eq:ch-02-BO-Ansatz} into \eqref{eq:ch-02-full-TISE} and neglecting the small $\nabla_A \psi_{\text{el}}$ term (the non-adiabatic coupling) gives two separate eigenvalue problems: an electronic one at each nuclear geometry $\mathbf R$,

\[\label{eq:ch-02-electronic} \hat H_{\text{el}}(\mathbf R)\, \psi_{\text{el},n}(\mathbf r; \mathbf R) \;=\; \varepsilon_n(\mathbf R)\, \psi_{\text{el},n}(\mathbf r; \mathbf R) ,\]

and a nuclear one in the electronic eigenvalue surface $\varepsilon_n(\mathbf R)$,

\[\label{eq:ch-02-nuclear} \Bigl[\,\hat T_N + \varepsilon_n(\mathbf R) + V_{NN}(\mathbf R)\Bigr]\, \chi_{\text{nuc},n\nu}(\mathbf R) \;=\; E_{n\nu}\, \chi_{\text{nuc},n\nu}(\mathbf R) .\]

Equation \eqref{eq:ch-02-electronic} is the electronic Schrödinger equation, and it is what we will solve for the rest of these notes. Equation \eqref{eq:ch-02-nuclear} is the nuclear Schrödinger equation; it is the equation of motion on the Born–Oppenheimer potential-energy surface $\varepsilon_n(\mathbf R)$, and is the starting point of vibrational spectroscopy (chapter 10) and of reaction-rate theory.

The electronic Hamiltonian appearing in \eqref{eq:ch-02-electronic} is exactly \eqref{eq:ch-02-hamiltonian}:

\[\label{eq:ch-02-Hel} \hat H_{\text{el}}(\mathbf R) \;=\; \underbrace{-\frac{1}{2}\sum_{i=1}^{N} \nabla_i^2}_{\hat T_e} \;\underbrace{-\sum_{i=1}^{N}\sum_{A=1}^{M} \frac{Z_A}{|\mathbf r_i - \mathbf R_A|}}_{\hat V_{en}} \;\underbrace{+\sum_{i<j}^{N} \frac{1}{|\mathbf r_i - \mathbf r_j|}}_{\hat V_{ee}} .\]

The $V_{NN}$ of \eqref{eq:ch-02-VNN} is the same in both \eqref{eq:ch-02-electronic} and the total energy, but in the electronic problem it is a constant (the $\mathbf R$ are parameters, not operators) and is added at the end.

Note (non-adiabatic coupling). We dropped the $\nabla_A \psi_{\text{el}}$ term that arises when the product \eqref{eq:ch-02-BO-Ansatz} is differentiated with respect to $\mathbf R_A$. This term is the source of non-adiabatic transitions between electronic states, and the Born–Oppenheimer approximation is exact only in the limit $M_A \to \infty$ (or, equivalently, $m_e/M_A \to 0$). In light molecules, in Jahn–Teller systems, at conical intersections, and whenever two electronic surfaces come close in energy, the non-adiabatic term is non-negligible and the Born–Oppenheimer picture breaks down. We will not need it for the ground-state molecules of these notes, but it is the conceptual reason Born–Oppenheimer molecular dynamics (the standard ab-initio MD) is a slightly different beast from non-adiabatic molecular dynamics (which carries surface-hopping or Ehrenfest corrections for the $\nabla_A \psi_{\text{el}}$ term).

2.2.2 Spin and the Pauli principle

A single electron has spin $s = 1/2$. The spin state lives in $\mathbb C^2$; the two basis vectors are conventionally labelled $\alpha$ (“spin up”) and $\beta$ (“spin down”),

\[\label{eq:ch-02-spin-basis} \alpha(\sigma) \;=\; \begin{pmatrix} 1 \\\\ 0 \end{pmatrix}, \qquad \beta(\sigma) \;=\; \begin{pmatrix} 0 \\\\ 1 \end{pmatrix}, \qquad \sigma \in \{-\tfrac{1}{2}, +\tfrac{1}{2}\}.\]

The combined spatial $\otimes$ spin Hilbert space of one electron is therefore $\mathcal H_1 = L^2(\mathbb R^3) \otimes \mathbb C^2$. A single-electron state is a spin-orbital

\[\label{eq:ch-02-spinorbital} \chi(\mathbf x) \;=\; \begin{cases} \phi(\mathbf r)\, \alpha(\sigma) & \text{(spin up)}, \\\\[4pt] \phi(\mathbf r)\, \beta(\sigma) & \text{(spin down)} , \end{cases}\]

where $\mathbf x = (\mathbf r, \sigma)$ is the combined space-spin coordinate and $\phi(\mathbf r) \in L^2(\mathbb R^3)$ is the spatial orbital. For a closed-shell atom or molecule with $N_\alpha = N_\beta = N/2$, we can pair up two spin-orbitals with the same spatial part $\phi(\mathbf r)$ — one $\alpha$, one $\beta$ — and call the pair a “spatial orbital” (sloppy but standard usage). This is what every quantum-chemistry code means when it talks about “$N/2$ doubly-occupied spatial orbitals”.

Postulate P6, in mathematics. The $N$-electron wavefunction $\Psi(\mathbf x_1, \dots, \mathbf x_N)$ must be totally antisymmetric under the exchange of any two combined coordinates $\mathbf x_i \leftrightarrow \mathbf x_j$:

\[\label{eq:ch-02-antisymmetry} \Psi(\mathbf x_1, \dots, \mathbf x_i, \dots, \mathbf x_j, \dots, \mathbf x_N) \;=\; -\,\Psi(\mathbf x_1, \dots, \mathbf x_j, \dots, \mathbf x_i, \dots, \mathbf x_N) .\]

The immediate corollary is the Pauli exclusion principle: if two electrons occupy the same spin-orbital, i.e. $\chi_i = \chi_j$, then the exchange gives $\Psi = -\Psi$, forcing $\Psi = 0$. Two electrons with the same spin ($\alpha\alpha$ or $\beta\beta$) cannot occupy the same spatial orbital; two electrons with opposite spins (one $\alpha$, one $\beta$) can. This is the origin of the Aufbau principle, of the Hund’s-rule multiplicity of free atoms, and of the column structure of the periodic table.

Tip. Antisymmetry is a fermionic property. Photons and $\,^4$He atoms are bosons and have symmetric wavefunctions; they can pile into the same state with no Pauli penalty. This is why a laser can be packed arbitrarily full of photons, while a metal can have at most two electrons per spatial orbital.

2.2.3 The Slater determinant

The simplest non-trivial antisymmetric wavefunction is a Slater determinant of $N$ orthonormal spin-orbitals ${\chi_i}_{i=1}^{N}$,

\[\label{eq:ch-02-slater} \Phi(\mathbf x_1, \dots, \mathbf x_N) \;=\; \frac{1}{\sqrt{N!}} \begin{vmatrix} \chi_1(\mathbf x_1) & \chi_2(\mathbf x_1) & \cdots & \chi_N(\mathbf x_1) \\\ \chi_1(\mathbf x_2) & \chi_2(\mathbf x_2) & \cdots & \chi_N(\mathbf x_2) \\\ \vdots & \vdots & \ddots & \vdots \\\ \chi_1(\mathbf x_N) & \chi_2(\mathbf x_N) & \cdots & \chi_N(\mathbf x_N) \end{vmatrix}.\]

The prefactor $1/\sqrt{N!}$ is the normalisation for orthonormal $\chi_i$; the determinant structure guarantees total antisymmetry — exchanging two columns of a determinant multiplies it by $-1$, and exchanging two rows of a matrix is the same as exchanging two columns of its transpose, so exchanging two electronic coordinates multiplies the determinant by $-1$.

It is worth checking that \eqref{eq:ch-02-slater} is correctly normalised. The square modulus, integrated over all $3N$ electronic coordinates and summed over all $2^N$ spin configurations, is

\[\label{eq:ch-02-norm} \langle \Phi \mid \Phi \rangle \;=\; \frac{1}{N!} \int d\mathbf x_1 \cdots d\mathbf x_N\; \left| \det\Bigl[\chi_i(\mathbf x_j)\Bigr] \right|^2 .\]

A classic result (sometimes called the Cauchy–Binet identity) reduces this multiple integral to a product of one-electron norms:

\[\label{eq:ch-02-cauchy-binet} \frac{1}{N!} \int d\mathbf x_1 \cdots d\mathbf x_N\; \Bigl| \det[\chi_i(\mathbf x_j)] \Bigr|^2 \;=\; \det\Bigl[ \langle \chi_i \mid \chi_j \rangle \Bigr]_{i,j=1}^{N} .\]

For orthonormal $\chi_i$, the overlap matrix on the right is the identity, and \eqref{eq:ch-02-norm} equals one. This is the reason the prefactor is $1/\sqrt{N!}$ and not, say, $1/N!$.

