Raman Scattering and Vibrational Quantization

Let us first review the classical description. For the normal mode $k$, let $Q_k$ denote the coordinate, $Q_{k0}$ the amplitude, and $\omega_k$ the characteristic angular frequency. Under an incident electric field $E_i(t)$ with angular frequency $\omega_i$, the molecular displacement and the field are expressed as follows.

Here we consider non-resonant Raman scattering, well away from electronic resonance, and treat the molecular vibration as a harmonic oscillator. Expanding the polarizability $\alpha(Q_k)$ in a Taylor series about the equilibrium position $Q_k=0$ to first order in displacement $Q_k$ yields the following form.

For simplicity in the subsequent discussion, we fix the polarization directions of the incident and scattered light, denoting the observed polarizability tensor component simply by $\alpha$.

Substituting these expressions into the induced dipole moment $\mu(t)=\alpha(Q_k) E_i(t)$ and applying standard trigonometric product-to-sum identities separates the vibration-dependent term into two distinct sidebands.

These are the Stokes component $\omega_s^{(\mathrm{St})}=\omega_i-\omega_k$ on the lower-frequency side and the anti-Stokes component $\omega_s^{(\mathrm{AS})}=\omega_i+\omega_k$ on the higher-frequency side. Thus, the spectral positions of the Raman lines, namely the frequency shifts themselves, are already explained clearly within classical electromagnetism. In classical theory, however, both sidebands share the identical dipole amplitude $\alpha^{\prime}Q_{k0}E_{i0}/2$, predicting equal vibrational intensity factors.

Because classical dipole radiation power scales with the fourth power of frequency $\omega_s^4$, a literal reading of classical theory would imply that higher-frequency anti-Stokes light should be stronger than Stokes light. In measured spectra, however, anti-Stokes light is substantially weaker even at room temperature and vanishes almost entirely at low temperatures. This failure of classical theory is resolved only by quantizing the vibration and recognizing the fundamental asymmetry between creating and annihilating a vibrational quantum.

The mechanical energy of a classical harmonic oscillator is proportional to the square of its amplitude $Q_{k0}$, which can be continuously reduced to zero. In quantum mechanics, however, even within the very same parabolic potential, the allowed energies are discrete and form an evenly spaced ladder.

Here the non-negative integer $v_k=0,1,2,\ldots$ is the vibrational quantum number.

The spacing between adjacent levels $\hbar\omega_k$ corresponds precisely to the energy of a single vibrational quantum. The lowest level is the vibrational ground state $v_k=0$, whose energy $E_0 = \frac{1}{2}\hbar\omega_k$ is known as the zero-point energy. Even at absolute zero, molecular vibration does not freeze entirely; the system perpetually retains this zero-point motion.

In the classical model, the Raman sidebands originated from the term $\alpha^{\prime}Q_k$, which modulates the polarizability proportionally to the vibrational coordinate. In quantum theory, this classical coordinate $Q_k$ is promoted to a quantum coordinate operator $\hat Q_k$. Using the standard formulation of the harmonic oscillator, $\hat Q_k$ is expressed in terms of the annihilation operator $\hat b_k$ and creation operator $\hat b_k^\dagger$.

Here $m_k$ is the effective mass associated with this normal mode, and $Q_{k,\mathrm{zpf}}$ is the quantum displacement scale of this vibrational mode.

The operator $\hat b_k$ is the annihilation operator, which removes one vibrational quantum, and $\hat b_k^\dagger$ is the creation operator, which adds one quantum. Their actions on an eigenstate $|v_k\rangle$ with vibrational quantum number $v_k$ follow standard ladder operations.

Consequently, acting on state $|v_k\rangle$ with the coordinate operator $\hat Q_k$ produces a linear combination of the state one step lower $|v_k-1\rangle$ and the state one step higher $|v_k+1\rangle$.

Classically, $Q_k(t)$ tracked the instantaneous geometric displacement of the molecule. In quantum mechanics, $\hat Q_k$ remains the displacement operator, but when viewed in the basis of vibrational eigenstates, it acts strictly as a transition operator coupling adjacent rungs of the ladder.

The interaction between light and vibration in the classical model proceeded through the first-order term $\alpha^{\prime}Q_k$. In quantum mechanics, the probability amplitude for an operator to drive a transition between initial and final states is governed by its matrix element. Here, the vibrational factor governing the transition from initial state $|v_k\rangle$ to final state $|v_k^{\prime}\rangle$ is the coordinate matrix element $\langle v_k^{\prime}|\hat Q_k|v_k\rangle$.

