quantum chemistry - How to calculate molecular dipole moment from a known wavefunction?

Say I have a molecular wavefunction as a set of molecular orbitals and want to calculate the molecule's dipole moment, but don't know how! I searched a lot but couldn't find any practical example.

$$\psi _i=\sum ^N_{i=1}C_i\mathrm e^{-\alpha _ir^2}$$

Answer

The dipole moment $\mu$ of a molecule is a measure of charge distribution in the molecule and the polarity formed by the nuclei and electron cloud.

We can perturb our system with an external electric field $\vec E$ and gauge the response of the electron cloud and nuclei by the polarisability, i.e how much the dipole moment changes. In practice the nuclei might be so heavy that their motion is not perturbed, while electrons being light are very mobile. If we imagine that the external field is caused by some other species, and that it itself is not changing, we can call this external constant electric field $\vec E$ at least over the volume of the molecule we are considering. Imagine for arbitrary book keeping that we point it down the $z$ axis. We could also investigate how the dipole moment changes with bond vibrations to discuss IR spectroscopy or if the polarisability changes during a vibration to give Raman spectroscopy.

We can use perturbation theory to expand the wavefunction and the molecular energy in terms of small perturbations of the field. We start by Taylor expanding the energy and molecular wavefunction in terms of the electric field which acts as the perturbation parameter.

$$\begin{equation} E(\vec E)=E^0+\bigg(\frac{\partial E}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2E}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3E}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots \end{equation}$$ $$\begin{equation} \psi(\vec E)=\psi^0+\bigg(\frac{\partial \psi}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2\psi}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3\psi}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots \end{equation}$$ If we use the notation that the wavefunction derivatives are given by; $$\begin{equation} \bigg(\frac {1}{i!}\bigg)\frac{\partial ^i\psi}{\partial \vec E^i}=\psi ^{(i)} \end{equation}$$ The Hamiltonian for such a system under the influence of an electric field in the $z$ direction is $\hat H(\vec E)$. $$\begin{equation} \hat H(\vec E)=\hat H^0-\vec E\hat \mu _z \end{equation}$$ Where $\hat \mu _z$ is the dipole moment operator that is a summation of the charges of the nuclei and electrons in the molecule. This is the result of Hellmann-Feynman theorem of the energy and the electric field, e.g. $\hat H(\vec E)=\hat H^0 +\hat H^1(\vec E)$, $$\begin{equation} \frac{dE}{d\vec E}=\bigg\langle \frac{d\hat H}{d\vec E}\bigg\rangle=\bigg\langle \frac{d(-\mu _z\vec E)}{d\vec E}\bigg\rangle \end{equation}$$

The time-independent Schrödinger equation is now,

$$\begin{equation} \hat H(\vec E)\psi(\vec E)=E(\vec E)\psi (\vec E) \end{equation}$$ With an energy; $$\begin{equation} E(\vec E)=\big \langle \psi (\vec E)\big|\hat H(\vec E)\big|\psi (\vec E)\big \rangle=\big\langle \psi ^{(0)}+\psi ^{(1)}\vec E+\psi ^2 \vec E^2+ \dots \big|\hat H-\vec E\hat \mu _z \big|\psi ^{(0)}+\psi ^{(1)}\vec E+\psi^2\vec E^2+\dots \big \rangle \end{equation}$$

With a little algebra and use of $E^{(0)}=\langle \psi ^{(0)}|\hat H^0 |\psi ^{(0)}\rangle$, as well as the Hermitian properties of the Hamiltonian.

$$\begin{equation} E(\vec E)=E^{(0)}+2\vec E\big \langle \psi ^{(1)}\big|\hat H^{(0)}\big|\psi ^{(0)}\rangle -\vec E\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big \rangle +\mathcal O(\vec E^2) \end{equation}$$ Using the Schrödinger equation $H\psi =E\psi$ and pulling the scalar energy out of the integral; $$\begin{equation} \big\langle \psi ^{(1)}\big|\hat H^0\big|\psi ^{(0)}\rangle =E^{(0)}\big\langle \psi ^{(1)}\big|\psi ^{(0)}\big \rangle \end{equation}$$ Since $\langle \phi |\psi \rangle=0$, $$\begin{equation} E(\vec E)=E^{(0)}-\vec E\big \langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle \end{equation}$$ Therefore the expectation value of the dipole moment along the $z$ axis for a molecular state $\psi ^{(0)}$ is $\langle \mu _z \rangle$; $$\begin{equation} \langle \mu _z\rangle =\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle \end{equation}$$ To understand the strength of the interaction that causes the transition between the states $\psi ^{(0)}$ and $\psi ^{(1)}$ we use the transition dipole moment which is basically exactly the same but there is a wavefunction from both of the states involved, initial and final!

If you were to repeat this process but retain higher orders (Messy!) you would get (2nd order) the polarisability of the molecule which in essence is the susceptibility of the electron cloud to change with respect to an external electric field (so how the dipole moment changes. Third order would give the hyperpolarisability etc., etc...

As I said, you could also approach this from a really different angle by interpreting the molecular orbital diagram and using computational chemistry (so variational principle etc) to find the molecular orbital coefficients! That would give you a good idea of what is going on!