Energies calculated by the Hartree-Fock method are typically in error by about $\frac{1}{2} \%$ for light atoms. On an absolute basis this is not much, but for the chemist it is too large. For example, the total energy of the carbon atom is about -1000 eV , and $\frac{1}{2} \%$ of this is 5 eV . Chemical single-bond energies run about 5 eV . Calculating a bond energy by taking the difference between Hartree-Fock molecular and atomic energies, which are in error by several electronvolts for light atoms, is an unreliable procedure. We must seek a way to improve Hartree-Fock wave functions and energies. (Our discussion will apply to molecules as well as atoms.)
A Hartree-Fock SCF wave function takes into account the interactions between electrons only in an average way. Actually, we must consider the instantaneous interactions between electrons. Since electrons repel each other, they tend to keep out of each other's way. For example, in helium, if one electron is close to the nucleus at a given instant, it is energetically more favorable for the other electron to be far from the nucleus at that instant. One sometimes speaks of a Coulomb hole surrounding each electron in an atom. This is a region in which the probability of finding another electron is small. The motions of electrons are correlated with each other, and we speak of electron correlation. We must find a way to introduce the instantaneous electron correlation into the wave function.
Actually, a Hartree-Fock wave function does have some instantaneous electron correlation. A Hartree-Fock function satisfies the antisymmetry requirement. Therefore [Eq. (10.20)], it vanishes when two electrons with the same spin have the same spatial coordinates. For a Hartree-Fock function, there is little probability of finding electrons of the same spin in the same region of space, so a Hartree-Fock function has some correlation of the motions of electrons with the same spin. This makes the Hartree-Fock energy lower than the Hartree energy. One sometimes refers to a Fermi hole around each electron in a Hartree-Fock wave function, thereby indicating a region in which the probability of finding another electron with the same spin is small.
The correlation energy $E{\text {corr }}$ is the difference between the exact nonrelativistic energy $E{\text {nonrel }}$ and the (nonrelativistic) Hartree-Fock energy $E_{\mathrm{HF}}$ :
\(
\begin{equation}
E{\mathrm{corr}} \equiv E{\mathrm{nonrel}}-E_{\mathrm{HF}} \tag{11.16}
\end{equation}
\)
where $E{\text {nonrel }}$ and $E{\mathrm{HF}}$ should both either include corrections for nuclear motion or omit these corrections. For the He atom, the (nonrelativistic) Hartree-Fock energy uncorrected for nuclear motion is $-2.86168\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$ [E. Clementi and C. Roetti, At. Data Nucl. Data Tables, 14, 177 (1974)] and variational calculations (Section 9.4) give the exact nonrelativistic energy uncorrected for nuclear motion as $-2.90372\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$. Therefore, $E{\text {corr, } \mathrm{He}}=-2.90372\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)+2.86168\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)=-0.04204\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$ $=-1.14 \mathrm{eV}$. For atoms and molecules where $E{\text {nonrel }}$ cannot be accurately calculated,
one combines the experimental energy with estimates for relativistic and nuclear-motion corrections to get $E{\text {nonrel }}$. For neutral ground-state atoms, $\left|E{\text {corr }}\right|$ has been found to increase roughly linearly with the number $n$ of electrons:
\(
E{\text {corr }} \approx-0.0170 n^{1.31}\left(e^{2} / 4 \pi \varepsilon{0} a_{0}\right)=-0.0170 n^{1.31}(27.2 \mathrm{eV})
\)
We have already indicated two of the ways in which we may provide for instantaneous electron correlation. One method is to introduce the interelectronic distances $r_{i j}$ into the wave function (Section 9.4).
Another method is configuration interaction. We found (Sections 9.3 and 10.4) the zeroth-order wave function for the helium-atom $1 s^{2}$ ground state to be $1 s(1) 1 s(2)[\alpha(1) \beta(2)-\beta(1) \alpha(2)] / \sqrt{2}$. We remarked that first- and higher-order corrections to the wave function will mix in contributions from excited configurations, producing configuration interaction (CI), also called configuration mixing (CM).
The most common way to do a configuration-interaction calculation on an atom or molecule uses the variation method. One starts by choosing a basis set of one-electron functions $\chi_{i}$. In principle, this basis set should be complete. In practice, one is limited to a basis set of finite size. One hopes that a good choice of basis functions will give a good approximation to a complete set. For atomic calculations, STOs [Eq. (11.14)] are often chosen as the basis functions.
The SCF atomic (or molecular) orbitals $\phi_{i}$ are written as linear combinations of the basis-set members [see (11.13)], and the Hartree-Fock equations (11.12) are solved to give the coefficients in these linear combinations. The number of atomic (or molecular) orbitals obtained equals the number of basis functions used. The lowest-energy orbitals are the occupied orbitals for the ground state. The remaining unoccupied orbitals are called virtual orbitals.
