# Experiment to theory#

## Isotropic molecular orientation#

When studying physical processes involving light–matter interactions, experiments are predominantly performed in the liquid phase. To connect theory to such experiments, it becomes necessary to take into account the random molecular orientation.

### Laboratory and molecular frames#

Let us consider a laboratory frame that is space fixed with Cartesian axes $$X$$, $$Y$$, and $$Z$$ and an arbitrarily oriented molecular frame with Cartesian axes $$x$$, $$y$$, and $$z$$. It is a common practice to relate these two frames of reference by means of a non-unique set of three Euler angles. We choose the elementary rotations to be intrinsic and referring to the axes sequence ($$z$$-$$x$$-$$z$$). The associated proper Euler angles are denoted by $$\phi$$, $$\theta$$, and $$\psi$$ and they are illustrated in the figure below.

The three unit vectors in the molecular frame relate to those in the laboratory frame according to

$\begin{split} (\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z) = (\mathbf{e}_X, \mathbf{e}_Y, \mathbf{e}_Z) \begin{pmatrix} c_\phi c_\psi - c_\theta s_\phi s_\psi & - c_\psi s_\phi - c_\phi c_\theta s_\psi & s_\theta s_\psi \\ c_\phi s_\psi + c_\theta c_\psi s_\phi & c_\phi c_\theta c_\psi - s_\phi s_\psi & - c_\psi s_\theta \\ s_\phi s_\theta & c_\phi s_\theta & c_\theta \end{pmatrix} \end{split}$

where $$c_\phi$$ is short for $$\cos \phi$$ and so on. The rotation matrix in the equation above is obtained from the three elemental rotation matrices

$R = R_z(\psi) R_x(\theta) R_z(\phi)$

We note that the elements of this rotation matrix are equal to the cosine of the angles in between the unit vectors, or, in other words,

$l_{i\lambda} = \mathbf{e}_i \cdot \mathbf{e}_\lambda$

with $$i \in \{X,Y,Z\}$$ and $$\lambda \in \{x,y,z\}$$. These elements are known as directional cosines.

A molecular property such as the dipole moment is a first-rank tensor for which we readily obtain the following transformation relation

$\mu_i = l_{i\lambda} \mu_\lambda$

where the Einstein summation convention has been adopted for the repeated index $$\lambda$$. The expression for the transformation of higher-order property tensors follow suit

$P_{i_1 \cdots i_n} = l_{i_1 \lambda_1} \cdots l_{i_n \lambda_n} P_{\lambda_1 \cdots \lambda_n}$

### Rotational averages#

In general, an expression for an observable $$T$$ takes the form

$\begin{equation*} T = A_{i_1 \cdots i_n} P_{i_1 \cdots i_n} \end{equation*}$

where the tensor $$P_{i_1 \cdots i_n}$$ contains the response of a molecule subject to external fields under conditions represented by the tensor $$A_{i_1 \cdots i_n}$$. The tensor components are given in the space-fixed laboratory frame, typically with an orientation chosen such that the polarization information can be straightforwardly expressed in $$A_{i_1 \cdots i_n}$$.

For a randomly oriented molecular system, the a calculation of the observable requires the calculation of rotationally averaged products of the aforementioned direction cosines

$\begin{equation*} I_{i_1 \cdots i_n ; \lambda_1 \cdots \lambda_n} = \frac{1}{8 \pi^2} \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{2 \pi} l_{i_1 \lambda_1} \cdots l_{i_n \lambda_n} \sin \theta \, d\phi d\theta d\psi \end{equation*}$

Since integrals of first order $$I_{i;\lambda}$$ are zero, the rotational averages of first-rank tensors (such as the dipole moment) vanish.

Induced moments, on the other hand, involve higher-order tensors that do not necessarily vanish upon rotational averaging. As an example, to linear order, the induced dipole moment in an electric field equals the product of the polarizability and the electric field, here assumed to be applied along the $$Z$$-axis in the laboratory frame,

$\mu_Z = \alpha_{ZZ} E_Z$

where

$\alpha_{ZZ} = I_{ZZ;\lambda_1 \lambda_2} \alpha_{\lambda_1 \lambda_2}$

The only nonzero integrals are those where $$\lambda_1 = \lambda_2$$ and one arrives at

$\alpha_{ZZ} = \frac{1}{3} \big( \alpha_{xx} + \alpha_{yy} + \alpha_{zz} \big)$

## Fields and polarization#

Experimental setups frequently use electromagnetic fields produced by lasers with control over wavelength and polarization. Such fields are purely solenoidal and perpendicular to the direction of propagation. We will here consider the case of a monochromatic light source and two common states of polarization namely linear and circular.

