Classical Theory of Raman Scattering

Light arrives as an electric field

First, let's put the incoming light into an equation. Light is an electromagnetic wave, and a laser is essentially monochromatic, so over the tiny region a molecule occupies we may treat the incident light as a single plane wave. Fix axes $x,y,z$ in space and write the incident electric-field vector at that point as $\boldsymbol{E}_{\mathrm{i}}(t)$. Writing the field amplitude as $E_{\mathrm{i}0}$, the unit vector giving the field's oscillation direction (the polarization vector) as $\boldsymbol{e}_{\mathrm{i}}$, and the incident angular frequency as $\omega_{\mathrm{i}}$,

$$\boldsymbol{E}_{\mathrm{i}}(t) = E_{\mathrm{i}0}\,\boldsymbol{e}_{\mathrm{i}}\cos\omega_{\mathrm{i}} t \tag{1}$$

The polarization vector has a component along each axis,

$$\boldsymbol{e}_{\mathrm{i}}=\begin{pmatrix} e_{\mathrm{i}x} \\ e_{\mathrm{i}y} \\ e_{\mathrm{i}z} \end{pmatrix} \tag{2}$$

When we later treat the light scattered back out, we write its polarization vector $\boldsymbol{e}_{\mathrm{s}}$ and angular frequency $\omega_{\mathrm{s}}$.

The field distorts the electron cloud

What does this field do to the molecule? Applying a field does not rearrange the molecule's shape, that is, the layout of its nuclei. What moves is the charge distribution. The positive and negative charges in the molecule feel opposite forces from the field: the positive charge shifts along the field, the negative electron cloud against it, each sliding to opposite sides. The centres of positive and negative charge thus separate, creating an imbalance of charge. This is the induced dipole, and since it slides farther the stronger the field, the dipole is proportional to the field.

Leaving the tangled question of direction for later, we look first at magnitude alone, as a scalar. Writing the magnitude of the induced dipole as $\mu(t)$,

$$\mu(t) = \alpha\,E_{\mathrm{i}0}\cos\omega_{\mathrm{i}} t$$

The constant $\alpha$ is the polarizability mentioned above; the more easily the electron cloud deforms, the larger $\alpha$ is.

Here, if the polarizability $\alpha$ is a constant $\alpha_0$ independent of time, the dipole can oscillate only at the incident angular frequency $\omega_{\mathrm{i}}$. And a dipole oscillating at one frequency radiates light only at that same frequency, so the scattered light keeps the incident colour unchanged. This is Rayleigh scattering. For the colour to shift, $\alpha$ itself must change with time, and what shakes it is the molecule's own vibration.

The vibrating polarizability

Now let us see how molecular vibration changes the polarizability. A molecule has several characteristic ways to vibrate. As its nuclei move away from equilibrium, the electron cloud reshapes with them, so the polarizability changes. In other words, the polarizability depends on the displacement of the nuclei.

Each of a molecule's fixed ways of oscillating is called a normal mode, numbered $k=1,2,\dots$. A normal mode is one of the independent oscillation patterns into which a molecule's tangled overall vibration is decomposed. For mode $k$, write the displacement of the nuclei along that oscillation from equilibrium as the normal coordinate $Q_k$, with $Q_k=0$ at rest. The polarizability is then a function of that displacement, $\alpha(Q_k)$.

Molecular vibrations have small amplitudes, so instead of needing the entire $\alpha(Q_k)$ curve, we focus on the neighbourhood of equilibrium, $Q_k=0$. Very close to a point, a curve nearly overlaps its tangent there. Thus $\alpha(Q_k)$ can be approximated using only its value and slope at equilibrium. This is the same as writing $f(x)\simeq f(0)+f'(0)x$ for a small $x$.

$$\alpha(Q_k)\simeq\alpha_0+\left(\frac{\partial\alpha}{\partial Q_k}\right)_{0}Q_k \tag{3}$$

$\alpha_0$ is the polarizability at equilibrium, and $(\partial\alpha/\partial Q_k)_0$ is the slope at $Q_k=0$. If the nuclei oscillate harmonically at angular frequency $\omega_k$ with amplitude $Q_{k0}$, the normal coordinate varies in time as

$$Q_k(t)=Q_{k0}\cos\omega_k t \tag{4}$$

Substituting this into Eq. (3), the polarizability varies periodically about its equilibrium value $\alpha_0$ at angular frequency $\omega_k$:

$$\alpha(t)=\alpha_0+\alpha_k\cos\omega_k t,\qquad \alpha_k=\left(\frac{\partial\alpha}{\partial Q_k}\right)_{0}Q_{k0} \tag{5}$$

Here we have bundled the width over which the polarizability swings into the constant $\alpha_k$. It is not a function of time but a constant giving how widely the polarizability swings about its equilibrium value, and its size is set by the equilibrium slope $(\partial\alpha/\partial Q_k)_0$. This slope will itself become the selection rule later.