Using the set of occupied and virtual spin-orbitals, one can form antisymmetric manyelectron functions that have different orbital occupancies. For example, for helium, one can form functions that correspond to the electron configurations $1 s^{2}, 1 s 2 s, 1 s 2 p, 2 s^{2}, 2 s 2 p, 2 p^{2}, 1 s 3 s$, and so on. Moreover, more than one function can correspond to a given electron configuration. Recall the functions (10.27) to (10.30) corresponding to the helium $1 s 2 s$ configuration. Each such many-electron function $\Phi{i}$ is a Slater determinant or a linear combination of a few Slater determinants. Use of more than one Slater determinant is required for certain openshell functions such as (10.44) and (10.45). Each $\Phi{i}$ is called a configuration state function or a configuration function or simply a "configuration." (This last name is unfortunate, since it leads to confusion between an electron configuration such as $1 s^{2}$ and a configuration function such as $|1 s \overline{1 s}|$.)
As we saw in perturbation theory, the true atomic (or molecular) wave function $\psi$ contains contributions from configurations other than the one that makes the main contribution to $\psi$, so we express $\psi$ as a linear combination of the configuration functions $\Phi_{i}$ :
\(
\begin{equation}
\psi=\sum{i} c{i} \Phi_{i} \tag{11.17}
\end{equation}
\)
We then regard (11.17) as a linear variation function (Section 8.5). Variation of the coefficients $c_{i}$ to minimize the variational integral leads to the equation
\(
\begin{equation}
\operatorname{det}\left(H{i j}-E S{i j}\right)=0 \tag{11.18}
\end{equation}
\)
where $H{i j} \equiv\left\langle\Phi{i}\right| \hat{H}\left|\Phi{j}\right\rangle$ and $S{i j} \equiv\left\langle\Phi{i} \mid \Phi{j}\right\rangle$. Commonly, the $\Phi_{i}$ functions are orthonormal, but if they are not orthogonal, they can be made so by the Schmidt method. [Only
configuration functions whose angular-momentum eigenvalues are the same as those of the state $\psi$ will contribute to the expansion (11.17); see Section 11.5.]
Because the many-electron configuration functions $\Phi{i}$ are ultimately based on a oneelectron basis set that is a complete set, the set of all possible configuration functions is a complete set for the many-electron problem: Any antisymmetric many-electron function (including the exact wave function) can be expressed as a linear combination of the $\Phi{i}$ functions. (For a proof of this, see Szabo and Ostlund, Section 2.2.7.) Therefore, if one starts with a complete one-electron basis set and includes all possible configuration functions, a CI calculation will give the exact atomic (or molecular) wave function $\psi$ for the state under consideration. In practice, one is limited to a finite, incomplete basis set, rather than an infinite, complete basis set. Moreover, even with a modest-size basis set, the number of possible configuration functions is extremely large, and one usually does not include all possible configuration functions. Part of the "art" of the CI method is choosing those configurations that will contribute the most.
Because it generally takes very many configuration functions to give a truly accurate wave function, configuration-interaction calculations for systems with more than a few electrons are time-consuming, even on supercomputers. Other methods for allowing for electron correlation are discussed in Chapter 16.
In summary, to do a CI calculation, we choose a one-electron basis set $\chi{i}$, iteratively solve the Hartree-Fock equations (11.12) to determine one-electron atomic (or molecular) orbitals $\phi{i}$ as linear combinations of the basis set, form many-electron configuration functions $\Phi{i}$ using the orbitals $\phi{i}$, express the wave function $\psi$ as a linear combination of these configuration functions, solve (11.18) for the energy, and solve the associated simultaneous linear equations for the coefficients $c_{i}$ in (11.17). [In practice, (11.18) and its associated simultaneous equations are solved by matrix methods; see Section 8.6.]
As an example, consider the ground state of beryllium. The Hartree-Fock SCF method would find the best forms for the $1 s$ and $2 s$ orbitals in the Slater determinant $|1 s \overline{1 s} 2 s \overline{2 s}|$ and use this for the ground-state wave function. [We are using the notation of Eq. (10.47).] Going beyond the Hartree-Fock method, we would include contributions from excited configuration functions (for example, $|1 s \bar{s} 3 s \overline{3 s}|$ ) in a linear variation function for the ground state. Bunge did a CI calculation for the beryllium ground state using a linear combination of 650 configuration functions [C. F. Bunge, Phys. Rev. A, 14, 1965 (1976)]. The Hartree-Fock energy is $-14.5730\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$, Bunge's CI result is $-14.6669\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$, and the exact nonrelativistic energy is $-14.6674\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$. Bunge was able to obtain $99.5 \%$ of the correlation energy.
A detailed CI calculation for the He atom is shown in Section 16.2.