In the Hartree-Fock approximation, the wave function of an atom (or molecule) is a Slater determinant or a linear combination of a few Slater determinants [for example, Eq. (10.44)]. A configuration-interaction wave function such as (11.17) is a linear combination of many Slater determinants. To evaluate the energy and other properties of atoms and molecules using Hartree-Fock or configuration-interaction wave functions, we must be able to evaluate integrals of the form $\left\langle D^{\prime}\right| \hat{B}|D\rangle$, where $D$ and $D^{\prime}$ are Slater determinants of orthonormal spin-orbitals and $\hat{B}$ is an operator.
Each spin-orbital $u{i}$ is a product of a spatial orbital $\theta{i}$ and a spin function $\sigma{i}$, where $\sigma{i}$ is either $\alpha$ or $\beta$. We have $u{i}=\theta{i} \sigma{i}$ and $\left\langle u{i}(1) \mid u{j}(1)\right\rangle=\delta{i j}$, where $\left\langle u{i}(1) \mid u{j}(1)\right\rangle$ involves a sum over the spin coordinate of electron 1 and an integration over its spatial coordinates. If $u{i}$ and $u{j}$ have different spin functions, then (10.12) ensures the orthogonality
of $u{i}$ and $u{j}$. If $u{i}$ and $u{j}$ have the same spin function, their orthogonality is due to the orthogonality of the spatial orbitals $\theta{i}$ and $\theta{j}$.
For an $n$-electron system, $D$ is
\(
D=\frac{1}{\sqrt{n!}}\left|\begin{array}{cccc}
u{1}(1) & u{2}(1) & \ldots & u{n}(1) \tag{11.76}\
u{1}(2) & u{2}(2) & \ldots & u{n}(2) \
\vdots & \vdots & \ddots & \vdots \
u{1}(n) & u{2}(n) & \ldots & u_{n}(n)
\end{array}\right|
\)
An example with $n=3$ is Eq. (10.40). $D^{\prime}$ has the same form as $D$ except that $u{1}, u{2}, \ldots, u{n}$ are replaced by $u{1}^{\prime}, u{2}^{\prime}, \ldots, u{n}^{\prime}$.
We shall assume that the columns of $D$ and $D^{\prime}$ are arranged so as to have as many as possible of their left-hand columns match. For example, if we were working with the Slater determinants $\left|1 s \overline{1 s} 2 s 3 p{0}\right|$ and $\left|1 s \overline{1 s} 3 p{0} 4 s\right|$, we would interchange the third and fourth columns of the first determinant (thereby multiplying it by -1 ) and let $D=\left|1 s \overline{1 s} 3 p{0} 2 s\right|$ and $D^{\prime}=\left|1 s \overline{1 s} 3 p{0} 4 s\right|$.
The operator $\hat{B}$ typically has the form
\(
\begin{equation}
\hat{B}=\sum{i=1}^{n} \hat{f}{i}+\sum{i=1}^{n-1} \sum{j>i} \hat{g}_{i j} \tag{11.77}
\end{equation}
\)
where the one-electron operator $\hat{f}{i}$ involves only coordinate and momentum operators of electron $i$ and the two-electron operator $\hat{g}{i j}$ involves electrons $i$ and $j$. For example, if $\hat{B}$ is the atomic Hamiltonian operator (11.1), then $\hat{f}{i}=-\left(\hbar^{2} / 2 m{e}\right) \nabla{i}^{2}-Z e^{2} / 4 \pi \varepsilon{0} r{i}$ and $\hat{g}{i j}=e^{2} / 4 \pi \varepsilon{0} r{i j}$.
Condon and Slater showed that the $n$-electron integral $\left\langle D^{\prime}\right| \hat{B}|D\rangle$ can be reduced to sums of certain one- and two-electron integrals. The derivation of these Condon-Slater formulas uses the determinant expression of Prob. 8.22 together with the orthonormality of the spin-orbitals. (See Parr, pp. 23-27 for the derivation.) Table 11.3 gives the Condon-Slater formulas.
