The first-order nuclear derivatives of properties other than the energy can be related to intensities in vibrational spectroscopies. More specifically, the derivative of the dipole moment and polarizability with respect to nuclear coordinates are needed to determine infrared (IR) and Raman intensities, respectively. Their calculation is described in more detail in the following.

IR intensities#

In order to calculate intensities in the IR spectrum, one needs to know how the electric dipole moment $$\boldsymbol{\mu} = (\mu_x, \mu_y, \mu_z)$$ changes along a normal mode. This means that nuclear derivatives of $$\boldsymbol{\mu}$$ have to be calculated, where each component $$\mu$$ can be decomposed into an electronic and a nuclear contribution, $$\mu = \mu_{\text{e}} + \mu_{\text{n}}$$. The nuclear part of the dipole moment $$\mu_{\text{n}}$$ is simply given by the classical expression

$\mu_{\text{n}} = \sum_{K} Z_K R_K$

with the charge $$Z_K$$ and Cartesian coordinate $$R_K$$ of nucleus $$K$$. The electronic part $$\mu_{\text{e}}$$ is calculed quantum-mechanically from the dipole moment integrals in AO basis, $$\mu_{\kappa \lambda}$$, and the one-particle density matrix $$\mathbf{P}$$,

$%:label: eq:electric_dipole_moment \mu_{\text{e}} = \sum_{\kappa \lambda} P_{\lambda \kappa} \mu_{\kappa \lambda} \, .$

The nuclear gradient of the dipole moment can again be calculated either numerically or analytically. Since the dipole moment itself as a first-order property can be considered a first derivative of the energy, its gradient corresponds to a mixed second derivative of the energy (once respect to an electric field and once with respect to a nuclear coordinate, and thus a second-order property), and its calculation thus resembles the molecular Hessian calculation, and the solution of the CPHF equations is required. While the analytic derivative of the nuclear contribution $$\mu_{\text{n}}$$ is trivial, the derivative of the electronic part $$\mu_{\text{e}}$$ with respect to a nuclear coordinate $$\chi$$ is given by

$%:label: eq:electronic_dipole_derivative \frac{\mathrm{d} \mu_{\text{e}}}{\mathrm{d} \chi} = \sum_{\kappa \lambda} \frac{\mathrm{d} P_{\lambda \kappa}}{\mathrm{d} \chi} \mu_{\kappa \lambda} + \sum_{\kappa \lambda} P_{\lambda \kappa} \frac{\mathrm{d} \mu_{\kappa \lambda}}{\mathrm{d} \chi} \, ,$

where the perturbed density and the derivatives of the dipole integrals are needed.

The IR transition dipole moment is then calculated by taking the dot product of the dipole moment gradient with the Cartesian normal modes $$\mathbf{l}^{\text{Cart}}$$, and the IR intensity as the square norm of corresponding transition moment. The intensities are successively converted to the unit of km mol$$^{-1}$$. For this, one has to multiply multiply the intensities, which are initially given in atomic unites times atomic mass units, with the Avogadro number, the conversion factor between the electron mass in atomic mass units, bohrs in kilometers, the fine structure constant squared, and with $$\pi/3$$.

Raman intensities#

Intensities in the vibrational Raman spectrum are calculated in an analogous manner, except that the nuclear derivative of the electric-dipole polarizability $$\boldsymbol{\alpha}(\omega)$$ along normal mode $$Q_k$$ is needed. Since the polarizability is already a second-order (or linear-response) property, this is more involved than IR intensities (in the time-independent regime, the polarizability gradient corresponds to a third derivative of the energy). However, it turns that out the analytical nuclear gradient of the polarizability can be calculated in manner analogous to the gradients of the TDHF or TDDFT schemes [RF07], which is similar to what has been discussed for here. The main differences include the following:

• all the density matrices ($$\boldsymbol{\gamma}$$ and $$\boldsymbol{\Gamma}$$) depend also on two Cartesian components, $$m,n \in \{ x,y,z \}$$, corresponding to the element $$\alpha_{mn}$$ of the polarizability tensor $$\boldsymbol{\alpha}$$, and need to be symmetrized with respect to $$m$$ and $$n$$.

• the right-hand side of the linear response equation needs to be included in the Lagrangian, corresponding in this case to the contraaction of response vector component $$m$$ with component $$n$$ of the dipole integrals (also to be symmetrized with respect to $$m$$ and $$n$$)

• in the dynamic case ($$\omega \neq 0$$), also the term $$\omega (\mathbf{X}^\dagger \mathbf{X} - \mathbf{Y}^\dagger \mathbf{Y})$$ needs to be taken into account (for all combinations of vector components $$m$$ and $$n$$)

For randomly oriented molecules and linearly polarized incident light, the Raman differenial cross-section is calculated in practice as [Gut19]

$\frac{\mathrm{d} \sigma_k}{\mathrm{d} \Omega} = \frac{\hbar \omega_{\text{L}} \omega_{\text{S}}^3}{32 \pi^2 \epsilon_0^2 c^4 \omega_k} \frac{S_{k}}{45} \, ,$

where $$\omega_{\text{L}}$$ is the frequency of the incident radiation, $$\omega_{\text{S}}$$ is the angular frequency of the scattered light, $$\omega_k$$ is the angular frequency of vibrational mode $$Q_k$$, and $$S_k$$ is the Raman activity of the mode, defined as [Gut19]

$S_k = 45 \bar{\alpha}_k^2 + 7 \bar{\gamma}_k^2 \, ,$

where $$\bar{\alpha}_k^2$$ and $$\bar{\gamma}_k^2$$ are Raman rotational invariants [Gut19],

$\begin{split} \bar{\alpha}_k^2 &= \frac19 \bigg( \frac{\mathrm{d} \alpha_{xx}}{\mathrm{d} Q_k} + \frac{\mathrm{d} \alpha_{yy}}{\mathrm{d} Q_k} +\frac{\mathrm{d} \alpha_{zz}}{\mathrm{d} Q_k} \bigg)^2 \\ \bar{\gamma}_k^2 &= 3 \bigg[ \bigg( \frac{\mathrm{d} \alpha_{xy}}{\mathrm{d} Q_k} + \frac{\mathrm{d} \alpha_{xz}}{\mathrm{d} Q_k} +\frac{\mathrm{d} \alpha_{yz}}{\mathrm{d} Q_k} \bigg)^2 \bigg] \\ &+\frac12 \bigg[ \bigg( \frac{\mathrm{d} \alpha_{xx}}{\mathrm{d} Q_k} - \frac{\mathrm{d} \alpha_{yy}}{\mathrm{d} Q_k} \bigg)^2 + \frac12 \bigg( \frac{\mathrm{d} \alpha_{xx}}{\mathrm{d} Q_k} - \frac{\mathrm{d} \alpha_{zz}}{\mathrm{d} Q_k} \bigg)^2 + \frac12 \bigg( \frac{\mathrm{d} \alpha_{yy}}{\mathrm{d} Q_k} - \frac{\mathrm{d} \alpha_{zz}}{\mathrm{d} Q_k} \bigg)^2 \bigg] \, .\end{split}$

The rotational invariants can further be used to calculate the parallel (or “polarized”) and perpendicular (or “depolarized”) intensities as $$I_{\text{pol}} = 45 \bar{\alpha}_k^2 + 4 \bar{\gamma}_k^2$$ and $$I_{\text{depol}} = 3 \bar{\gamma}_k^2$$, respectively, from which the so-called depolarization ratio $$\rho$$ can be calculated as

$\rho = \frac{I_{\text{depol}}}{I_{\text{pol}}} \, .$