### Linear polarization#

In a state of linear polarization, the electric field takes the form

$\mathbf{E}_\mathrm{lp} = E_0 \left( \cos\theta \, \mathbf{e}_X + \sin\theta \, \mathbf{e}_Y \right) \cos(k z - \omega t + \delta)$

where $$\mathbf{k} = k \mathbf{e}_Z$$ is the wave vector pointing along the direction of propagation. The associated magnetic field is perpendicular to the electric field and given by

$\mathbf{B}_\mathrm{lp} = \frac{E_0}{c} \left( -\sin\theta \, \mathbf{e}_X + \cos\theta \, \mathbf{e}_Y \right) \cos(k z - \omega t + \delta)$

We note that the magnetic field strength is equal to that of that electric field divided by the speed of light. In atomic units with $$c=137.036$$ a.u., the magnetic field is thus two orders of magnitude smaller than the electric field and depending on the spectroscopy at hand, it can often be ignored in calculations of induced dipole moments.

### Circular polarization#

In a state of circular polarization, the electric field takes the form

$\mathbf{E}_\mathrm{cp}(t) = E_0 \left( \cos\left(k Z -\omega t + \delta\right) \, \mathbf{e}_x -(-1)^{m}\sin\left(k Z - \omega t + \delta\right) \, \mathbf{e}_y \right)$

where $$m$$ is an integer. At a certain instant in time, say for simplicity $$t_0 = \delta / \omega$$, the electric field with $$m$$ being even equals

$\mathbf{E}_\mathrm{cp}(t_0) = E_0 \left( \cos k Z \, \mathbf{e}_X - \sin k Z \, \mathbf{e}_Y \right)$

This field is illustrated in the figure below.

In this time frozen spatial view, this electric field forms a left-handed helix and we refer to this state of polarization as left circularly polarized light. If we instead time trace the electric field vector at a fixed point in space, say $$Z =0$$, it is seen to circle in an anti-clockwise direction when viewed in the direction of the source.

### Field strengths and intensities#

For a monochromatic electromagnetic wave, the intensity relates to the electric field strength according to

$I = \frac{1}{2} \varepsilon_0 c E_0^2$

The intensity depends not only on the laser at hand but also on the experimental setup. In a test bed for optical power limiting materials that depend critically on nonlinear absorption, a laser that delivered 1 mJ pulses of 10 ns duration was used. These pulses were focused to a spot size (or circle radius) of 100 $$\mu$$m. The associated intensity becomes

$I = \frac{E_\mathsf{pulse}}{t_\mathsf{pulse} \cdot A_\mathsf{spot}} \approx 0.3\; \mathsf{GW/cm}^2.$

Combining these two equation, we are able to determine the electric field strength for this experimental setup that is designed for the characterization of molecular materials without causing (too frequent) damage. Still, on the atomic scale, this electric field is weak and in the order of $$10^{-5}$$ a.u., which very much is the foundation for our approach of identifying molecular properties from perturbational field expansions of induced moments.

### Spatial averaging of high order properties#

In the perturbative expansion of the electric dipole polarization - the expectation value of the induced electric dipole operator where nonvanishing permanent dipole is subtracted - one often encounters situations where the general expression of the molecular contribution to a particular term of the expansion takes the form

$\sum_j \sum_k \cdots \sum_l \sum_m \frac{ (\mathbf{A}^{f j} \cdot \mathbf{R}_{-\omega_\sigma}) (\mathbf{B}^{j k} \cdot \mathbf{S}_{\omega_n}) \cdots (\mathbf{L}^{l m} \cdot \mathbf{W}_{\omega_2}) (\mathbf{M}^{m i} \cdot \mathbf{Z}_{\omega_1}) }{ \Delta (\omega_1, \omega_2, \cdots , +\omega_n, \omega_\sigma, \omega^{o p}) }$

The contribution of this term along the $$\mathbf{e}_q$$ ($$q=X$$, $$Y$$ or $$Z$$) laboratory cartesian axe is then

$\sum_j \sum_k \cdots \sum_l \sum_m \frac{ (\mathbf{A}^{f j} \cdot \mathbf{e}_q) (\mathbf{B}^{j k} \cdot \mathbf{S}_{\omega_n}) \cdots (\mathbf{L}^{l m} \cdot \mathbf{W}_{\omega_2}) (\mathbf{M}^{m i} \cdot \mathbf{Z}_{\omega_1}) }{ \Delta (\omega_1, \omega_2, \cdots , +\omega_n, \omega_\sigma, \omega^{o p}) }$