Tip. The Slater determinant \eqref{eq:ch-02-slater} is the “determinantal state” of second quantisation. In the occupation-number representation it is the $|\chi_1, \chi_2, \dots, \chi_N\rangle$ state — the one in which the first $N$ spin-orbitals are occupied and the rest are empty. Most of modern quantum chemistry is built on top of second quantisation (see Helgaker, Jorgensen, Olsen, Molecular Electronic-Structure Theory, §11), but for these notes the first-quantised determinant \eqref{eq:ch-02-slater} is enough.

The one-particle density associated with a single determinant is

\[\label{eq:ch-02-1pdm} \rho(\mathbf r) \;=\; \sum_{i=1}^{N} \sum_{\sigma \in \{\uparrow,\downarrow\}} \Bigl| \chi_i(\mathbf r, \sigma) \Bigr|^2 .\]

For a closed-shell determinant with $N/2$ doubly-occupied spatial orbitals ${\phi_a}_{a=1}^{N/2}$, this reduces to the spin-summed density

\[\label{eq:ch-02-rho-closed} \rho(\mathbf r) \;=\; 2 \sum_{a=1}^{N/2} |\phi_a(\mathbf r)|^2 .\]

Equations \eqref{eq:ch-02-1pdm} and \eqref{eq:ch-02-rho-closed} are the only properties of a Slater determinant that survive the single-determinant approximation and are exact for the non-interacting (mean-field) problem. They are also the ingredients of density-functional theory — see chapter 04.

2.2.4 Many Slater determinants: the FCI wavefunction

A single Slater determinant is the exact ground state of a non-interacting system of $N$ fermions in $N$ spin-orbitals — that is, a system in which the electron–electron repulsion $\hat V_{ee}$ is absent. For an interacting system, the ground state is not in general a single determinant. The standard move is to allow a linear combination of determinants.

Choose a finite orthonormal basis of $K$ spin-orbitals ${\chi_p}_{p=1}^{K}$. An $N$-electron Slater determinant is specified by choosing which $N$ of the $K$ spin-orbitals are occupied. The number of such choices factors into the $\alpha$ and $\beta$ sectors,

\[\label{eq:ch-02-ndet} N_{\text{det}}(K, N_\alpha, N_\beta) \;=\; \binom{K}{N_\alpha} \binom{K}{N_\beta} .\]

The full-CI wavefunction is the most general linear combination of all of them:

\[\label{eq:ch-02-fci-explicit} \Psi_0(\mathbf x_1, \dots, \mathbf x_N) \;=\; \sum_{I=1}^{N_{\text{det}}} c_I\, \Phi_I(\mathbf x_1, \dots, \mathbf x_N) .\]

Equation \eqref{eq:ch-02-fci-explicit} is the same as \eqref{eq:ch-02-fci}; we are simply being explicit about the $\binom{K}{N_\alpha}\binom{K}{N_\beta}$ basis. The coefficients $\{ c_I \}$ are determined by minimising the energy expectation value

\[\label{eq:ch-02-var} E_0 \;=\; \min_{\\{ c_I \\}} \frac{\sum_{I,J} c_I^*\, c_J\, \langle \Phi_I \mid \hat H_{\text{el}} \mid \Phi_J \rangle} {\sum_{I} |c_I|^2} .\]

This is the Rayleigh–Ritz variational principle applied to the finite-dimensional subspace spanned by the determinants. In the $K \to \infty$ limit the span becomes dense in the full electronic Hilbert space, and $E_0$ converges to the exact ground-state energy in the same limit. Equation \eqref{eq:ch-02-fci-explicit} is exact in the basis-set limit; the only remaining error is the basis-set incompleteness, which we treat in chapter 06.

The equation for the optimal $\{ c_I \}$ is the standard eigenvalue problem in the determinant basis,

\[\label{eq:ch-02-ci-eig} \sum_{J} \langle \Phi_I \mid \hat H_{\text{el}} \mid \Phi_J \rangle c_J \;=\; E\, c_I ,\]

i.e. $\mathbf H \mathbf c = E \mathbf c$ with $H_{IJ} = \langle \Phi_I \mid \hat H_{\text{el}} \mid \Phi_J \rangle$. The matrix elements of $\hat H_{\text{el}}$ between two $N$-electron Slater determinants are given by the Slater–Condon rules; the section 2.2.6 below writes them out for the cases needed in the worked example.

Note. Equation \eqref{eq:ch-02-ci-eig} is a general eigenvalue problem in a finite, exponentially-large basis. The size of $\mathbf H$ is $N_{\text{det}} \times N_{\text{det}}$ and grows as $\binom{K}{N_\alpha}\binom{K}{N_\beta}$ — equation \eqref{eq:ch-02-exp-wall}. Even for $K = 30$ spatial orbitals and $N = 10$ electrons ($N_\alpha = N_\beta = 5$), the dimension is $\binom{30}{5}^2 = 142{,}506$, a $200\,\text{kB}$ Hamiltonian matrix whose lowest eigenvalue is exact for the model. For $K = 50$ and $N = 20$, the dimension is $\binom{50}{10}^2 \approx 1.0 \times 10^{14}$, which is a matrix of $8 \times 10^{19}$ complex entries — about $160$ petabytes at 16 bytes per entry. Storing the matrix, never mind diagonalising it, is out of reach.

2.2.5 The exponential wall

Equation \eqref{eq:ch-02-exp-wall} is a fact, not a conjecture. The combinatorial identity

\[\label{eq:ch-02-binom-stirling} \binom{K}{N} \;=\; \frac{K!}{N!\,(K-N)!} \;\xrightarrow[K \gg N]{}\; \frac{K^N}{N!} \;\xrightarrow[K \to \infty]{}\; e^{N \ln K - \ln N!}\]

shows that the determinant count grows exponentially in the number of electrons $N$, with a base that depends on the basis size $K$. Three concrete settings illustrate the wall:

System $K$ $N$ $N_{\text{det}}$ Notes
$\mathrm H_2$ / STO-3G 2 2 $\binom{2}{1}^2 = 4$ Closed-shell, full CI is 4 dets
$\mathrm H_2$ O / 6-31G 13 10 $\binom{13}{5}^2 = 128{,}497$ Full CI is feasible, ~minute
Benzene / `6-31G* 36 30 $\binom{36}{15}^2 \approx 1.6 \times 10^{16}$ Not feasible — *petabytes
Benzene / cc-pVDZ 114 30 $\binom{114}{15}^2 \approx 2.5 \times 10^{26}$ Out of reach of any conceivable computer

The numbers in the third and fourth rows of the table are what the phrase “exponential wall” refers to. No known algorithm defeats it in general. The orbital-optimised approaches of modern quantum Monte Carlo (DMC, AFQMC, FCIQMC) and the selected CI methods (CIPSI, ASCI, Heat-bath CI) attack the wall by sampling the determinant space rather than enumerating it; the density matrix renormalisation group (DMRG) and tensor network methods attack it by factorising the wavefunction into a product of small-tensor pieces. All of them work in regimes; none of them works in general. Density-functional theory, the subject of the rest of these notes, reformulates the problem so that the exponential wall never appears — at the cost of introducing an unknown exchange–correlation functional.

Tip. The phrase “exponential wall” was coined by W. Kohn in his 1996 Nobel lecture, in the specific context of wavefunction methods. DFT is the work-around: the basic variables of the theory — the one-electron density $\rho(\mathbf r)$ — live in a three-dimensional space that is independent of $N$. The price is that the exchange–correlation functional is unknown and has to be approximated.

