We now show how to calculate molecular dipole moments from wave functions.
The classical expression for the electric dipole moment $\boldsymbol{\mu}{\mathrm{cl}}$ of a set of discrete charges $Q{i}$ is
\(
\begin{equation}
\boldsymbol{\mu}{\mathrm{cl}}=\sum{i} Q{i} \mathbf{r}{i} \tag{14.9}
\end{equation}
\)
where $\mathbf{r}_{i}$ is the position vector from the origin to the $i$ th charge [Eq. (5.33)]. The electric dipole moment is a vector whose $x$ component is
\(
\begin{equation}
\mu{x, \mathrm{cl}}=\sum{i} Q{i} x{i} \tag{14.10}
\end{equation}
\)
with similar expressions for the other components. For a continuous charge distribution with charge density $\rho{Q}(x, y, z), \boldsymbol{\mu}{\mathrm{cl}}$ is found by summing over the infinitesimal elements of charge $d Q{i}=\rho{Q}(x, y, z) d x d y d z$ :
\(
\begin{equation}
\boldsymbol{\mu}{\mathrm{cl}}=\int \rho{Q}(x, y, z) \mathbf{r} d x d y d z, \quad \text { where } \quad \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \tag{14.11}
\end{equation}
\)
Now consider the quantum-mechanical definition of the electric dipole moment. Suppose we apply a uniform external electric field $\mathbf{E}$ to an atom or molecule and ask for the effect on the energy of the system. To form the Hamiltonian operator, we first need the classical expression for the energy. The electric field strength $\mathbf{E}$ is defined as $\mathbf{E} \equiv \mathbf{F} / Q$, where $\mathbf{F}$ is the force the field exerts on a charge $Q$. We take the $z$ direction as the direction of the applied field: $\mathbf{E}=\mathscr{E}_{z} \mathbf{k}$. The potential energy $V$ is [Eq. (4.24)]
\(
d V / d z=-F{z}=-Q \mathscr{E}{z} \quad \text { and } \quad V=-Q \mathscr{C}_{z} z
\)
This is the potential energy of a single charge in the field. For a system of charges,
\(
\begin{equation}
V=-\mathscr{E}{z} \sum{i} Q{i} z{i} \tag{14.12}
\end{equation}
\)
where $z{i}$ is the $z$ coordinate of charge $Q{i}$. The extension of (14.12) to the case where the electric field points in an arbitrary direction follows from (4.24) and is
\(
\begin{equation}
V=-\mathscr{E}{x} \sum{i} Q{i} x{i}-\mathscr{E}{y} \sum{i} Q{i} y{i}-\mathscr{E}{z} \sum{i} Q{i} z{i}=-\mathbf{E} \cdot \boldsymbol{\mu}_{\mathrm{cl}} \tag{14.13}
\end{equation}
\)
This is the classical-mechanical expression for the energy of an electric dipole in a uniform applied electric field.
To calculate the quantum-mechanical expression, we use perturbation theory. The perturbation operator $\hat{H}^{\prime}$ corresponding to (14.13) is $\hat{H}^{\prime}=-\mathbf{E} \cdot \hat{\boldsymbol{\mu}}$, where the electric dipole-moment operator $\hat{\boldsymbol{\mu}}$ is
\(
\begin{gather}
\hat{\boldsymbol{\mu}}=\sum{i} Q{i} \hat{\mathbf{r}}{i}=\mathbf{i} \hat{\mu}{x}+\mathbf{j} \hat{\mu}{y}+\mathbf{k} \hat{\mu}{z} \tag{14.14}\
\hat{\mu}{x}=\sum{i} Q{i} x{i}, \quad \hat{\mu}{y}=\sum{i} Q{i} y{i}, \quad \hat{\mu}{z}=\sum{i} Q{i} z{i} \tag{14.15}
\end{gather}
\)
The first-order correction to the energy is [Eq. (9.22)]
\(
\begin{equation}
E^{(1)}=-\mathbf{E} \cdot \int \psi^{(0) } \hat{\boldsymbol{\mu}} \psi^{(0)} d \tau \tag{14.16}
\end{}
\)
where $\psi^{(0)}$ is the unperturbed wave function. Comparison of (14.16) and (14.13) shows that the quantum-mechanical quantity that corresponds to $\boldsymbol{\mu}_{\mathrm{cl}}$ is the integral
\(
\begin{equation}
\boldsymbol{\mu}=\int \psi^{(0)} \hat{\boldsymbol{\mu}} \psi^{(0)} d \tau \tag{14.17}
\end{}
\)
$\boldsymbol{\mu}$ in (14.17) is the quantum-mechanical electric dipole moment of the system.