Above $$\Delta (\omega_1, \omega_2, \cdots , +\omega_n, \omega_\sigma, \omega^{o p})$$ is a general function involving $$\omega^{o p}$$ - the collection of transition frequencies between states $$o$$ and $$p$$, $$\omega_1, \cdots, \omega_n$$ - the angular frequencies of the radiation, and $$\omega_\sigma = \omega_1+\cdots+\omega_n$$. Summations run over all states of the molecule. Furthermore, $$\mathbf{A}, \mathbf{B}, \cdots, \mathbf{M}$$ are operators such as $$\hat{\boldsymbol{\mu}}$$ or $$\hat{\mathbf{m}}$$, yielding transition moments in the molecular frame, whereas $$\mathbf{R}$$ is a scalar or - see above - a unit vector while $$\mathbf{S}, \cdots, \mathbf{Z}$$ represent field quantities aligned with the laboratory frame. Usually the terms within brackets in the numerator are elements of the multipolar expansion of the hamiltonian describing the interaction of the molecule with the electromagnetic interaction, for example

$(\mathbf{B}^{j k} \cdot \mathbf{S}^{\omega_n}) = (\hat{\boldsymbol{\mu}}^{j k} \cdot \mathbf{E}^{\omega_n})$

When the perturbation includes electromagnetic field gradients (i.e. tensorial physical quantities of rank two or more, as for instance terms like ($$\hat{\mathbf{q}} : \boldsymbol{\nabla} \mathbf{E}$$), the scalar product turns into general tensor contractions, f.ex.

$(\mathbf{B}^{j k} \cdot \mathbf{S}^{\omega_n}) \rightarrow (\hat{\mathbf{q}}^{j k} : \boldsymbol{\nabla} \mathbf{E}^{\omega_n}) = \sum_\alpha \sum_\beta \hat{\mathbf{q}}^{j k}_{\alpha \beta} \boldsymbol{\nabla}_\alpha \mathbf{E}_\beta^{\omega_n} \rightarrow [\hat{\boldsymbol{\mu}} \cdot ( \hat{\boldsymbol{\mu}} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_n}]^{j k}$

ultimately involving combinations of unit vector representing the cartesian directions of the laboratory frame. Note that for sake of symplicity we use the traced form of the quadrupole $$\mathbf{q}$$ in place of the more popolar traceless form $$\boldsymbol{\Theta}$$ throughout the rest of these notes, and that the second $$\leftarrow$$ (in place of the equal sign) above highlights the existence of a an electron charge factor implied at the denominator.

As a specific example of what we are going to discuss in this section, let’s concentrate on the perturbative expansion of the electric dipole polarization, a vector, in particular on the all electric dipole second order contribution $$\boldsymbol{\mu}(\omega_\sigma,t)$$ describing the sum frequency generation process when two beams propagating in the same direction and of angular frequency $$\omega_1$$ and $$\omega_2$$ impinge into a molecular system. The structure of one of the six terms entering in the general case of different optical angular frequencies this contribution is

$\sum_k \sum_l \frac{\hat{\boldsymbol{\mu}}^{il}(\hat{\boldsymbol{\mu}}^{l k} \cdot \mathbf{E}_{\omega_2})(\hat{\boldsymbol{\mu}}^{k i} \cdot \mathbf{E}_{\omega_1})} {\hbar^2 (\omega^{l i} + \omega_\sigma)(\omega^{k i} + \omega_1)}$

The other five terms are obtained by permuting optical frequencies $$\omega_1, \omega_2, \omega_\sigma$$. For our further discussion, only the numerators of the above equation are essential. Following the example of sum frequency generation, we concentrate on the numerator

$\hat{\boldsymbol{\mu}}^{i l} (\hat{\boldsymbol{\mu}}^{l k} \cdot \mathbf{E}_{\omega_2}) (\hat{\boldsymbol{\mu}}^{k i} \cdot \mathbf{E}_{\omega_1})$

The vectors representing field quantities are usually reduced to the product of a scalar field strength times linear (in general complex) combinations of the unit cartesian vectors defining the laboratory frame, see definitions given in the previous section. Indeed, the matrix elements of the electric dipole operator are fixed on the molecule whereas the radiation field vectors are defined in the frame of the laboratory. If the molecule is rigidly oriented with respect to the laboratory, we can choose to align its axes along the laboratory frame. Without loss of generality, in the particular case of linearly polarized fields propagating along the $$Z$$ direction, we can then align the first field along the $$X$$ direction, and the second on the $$X,Y$$ plane. For this example, let’s choose as direction of polarization for the second field directly the $$Y$$ axis. Then $$\mathbf{E}^{\omega_1} = E_X^{\omega_1} \mathbf{e}_X$$ and $$\mathbf{E}^{\omega_2} = E_Y^{\omega_2} \mathbf{e}_Y$$, where the dependence on time and angular frequency is in the scalar field strengths $$E_X^{\omega_1}$$ and $$E_Y^{\omega_2}$$. The contribution to the polarization vector along the $$Z$$-direction (i.e. $$q=Z$$ above), restricting our focus on the numerator will then be of the form