2.2.6 The Slater–Condon rules

To turn the eigenvalue problem \eqref{eq:ch-02-ci-eig} into a computable recipe, we need a closed-form expression for the matrix elements $\langle \Phi_I \mid \hat H_{\text{el}} \mid \Phi_J \rangle$. The Slater–Condon rules give exactly that, by exploiting the determinant structure of the $\Phi$’s. The rules express every matrix element in terms of one-electron integrals

\[\label{eq:ch-02-h1} h_{pq} \;=\; \langle \chi_p \mid \hat h \mid \chi_q \rangle, \qquad \hat h \;=\; -\tfrac{1}{2}\nabla^2 -\sum_{A=1}^{M} \frac{Z_A}{|\mathbf r - \mathbf R_A|},\]

and two-electron integrals

\[\label{eq:ch-02-v2} \langle pq \mid rs \rangle \;=\; \int\!\!\int \frac{\chi_p^*(\mathbf x_1)\, \chi_q(\mathbf x_1)\, \chi_r^*(\mathbf x_2)\, \chi_s(\mathbf x_2)} {|\mathbf r_1 - \mathbf r_2|}\; d\mathbf x_1\, d\mathbf x_2 .\]

The notation $\langle pq \mid rs \rangle$ is physicists’ notation — the bar separates the bra/ket for electron 1 from the bra/ket for electron 2. (The equivalent chemists’ notation is $[pq|rs]$; they differ by a complex conjugation of the second pair, and the two are equal for real orbitals.)

Note (real orbitals). In most of these notes the spin-orbitals are real, so the distinction between physicists’ and chemists’ notation is moot. In the general case the conversion is $$ \begin{equation} \label{eq:ch-02-notation-conv} \langle pq \mid rs \rangle \;=\; [pr|qs] , \end{equation> where the chemist integral is $[pr|qs] = \int!!\int \chi_p(\mathbf r_1)\chi_r(\mathbf r_2)\, \chi_q(\mathbf r_1)\chi_s(\mathbf r_2)/r_{12}\, d\mathbf r_1 d\mathbf r_2$.

The Slater–Condon rules state that the matrix element of $\hat H_{\text{el}}$ between two determinants is a sum of one- and two-electron contributions, with the structure determined by how the two determinants differ in their occupation of the spin-orbital basis:

The proof of these rules is in any quantum-chemistry textbook (Szabo & Ostlund §4.2; Helgaker–Jorgensen–Olsen §11.2). The key fact is that the antisymmetrised matrix element between two $N$-electron determinants factors into a sum of $N$-electron cofactors, and most of those cofactors vanish unless the occupation patterns of $I$ and $J$ differ in at most two places. The Coulomb integral $\langle ij \mid ij \rangle$ is the classical repulsion of two charge densities $|\chi_i|^2$ and $|\chi_j|^2$; the exchange integral $\langle ij \mid ji \rangle$ is a purely quantum effect with no classical analogue, and vanishes identically for $i, j$ of opposite spin (the exchange integral is a property of identical fermions only).

Tip. The “1/2” in the diagonal \eqref{eq:ch-02-sc-diag} is the price of double-counting: the sum over ordered pairs $(i, j)$ with $i, j \in I$ counts each unordered pair ${i, j}$ twice. It is the same combinatorial factor as the $N(N-1)/2$ in the definition of the variance of a uniform distribution on $N$ points.

2.2.7 Why a single determinant is not enough: the correlation energy

A single Slater determinant is a natural variational Ansatz — it satisfies Pauli automatically, it is a “determinantal state” of second quantisation, and for non-interacting fermions it gives the exact ground state. The natural question is: how close is the best single determinant to the true ground state of an interacting system?

The answer is given by the Hartree–Fock method of chapter 03: the restricted Hartree–Fock (RHF) energy is the minimum of $\langle \Phi \mid \hat H_{\text{el}} \mid \Phi \rangle$ over all single Slater determinants $\Phi$ built from a chosen basis. The correlation energy is the leftover,

\[\label{eq:ch-02-ecorr} E_{\text{corr}} \;\equiv\; E_0 - E_{\text{HF}} ,\]

where $E_0$ is the exact non-relativistic ground-state energy (in the same one-electron basis) and $E_{\text{HF}}$ is the RHF energy. By construction $E_{\text{corr}} \le 0$ — the exact wavefunction is always at least as low in energy as the best single determinant, because the determinant is one of the determinants in the FCI expansion \eqref{eq:ch-02-fci-explicit} and the variational principle guarantees that adding more determinants can only lower the energy.

Equation \eqref{eq:ch-02-ecorr} hides two effects, and it is worthwhile to disentangle them.

For atoms and small molecules near equilibrium, the magnitude of $E_{\text{corr}}$ is small compared to the total energy: for water in a triple-zeta basis, $E_{\text{HF}} \approx -76.0\,E_h$ and $E_{\text{corr}} \approx -0.3\,E_h$, i.e. about 0.4% of the total. But $E_{\text{corr}}$ is the only quantity that distinguishes different molecules’ bond energies reliably. A calculation that gets $E_{\text{HF}}$ to within 0.1% but misses the correlation energy entirely will not predict reaction energies to within chemical accuracy ($\sim 1\,\text{kcal/mol} \approx 1.6 \times 10^{-3}\,E_h$). This is the practical motivation for the hierarchy of post-Hartree–Fock methods below.

Note (the Coulson–Fischer point). For molecular geometries far from equilibrium — stretched bonds, near-dissociation limits, transition states for radical reactions — a single determinant can become qualitatively wrong, not just slightly wrong. The Coulson–Fischer point is the geometry at which the restricted Hartree–Fock solution spontaneously breaks spin symmetry and becomes unstable to a different (lower) determinant. Beyond this point, RHF is qualitatively wrong: it predicts fractional bond orders, spurious ionic contributions, and the wrong dissociation limit. This is the regime of strong (or static) correlation, and it is the regime in which the determinant basis of \eqref{eq:ch-02-fci-explicit} is needed at near-degenerate weights.

2.2.8 The hierarchy of post-HF methods

Once we accept that a single determinant is approximate, the next question is: how do we add correlation back in systematically? The standard answer is a hierarchy of methods, each with a known accuracy and a known cost. The methods are introduced in roughly increasing order of accuracy (and cost):

Method Ansatz Cost (leading) Typical error (kcal/mol)
HF Single determinant $\mathcal O(K^2 N)$ 1 – 10 (relative)
MP2 2nd-order Møller–Plesset perturbation theory $\mathcal O(K^4)$ 0.5 – 2
CISD Singles + doubles CI $\mathcal O(K^6)$ 0.5 – 2
CCSD Coupled-cluster singles + doubles $\mathcal O(K^6)$ 0.2 – 1
CCSD(T) CCSD + perturbative triples $\mathcal O(K^7)$ 0.1 – 0.5
Full CI Exact (in the basis) $\mathcal O(K^N)$ 0 (in basis)

All of the methods below “Full CI” are systematic approximations to the full-CI sum. They can be derived in three standard ways:

Tip. “Cost (leading)” in the table is the leading-order polynomial scaling with the basis-set size $K$. All of the methods have a steep prefactor (especially CCSD(T) and full CI), and the scaling is what makes them intractable in practice. A useful rule of thumb: CCSD(T) on a 20-atom organic molecule with a triple-zeta basis is feasible (a few days on a workstation); CCSD(T) on a 100-atom protein is not (it would take longer than the age of the universe). Full CI on more than about 12 electrons in a double-zeta basis is not (it would take longer than 9 million years, as the worked example in section 2.5 makes concrete).

2.2.9 Size consistency and size extensivity

The full-CI wavefunction \eqref{eq:ch-02-fci-explicit} is exact in the basis-set limit, and it has a property that no truncated method preserves without further consideration: the energy of a super-molecule made of two non-interacting fragments is the sum of the energies of the fragments. This property has two names, which are sometimes used interchangeably and sometimes not.

Size consistency (Pople, 1973). A method is size consistent if, for two non-interacting fragments $A$ and $B$ at infinite separation, the total energy equals the sum of the fragment energies:

\[\label{eq:ch-02-size-cons} E_{AB} \;=\; E_A + E_B \qquad \text{(fragments at } R \to \infty\text{)} .\]

The “non-interacting” caveat is essential: size consistency is defined for a vanishing coupling between the fragments. A method that satisfies \eqref{eq:ch-02-size-cons} for any two non-interacting fragments at any separation is size consistent.

Size extensivity (Bartlett, 1981). A method is size extensive if the energy scales linearly with the number of particles $N$ for a homogeneous system:

\[\label{eq:ch-02-size-ext} E_N \;=\; \mathcal O(N) \qquad (N \to \infty) .\]

This is a stronger property than size consistency: it requires the energy to be proportional to the number of particles, not just additive at infinite separation. For a uniform electron gas, the HF energy per electron approaches a constant, the MP2 energy per electron approaches a constant, and the CCSD(T) energy per electron approaches a constant — so all of these methods are size extensive. CISD, on the other hand, gives an energy that grows less than linearly with $N$, because the truncated CI misses the disconnected contributions to the energy (the products of excitations on different fragments) and the missing piece grows with $N$.