In Table 11.3, each matrix element of $\hat{g}{12}$ involves summation over the spin coordinates of electrons 1 and 2 and integration over the full range of the spatial coordinates of electrons 1 and 2. Each matrix element of $\hat{f}{1}$ involves summation over the spin coordinate of electron 1 and integration over its spatial coordinates. The variables in the sums and definite integrals are dummy variables.
TABLE 11.3 The Condon-Slater Rules
| $D$ and $D^{\prime}$ differ by | $\left\langle D^{\prime}\right\ | \sum{i=1}^{n} \hat{f}{i}\ | D\rangle$ | $\left\langle D^{\prime}\right\ | \sum{i=1}^{n-1} \sum{j>1} \hat{g}_{i j}\ | D\rangle$ | ||
|---|---|---|---|---|---|---|---|---|
| no spin-orbitals | $\sum{i=1}^{n}\left\langle u{i}(1)\right\ | \hat{f}_{1}\left\ | u_{i}(1)\right\rangle$ | $\begin{aligned} \sum{i=1}^{n-1} \sum{j>1}[ & \left\langle u{i}(1) u{j}(2)\right\ | \hat{g}_{12}\left\ | u{i}(1) u{j}(2)\right\rangle \ & \left.-\left\langle u{i}(1) u{j}(2)\right\ | \hat{g}_{12}\left\ | u{j}(1) u{i}(2)\right\rangle\right] \end{aligned}$ |
| one spin-orbital $u{n}^{\prime} \neq u{n}$ | $\left\langle u_{n}^{\prime}(1)\right\ | \hat{f}_{1}\left\ | u_{n}(1)\right\rangle$ | $\begin{aligned} & \sum{j=1}^{n-1}\left[\left\langle u{n}^{\prime}(1) u_{j}(2)\right\ | \hat{g}_{12}\left\ | u{n}(1) u{j}(2)\right\rangle\right. \ & \left.\quad-\left\langle u{n}^{\prime}(1) u{j}(2)\right\ | \hat{g}_{12}\left\ | u{j}(1) u{n}(2)\right\rangle\right] \end{aligned}$ |
| two spin-orbitals $u{n}^{\prime} \neq u{n}, u{n-1}^{\prime} \neq u{n-1}$ | 0 | $\begin{aligned} & \left\langle u{n}^{\prime}(1) u{n-1}^{\prime}(2)\right\ | \hat{g}_{12}\left\ | u{n}(1) u{n-1}(2)\right\rangle \ & \quad-\left\langle u{n}^{\prime}(1) u{n-1}^{\prime}(2)\right\ | \hat{g}_{12}\left\ | u{n-1}(1) u{n}(2)\right\rangle \end{aligned}$ | ||
| three or more spin-orbitals | 0 | 0 |
If the operators $\hat{f}{i}$ and $\hat{g}{i j}$ do not involve spin, the expressions in Table 11.3 can be further simplified. We have $u{i}=\theta{i} \sigma_{i}$ and
\(
\begin{aligned}
\left\langle u{i}(1)\right| \hat{f}{1}\left|u{i}(1)\right\rangle & =\int \theta{i}^{}(1) \hat{f}{1} \theta{i}(1) d v{1} \sum{m{s 1}} \sigma{i}^{}(1) \sigma{i}(1) \
& =\int \theta{i}^{*}(1) \hat{f}{1} \theta{i}(1) d v{1}=\left\langle\theta{i}(1)\right| \hat{f}{1}\left|\theta{i}(1)\right\rangle
\end{aligned}
\)
since $\sigma{i}$ is normalized. Using this result and the orthonormality of $\sigma{i}$ and $\sigma_{j}$, we get for the case $D=D^{\prime}$ (Prob. 11.37)
\(
\begin{equation}
\langle D| \sum{i=1}^{n} \hat{f}{i}|D\rangle=\sum{i=1}^{n}\left\langle\theta{i}(1)\right| \hat{f}{1}\left|\theta{i}(1)\right\rangle \tag{11.78}
\end{equation}
\)
$\langle D| \sum{i=1}^{n-1} \sum{j>1} \hat{g}{i j}|D\rangle=$
$\sum{i=1}^{n-1} \sum{j>1}\left[\left\langle\theta{i}(1) \theta{j}(2)\right| \hat{g}{12}\left|\theta{i}(1) \theta{j}(2)\right\rangle-\delta{m{s, i} m{s, j}}\left\langle\theta{i}(1) \theta{j}(2)\right| \hat{g}{12}\left|\theta{j}(1) \theta{i}(2)\right\rangle\right]$
where $\delta{m{s i,} m{s, j}}$ is 0 or 1 according to whether $m{s, i} \neq m{s, j}$ or $m{s, i}=m_{s, j}$. Similar equations hold for the other integrals.