$\hat{{\mu}}_Z^{i l} \hat{{\mu}}^{l k}_X \hat{{\mu}}^{k i}_Y E_Y^{\omega_2} E_X^{\omega_1}$

The general molecular property term therefore (restricting again to the numerator) takes the form

$(\mathbf{A} \cdot \mathbf{e}_{i_1}) (\mathbf{B} \cdot \mathbf{e}_{i_2}) \cdots (\mathbf{M} \cdot \mathbf{e}_{i_n}) = A_{i_1} B_{i_2} \cdot M_{i_n}$

Under rotation this product transforms as a tensor with a rank equal to the number of dots enclosed within parentheses (left) or components (right). Therefore if $$n=3$$ in the last equation, the product transforms as a tensor of rank three. If we swap f.ex. $$(\mathbf{B} \cdot \mathbf{e}_{i_2}) \rightarrow (\hat{\mathbf{q}} : \mathbf{e}_{i_2}\mathbf{e}_{i_3})$$, or equally $$B_{i_2} \rightarrow B_{i_2 i_3}$$ then the product transforms as a tensor of rank four. The rotational averaging of such higher-rank Cartesian tensors has been treated in the literature by several authors, see e.g. [Bar04, FBR14]. We follow here the approach taken by Wagnière . We indicate the isotropic average of the sequence $$A_{i_1} B_{i_2} \cdots M_{i_n}$$ over the solid angle $$\Omega$$ as $$\langle A_{i_1} B_{i_2} \cdots M_{i_n} \rangle_\Omega$$.

For sum frequency generation, performing the necessary integrals involving directional cosines introduced above one gets easily to the following expression

$\langle \hat{\boldsymbol{\mu}}^{il}(\hat{\boldsymbol{\mu}}^{l k} \cdot \mathbf{E}^{\omega_2})(\hat{\boldsymbol{\mu}}^{k i} \cdot \mathbf{E}^{\omega_1}) \rangle_\Omega = \frac{1}{6} \hat{\boldsymbol{\mu}}^{il} \cdot (\hat{\boldsymbol{\mu}}^{l k} \times \hat{\boldsymbol{\mu}}^{k i})(\mathbf{E}^{\omega_2} \times \mathbf{E}^{\omega_1})$

yielding for the averaged second order contribution to the electric dipole polarization describing a generic sum frequency generation process the expression

$\frac{1}{6} \chi_{av}(-\omega_\sigma;\omega_2,\omega_1) (\mathbf{E}^{\omega_2} \times \mathbf{E}^{\omega_1})$

where (reintroducing the denominators, and gathering all terms obtained by permuting transition moments and associated angular frequencies)

$\chi_{av}(-\omega_\sigma;\omega_2,\omega_1) = P \sum_k \sum_l \frac{ \hat{\boldsymbol{\mu}}^{il} \cdot (\hat{\boldsymbol{\mu}}^{l k} \times \hat{\boldsymbol{\mu}}^{k i}) }{\hbar^2 (\omega^{l i} + \omega_\sigma)(\omega^{k i} + \omega_1)}$

We first notice that the averaging process decouples molecule fixed vectors/tensors from laboratory-fixed field quantities. Furthermore, the vector products determining the molecular response $$\chi_{av}(-\omega_\sigma;\omega_2,\omega_1)$$ yield a nonvanishing pseudoscalar only for systems belonging to point groups where a center of inversion (or more rigorously an improper axis of rotation) is included in the symmetry group generator, e.g. in molecules exhibiting chirality. Equally important is the fact that the same pseudoscalar vanishes in all isotropic fluid media when $$\omega_1=\omega_2$$ (optical second harmonic generation) and where it is nonvanishing, the corresponding induced polarization can in principle be detected only along the direction perpendicular to the plane containing the polarization directions of the two electric fields $$\mathbf{E}^{\omega_1}$$ and $$\mathbf{E}^{\omega_2}$$, directions that cannot coincide. All other contributions vanish. Thus the optical sum frequency generation response (due to the interaction of the molecules solely with the optical electric fields composing the impinging radiation) can be seen only in chiral fluid, with an experimental setup with crossed optical electric fields of different angular frequency and aligning the detector along the direction perpendicular to the plane defined by the polarization vectors (transversal). Below we will extend the analysis to the effects of quadrupolar (electric field gradients) interactions.