The distinction between the two properties is most easily seen in a dimer calculation. Take two non-interacting copies of $\mathrm{H_2O}$ at infinite separation, and compare the size- consistent prediction of the energy to the size-inconsistent one:

Method $E_A$ (per $\mathrm{H_2O}$) $E_{AB}$ (two $\mathrm{H_2O}$) Size-consistent? Size-extensive?
HF $-76.0\,E_h$ $-152.0\,E_h$ yes yes
MP2 $-76.3\,E_h$ $-152.6\,E_h$ yes yes
CISD $-76.3\,E_h$ $-152.4\,E_h$ no ($\Delta = 0.2\,E_h$) no
CCSD $-76.3\,E_h$ $-152.6\,E_h$ yes yes
CCSD(T) $-76.32\,E_h$ $-152.64\,E_h$ yes yes
Full CI $-76.32\,E_h$ $-152.64\,E_h$ yes (by construction) yes (by construction)

The numbers in the table are illustrative (rounded, not from a real calculation), but the qualitative pattern is correct. CISD gives the same energy per fragment as CCSD, but it underestimates the energy of the dimer: it misses $\sim 0.2\,E_h$ of correlation energy, which is exactly the disconnected contribution of excitations on the two fragments acting in concert. The same $0.2\,E_h$ is also the difference between “size consistent” and “size inconsistent” for two $\mathrm{H_2O}$ fragments. This is the practical reason CCSD has replaced CISD in essentially every production code: it has the same asymptotic cost and is size extensive, while CISD is not.

Note (the linked diagram theorem). The formal reason CCSD is size extensive is the linked diagram theorem: when the CC equations are derived in the language of many-body perturbation theory, only connected (linked) diagrams appear. Connected diagrams scale linearly with the number of particles, while disconnected diagrams scale as products of subsystem contributions. Truncated CI includes both linked and unlinked diagrams; the unlinked pieces do not vanish in general, and they make the truncated CI miss the size-extensive part of the energy. The exponential Ansatz $e^{\hat T}$ of coupled-cluster theory is the only non-trivial combination of excitations that automatically drops the unlinked diagrams, which is why the linked-diagram theorem makes CC methods size extensive and CI methods not.

2.2.10 The road ahead

Three avenues of escape from the exponential wall are visible from where we stand:

  1. Systematically improvable wavefunctions. The MP / CI / CC hierarchy of section 2.2.8, the selected CI family (CIPSI, ASCI, Heat-bath CI), quantum Monte Carlo (VMC, DMC, FCIQMC, AFQMC), and tensor network methods (DMRG, PEPS, MERA). Each has a regime where it works and a regime where it does not. These methods converge to the full-CI answer in the limit of their respective control parameters (perturbation order, number of selected determinants, walker population, bond dimension), and the convergence is the proof that the method is doing exactly the calculation it claims to.
  2. Density-functional reformulation. Throw out the wavefunction, reformulate the problem in terms of the one-electron density $\rho(\mathbf r)$ of \eqref{eq:ch-02-1pdm}, and accept that the exchange–correlation functional is unknown. The basic variables of the theory now live in $\mathbb R^3$, which is independent of $N$. This is the path that Kohn–Sham DFT takes, and it is the subject of chapter 04 and chapter 05.
  3. Avoid the problem entirely. Pick a system that is not strongly correlated, pick a property that is not a small energy difference, and use a method that is known to be accurate for the regime. Tight-binding, semi-empirical methods, force fields, and machine-learning potentials are all of this kind; they work because they avoid the exponentially-hard problem rather than solving it.

The rest of these notes is almost entirely about option 2. We take a short detour through Hartree–Fock first, because the HF orbitals and eigenvalues are the language in which Kohn–Sham DFT is written, and because the Fock operator is the cleanest example of a mean-field theory.

2.3 The code

The “many-body” character of the problem is hard to see in a production code — the production code just diagonalises a huge matrix — but it is easy to make visible in a toy problem. We build a one-dimensional model of two electrons in a box of length $L = 4$ bohr, with a soft short-range repulsion between them, and solve the model in a tiny basis by hand. The script dft_notes/python_codes/chapter_02/01-two-electron-box.py does exactly this.

# dft_notes/python_codes/chapter_02/01-two-electron-box.py
"""
Two electrons in a 1-D box with a soft short-range repulsion.

A toy many-body problem solved by direct full CI in a sine basis.

  - One-electron basis: the first K sin(n pi x / L) functions on
    [0, L] with the particle-in-a-box Hamiltonian h_1. - Two-electron interaction: a soft Gaussian repulsion
    V(x1, x2) = lambda * exp(-(x1 - x2)^2 / sigma^2) with
    lambda = 1.0 Hartree, sigma = 0.5 bohr.
  - N_alpha = N_beta = 1, so N_det = K^2 (spin-adapted CI:  K_orb
    singlet and triplet combinations).
  - Build the Slater-Condon matrix elements, diagonalise, plot the
    first few eigenvalues vs. K to show convergence.

This is the smallest problem that has *all* the ingredients of a
real many-body calculation: a one-electron basis, a two-electron
integrals tensor, the Slater-Condon rules, and an exponential
growth of the determinant count.
"""
import os
import numpy as np
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# --- Parameters ---------------------------------------------------------
L       = 4.0       # box length, bohr
N_e     = 2         # total electrons
N_alpha = 1
N_beta  = 1
LAMBDA  = 1.0       # repulsion strength, Hartree
SIGMA   = 0.5       # repulsion range,    bohr
N_GRID  = 400       # spatial grid for integrals

# --- One-electron grid --------------------------------------------------
x  = np.linspace(0.0, L, N_GRID)
dx = x[1] - x[0]

# --- One-electron basis: sine functions ---------------------------------
def sin_basis(K, L, x):
    """Return the K lowest sin(n pi x / L) functions on a grid."""
    n = np.arange(1, K + 1)[:, None]
    return np.sqrt(2.0 / L) * np.sin(n * np.pi  x / L)

# --- One-electron Hamiltonian matrix -----------------------------------
def one_electron_hamiltonian(K, L, x, dx):
    """<chi_p | -1/2 d^2/dx^2 | chi_q> on a sine basis."""
    chi = sin_basis(K, L, x)            # (K, N_GRID)
    lap = (np.roll(chi, -1, axis=1)
         - 2.0 * chi
         + np.roll(chi,  1, axis=1)) / dx**2
    lap[:,  0] = 0.0
    lap[:, -1] = 0.0
    return -0.5 * chi @ lap.T * dx       # (K, K)

# --- Two-electron integrals <pq|rs> -------------------------------------
def two_electron_integrals(K, L, x, dx, lam, sig):
    """
    Return the four-index tensor <pq|rs> in physicists' notation,
    using a soft Gaussian repulsion
        V(x1, x2) = lam * exp(-(x1 - x2)^2 / sig^2).
    """
    chi = sin_basis(K, L, x)            # (K, N_GRID)
    g   = np.exp(-(x[:, None] - x[None, :])**2 / sig**2)  # (N_GRID, N_GRID)
    # (pq|rs) = sum_{x1 x2} chi_p(x1) chi_q(x1) V(x1,x2) chi_r(x2) chi_s(x2) dx1 dx2
    V    = lam * g * dx  dx
    chiV = chi @ V                       # (K, N_GRID, N_GRID) -> einsum
    return np.einsum("px,qx,rx,sx,pqrs->pqrs", chi, chi, chi, chi, V)
    # actually do it step by step to keep the storage small

The complete script (with the Slater–Condon rules, the diagonalisation, and the convergence plot) is in dft_notes/python_codes/chapter_02/01-two-electron-box.py; the run command from the repo root is ‘python dft_notes/python_codes/chapter_02/01-two-electron-box.py’. The script produces two plots: a convergence plot of the first few eigenvalues vs. basis size $K$, and a density plot of the ground-state wavefunction $\Psi(x_1, x_2)$. The point of the exercise is not the physics (the model is a toy) but the code structure*: the same eight lines of Slater–Condon rules are what every quantum-chemistry package uses internally, with $K$ replaced by 100+ and $N$ replaced by 10+.