Let us apply these equations to evaluate $\langle D| \hat{H}|D\rangle$, where $\hat{H}$ is the Hamiltonian of an $n$-electron atom with spin-orbit interaction neglected and $D$ is a Slater determinant of $n$ spin-orbitals. We have $\hat{H}=\sum{i} \hat{f}{i}+\sum{i} \sum{j>i} \hat{g}{i j}$, where $\hat{f}{i}=-\left(\hbar^{2} / 2 m{e}\right) \nabla{i}^{2}-Z e^{2} / 4 \pi \varepsilon{0} r{i}$ and $\hat{g}{i j}=e^{2} / 4 \pi \varepsilon{0} r_{i j}$. Introducing the Coulomb and exchange integrals (9.99) and (9.100) and using (11.78) and (11.79), we have
\(
\begin{gather}
\langle D| \hat{H}|D\rangle=\sum{i=1}^{n}\left\langle\theta{i}(1)\right| \hat{f}{1}\left|\theta{i}(1)\right\rangle+\sum{i=1}^{n-1} \sum{j>1}\left(J{i j}-\delta{m{s, i}, m{s, j}} K{i j}\right) \tag{11.80}\
J{i j}=\left\langle\theta{i}(1) \theta{j}(2)\right| e^{2} / 4 \pi \varepsilon{0} r{12}\left|\theta{i}(1) \theta{j}(2)\right\rangle, K{i j}=\left\langle\theta{i}(1) \theta{j}(2)\right| e^{2} / 4 \pi \varepsilon{0} r{12} \mid \theta{j}(1) \theta{i}(2) \tag{11.81}\
\hat{f}{1}=-\left(\hbar^{2} / 2 m{e}\right) \nabla{1}^{2}-Z e^{2} / 4 \pi \varepsilon{0} r{1} \tag{11.82}
\end{gather}
\)
The Kronecker delta in (11.80) results from the orthonormality of the one-electron spin functions.
As an example, consider Li. The SCF approximation to the ground-state $\psi$ is the Slater determinant $D=|1 s \overline{1 s} 2 s|$. The spin-orbitals are $u{1}=1 s \alpha, u{2}=1 s \beta$, and $u{3}=2 s \alpha$. The spatial orbitals are $\theta{1}=1 s, \theta{2}=1 s$, and $\theta{3}=2 s$. We have $J{12}=J{1 s 1 s}$ and $J{13}=J{23}=J{1 s 2 s}$. Since $m{s 1} \neq m{s 2}$ and $m{s 2} \neq m{s 3}$, the only exchange integral that appears in the energy expression is $K{13}=K_{1 s 2}$. We get exchange integrals only between spin-orbitals with the same spin. Equation (11.80) gives the SCF energy as
\(
E=\langle D| \hat{H}|D\rangle=2\langle 1 s(1)| \hat{f}{1}|1 s(1)\rangle+\langle 2 s(1)| \hat{f}{1}|2 s(1)\rangle+J{1 s 1 s}+2 J{1 s 2 s}-K_{1 s 2 s}
\)
The terms involving $\hat{f}_{1}$ are hydrogenlike energies, and their sum equals $E^{(0)}$ in Eq. (10.49). The remaining terms equal $E^{(1)}$ in Eq. (10.51). As noted at the beginning of Section 9.4, $E^{(0)}+E^{(1)}$ equals the variational integral $\langle D| \hat{H}|D\rangle$, so the Condon-Slater rules have been checked in this case.