### Formulae for rank two to rank five tensor properties#

The procedure to obtain general formulae for rotational averages of tensors of any order is a straightforward, albeit increasingly mathematically intense as the order rises, generalization of that discussed above. The resulting compact expressions for tensors of rank from two to five are

Rank Two:

$\begin{equation*} \langle(\mathbf{A} \cdot \mathbf{N}) (\mathbf{B} \cdot \mathbf{O}) \rangle_\Omega = \frac{1}{3} (\mathbf{A} \cdot \mathbf{B}) (\mathbf{N} \cdot \mathbf{O}) \end{equation*}$

Rank Three:

$\begin{equation*} \langle(\mathbf{A} \cdot \mathbf{N}) (\mathbf{B} \cdot \mathbf{O}) (\mathbf{C} \cdot \mathbf{P}) \rangle_\Omega = \frac{1}{6} (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C} ) (\mathbf{N} \cdot \mathbf{O} \times \mathbf{P}) \end{equation*}$

Rank Four:

$\begin{eqnarray*} & & \langle(\mathbf{A} \cdot \mathbf{N}) (\mathbf{B} \cdot \mathbf{O}) (\mathbf{C} \cdot \mathbf{P}) (\mathbf{D} \cdot \mathbf{Q}) \rangle_\Omega = \\ & & \Big [ \frac{2}{15} (\mathbf{A} \cdot \mathbf{B}) (\mathbf{C} \cdot \mathbf{D}) -\frac{1}{30} (\mathbf{A} \cdot \mathbf{C}) (\mathbf{B} \cdot \mathbf{D}) -\frac{1}{30} (\mathbf{A} \cdot \mathbf{D}) (\mathbf{B} \cdot \mathbf{C}) \Big ] (\mathbf{M} \cdot \mathbf{N}) (\mathbf{P} \cdot \mathbf{Q}) \\ & & \Big [ -\frac{1}{30} (\mathbf{A} \cdot \mathbf{B}) (\mathbf{C} \cdot \mathbf{D}) +\frac{2}{15} (\mathbf{A} \cdot \mathbf{C}) (\mathbf{B} \cdot \mathbf{D}) -\frac{1}{30} (\mathbf{A} \cdot \mathbf{D}) (\mathbf{B} \cdot \mathbf{C}) \Big ] (\mathbf{M} \cdot \mathbf{P}) (\mathbf{N} \cdot \mathbf{Q}) \\ & & \Big [ -\frac{1}{30} (\mathbf{A} \cdot \mathbf{B}) (\mathbf{C} \cdot \mathbf{D}) -\frac{1}{30} (\mathbf{A} \cdot \mathbf{C}) (\mathbf{B} \cdot \mathbf{D}) +\frac{2}{15} (\mathbf{A} \cdot \mathbf{D}) (\mathbf{B} \cdot \mathbf{C}) \Big ] (\mathbf{M} \cdot \mathbf{Q}) (\mathbf{N} \cdot \mathbf{P}) \end{eqnarray*}$

Rank Five

$\begin{eqnarray*} & & \langle(\mathbf{A} \cdot \mathbf{N}) (\mathbf{B} \cdot \mathbf{O}) (\mathbf{C} \cdot \mathbf{P}) (\mathbf{D} \cdot \mathbf{Q}) (\mathbf{E} \cdot \mathbf{R})\rangle_\Omega = \\ & & \frac{1}{30} \Big [ +(\mathbf{A} \cdot \mathbf{B}) (\mathbf{C} \cdot \mathbf{D} \times \mathbf{E} ) (\mathbf{M} \cdot \mathbf{N}) (\mathbf{P} \cdot \mathbf{Q} \times \mathbf{R} ) + (\mathbf{A} \cdot \mathbf{C}) (\mathbf{B} \cdot \mathbf{D} \times \mathbf{E} ) (\mathbf{M} \cdot \mathbf{P}) (\mathbf{N} \cdot \mathbf{Q} \times \mathbf{R} ) \\ & & +(\mathbf{A} \cdot \mathbf{D}) (\mathbf{B} \cdot \mathbf{C} \times \mathbf{E} ) (\mathbf{M} \cdot \mathbf{Q}) (\mathbf{N} \cdot \mathbf{P} \times \mathbf{R} ) + (\mathbf{A} \cdot \mathbf{E}) (\mathbf{B} \cdot \mathbf{C} \times \mathbf{D} ) (\mathbf{M} \cdot \mathbf{R}) (\mathbf{N} \cdot \mathbf{P} \times \mathbf{Q} ) \\ & & +(\mathbf{B} \cdot \mathbf{C}) (\mathbf{A} \cdot \mathbf{D} \times \mathbf{E} ) (\mathbf{N} \cdot \mathbf{P}) (\mathbf{M} \cdot \mathbf{Q} \times \mathbf{R} ) + (\mathbf{B} \cdot \mathbf{D}) (\mathbf{A} \cdot \mathbf{C} \times \mathbf{E} ) (\mathbf{N} \cdot \mathbf{Q}) (\mathbf{M} \cdot \mathbf{P} \times \mathbf{R} ) \\ & & +(\mathbf{B} \cdot \mathbf{E}) (\mathbf{A} \cdot \mathbf{C} \times \mathbf{D})(\mathbf{N} \cdot \mathbf{R}) (\mathbf{M} \cdot \mathbf{P} \times \mathbf{Q}) +(\mathbf{C} \cdot \mathbf{D}) (\mathbf{A} \cdot \mathbf{B} \times \mathbf{E})(\mathbf{P} \cdot \mathbf{Q}) (\mathbf{M} \cdot \mathbf{N} \times \mathbf{R}) \\ & & +(\mathbf{C} \cdot \mathbf{E}) (\mathbf{A} \cdot \mathbf{B} \times \mathbf{D})(\mathbf{P} \cdot \mathbf{R}) (\mathbf{M} \cdot \mathbf{N} \times \mathbf{Q}) +(\mathbf{D} \cdot \mathbf{E}) (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C}) (\mathbf{Q} \cdot \mathbf{R}) (\mathbf{M} \cdot \mathbf{N} \times \mathbf{P}) \Big ] \end{eqnarray*}$