Note (storage). The full four-index tensor $\langle pq \mid rs \rangle$ has $K^4$ entries. For $K = 30$, that is $30^4 = 810{,}000$ floats ($\sim 6$ MB). For $K = 100$, it is $10^8$ floats ($\sim 800$ MB). For $K = 300$, it is $8.1 \times 10^9$ floats ($\sim 64$ GB) — already past the memory of a workstation. This is the practical reason every production code uses density fitting (a.k.a. resolution of the identity) to compress the ERI tensor from $\mathcal O(K^4)$ to $\mathcal O(K^3)$. We will not pursue this further.

2.4 The diagram

The flow of information in a full-CI calculation is a useful structural picture. The diagram below is the one to internalise before reading any code.

%%{init: {'flowchart': {'htmlLabels': true, 'curve': 'basis'}}}%%
graph TD
  GEOM["Molecular geometry<br/><small>{R_A, Z_A}</small>"] --> AOINT["One-electron<br/>integrals h_pq<br/>+ two-electron ERI<br/>tensor &lt;pq|rs&gt;"]
  AOINT --> DETS["Build the determinant basis<br/><small>choose N_alpha of K<br/>+ N_beta of K</small>"]
  DETS -->|"N_det = C(K,N_a) * C(K,N_b)<br/>(exponential in N)"| HCORE["Compute the CI matrix<br/>H_IJ = &lt;Phi_I|H|Phi_J&gt;<br/>via Slater-Condon rules"]
  HCORE --> EIG["Diagonalise H<br/><small>np.linalg.eigh</small>"]
  EIG --> E0["Lowest eigenvalue<br/>E_0 = exact (in basis)"]
  EIG --> C0["Lowest eigenvector<br/>c_I = full-CI coefficients"]
  C0 --> RHO["One-particle density<br/>rho(r) = sum_I c_I^2 ...<br/>(eq. 2.13)"]
  RHO --> NEXT["Chapter 04:<br/>Kohn-Sham DFT<br/>avoids the determinant basis"]

  classDef input fill:#faf9f5,stroke:#6c6a64,stroke-width:1px;
  classDef compute fill:#e3e6d8,stroke:#3a342a,stroke-width:1px;
  classDef explosion fill:#cc785c,stroke:#ffffff,stroke-width:2px;
  classDef output fill:#d6dcc8,stroke:#3a4031,stroke-width:1px;
  classDef next fill:#1c1f17,stroke:#f0ebde,stroke-width:1px;

  class GEOM input
  class AOINT compute
  class DETS compute
  class HCORE compute
  class EIG compute
  class DETS explosion
  class E0 output
  class C0 output
  class RHO output
  class NEXT next

The “explosion” node (coloured coral, in the centre of the diagram) is the only place where the method becomes intractable: the determinant basis grows exponentially with the number of electrons, and the diagonalisation cost is the cube of that. Every approximation in the post-HF hierarchy is a strategy for replacing or reducing this node:

The replacement of the explosion node is what the rest of these notes is about. In Kohn–Sham DFT (chapter 04) the explosion is removed entirely: the wavefunction is gone, and only the density $\rho(\mathbf r)$ remains.

2.5 Worked example — full CI on the 2-electron STO-3G H₂ molecule

We now make the machinery of §2.2 concrete on the smallest non-trivial molecule, $\mathrm{H}_2$, in a minimal basis. The worked example reproduces the Szabo–Ostlund benchmark (Szabo & Ostlund, Modern Quantum Chemistry, §3.8.5 and Table 3.5) for minimal-basis full CI on $\mathrm{H}_2$ at the equilibrium bond length, and every number below is reproduced by dft_notes/python_codes/chapter_02/03-h2-full-ci-toy.py.

2.5.1 The system

Take $\mathrm{H}_2$ with one STO-3G contracted $s$-function on each hydrogen (the same set used in chapter 06). The bond axis is the $z$-axis, the two nuclei are at $\mathbf A = (0, 0, -R/2)$ and $\mathbf B = (0, 0, +R/2)$ with $R = 1.4\,a_0$, and the raw basis is

\[\label{eq:ch-02-h2-basis} \chi_1(\mathbf r) \;=\; \chi(\mathbf r - \mathbf A), \qquad \chi_2(\mathbf r) \;=\; \chi(\mathbf r - \mathbf B).\]

With $K = 2$ spatial orbitals and $N = 2$ electrons ($N_\alpha = N_\beta = 1$), the full-CI basis of \eqref{eq:ch-02-fci-explicit} has $\binom{2}{1}\binom{2}{1} = 4$ determinants.

A trap to avoid: the raw AOs are not orthonormal. The two centred functions overlap, $S_{12} = \langle \chi_1 \mid \chi_2 \rangle = 0.6593$, so determinants built directly from ${\chi_1, \chi_2}$ are not orthonormal either — $\langle \Phi_I \mid \Phi_J \rangle \neq \delta_{IJ}$ — and the Slater–Condon rules of §2.2.6 do not apply in that raw basis. (Applying them anyway produces a Hamiltonian whose ground-state energy lies above the HF energy, violating the variational principle.) Every determinant method — from this toy example to production FCI — therefore works in an orthonormal orbital basis. We use the canonical RHF orbitals: the bonding combination $\sigma_g$ and antibonding $\sigma_u$ with orbital energies $\varepsilon_g = -0.5782$ and $\varepsilon_u = +0.6703\ E_h$. (Symmetric Löwdin orthogonalisation gives the same physics; see Problem 4.)

In terms of the orthonormal MOs the four determinants are

\[\label{eq:ch-02-h2-dets} \Phi_1 = | \sigma_g \bar\sigma_g \rangle, \quad \Phi_2 = | \sigma_g \bar\sigma_u \rangle, \quad \Phi_3 = | \sigma_u \bar\sigma_g \rangle, \quad \Phi_4 = | \sigma_u \bar\sigma_u \rangle .\]

(Here $\sigma_i$ and $\bar\sigma_i$ denote the spin-orbitals with $\sigma_i$ in the $\alpha$ and $\beta$ spin sectors, respectively.)

Note carefully what spin symmetry says about this basis. All four determinants have $M_S = 0$: each contains exactly one $\alpha$ and one $\beta$ electron. What $M_S$ conservation does not give us is a block-diagonal matrix. Instead, the clean statement is in terms of spin-adapted combinations of $\Phi_2$ and $\Phi_3$ (§2.5.4): the open-shell singlet and the open-shell triplet decouple from each other, and the triplet decouples from everything else.

2.5.2 The integrals

For reference, the raw AO-basis integrals at $R = 1.4\,a_0$ (Szabo–Ostlund Table 4.1; computed by chapter_06/01-sto-3g-h2.py, in Hartree) are

\[\label{eq:ch-02-h2-h} h_{11} = h_{22} = -1.1204, \qquad h_{12} = h_{21} = -0.9584 ,\] \[\label{eq:ch-02-h2-v} \begin{aligned} (\,11 \mid 11\,) &= 0.7746, & (\,11 \mid 22\,) &= 0.5697, \\ (\,12 \mid 12\,) &= 0.2970, & (\,11 \mid 12\,) &= 0.4441 . \end{aligned}\]

Transforming to the canonical MOs (one-electron integrals rotate as $h_{pq} \to \sum_{\mu\nu} C_{\mu p} h_{\mu\nu} C_{\nu q}$, two-electron integrals as $(pq \mid rs) \to \sum C_{\mu p} C_{\nu r} C_{\lambda s} C_{\sigma q}\,(\mu\nu \mid \lambda\sigma)$) gives the MO-basis set we actually need:

\[\label{eq:ch-02-h2-mo-ints} h_{gg} = -1.2528, \quad h_{uu} = -0.4756; \qquad J_{gg} = 0.6746, \;\; J_{uu} = 0.6975,\] \[\label{eq:ch-02-h2-mo-ints2} J_{gu} = (\,gg \mid uu\,) = 0.6636, \qquad K_{gu} = (\,gu \mid gu\,) = 0.1813 .\]

The nuclear repulsion $V_{NN} = 1/R = 0.7143\,E_h$ is added at the end.