For an atom with closed subshells only (for example, ground-state Be with a $1 s^{2} 2 s^{2}$ configuration), the $n$ electrons reside in $n / 2$ different spatial orbitals, so $\theta{1}=\theta{2}, \theta{3}=\theta{4}$, and so on. Let $\phi{1} \equiv \theta{1}=\theta{2}, \phi{2} \equiv \theta{3}=\theta{4}, \ldots, \phi{n / 2} \equiv \theta{n-1}=\theta_{n}$. If one rewrites
(11.80) using the $\phi$ 's instead of the $\theta$ 's, one finds (Prob. 11.38) for the SCF energy of the ${ }^{1} S$ term produced by a closed-subshell configuration
\(
\begin{equation}
E=\langle D| \hat{H}|D\rangle=2 \sum{i=1}^{n / 2}\left\langle\phi{i}(1)\right| \hat{f}{1}\left|\phi{i}(1)\right\rangle+\sum{j=1}^{n / 2} \sum{i=1}^{n / 2}\left(2 J{i j}-K{i j}\right) \tag{11.83}
\end{equation}
\)
where $\hat{f}{1}$ is given by (11.82) and where $J{i j}$ and $K{i j}$ have the forms in (11.81) but with $\theta{i}$ and $\theta{j}$ replaced by $\phi{i}$ and $\phi_{j}$. Each sum in (11.83) goes over all the $n / 2$ different spatial orbitals.
For example, consider the $1 s^{2} 2 s^{2}$ electron configuration. We have $n=4$ and the two different spatial orbitals are $1 s$ and $2 s$. The double sum in Eq. (11.83) is equal to $2 J{1 s 1 s}-K{1 s 1 s}+2 J{1 s 2 s}-K{1 s 2 s}+2 J{2 s 1 s}-K{2 s 1 s}+2 J{2 s 2 s}-K{2 s 2 s}$. From the definition (11.81), it follows that $J{i i}=K{i i}$. The labels 1 and 2 in (11.81) are dummy variables, and interchanging them can have no effect on the integrals. Interchanging 1 and 2 in $J{i j}$ converts it to $J{j i}$; therefore, $J{i j}=J{j i}$. The same reasoning gives $K{i j}=K{j i}$. Thus
\(
\begin{equation}
J{i i}=K{i i}, \quad J{i j}=J{j i}, \quad K{i j}=K{j i} \tag{11.84}
\end{equation}
\)
Use of (11.84) gives the Coulomb- and exchange-integrals expression for the $1 s^{2} 2 s^{2}$ configuration as $J{1 s 1 s}+J{2 s 2 s}+4 J{1 s 2 s}-2 K{1 s 2 s}$. Between the two electrons in the $1 s$ orbital, there is only one Coulombic interaction, and we get the term $J{1 s 1 s}$. Each $1 s$ electron interacts with two $2 s$ electrons, for a total of four $1 s-2 s$ interactions, and we get the term $4 J{1 s 2 s}$. As noted earlier, exchange integrals occur only between spin-orbitals of the same spin. There is an exchange integral between the $1 s \alpha$ and $2 s \alpha$ spin-orbitals and an exchange integral between the $1 s \beta$ and $2 s \beta$ spin-orbitals, which gives the $-2 K_{1 s 2 s}$ term.
The magnitude of the exchange integrals is generally much less than the magnitude of the Coulomb integrals [for example, see Eq. (9.111)].