### Example of application: electric dipole polarizability#

In the section above, the second equation (rank three) is a generalization of the results obtained in the application of the averaging process to the case of sum frequency generation. Let’s use the first equation, involving rank two tensors, and apply it to the case of the electric dipole polarizability $$\alpha(-\omega;\omega)$$, when $$\mathbf{A} = \hat{\boldsymbol{\mu}}^{ik}$$, $$\mathbf{B} = \hat{\boldsymbol{\mu}}^{ki}$$,

$\langle \hat{\boldsymbol{\mu}}^{il} (\hat{\boldsymbol{\mu}}^{li} \cdot \mathbf{E}^{\omega}) \rangle_\Omega = \frac{1}{3} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{li}) \mathbf{E}^\omega$

or, turning to the linear optical electric dipole contribution to the induced dipole polarization

$\begin{eqnarray*} \langle \boldsymbol{\mu}(\omega) \rangle_\Omega (t) & = & \frac{1}{3} P \sum_l \frac{\hat{\boldsymbol{\mu}}^{i l} \cdot \hat{\boldsymbol{\mu}}^{l i}} {\hbar (\omega^{l i} + \omega)} \mathbf{E}^\omega = \\ & = & \frac{1}{3} P \sum_l \frac{(\hat{\mu_{X}}^{i l} \hat{\mu_{X}}^{l i} + \hat{\mu_{Y}}^{i l} \hat{\mu_{Y}}^{l i} + \hat{\mu_{Z}}^{i l} \hat{\mu_{Z}}^{l i})} {\hbar (\omega^{l i} + \omega)} \mathbf{E}^\omega = \\ & = & \frac{1}{3} (\alpha_{XX} + \alpha_{YY} + \alpha_{ZZ}) \mathbf{E}^\omega = \\ & = & \alpha_{\mathrm{av}}(-\omega;\omega) \mathbf{E}^\omega \end{eqnarray*}$

This coincides with the expression obtained above where we introduced the matter of rotational averages.

## Example of application: quadrupolar contributions#

As a further useful example here, let’s discuss cases where terms like $$\hat{\mathbf{q}} : \boldsymbol{\nabla} \mathbf{E}$$ appear in the averaging process.

### Linear response#

We focus first on the simple example of the (vanishing) quadrupolar contribution to natural electronic circular dichroism (ECD), where terms like

$(\hat{\boldsymbol{\mu}}^{i l} \cdot \mathbf{E}^{\omega_1}) (\hat{\mathbf{q}}^{l i} : \boldsymbol{\nabla} \mathbf{E}^{\omega_2}) \rightarrow (\hat{\boldsymbol{\mu}}^{i l} \cdot \mathbf{E}^{\omega_1})[\hat{\mathbf{r}} \cdot ( \hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_2}]^{l i}$

appear. The operator $$\boldsymbol{\nabla}$$ implies a differentiation applying only to the $$\mathbf{E}^{\omega_2}$$ field vector(s) eventually appearing on the right. This is an important remark since in the process of the isotropic averaging the $$\boldsymbol{\nabla}$$ vector operator behaves as any other space-fixed vectors in the application of the rules given above for tensors of ranks from two to five and it moves in principle around in the corresponding expression. In our specific case, since the transformation is here that of a rank three tensor, we apply the same procedure seen above for sum frequency generation and write