Frequencies contained in the dipole

The field oscillates at $\omega_{\mathrm{i}}$, while the polarizability oscillates at $\omega_k$. The induced dipole is their product, $\mu=\alpha E$. Substituting Eq. (5) into the dipole equation gives

$$\mu(t)=\big(\alpha_0+\alpha_k\cos\omega_k t\big)E_{\mathrm{i}0}\cos\omega_{\mathrm{i}} t \tag{6}$$

A product of two cosines of different angular frequency has appeared. We want to extract which frequencies hide in this product, so we use the product-to-sum identity

$$\cos\omega_k t\,\cos\omega_{\mathrm{i}} t = \tfrac12\left[\cos(\omega_{\mathrm{i}}-\omega_k)t + \cos(\omega_{\mathrm{i}}+\omega_k)t\right]$$

to obtain

$$\mu(t)=\underbrace{\alpha_0E_{\mathrm{i}0}\cos\omega_{\mathrm{i}} t}_{\text{Rayleigh}}+\tfrac12\alpha_k E_{\mathrm{i}0}\Big[\underbrace{\cos(\omega_{\mathrm{i}}-\omega_k)t}_{\text{Stokes}}+\underbrace{\cos(\omega_{\mathrm{i}}+\omega_k)t}_{\text{anti-Stokes}}\Big] \tag{7}$$

The first term oscillates at the incident frequency $\omega_{\mathrm{i}}$ and gives Rayleigh scattering. The remaining two have the new frequencies $\omega_{\mathrm{i}}-\omega_k$ and $\omega_{\mathrm{i}}+\omega_k$. The lower one is Stokes light and the higher one is anti-Stokes light. The vibration of the polarizability modulates the incident light and creates sum and difference sidebands. This is the connection with AM radio mentioned above. The factor $\tfrac12$ in the sideband amplitudes comes from the product-to-sum identity.

Including direction: the Raman tensor

So far we have treated the polarizability as a single number. That was enough to see where the shifted frequencies come from. In a real molecule, however, the electron cloud does not deform equally in every direction, and a field in one direction can produce a dipole component in another. To keep track of these directions, we write the polarizability as a second-rank tensor $\boldsymbol{\alpha}$. Whereas the scalar relation $\mu=\alpha E$ multiplies by one number, the tensor relates an electric-field vector to an induced-dipole vector.

So what does that operation look like in components? A field applied along $x$ can produce not only an $x$ dipole but $y$ and $z$ dipoles as well. Conversely, the $x$ component of the dipole, $\mu_x$, picks up contributions not only from the $x$ field but from the $y$ and $z$ fields too. Each contribution, like the scalar $\mu=\alpha E$, is a product of a polarizability component and a field component. Each component of the incident field splits into the amplitude $E_{\mathrm{i}0}$ and a direction $e_{\mathrm{i}\sigma}$, so the $x$ field component is $E_{\mathrm{i}0}e_{\mathrm{i}x}$, the $y$ component $E_{\mathrm{i}0}e_{\mathrm{i}y}$, and so on. Adding the three contributions,

$$\mu_x = E_{\mathrm{i}0}\left(\alpha_{xx}\,e_{\mathrm{i}x} + \alpha_{xy}\,e_{\mathrm{i}y} + \alpha_{xz}\,e_{\mathrm{i}z}\right)\cos\omega_{\mathrm{i}} t$$

Here $\alpha_{xx}$ is how effectively an $x$ field makes an $x$ dipole, and $\alpha_{xy}$ how effectively a $y$ field makes an $x$ dipole. $\mu_y$ and $\mu_z$ take the same form. Writing all three out each time is tedious, so we introduce indices $\rho,\sigma$, each standing for one of $x,y,z$, and collapse them into one:

$$\mu_\rho(t) = \sum_\sigma \alpha_{\rho\sigma}\,E_{\mathrm{i}0}\,e_{\mathrm{i}\sigma}\cos\omega_{\mathrm{i}} t \tag{8}$$

The sum $\sum_\sigma$ means summing over $\sigma=x,y,z$. The components satisfy the symmetry $\alpha_{\rho\sigma}=\alpha_{\sigma\rho}$, leaving six independent components.