Taking the inner product of $\hat Q_k|v_k\rangle$ from Eq. (8) with the final state $\langle v_k^{\prime}|$ yields:

Because distinct vibrational eigenstates are mutually orthogonal, this matrix element is non-zero only when $v_k^{\prime}=v_k-1$ or $v_k^{\prime}=v_k+1$. Therefore, the selection rule governing the change in vibrational quantum number $\Delta v_k$ is:

This is the vibrational selection rule established under the harmonic oscillator approximation and first-order polarizability expansion. For fundamental scattering to occur in this framework, the polarizability derivative must also be non-zero, $\alpha^{\prime}\neq0$. The classical requirement that molecular vibration must alter the polarizability carries over intact into quantum mechanics.

Stokes Scattering and Energy Transfer to the Molecule

In the transition $v_k \to v_k+1$, the vibrational energy of the molecule increases by $\hbar\omega_k$. Formulating energy conservation for the combined light-molecule system gives:

The incident light imparts one quantum of vibrational energy to the molecule, decreasing the angular frequency of the scattered photon by $\omega_k$ and emitting red-shifted Stokes light.

Anti-Stokes Scattering and Energy Transfer to Light

Conversely, in the transition $v_k \to v_k-1$, the molecule sheds vibrational energy $\hbar\omega_k$. Because this released energy is transferred to the scattered photon, conservation of energy demands:

The two sideband frequencies $\omega_i \mp \omega_k$, derived formally from trigonometric identities in classical theory, now gain a clear physical interpretation: the exchange of exactly one vibrational quantum $\hbar\omega_k$ transferred to or taken from the molecule.

In quantum mechanics, transition probabilities scale with the squared magnitude of the transition amplitude. The vibrational contribution to the scattering intensity is therefore proportional to the squared coordinate matrix element $|\langle v_k^{\prime}|\hat Q_k|v_k\rangle|^2$. Evaluating this quantity from Eq. (9) for Stokes ($v_k \to v_k+1$) and anti-Stokes ($v_k \to v_k-1$) transitions yields:

Setting aside the $\omega_s^4$ radiation factor and the common constant $|\alpha^{\prime}|^2Q_{k,\mathrm{zpf}}^2$, the vibrational intensity factor per molecule in initial state $v_k$ simplifies to an elegant result:

Remarkably, even when starting from the identical initial state $v_k$, the Stokes factor $v_k+1$ and anti-Stokes factor $v_k$ differ by strictly one quantum. Whereas classical theory predicted equal amplitudes $Q_{k0}/2$ for both sidebands, quantum mechanics introduces an intrinsic asymmetry into the transition probabilities, directly reflecting the commutator $[\hat b_k, \hat b_k^\dagger] = 1$.

Scattering from the Vibrational Ground State

Consider a molecule in the vibrational ground state $v_k=0$. As shown in Eq. (15), the molecule can absorb energy from the incident light and transition to $v_k=1$, a Stokes process with finite matrix element $Q_{k,\mathrm{zpf}}$. In contrast, because no vibrational quanta exist to be removed, meaning $\hat b_k|0\rangle=0$, the anti-Stokes transition probability is strictly zero. Thus, creating a vibrational quantum requires no pre-existing thermal excitation, whereas annihilating a quantum cannot occur without prior excitation. This fundamental distinction explains why the Stokes line remains fully observable even upon cooling to zero temperature.

Within a macroscopic sample, countless molecules are distributed statistically across various vibrational levels by thermal energy. While all molecules, including those in the ground state, can contribute to Stokes scattering, anti-Stokes scattering is restricted to molecules already occupying excited levels $v_k \ge 1$. As temperature drops, the fraction of molecules in these excited states diminishes exponentially.

Boltzmann Distribution and Mean Quantum Number

In thermal equilibrium at temperature $T$, the probability $p_{v_k}$ of finding a molecule in vibrational level $v_k$ obeys Maxwell-Boltzmann statistics. With $k_B$ denoting the Boltzmann constant, this probability is expressed via the partition function:

Total Scattering Intensity by Thermal Averaging

Taking the ratio of these vibrational factors and inserting Eq. (18) recovers the Boltzmann factor directly:

In the low-temperature regime where $k_B T \ll \hbar\omega_k$, meaning $\bar v_k \to 0$, the anti-Stokes factor vanishes while the Stokes factor approaches 1. The striking spectral asymmetry highlighted at the outset simply reflects the extreme statistical scarcity of molecules possessing an available quantum to be removed.

Radiation Efficiency and the Total Intensity Ratio

Finally, we incorporate the $\omega_s^4$ factor governing classical dipole radiation efficiency. Under identical incident intensity and collection geometry, the total scattered intensities are given by:

Taking their ratio yields the celebrated intensity ratio equation:

Here $\omega_i>\omega_k$, and we utilize the same polarizability derivative $\alpha^{\prime}$ for both components under non-resonant conditions. Equation (22) carries profound practical value: by measuring the intensity ratio of Stokes and anti-Stokes lines, one can determine the local absolute temperature $T$ of a sample non-invasively and in real time without external thermometers.