$\langle (\hat{\boldsymbol{\mu}}^{i l} \cdot \mathbf{E}^{\omega_1})[\hat{\mathbf{r}} \cdot ( \hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_2}]^{l i} \rangle_\Omega = \frac{1}{6} ( \hat{\boldsymbol{\mu}}^{i l} \cdot [ \hat{\boldsymbol{\mu}} \times \hat{\boldsymbol{\mu}} ]^{l i}) (\mathbf{E}^{\omega_1} \cdot \boldsymbol{\nabla} \times \mathbf{E}^{\omega_2}) = 0$

The last equal sign stems directly from the structure of the $$( \hat{\boldsymbol{\mu}}^{i l} \cdot [ \hat{\boldsymbol{\mu}} \times \hat{\boldsymbol{\mu}}]^{l i})$$ product. The same applies to the contribution obtained permuting the two operator with associated angular frequencies.

Finally, let’s work out the expression of the spatially isotropically averaged contribution to the circularly difference effect in an optically active system for sum frequency generation due to the quadrupolar interaction. For instance, we concentrate on the term (one of the six arising from the permutation of operators and associated frequencies, and we assume this as understood when we write $$\langle \hat{\mathbf{\mu}}^{il} (\hat{\mathbf{q}}^{lk} : \boldsymbol{\nabla} \mathbf{E}^{\omega_1})(\hat{\boldsymbol{\mu}}^{ki} \cdot \mathbf{E}^{\omega_2}) \rangle_\Omega = \ldots$$)

$\langle \hat{\boldsymbol{\mu}}^{il} (\hat{\mathbf{q}}^{lk} : \boldsymbol{\nabla} \mathbf{E}^{\omega_1})(\hat{\boldsymbol{\mu}}^{ki} \cdot \mathbf{E}^{\omega_2}) \rangle_\Omega = \langle \hat{\boldsymbol{\mu}}^{il} [\hat{\boldsymbol{\mu}} \cdot ( \hat{\boldsymbol{\mu}} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_1}]^{l k}(\hat{\boldsymbol{\mu}}^{ki} \cdot \mathbf{E}^{\omega_2}) \rangle_\Omega$

This requires a rank four transformation which can be worked out setting in the appropriate equation

$\begin{eqnarray*} \mathbf{A} \rightarrow \hat{\boldsymbol{\mu}}^{il}; \mathbf{B} \rightarrow \hat{\boldsymbol{\mu}}^{lk} & ; & \mathbf{C} \rightarrow \hat{\boldsymbol{\mu}}^{lk}; \mathbf{D} \rightarrow \hat{\boldsymbol{\mu}}^{ki} \\ \mathbf{M} \rightarrow 1 ({\mathrm{scalar}}); \mathbf{N} \rightarrow \boldsymbol{\nabla} & ; & \mathbf{P} \rightarrow \mathbf{E}^{\omega_1}; \mathbf{Q} \rightarrow \mathbf{E}^{\omega_2} \\ \end{eqnarray*}$

Note that the vectors (rank one tensors) $$\hat{\boldsymbol{\mu}}$$ arising from the (rank two) tensor $$\hat{\mathbf{q}}$$ both carry the indices $$lk$$ and must be reassociated after the application of the averaging. In short, f.ex., a $$\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{lk}$$ turns into

$(\hat{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{\mu}})^{lk} = (\hat{\boldsymbol{\mu}}_X \hat{\boldsymbol{\mu}}_X + \hat{\boldsymbol{\mu}}_Y \hat{\boldsymbol{\mu}}_Y + \hat{\boldsymbol{\mu}}_Z \hat{\boldsymbol{\mu}}_Z)^{lk} \rightarrow (\hat{\mathbf{q}}_{XX}^{lk}+\hat{\mathbf{q}}_{YY}^{lk} + \hat{\mathbf{q}}_{ZZ}^{lk})$

whereas $$\hat{\boldsymbol{\mu}}^{lk} \hat{\boldsymbol{\mu}}^{lk}$$ has to be rewritten as $$(\hat{\boldsymbol{\mu}} \hat{\boldsymbol{\mu}})^{lk} \rightarrow \hat{\mathbf{q}}^{lk}$$. One immediately sees that the resulting field arrangements are

