How is the wave function of a many-electron molecule related to the electron probability density? We want to find the probability of finding an electron in the rectangular volume element located at point $(x, y, z)$ in space with edges $d x, d y, d z$. The electronic wave function $\psi$ is a function of the spatial and spin coordinates of the $n$ electrons. (For simplicity the parametric dependence on the nuclear configuration will not be explicitly indicated.) We know that
\(
\begin{equation}
\left|\psi\left(x{1}, \ldots, z{n}, m{s 1}, \ldots, m{s n}\right)\right|^{2} d x{1} d y{1} d z{1} \cdots d x{n} d y{n} d z{n} \tag{14.1}
\end{equation}
\)
is the probability of simultaneously finding electron 1 with spin $m{s 1}$ in the volume element $d x{1} d y{1} d z{1}$ at $\left(x{1}, y{1}, z{1}\right)$, electron 2 with spin $m{s 2}$ in the volume element $d x{2} d y{2} d z{2}$ at $\left(x{2}, y{2}, z{2}\right)$, and so on. Since we are not interested in what spin the electron we find at ( $x, y, z$ ) has, we sum the probability (14.1) over all possible spin states of all electrons to give the probability of simultaneously finding each electron in the appropriate volume element with no regard for spin:
\(
\begin{equation}
\sum{m{s 1}} \cdots \sum{m{s n}}|\psi|^{2} d x{1} \cdots d z{n} \tag{14.2}
\end{equation}
\)
Suppose we want the probability of finding electron 1 in the volume element $d x d y d z$ at $(x, y, z)$. For this probability we do not care where electrons 2 through $n$ are. We therefore add the probabilities for all possible locations for these electrons. This amounts to integrating (14.2) over the coordinates of electrons $2,3, \ldots, n$ :
\(
\begin{equation}
\left[\sum{\text {all } m{s}} \int \cdots \int\left|\psi\left(x, y, z, x{2}, y{2}, z{2}, \ldots, x{n}, y{n}, z{n}, m{s 1}, \ldots, m{s n}\right)\right|^{2} d x{2} \cdots d z{n}\right] d x d y d z \tag{14.3}
\end{equation}
\)
where there is a $(3 n-3)$-fold integration over $x{2}$ through $z{n}$.
Now suppose we ask for the probability of finding electron 2 in the volume element $d x d y d z$ at $(x, y, z)$. By analogy to (14.3), this is
\(
\begin{equation}
\left[\sum{\mathrm{all} m{s}} \int \cdots \int\left|\psi\left(x{1}, y{1}, z{1}, x, y, z, x{3}, \ldots, z{n}, m{s 1}, \ldots, m{s n}\right)\right|^{2} d x{1} d y{1} d z{1} d x{3} \cdots d z{n}\right] d x d y d z \tag{14.4}
\end{equation}
\)
Of course, electrons do not come with labels, and this indistinguishability (Section 10.3) means that the probabilities (14.3) and (14.4) must be equal. This equality is readily proved. The wave function $\psi$ is antisymmetric with respect to electron exchange, so $|\psi|^{2}$ is unchanged by an electron exchange. Interchanging the spatial and spin coordinates of electrons 1 and 2 in $\psi$ in (14.4) and doing some relabeling of dummy variables, we see that (14.4) is equal to (14.3). Thus (14.3) gives the probability of finding any one particular electron in $d x d y d z$. Since the system has $n$ electrons, the probability of finding an electron in $d x d y d z$ is $n$ times (14.3). (In drawing this conclusion, we assume that the probability of finding more than one electron in the infinitesimal region $d x d y d z$ is negligible compared with the probability of finding one electron. This is certainly valid since the probability of finding two electrons will involve the product of six infinitesimal quantities as compared with the product of three infinitesimal quantities for the probability of finding one electron.)
Thus the probability density $\rho$ for finding an electron in the neighborhood of point $(x, y, z)$ is
\(
\begin{align}
\rho(x, y, z) & =n \sum{\text {all } m{s}} \int \cdots \int\left|\psi\left(x, y, z, x{2}, \ldots, z{n}, m{s 1}, \ldots, m{s n}\right)\right|^{2} d x{2} \cdots d z{n} \
\rho(\mathbf{r}) & =n \sum{\text {all } m{s}} \int \cdots \int\left|\psi\left(\mathbf{r}, \mathbf{r}{2}, \ldots, \mathbf{r}{n}, m{s 1}, \ldots, m{s n}\right)\right|^{2} d \mathbf{r}{2} \cdots d \mathbf{r}{n} \tag{14.5}
\end{align}
\)
where the vector notation for spatial variables (Section 5.2) is used. The atomic units of $\rho$ are electrons $/$ bohr $^{3}$.
$\rho$ is the electron probability density. The corresponding electronic charge density averaged over time is equal to $-e \rho(x, y, z)$, where $-e$ is the charge on an electron. In atomic units, the electronic charge density is $-\rho$. In addition, there are the positive charges of the nuclei. The term electronic charge density is commonly shortened to charge density.