2.5.3 The CI matrix

The Slater–Condon rules \eqref{eq:ch-02-sc-doubles} now apply directly, because the MOs are orthonormal. Element by element:

Putting it all together,

\[\label{eq:ch-02-h2-ci-matrix} \mathbf H_{\text{el}} \;=\; \begin{pmatrix} -1.8310 & 0 & 0 & 0.1813 \\ 0 & -1.0648 & -0.1813 & 0 \\ 0 & -0.1813 & -1.0648 & 0 \\ 0.1813 & 0 & 0 & -0.2537 \end{pmatrix}\; E_h .\]

The structure is now transparent: a $2\times 2$ closed-shell block ${\Phi_1, \Phi_4}$ coupled through $K_{gu}$, and a $2\times 2$ open-shell block ${\Phi_2, \Phi_3}$ coupled through $-K_{gu}$, with Brillouin’s theorem keeping the blocks independent.

2.5.4 Spin adaptation and the block eigenvalues

Define the spin-adapted combinations

\[\label{eq:ch-02-h2-csf} T_0 \;=\; \tfrac{1}{\sqrt 2}\bigl(\Phi_2 + \Phi_3\bigr), \qquad S_{gu} \;=\; \tfrac{1}{\sqrt 2}\bigl(\Phi_2 - \Phi_3\bigr),\]

which are the $M_S = 0$ components of the open-shell triplet ($S = 1$) and open-shell singlet ($S = 0$), respectively. In the ${T_0, S_{gu}}$ block the Hamiltonian diagonalises into

\[E(T_0) \;=\; H_{22} + H_{23} \;=\; -1.2461\; E_h \quad\text{(electronic)},\] \[E(S_{gu}) \;=\; H_{22} - H_{23} \;\phantom{=}\; -0.8835\; E_h \quad\text{(electronic)}.\]

Adding $V_{NN}$: the triplet $^3\Sigma_u$ state lies at $-0.5318\,E_h$ total and the open-shell singlet $^1\Sigma_u$ at $-0.1693\,E_h$ total. The triplet lies below the corresponding singlet by $2K_{gu} = 0.363\,E_h$ — the same Hund-rule exchange splitting seen in atomic spectra — but both lie well above the ground state.

The closed-shell block ${\Phi_1, \Phi_4}$ is the classic two-configuration problem, with eigenvalues

\[\label{eq:ch-02-h2-closed-evals} E_\pm \;=\; \frac{H_{11} + H_{44}}{2} \;\pm\; \sqrt{\Bigl(\frac{H_{11} - H_{44}}{2}\Bigr)^{2} + H_{14}^{2}}\] \[\label{eq:ch-02-h2-closed-numbers} E_- \;=\; -1.8516\; E_h \;\xrightarrow{\;+\,V_{NN}\;}\; -1.1373\; E_h, \qquad E_+ \;=\; -0.2331\; E_h \;\to\; +0.4811\; E_h .\]

2.5.5 The ground state and its energy

Animation 2.1 — the two-configuration problem solved live: every matrix element and every arithmetic step of the eigenvalue evaluation is shown explicitly. Rendered with Manim; source script in chapter 02's animation folder.

The four full-CI energies of \eqref{eq:ch-02-h2-ci-matrix}, including $V_{NN}$, are

\[\label{eq:ch-02-h2-fci-evals} E_0 = -1.1373\; E_h \;\;(^1\Sigma_g), \quad E(T_0) = -0.5318\; E_h \;\;(^3\Sigma_u),\] \[E(S_{gu}) = -0.1693\; E_h \;\;(^1\Sigma_u), \quad E_+ = +0.4811\; E_h \;\;(^1\Sigma_g, \text{doubly excited}) .\]

The ground state is the lower root of the two-configuration problem, with eigenvector

\[\label{eq:ch-02-h2-gs} \Psi_0 \;=\; 0.9936\,\Phi_1 \;-\; 0.1128\,\Phi_4 \;=\; 0.9936\,|\sigma_g \bar\sigma_g\rangle - 0.1128\,|\sigma_u \bar\sigma_u\rangle .\]

This is the textbook picture of left–right correlation: the exact wavefunction is dominated by the HF configuration but mixes in a small amount of the doubly excited configuration, which shifts charge density away from the bond midpoint toward the two nuclei. The weights (98.7% and 1.3%) quantify how “good” the HF approximation is for $\mathrm{H}_2$ near equilibrium — good, but not exact.

The correlation energy is

\[\label{eq:ch-02-h2-ecorr} E_{\text{corr}} \;=\; E_0 - E_{\text{HF}} \;=\; -1.1373 - (-1.1167) \;=\; -0.0206\; E_h .\]

By convention $E_{\text{corr}} \le 0$: correlation always lowers the energy relative to HF, because FCI minimises over a strictly larger variational space than any single determinant. (If your “correlation energy” comes out positive, the calculation has an error — most commonly, inconsistent orbital bases between the two numbers, as Problem 4 illustrates.)

2.5.6 The 1-particle density

The one-particle density of the ground state follows from \eqref{eq:ch-02-1pdm}. Because $\Psi_0$ contains no $\Phi_2/\Phi_3$ character, the spin-summed 1-RDM is diagonal in the MO basis, with occupations

\[\gamma_{gg} = 2\,(0.9936)^2 = 1.9746, \qquad \gamma_{uu} = 2\,(0.1128)^2 = 0.0254 ,\]

so the density is

\[\label{eq:ch-02-h2-density} \rho(\mathbf r) \;=\; 1.9746\,\lvert\sigma_g(\mathbf r)\rvert^2 \;+\; 0.0254\,\lvert\sigma_u(\mathbf r)\rvert^2 ,\]

which integrates to exactly $N = 2$. The numbers $(1.9746,\ 0.0254)$ are the natural occupation numbers — the eigenvalues of the 1-RDM — and they quantify the correlation: an uncorrelated closed-shell state has occupations $(2, 0)$, while the deviation $(1.9746,\ 0.0254)$ measures how much the doubly excited configuration has smeared electrons out of the bonding orbital.

2.5.7 What this tells us

The full-CI calculation on $\mathrm{H}_2$ in a minimal basis is small enough to do by hand in a few minutes, and it contains every ingredient of a real many-body calculation: an orthonormal orbital basis, a two-electron integrals tensor, the Slater–Condon rules, Brillouin’s theorem, spin adaptation, a $4 \times 4$ Hamiltonian matrix in the determinant basis, a numerical diagonalisation, and an interpretation of the resulting eigenvector in terms of a physically meaningful wavefunction. The only difference between this calculation and a real production calculation is the size of the basis and the number of electrons.

The size is what gets us into trouble. A minimal-basis full CI on benzene in a 6-31G* basis has $K = 36$ spatial orbitals and $N = 30$ electrons, so $N_{\text{det}} = \binom{36}{15}^2 \approx 1.6 \times 10^{16}$. A full CI on water in cc-pVTZ has $K = 58$ spatial orbitals and $N = 10$ electrons, so $N_{\text{det}} = \binom{58}{5}^2 \approx 4.2 \times 10^{10}$. The first calculation is hopeless; the second is at the edge of feasibility (a $4.2 \times 10^{10}$ complex Hamiltonian is $\sim 7$ PB at 16 bytes per entry, and its diagonalisation would take more than the age of the universe on a laptop). Every practical method is a strategy for not having to build this matrix.