$\begin{eqnarray*} {\mathrm{From}} (\mathbf{M} \cdot \mathbf{N})(\mathbf{P} \cdot \mathbf{Q}): \ \ \ \ & \rightarrow & \boldsymbol{\nabla} ( \mathbf{E}^{\omega_1} \cdot \mathbf{E}^{\omega_2}) \\ {\mathrm{From}} (\mathbf{M} \cdot \mathbf{P}) (\mathbf{N} \cdot \mathbf{Q}): \ \ \ \ & \rightarrow & \mathbf{E}^{\omega_1} (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_2}) \rightarrow (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_2}) \mathbf{E}^{\omega_1} \rightarrow (\mathbf{E}^{\omega_2} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_1} \\ {\mathrm{From}} (\mathbf{M} \cdot \mathbf{Q})(\mathbf{N} \cdot \mathbf{P}): \ \ \ \ & \rightarrow & \mathbf{E}^{\omega_2} (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_1}) \end{eqnarray*}$

where we have used the fact that the differentiation applies only to the $$\mathbf{E}^{\omega_1}$$ field and it moves further to the right when it is in front of $$\mathbf{E}^{\omega_2}$$.

Let’s the concentrate on the molecular response conterpart (i.e. the arrangement of the $$\mathbf{A}$$, $$\mathbf{B}$$, $$\mathbf{C}$$ and $$\mathbf{D}$$ terms). From both the first and second row of the rank four formula given above the quantity associated to the sum of the $$\boldsymbol{\nabla} ( \mathbf{E}^{\omega_1} \cdot \mathbf{E}^{\omega_2})$$ and $$(\mathbf{E}^{\omega_2} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_1}$$ field factors is seen to be

$\begin{eqnarray*} & & \frac{4}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}) -\frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}) -\frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{ki})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{lk}) = \\ & & = \frac{1}{10} [\hat{\boldsymbol{\mu}}^{il} \cdot (\hat{\boldsymbol{\mu}} \hat{\boldsymbol{\mu}})^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}] -\frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk}) (\hat{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{\mu}})^{lk} \end{eqnarray*}$

whereas from the third row we obtain for the quantity in front o the field factor $$\mathbf{E}^{\omega_2} (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_1})$$

$\begin{eqnarray*} & & -\frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}) -\frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}) +\frac{4}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{ki})(\hat{\boldsymbol{\mu}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{lk}) = \\ & & = -\frac{1}{15} [ \hat{\boldsymbol{\mu}}^{il} \cdot (\hat{\boldsymbol{\mu}} \hat{\boldsymbol{\mu}})^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki} ] +\frac{2}{15} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{\mu}})^{lk} \end{eqnarray*}$

The final expression is therefore

$\begin{eqnarray*} & & \langle \hat{\boldsymbol{\mu}}^{il} (\hat{\mathbf{q}}^{lk} : \boldsymbol{\nabla} \mathbf{E}^{\omega_1})(\hat{\boldsymbol{\mu}}^{ki} \cdot \mathbf{E}^{\omega_2}) \rangle_\Omega = \\ & & = \big[ \frac{1}{10} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\mathbf{q}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki} ) - \frac{1}{30} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\hat{\boldsymbol{\mu}}}}^{lk})(\hat{\mathbf{q}}_{XX}^{lk}+\hat{\mathbf{q}}_{YY}^{lk}+\hat{\mathbf{q}}_{ZZ}^{lk}) \big] [ \boldsymbol{\nabla} ( \mathbf{E}^{\omega_1} \cdot \mathbf{E}^{\omega_2}) + (\mathbf{E}^{\omega_2} \cdot \boldsymbol{\nabla}) \mathbf{E}^{\omega_1} ]+ \\ & & + \big[ -\frac{1}{15} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\mathbf{q}}^{lk} \cdot \hat{\boldsymbol{\mu}}^{ki}) +\frac{2}{15} (\hat{\boldsymbol{\mu}}^{il} \cdot \hat{\boldsymbol{\mu}}^{lk})(\hat{\mathbf{q}}_{XX}^{lk}+\hat{\mathbf{q}}_{YY}^{lk}+\hat{\mathbf{q}}_{ZZ}^{lk}) \big] [\mathbf{E}^{\omega_2} (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_1})] \end{eqnarray*}$

Once again, this is only one of the terms coming from permutations of operators and associated angular frequencies. Moreover, the process has contributions also for the term where the quadrupolar contribution is associated to the angular frequency $$\omega_2$$. Note in the last expression that the contribution coming from the field factor $$\mathbf{E}^{\omega_2} (\boldsymbol{\nabla} \cdot \mathbf{E}^{\omega_1})$$ vanishes due to the presence of the divergence of the electric field $$\mathbf{E}^{\omega_1}$$. The expression holds for all arrangement of the fields and for every choice of polarization, and again it yields an immediate information of how fields (mutual direction of propagation, choice of the circular frequencies and of their polarization) should be arranged in the experiment to allow detection of a given high order optical effect in freely rotating assemblies of absorbers.