Slater pointed out in 1929 that a determinant of the form (10.40) satisfies the antisymmetry requirement for a many-electron atom. A determinant like (10.40) is called a Slater determinant. All the elements in a given column of a Slater determinant involve the same spin-orbital, whereas elements in the same row all involve the same electron. (Since interchanging rows and columns does not affect the value of a determinant, we could write the Slater determinant in another, equivalent form.)
Consider how the zeroth-order helium wave functions that we found previously can be written as Slater determinants. For the ground-state configuration $1 s^{2}$, we have the spin-orbitals $1 s \alpha$ and $1 s \beta$, giving the Slater determinant
\(
\frac{1}{\sqrt{2}}\left|\begin{array}{ll}
1 s(1) \alpha(1) & 1 s(1) \beta(1) \tag{10.41}\
1 s(2) \alpha(2) & 1 s(2) \beta(2)
\end{array}\right|=1 s(1) 1 s(2) \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]
\)
which agrees with (10.26). For the states corresponding to the excited configuration $1 s 2 s$, we have the possible spin-orbitals $1 s \alpha, 1 s \beta, 2 s \alpha, 2 s \beta$, which give the four Slater determinants
\(
\begin{array}{ll}
D{1}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll}
1 s(1) \alpha(1) & 2 s(1) \alpha(1) \
1 s(2) \alpha(2) & 2 s(2) \alpha(2)
\end{array}\right| & D{2}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll}
1 s(1) \alpha(1) & 2 s(1) \beta(1) \
1 s(2) \alpha(2) & 2 s(2) \beta(2)
\end{array}\right| \
D{3}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll}
1 s(1) \beta(1) & 2 s(1) \alpha(1) \
1 s(2) \beta(2) & 2 s(2) \alpha(2)
\end{array}\right| & D{4}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll}
1 s(1) \beta(1) & 2 s(1) \beta(1) \
1 s(2) \beta(2) & 2 s(2) \beta(2)
\end{array}\right|
\end{array}
\)
Comparison with (10.27) to (10.30) shows that the $1 s 2 s$ zeroth-order wave functions are related to these four Slater determinants as follows:
\(
\begin{gather}
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] \alpha(1) \alpha(2)=D{1} \tag{10.42}\
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] \beta(1) \beta(2)=D{4} \tag{10.43}\
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] 2^{-1 / 2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)]=2^{-1 / 2}\left(D{2}+D{3}\right) \tag{10.44}\
2^{-1 / 2}[1 s(1) 2 s(2)+2 s(1) 1 s(2)] 2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]=2^{-1 / 2}\left(D{2}-D{3}\right) \tag{10.45}
\end{gather}
\)
(To get a zeroth-order function that is an eigenfunction of the spin and orbital angularmomentum operators, we sometimes have to take a linear combination of the Slater determinants of a configuration; see Chapter 11.)
Next, consider some notations used for Slater determinants. Instead of writing $\alpha$ and $\beta$ for spin functions, one often puts a bar over the spatial function to indicate the spin function $\beta$, and a spatial function without a bar implies the spin factor $\alpha$. With this notation, (10.40) is written as
\(
\psi^{(0)}=\frac{1}{\sqrt{6}}\left|\begin{array}{lll}
1 s(1) & \overline{1 s}(1) & 2 s(1) \tag{10.46}\
1 s(2) & \overline{1 s}(2) & 2 s(2) \
1 s(3) & \overline{1 s}(3) & 2 s(3)
\end{array}\right|
\)
Given the spin-orbitals occupied by the electrons, we can readily construct the Slater determinant. Therefore, a shorthand notation for Slater determinants that simply specifies the spin-orbitals is often used. In this notation, (10.46) is written as
\(
\begin{equation}
\psi^{(0)}=|1 s \overline{1 s} 2 s| \tag{10.47}
\end{equation}
\)
where the vertical lines indicate formation of the determinant and multiplication by $1 / \sqrt{6}$.
We showed that the factor $1 / \sqrt{6}$ normalizes a third-order Slater determinant constructed of orthonormal functions. The expansion of an $n$ th-order determinant has $n$ ! terms (Prob. 8.20). For an $n$ th-order Slater determinant of orthonormal spin-orbitals, the same reasoning used in the third-order case shows that the normalization constant is $1 / \sqrt{n!}$. We always include a factor $1 / \sqrt{n!}$ in defining a Slater determinant of order $n$.