2.6 Problems

Problem 1 (easy) — Antisymmetry of the Slater determinant for He Write down the two-electron Slater determinant for the helium atom in a minimal basis, with both electrons in the same spatial $1s$ orbital $\phi_{1s}(\mathbf r)$ and opposite spins (one $\alpha$, one $\beta$). Show explicitly that the wavefunction is antisymmetric under exchange of the two combined coordinates $(\mathbf r_1, \sigma_1) \leftrightarrow (\mathbf r_2, \sigma_2)$, and that it vanishes when the two electrons have the same spin and the same spatial orbital. Then compute the spin-summed one-particle density $\rho(\mathbf r)$ of this state, and show that it is normalised to $N = 2$.
Show answer The Slater determinant for the He atom's ground state is $$ \Phi_{\text{He}}(\mathbf x_1, \mathbf x_2) \;=\; \frac{1}{\sqrt 2} \begin{vmatrix} \phi_{1s}(\mathbf r_1)\,\alpha(\sigma_1) & \phi_{1s}(\mathbf r_1)\,\beta(\sigma_1) \\\ \phi_{1s}(\mathbf r_2)\,\alpha(\sigma_2) & \phi_{1s}(\mathbf r_2)\,\beta(\sigma_2) \end{vmatrix}. $$ Expanding the determinant, $$ \Phi_{\text{He}}(\mathbf x_1, \mathbf x_2) \;=\; \frac{1}{\sqrt 2}\, \phi_{1s}(\mathbf r_1)\phi_{1s}(\mathbf r_2) \Bigl[\alpha(\sigma_1)\beta(\sigma_2) -\beta(\sigma_1)\alpha(\sigma_2)\Bigr] . $$ **Antisymmetry.** Exchange $\mathbf x_1 \leftrightarrow \mathbf x_2$. The spatial part $\phi_{1s}(\mathbf r_1)\phi_{1s}(\mathbf r_2)$ is symmetric in the exchange (the two factors are identical), and the spin part $\alpha(1)\beta(2) - \beta(1)\alpha(2)$ is antisymmetric (it is the spin singlet, which flips sign under exchange of the *spin* labels). The product is therefore antisymmetric in $(\mathbf r, \sigma) \leftrightarrow (\mathbf r', \sigma')$, which is the content of postulate P6. **Vanishing for same spin and same orbital.** Suppose both electrons are $\alpha$ and both in $\phi_{1s}$: $$ \Psi(\mathbf x_1, \mathbf x_2) \;=\; \frac{1}{\sqrt 2} \begin{vmatrix} \phi_{1s}(\mathbf r_1)\alpha(\sigma_1) & \phi_{1s}(\mathbf r_1)\alpha(\sigma_1) \\\ \phi_{1s}(\mathbf r_2)\alpha(\sigma_2) & \phi_{1s}(\mathbf r_2)\alpha(\sigma_2) \end{vmatrix} \;=\; 0 , $$ because the two columns are *identical* and the determinant of a matrix with two equal columns is zero. This is the Pauli exclusion principle: two electrons with the same spin cannot occupy the same spatial orbital. **Spin-summed density.** From \eqref{eq:ch-02-rho-closed} (or directly from the wavefunction above, summing $|\Phi_{\text{He}}|^2$ over $\sigma_1, \sigma_2 \in \{\uparrow, \downarrow\}$): $$ \rho(\mathbf r) \;=\; 2\,|\phi_{1s}(\mathbf r)|^2 . $$ The factor 2 is the *spin* factor: both the $\alpha$ and the $\beta$ sectors contribute the same $|\phi_{1s}|^2$. Normalisation: $$ \int \rho(\mathbf r)\, d\mathbf r \;=\; 2 \int |\phi_{1s}(\mathbf r)|^2\, d\mathbf r \;=\; 2 \cdot 1 \;=\; N \;=\; 2 . $$ The last equality uses the normalisation $\int |\phi_{1s}|^2\, d\mathbf r = 1$ of the spatial orbital.
Problem 2 (medium) — Slater–Condon rules for a 2-electron system For a 2-electron system with one-electron integrals $\{h_{ii}\}$ and two-electron integrals $\{\langle ii \mid ii \rangle, \langle ii \mid jj \rangle, \langle ij \mid ij \rangle\}$ (no other distinct integrals, by the symmetry of real orbitals in 2 electrons), show by direct evaluation of the 2 × 2 Slater determinants that the Slater–Condon rules \eqref{eq:ch-02-sc-diag} and \eqref{eq:ch-02-sc-singles} give exactly the matrix elements of the 2 × 2 CI problem. Concretely, take the basis $\{\Phi_1, \Phi_2\} = \{|1\bar 1\rangle, |2\bar 2\rangle\}$ for a closed-shell 2-electron system with $K = 2$ spatial orbitals. Show that (a) the diagonal element $\langle \Phi_1 \mid \hat H_{\text{el}} \mid \Phi_1 \rangle = 2 h_{11} + \langle 11 \mid 11 \rangle$; (b) the off-diagonal element $\langle \Phi_1 \mid \hat H_{\text{el}} \mid \Phi_2 \rangle = \langle 11 \mid 22 \rangle - \langle 12 \mid 21 \rangle$.
Show answer **Part (a).** The wavefunction $\Phi_1 = |1 \bar 1\rangle = \phi_1(\mathbf r_1)\alpha(\sigma_1) \phi_1(\mathbf r_2)\beta(\sigma_2)$, normalised. Apply $\hat H_{\text{el}}$ to $\Phi_1$, written as a sum of one- and two-electron operators: $$ \hat H_{\text{el}} \Phi_1 \;=\; (\hat h_1 + \hat h_2 + \hat v_{12})\,\Phi_1 \;=\; \hat h_1 \Phi_1 + \hat h_2 \Phi_1 + \hat v_{12} \Phi_1 . $$ Project onto $\Phi_1$ and integrate: - **$\hat h_1$:** $\langle \Phi_1 \mid \hat h_1 \mid \Phi_1 \rangle = h_{11}$ by the spin orthogonality $\langle \alpha \mid \alpha \rangle = 1$, $\langle \beta \mid \alpha \rangle = 0$. The electron-1 factor is $\phi_1(\mathbf r_1)\alpha(\sigma_1)$ on both sides, so the spin integral is 1, and the spatial integral is $h_{11}$. - **$\hat h_2$:** $\langle \Phi_1 \mid \hat h_2 \mid \Phi_1 \rangle = h_{11}$ by the same argument with electron 2. - **$\hat v_{12}$:** $\langle \Phi_1 \mid \hat v_{12} \mid \Phi_1 \rangle$ involves the two-electron operator $1/|\mathbf r_1 - \mathbf r_2|$. The spin factors are still $\langle \alpha \mid \alpha \rangle \langle \beta \mid \beta \rangle = 1$, and the spatial integral is exactly $\langle 11 \mid 11 \rangle$ as defined in \eqref{eq:ch-02-v2}. Adding, $$ \langle \Phi_1 \mid \hat H_{\text{el}} \mid \Phi_1 \rangle \;=\; 2 h_{11} + \langle 11 \mid 11 \rangle . $$ This is the Slater–Condon rule \eqref{eq:ch-02-sc-diag} for the 2-electron case, with the exchange term $-\langle 11 \mid 11 \rangle$ cancelling against the 1/2 in front of the double sum (because the only pair is $i = j = 1$, so the 1/2 is irrelevant for a single-pair diagonal). **Part (b).** The two determinants differ by *two* spin-orbitals ($\Phi_1$ has $\{1\alpha, 1\beta\}$, $\Phi_2$ has $\{2\alpha, 2\beta\}$), so the only non-zero contribution to $\langle \Phi_1 \mid \hat H_{\text{el}} \mid \Phi_2 \rangle$ comes from the two-electron operator — the one-electron operator $\hat h_1$ would need $\langle 1 \mid \hat h \mid 2 \rangle$ in the spatial sector, but the spin part of $\Phi_1$ is $\alpha(1)\beta(2)$ and that of $\Phi_2$ is $\alpha(1)\beta(2)$ too, so the spin integral is still 1, but the spatial integral is $h_{12}$, *not* the 1-electron integral $h_{11}$. However, the antisymmetrisation in the determinant gives a *plus* sign for $\Phi_1$ and a minus sign for $\Phi_2$ in the same-row interchange, so the two terms cancel exactly. This is a 2-electron version of the fact that the matrix element of a 1-electron operator between two determinants that differ in two spin-orbitals vanishes. The two-electron piece: $$ \langle \Phi_1 \mid \hat v_{12} \mid \Phi_2 \rangle \;=\; \int d\mathbf x_1 d\mathbf x_2\; \phi_1(\mathbf r_1)\alpha(\sigma_1)\phi_1(\mathbf r_2)\beta(\sigma_2) \cdot \frac{1}{r_{12}} \cdot \phi_2(\mathbf r_1)\alpha(\sigma_1)\phi_2(\mathbf r_2)\beta(\sigma_2) . $$ Spin part: $\langle \alpha \mid \alpha \rangle \langle \beta \mid \beta \rangle = 1$. Spatial part: the Slater–Condon rule \eqref{eq:ch-02-sc-doubles} for two-electron systems gives $$ \int d\mathbf r_1 d\mathbf r_2\; \phi_1(\mathbf r_1)\phi_1(\mathbf r_2)\, \frac{1}{r_{12}}\, \phi_2(\mathbf r_1)\phi_2(\mathbf r_2) \;=\; \langle 11 \mid 22 \rangle . $$ However, the *antisymmetrised* matrix element picks up an *exchange* correction because the determinant structure reverses the sign of the two rows in a way that is sensitive to the identity of the orbitals. Concretely, expanding the determinants and projecting, $$ \langle \Phi_1 \mid \hat v_{12} \mid \Phi_2 \rangle \;=\; \langle 11 \mid 22 \rangle - \langle 12 \mid 21 \rangle . $$ For real orbitals, $\langle 12 \mid 21 \rangle = \langle 12 \mid 12 \rangle$, so the result is the *exchange-corrected* Coulomb integral $$ \boxed{H_{12} \;=\; \langle 11 \mid 22 \rangle - \langle 12 \mid 12 \rangle .} $$ For a 2-electron 2-orbital problem with real orbitals, the exchange term is the only correction to the bare Coulomb integral. This is the result we used in \eqref{eq:ch-02-h2-h14}.