The reasoning for why the polarizability varies in time is the same as in the scalar case, but the equilibrium slope takes a different value for each direction pair $(\rho,\sigma)$. The array of these component-by-component slopes is the Raman tensor $\boldsymbol{R}_k$ of mode $k$: a record of in which direction, and by how much, the vibration changes the polarizability.

$$(R_k)_{\rho\sigma} = \left(\frac{\partial\alpha_{\rho\sigma}}{\partial Q_k}\right)_0 \tag{9}$$

Since several modes may vibrate at once, the time dependence of each tensor component sums the contributions of all modes,

$$\alpha_{\rho\sigma}(t) = (\alpha_0)_{\rho\sigma} + \sum_k (\alpha_k)_{\rho\sigma}\cos\omega_k t,\qquad (\alpha_k)_{\rho\sigma} = (R_k)_{\rho\sigma}\,Q_{k0} \tag{10}$$

Substituting this into Eq. (8) and using the same product-to-sum identity as in the scalar case,

$$\begin{aligned}\mu_\rho={}&\underbrace{\sum_\sigma(\alpha_0)_{\rho\sigma}E_{\mathrm{i}0} e_{\mathrm{i}\sigma}\cos\omega_{\mathrm{i}} t}_{\text{Rayleigh}}\\&+\tfrac12\sum_\sigma\sum_k(\alpha_k)_{\rho\sigma}E_{\mathrm{i}0} e_{\mathrm{i}\sigma}\Big[\underbrace{\cos(\omega_{\mathrm{i}}-\omega_k)t}_{\text{Stokes}}+\underbrace{\cos(\omega_{\mathrm{i}}+\omega_k)t}_{\text{anti-Stokes}}\Big]\end{aligned} \tag{11}$$

The equation has grown long, but its structure is exactly that of the scalar Eq. (7). The only difference is that it now carries the information of direction and polarization.

Radiation: from dipole to scattered intensity

So far we know at which frequencies the induced dipole oscillates. Next we want to know how strongly that dipole radiates light. The three terms of Eqs. (7) and (11) radiate at $\omega_{\mathrm{i}}$, $\omega_{\mathrm{i}}-\omega_k$, and $\omega_{\mathrm{i}}+\omega_k$ respectively. Let us write the scattered angular frequency of whichever term we are looking at collectively as $\omega_{\mathrm{s}}$.

What radiates an electromagnetic wave is the acceleration of the shaken charge. When the dipole oscillates as $\cos\omega_{\mathrm{s}} t$, its acceleration is the second time derivative, which brings down a factor $\omega_{\mathrm{s}}^2$. In dipole radiation the radiated field is proportional to this acceleration, so a factor $\omega_{\mathrm{s}}^2$ appears in the field as well.

A wave from a point source spreads over a sphere. Because the sphere's surface area grows as $r^2$, the intensity per unit area falls as $1/r^2$. Intensity is proportional to the square of the electric-field amplitude, so the field amplitude itself falls as $1/r$. Light also takes a time $r/c$ to travel a distance $r$, so the wave at the observation point is delayed in phase by that amount.

We want to gather the polarization connections into a single quantity, so we introduce the scattering tensor $\boldsymbol{a}$. It bundles the relation among the incident polarization $\boldsymbol{e}_{\mathrm{i}}$, the scattered polarization $\boldsymbol{e}_{\mathrm{s}}$, and the scattering amplitude; comparing with Eqs. (7) and (11), for the Rayleigh term $\boldsymbol{a}=\boldsymbol{\alpha}_0$, and for the Raman sideband terms, including the $\tfrac12$ from before, $\boldsymbol{a}=\boldsymbol{\alpha}_k/2$. With it, the scattered field $E_{\mathrm{s}}$ at an observation point a distance $r$ away is

$$E_{\mathrm{s}} \propto \frac{\omega_{\mathrm{s}}^2}{c^2 r}\,(\boldsymbol{e}_{\mathrm{s}}\!\cdot\boldsymbol{a}\cdot\boldsymbol{e}_{\mathrm{i}})\,E_{\mathrm{i}0}\cos\!\Big(\omega_{\mathrm{s}} t-\frac{\omega_{\mathrm{s}}}{c}r\Big) \tag{12}$$