An objection to taking (14.17) as the dipole moment is that we considered only the first-order energy correction. If we had included $E^{(2)}$ in (14.16), the comparison with (14.13) would not have given (14.17) as the dipole moment. Actually, (14.17) is the dipole
moment of the system in the absence of an applied electric field and is the permanent electric dipole moment. Application of the field distorts the wave function from $\psi^{(0)}$, giving rise to an induced electric dipole moment in addition to the permanent dipole moment. The induced dipole moment corresponds to the energy correction $E^{(2)}$. (For the details, see Merzbacher, Section 17.4.) The induced dipole moment $\boldsymbol{\mu}_{\text {ind }}$ is related to the applied electric field $\mathbf{E}$ by
\(
\begin{equation}
\boldsymbol{\mu}_{\text {ind }}=\alpha \mathbf{E} \tag{14.18}
\end{equation}
\)
where $\alpha$ is the polarizability of the atom or molecule. The greater the polarizability of molecule B, the greater the London dispersion force (Section 13.7) between two $B$ molecules.
The shift in the energy of a quantum-mechanical system caused by an applied electric field is called the Stark effect. The first-order (or linear) Stark effect is given by (14.16), and from (14.17) it vanishes for a system with no permanent electric dipole moment. The second-order (or quadratic) Stark effect is given by the energy correction $E^{(2)}$ and is proportional to the square of the applied field.
The electric dipole-moment operator (14.14) is an odd function of the coordinates. If the wave function in (14.17) is either even or odd, then the integrand in (14.17) is an odd function, and the integral over all space vanishes. We conclude that the permanent electric dipole moment $\boldsymbol{\mu}$ is zero for states of definite parity.
The permanent electric dipole moment of a molecule in electronic state $\psi_{\mathrm{el}}$ is
\(
\begin{equation}
\boldsymbol{\mu}=\int \psi_{\mathrm{el}}^{} \hat{\boldsymbol{\mu}} \psi{\mathrm{el}} d \tau{\mathrm{el}} \tag{14.19}
\end{}
\)
The electronic wave functions of homonuclear diatomic molecules can be classified as $g$ or $u$, according to their parity. Hence, a homonuclear diatomic molecule has a zero permanent electric dipole moment, a result that is not too astonishing. The same holds true for any molecule with a center of symmetry. The electric dipole-moment operator for a molecule includes summation over both the electronic and nuclear charges:
\(
\begin{equation}
\hat{\boldsymbol{\mu}}=\sum{i}\left(-e \mathbf{r}{i}\right)+\sum{\alpha} Z{\alpha} e \mathbf{r}_{\alpha} \tag{14.20}
\end{equation}
\)
where $\mathbf{r}{\alpha}$ is the vector from the origin to the nucleus of atomic number $Z{\alpha}$, and $\mathbf{r}{i}$ is the vector to electron $i$. Since both the dipole-moment operator (14.20) and the electronic wave function depend on the parameters defining the nuclear configuration, the molecular electronic dipole moment $\boldsymbol{\mu}$ depends on the nuclear configuration. To indicate this, the quantity (14.19) can be called the dipole-moment function of the molecule. In writing (14.19), we ignored the nuclear motion. When the dipole moment of a molecule is experimentally determined, what is measured is the quantity (14.19) averaged over the zero-point vibrations (assuming the temperature is not high enough for there to be appreciable population of higher vibrational levels). We might use $\boldsymbol{\mu}{0}$ and $\boldsymbol{\mu}_{e}$ to indicate the dipole moment averaged over zero-point vibrations and the dipole moment at the equilibrium nuclear configuration, respectively.
Since the second sum in (14.20) is independent of the electronic coordinates, we have
\(
\begin{aligned}
\boldsymbol{\mu} & =\int \psi{\mathrm{el}}^{*} \sum{i}\left(-e \mathbf{r}{i}\right) \psi{\mathrm{el}} d \tau{\mathrm{el}}+\sum{\alpha} Z{\alpha} e \mathbf{r}{\alpha} \int \psi{\mathrm{el}}^{*} \psi{\mathrm{el}} d \tau{\mathrm{el}} \
& =-e \int\left|\psi{\mathrm{el}}\right|^{2} \sum{i} \mathbf{r}{i} d \tau{\mathrm{el}}+e \sum{\alpha} Z{\alpha} \mathbf{r}{\alpha}
\end{aligned}
\)
Using (14.8), we have
\(
\begin{equation}
\boldsymbol{\mu}=-e \iiint \rho(x, y, z) \mathbf{r} d x d y d z+e \sum{\alpha} Z{\alpha} \mathbf{r}_{\alpha} \tag{14.21}
\end{equation}
\)
where $\rho$ is the electron probability density. Equation (14.21) is what would be obtained if we pretended that the electrons were smeared out into a continuous charge distribution whose charge density is given by $-e \rho(x, y, z)$ and we used the classical equation (14.11) to calculate $\boldsymbol{\mu}$.