Problem 3 (hard) — Correlation energy of H₂ in a minimal basis The full-CI ground-state energy of $\mathrm{H}_2$ in a STO-3G minimal basis at $R = 1.4\,a_0$ is $E_0 = -1.1373\,E_h$ (§2.5.5). The restricted Hartree–Fock ground-state energy in the same basis is $E_{\text{HF}} = -1.1167\,E_h$ ([chapter 06](/DFT-notes/dft-notes/chapter-06/), §6.9). Carry out the following calculation, step by step, *by hand* (with a calculator — no computer): 1. Compute the correlation energy $E_{\text{corr}} = E_0 - E_{\text{HF}}$. 2. Express $E_{\text{corr}}$ in **eV** and in **kcal/mol** (using $1\,E_h = 27.2114\,\text{eV} = 627.509\,\text{kcal/mol}$). 3. The ground state \eqref{eq:ch-02-h2-gs} is $\Psi_0 = 0.9936\,\Phi_1 - 0.1128\,\Phi_4$. Compute the *energy expectation value* of $\hat H_{\text{el}} + V_{NN}$ in this state directly from the matrix elements \eqref{eq:ch-02-h2-h11}, \eqref{eq:ch-02-h2-h44} and \eqref{eq:ch-02-h2-h14}, and verify that it equals $E_0$. 4. Compare with the single-determinant HF energy. Show that the variational principle guarantees $\langle\Psi_0|\hat H|\Psi_0\rangle \le E_{\text{HF}}$, and explain in one sentence why a calculation that reports $E_0 > E_{\text{HF}}$ in the same orbital basis has an error.
Show answer **Part 1.** $$ E_{\text{corr}} \;=\; E_0 - E_{\text{HF}} \;=\; -1.1373 - (-1.1167) \;=\; -0.0206\,E_h . $$ The sign is *negative*: correlation always lowers the energy, because FCI minimises over a strictly larger variational space than any single determinant. **Part 2.** Using the conversions, $$ \begin{aligned} E_{\text{corr}} \;=\; -0.0206 \times 27.2114\;\text{eV} \;\approx\; -0.561\;\text{eV}, \\ E_{\text{corr}} \;=\; -0.0206 \times 627.509\;\text{kcal/mol} \;\approx\; -12.93\;\text{kcal/mol}. \end{aligned} $$ This is a substantial fraction of the $\mathrm H_2$ bond energy ($D_e \approx 0.174\,E_h \approx 109$ kcal/mol): even in this tiny system, left–right correlation contributes at the several-kcal/mol level, which matters for thermochemistry. **Part 3.** For $\Psi_0 = c_1\Phi_1 + c_4\Phi_4$ with $c_1 = 0.9936$ and $c_4 = -0.1128$, the expectation value is $$ \langle \Psi_0 | \hat H | \Psi_0 \rangle \;=\; c_1^2 H_{11} + c_4^2 H_{44} + 2 c_1 c_4 H_{14}, $$ $$ \;=\; (0.9873)(-1.8310) + (0.0127)(-0.2537) + 2(0.9936)(-0.1128)(0.1813) $$ $$ \;=\; -1.8078 - 0.0032 - 0.0407 \;=\; -1.8517\; E_h \quad(\text{electronic}), $$ and adding $V_{NN} = 0.7143\,E_h$ gives $-1.1374\,E_h \approx E_0$ (the last digit differs by rounding the coefficients to four decimals). This is just the statement that $\Psi_0$ is an eigenvector: its expectation value equals the eigenvalue. **Part 4.** The variational principle gives $\langle\Psi_0|\hat H|\Psi_0\rangle \le E_{\text{HF}}$ because the variational space of FCI contains the HF determinant as one of its basis states ($\Phi_1$ *is* the HF determinant), so minimising over the larger space can only stay equal or go lower. A reported $E_0 > E_{\text{HF}}$ therefore signals an inconsistency — most commonly, building the CI matrix in a non-orthonormal determinant basis (see Problem 4) or comparing energies from two different orbital bases.
Problem 4 (medium) — Why the orbitals must be orthonormal Suppose you build the $4\times 4$ CI matrix of §2.5 directly in the *raw* AO basis $\{\chi_1, \chi_2\}$ — i.e. determinants $|1\bar 1\rangle, |1\bar 2\rangle, |2\bar 1\rangle, |2\bar 2\rangle$ with $\langle\chi_1|\chi_2\rangle = S_{12} = 0.6593$ — while using the ordinary Slater–Condon rules, which assume orthonormal spin-orbitals. 1. Compute $\langle 1\bar 1 \mid 1\bar 2\rangle$ for these raw-AO determinants and show that the determinant basis is not orthonormal. 2. Using the raw AO integrals of \eqref{eq:ch-02-h2-h} and \eqref{eq:ch-02-h2-v}, build the naive $4\times 4$ matrix (diagonal elements $2h_{11} + (11|11)$ etc., coupling $(11|22) - (12|21)$ for the double excitation), diagonalise it, and compare the lowest root with $E_\mathrm{RHF} = -1.1167$. 3. Explain why the result violates the variational principle, and fix it by repeating the calculation in the Löwdin-orthogonalised basis $\phi_i = \sum_j (S^{-1/2})_{ji}\chi_j$. Verify that you recover the correct FCI energy $-1.1373\,E_h$.
Show answer **Part 1.** For two determinants built from non-orthogonal spin-orbitals, $$ \langle 1\bar 1 \mid 1\bar 2 \rangle \;=\; \langle \chi_1|\chi_1\rangle \langle \bar\chi_1|\bar\chi_2\rangle \;=\; 1 \times S_{12} \;=\; 0.6593 \neq 0 . $$ The four determinants are linearly independent but far from orthogonal, so they do not form an orthonormal basis and the usual eigenvalue problem $\mathbf H\mathbf c = E\mathbf c$ does not apply. **Part 2.** Applying the orthonormal-basis Slater–Condon rules to the raw integrals gives $$ H_{11} = H_{44} = 2(-1.1204) + 0.7746 = -1.4662, \qquad H_{14} = (11|22) - (12|21) = 0.5697 - 0.2970 = 0.2727, $$ (and analogously for the open-shell block). Diagonalising the resulting $4\times 4$ matrix yields a lowest root near $-1.59\,E_h$, which lies *below* both the correct answer $-1.1373\,E_h$ and the exact non-relativistic $\mathrm H_2$ energy $-1.1745\,E_h$ — an impossible outcome. **Part 3.** The violation happens because the generalised eigenproblem for a non-orthogonal basis is $\mathbf H\mathbf c = E\,\mathbf S\mathbf c$ with the determinant overlap matrix $\mathbf S_{IJ} = \langle\Phi_I|\Phi_J\rangle$; ignoring $\mathbf S$ amounts to solving the wrong problem, and the resulting energies are not variational upper bounds. With the Löwdin orthogonalisation $\boldsymbol\phi = \boldsymbol\chi\,\mathbf S^{-1/2}$, where $$ \mathbf X = \mathbf S^{-1/2} = \begin{pmatrix} 1.2448 & -0.4685 \\ -0.4685 & 1.2448 \end{pmatrix}, $$ the transformed integrals reproduce the canonical-MO results of §2.5 up to a rotation among the degenerate-symmetry orbitals, and the lowest root becomes exactly the full-CI energy $-1.1373\,E_h$. Moral: **second quantisation and the Slater–Condon rules presuppose an orthonormal spin-orbital basis — always orthogonalise first.**

2.7 Cross-references

Backward (what we used from earlier chapters):

Forward (where this material is used):

External references (for the reader who wants to go deeper):

2.8 What we left out

This chapter is the setup for everything else; it is necessarily a summary of an enormous literature. The topics we have not covered, and the reader should know are missing, are:

Next: Chapter 03 — Hartree–Fock: the mean-field approximation to the many-body problem, the Fock operator, the self-consistent field loop, and the only wavefunction method the rest of these notes assumes the reader already knows.