The $r/c$ inside the phase is, as we just saw, the time for the wave to travel the distance $r$. The observed scattered intensity $I_{\mathrm{s}}$ is proportional to the square of the field amplitude, so writing the incident intensity as $I_{\mathrm{i}}$ and squaring both sides,

$$I_{\mathrm{s}}\,r^2 \propto \frac{\omega_{\mathrm{s}}^4}{c^4}\,\big|\boldsymbol{e}_{\mathrm{s}}\!\cdot\boldsymbol{a}\cdot\boldsymbol{e}_{\mathrm{i}}\big|^2\,I_{\mathrm{i}} \tag{13}$$

The intensity is proportional to the incident intensity $I_{\mathrm{i}}$, to the square of the scattering tensor, and to the fourth power of the scattered angular frequency $\omega_{\mathrm{s}}$. The same fourth-power law that explains why the sky is blue shows up here too.

Which vibrations are visible? Selection rules

Now that we have reached the intensity, we can read off the condition for a vibration to appear as Raman-scattered light. The sideband amplitude is carried by $(\alpha_k)_{\rho\sigma}$, that is, the Raman tensor $(R_k)_{\rho\sigma}$, which was proportional to the slope $(\partial\alpha/\partial Q_k)_0$ we met in the scalar Eq. (5). That slope measures how much the polarizability changes when the molecule moves slightly along the vibration, so its being non-zero is nothing other than the deformability of the electron cloud changing as the molecule vibrates.

Thus, if for mode $k$ the slope is zero for every direction pair $(\rho,\sigma)$, the sideband amplitude vanishes completely. Put the other way, the condition to be seen in Raman scattering is that the slope be non-zero for at least one component:

$$(R_k)_{\rho\sigma}=\left(\frac{\partial\alpha_{\rho\sigma}}{\partial Q_k}\right)_{0}\neq 0\quad\text{for some }(\rho,\sigma) \tag{14}$$

In short, Raman scattering sees only vibrations that change how easily the electron cloud deforms. This is the selection rule. How strongly a mode is actually observed also depends on the polarizations of the incident and scattered light.

In the symmetric stretch, the two oxygens approach and recede from the carbon together. The $\alpha(Q)$ curve has a non-zero slope at equilibrium, so the polarizability changes with the vibration and sidebands appear. In the antisymmetric stretch, one bond lengthens while the other shortens. In bending, the straight molecule bends. For these last two motions, the slope of $\alpha(Q)$ is zero at equilibrium in the figure, so no sidebands appear in the first-order approximation used here.

The zero slope becomes intuitive if we compare the two sides of equilibrium. In the antisymmetric stretch, a shape with the left bond longer and one with the right bond longer differ only by exchanging left and right. In bending, the upward- and downward-bent shapes are mirror images. For $Q$ and $-Q$, the electron cloud is only exchanged or reflected, so it is equally easy to deform. The $\alpha(Q)$ curve is therefore symmetric about equilibrium and has zero slope at its centre. In the symmetric stretch, however, the shape with both bonds longer is not equivalent to the one with both bonds shorter. The ease of deforming the electron cloud changes, leaving a non-zero slope.

What we have found

A monochromatic electric field deforms the electron cloud and creates an induced dipole. If the polarizability is constant, the dipole oscillates only at the incident frequency $\omega_{\mathrm{i}}$, producing Rayleigh scattering. If molecular vibration changes the polarizability at $\omega_k$, the dipole gains components at $\omega_{\mathrm{i}}-\omega_k$ and $\omega_{\mathrm{i}}+\omega_k$, producing Stokes and anti-Stokes light. The frequency difference $\omega_k$ reveals the molecular vibration. A vibration appears in Raman scattering when the polarizability has a non-zero slope at equilibrium, and the scattered intensity is proportional to $\omega_{\mathrm{s}}^4$.

Where the classical picture fails

Eqs. (7) and (11) give the Stokes and anti-Stokes sidebands the same dipole amplitude. The radiation formula (13) then adds the $\omega_{\mathrm{s}}^4$ dependence. A classical estimate therefore predicts that the higher-frequency anti-Stokes light should be slightly stronger.

Experiments show the opposite: anti-Stokes light is weaker, and it becomes stronger as the temperature rises. This difference cannot be explained by classical theory. The absolute scattering intensity and resonance Raman also lie beyond this classical model.