A molecule's $\rho$ is an experimentally observable quantity that can be found from measured x-ray diffraction intensities of molecular crystals or electron-diffraction intensities of gases. See P. Coppens and M. B. Hall (eds.), Electron Distributions and the Chemical Bond, Plenum, 1982; D. A. Kohl and L. S. Bartell, J. Chem. Phys., 51, 2891, 2896 (1969); P. Coppens, J. Phys. Chem., 93, 7979 (1989); P. Coppens, Annu. Rev. Phys. Chem., 43, 663 (1992); C. Gatti and P. Macchi (eds.), Modern Charge-Density Analysis, Springer, 2012.
To illustrate (14.5), consider the electron density for the simple VB and MO groundstate $\mathrm{H}{2}$ wave functions. The wave function is a product of a spatial factor and the spin function (11.60). (For more than two electrons, $\psi$ cannot be factored into a product of a spatial part and a spin part; see Chapter 10.) Summation of (11.60) over $m{s 1}$ and $m{s 2}$ gives one (Section 10.4). Thus (14.5) becomes for $\mathrm{H}{2}$
\(
\rho(x, y, z)=2 \iiint\left|\phi\left(x, y, z, x{2}, y{2}, z{2}\right)\right|^{2} d x{2} d y{2} d z{2}
\)
where $\phi$ is the spatial factor. When $\phi$ is taken as the spatial factor in the VB function (13.100) and Prob. 13.33 or the MO function (13.94), we get (Prob. 14.1)
\(
\begin{equation}
\rho{\mathrm{VB}}=\frac{1 s{a}^{2}+1 s{b}^{2}+2 S{a b} 1 s{a} 1 s{b}}{1+S{a b}^{2}}, \quad \rho{\mathrm{MO}}=\frac{1 s{a}^{2}+1 s{b}^{2}+2\left(1 s{a} 1 s{b}\right)}{1+S_{a b}} \tag{14.6}
\end{equation}
\)
One finds (Prob. 14.2) that $\rho{\mathrm{MO}}>\rho{\mathrm{VB}}$ at the midpoint of the bond, so the MO function (which underestimates electron correlation) piles up more charge between the nuclei than the VB function.
The MO probability density in (14.6) is twice $\rho$ for the $\mathrm{H}{2}^{+}$-like $1 s{\mathrm{A}}+1 s_{\mathrm{B}}$ MO [Eq. (13.65)]. One can prove that, for a many-electron MO wave function, $\rho$ is found by multiplying the probability-density function of each MO by the number of electrons occupying it and summing the results:
\(
\begin{equation}
\rho(x, y, z)=\sum{j} n{j}\left|\phi_{j}\right|^{2} \tag{14.7}
\end{equation}
\)
where the sum is over the different orthogonal spatial MOs, and $n{j}$ (whose possible values are 0,1 , or 2 ) is the number of electrons in the $\mathrm{MO} \phi{j}$. [We used (14.7) in Eq. (11.11).]
Calculations of $\rho$ from high-quality wave functions show that for nearly all molecules, local maxima in $\rho$ occur only at the nuclei. One of the few exceptions is the ground electronic state of $\mathrm{Li}_{2}$, for which $\rho$ has a small local maximum at the bond midpoint [Bader, Section E2.1; G. I. Bersuker et al., J. Phys. Chem., 97, 9323 (1993)].
Let $B\left(\mathbf{r}{i}\right)$ be a function of the spatial coordinates $x{i}, y{i}, z{i}$ of electron $i$, where the notation of Section 5.2 is used. For an $n$-electron molecule, consider the average value
\(
\langle\psi| \sum{i=1}^{n} B\left(\mathbf{r}{i}\right)|\psi\rangle=\int \psi^{*} \sum{i=1}^{n} B\left(\mathbf{r}{i}\right) \psi d \tau=\sum{i=1}^{n} \int|\psi|^{2} B\left(\mathbf{r}{i}\right) d \tau
\)
where $\psi$ is the electronic wave function. Since the electrons are indistinguishable, each term in the sum $\sum{i} \int|\psi|^{2} B d \tau$ must have the same value. Therefore we have $\langle\psi| \sum{i=1}^{n} B\left(\mathbf{r}{i}\right)|\psi\rangle=\int n|\psi|^{2} B\left(\mathbf{r}{1}\right) d \tau$. Since $B\left(\mathbf{r}{1}\right)$ depends only on $x{1}, y{1}, z{1}$, before we integrate over $x{1}, y{1}, z{1}$, we can integrate $n|\psi|^{2}$ over the spatial coordinates of electrons 2 to $n$ and sum over all the spin coordinates. From Eq. (14.5), this produces the electron probability density $\rho\left(\mathbf{r}{1}\right)$. Therefore, $\langle\psi| \sum{i=1}^{n} B\left(\mathbf{r}{i}\right)|\psi\rangle=\int \rho\left(\mathbf{r}{1}\right) B\left(\mathbf{r}{1}\right) d \mathbf{r}_{1}$. The subscript 1 on the integration variables is not needed, and the final result is
\(
\begin{equation}
\int \psi^{} \sum{i=1}^{n} B\left(\mathbf{r}{i}\right) \psi d \tau=\int \rho(\mathbf{r}) B(\mathbf{r}) d \mathbf{r} \tag{14.8}
\end{}
\)
where the integration is over the three spatial coordinates $x, y, z$. This result will be used later in this chapter and in Chapter 15.