Linear Triatomic Molecule: Modes of Vibration

Normal Modes of Vibration of a Linear Triatomic Molecule

We consider a linear triatomic molecule consisting of three atoms arranged in a straight line. The two outer atoms have equal mass $m$, while the central atom has mass $M$. The atoms are connected by identical springs of force constant $k$.

Let $x_1$, $x_2$, and $x_3$ be the small displacements of the three atoms from their equilibrium positions along the line of the molecule.

Assumptions

To simplify the analysis, we make the following standard assumptions:

The oscillations are small.
The restoring forces obey Hooke’s law.
The forces are linear functions of displacement.
The resulting motion is simple harmonic.

Under these assumptions, the problem of molecular vibration reduces to a problem in linear algebra.

Kinetic Energy of the System

The total kinetic energy of the system is the sum of the kinetic energies of the three atoms. Since each atom moves independently along the line of the molecule, there are no cross terms in the velocities.

\[ T = \frac{1}{2} m \dot{x}_1^2 + \frac{1}{2} M \dot{x}_2^2 + \frac{1}{2} m \dot{x}_3^2 \]

There are no terms like $\dot{x}_1 \dot{x}_2$ because kinetic energy depends only on individual velocities, not on relative motion between particles.

Vector Representation

We now introduce the displacement vector

\[ \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \]

and the corresponding velocity vector

\[ \dot{\mathbf{X}} = \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} \]

The kinetic energy can be written in compact matrix form as

\[ T = \frac{1}{2} \, \dot{\mathbf{X}}^{T} \, \mathbf{T} \, \dot{\mathbf{X}} \]

where $\mathbf{T}$ is called the kinetic energy matrix.

Derivation of the T-Matrix

Writing the quadratic form explicitly,

\[ T = \frac{1}{2} \begin{bmatrix} \dot{x}_1 & \dot{x}_2 & \dot{x}_3 \end{bmatrix} \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} & T_{23} \\ T_{31} & T_{32} & T_{33} \end{bmatrix} \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} \]

Expanding this expression gives

\[ \begin{aligned} T = \frac{1}{2} (& T_{11}\dot{x}_1^2 + T_{22}\dot{x}_2^2 + T_{33}\dot{x}_3^2 \\ &+ T_{12}\dot{x}_1\dot{x}_2 + T_{21}\dot{x}_2\dot{x}_1 \\ &+ T_{13}\dot{x}_1\dot{x}_3 + T_{31}\dot{x}_3\dot{x}_1 \\ &+ T_{23}\dot{x}_2\dot{x}_3 + T_{32}\dot{x}_3\dot{x}_2 ) \end{aligned} \]

Comparing this with the physical expression for kinetic energy, we see that only the diagonal terms survive.

\[ \mathbf{T} = \begin{bmatrix} m & 0 & 0 \\ 0 & M & 0 \\ 0 & 0 & m \end{bmatrix} \]

Thus, the T-matrix is simply a convenient way of recording the masses of the atoms. It is diagonal and symmetric.

Once the kinetic energy and potential energy matrices are constructed, the problem of molecular vibration reduces to solving a matrix eigenvalue equation. The resulting eigenvalues give the normal mode frequencies, and the eigenvectors describe the corresponding normal modes of vibration.

Potential Energy of the System

The molecule has two springs: Spring 1 between masses $(m, M)$ and Spring 2 between masses $(M, m)$. For small oscillations, each spring stores potential energy according to Hooke’s law. The important point is that a spring “feels” the relative (effective) displacement of its endpoints.

Potential energy in each spring

For Spring 1, the extension (effective displacement) is $(x_2 - x_1)$, so

\[ V_1 = \frac{1}{2}k(x_2-x_1)^2 \]

For Spring 2, the extension is $(x_3 - x_2)$, so

\[ V_2 = \frac{1}{2}k(x_3-x_2)^2 \]

Therefore, the total potential energy is the sum:

\[ V = V_1 + V_2 \]

Expanding and collecting terms

Expanding each square gives

\[ \begin{aligned} V &= \frac{1}{2}k(x_2-x_1)^2+\frac{1}{2}k(x_3-x_2)^2 \\ &= \frac{1}{2}k(x_2^2-2x_1x_2+x_1^2) + \frac{1}{2}k(x_3^2-2x_3x_2+x_2^2) \end{aligned} \]

Now combine like terms:

\[ V=\frac{1}{2}k\Big[x_1x_1 + 2x_2x_2 + x_3x_3 - 2x_1x_2 - 2x_3x_2\Big] \]

Rearranging into a quadratic (matrix) form

We rewrite the same expression by arranging it as a symmetric quadratic form:

\[ V=\frac{1}{2}k \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \]

So, in compact matrix notation,

\[ V=\frac{1}{2}\mathbf{X}^T\,\mathbf{V}\,\mathbf{X} \]

Here, the V-matrix records how the displacements are coupled through the springs. Unlike the kinetic energy matrix (which was diagonal), the potential energy matrix has off-diagonal terms because springs depend on relative displacement.

Connecting Energies to Equations of Motion (Lagrange’s Equation)

Once the kinetic energy $T$ and potential energy $V$ are known, the equations of motion can be obtained using Lagrange’s equation.

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)-\frac{\partial L}{\partial q_j}=0 \]

where the Lagrangian is

\[ L=T-V \]

In this problem, we may take the generalized coordinates as the displacements themselves:

\[ q_j = x_j \quad (\text{in this case}) \]

Then Lagrange’s equation can be written in the form

\[ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)+\frac{\partial V}{\partial q_j}=0 \]

With the matrix forms,

\[ T=\frac{1}{2}\dot{x}_i\,T_{ij}\,\dot{x}_j, \qquad V=\frac{1}{2}x_i\,V_{ij}\,x_j \]

This is the key point: writing $T$ and $V$ in matrix form prepares the problem for a clean linear-algebra treatment, which naturally leads to the normal-mode (eigenvalue) formulation in the next step.

Matrix Equation of Motion

From the Lagrange equations written in matrix form, the equations of motion can be expressed as

\[ \mathbf{T}\,\ddot{\mathbf{X}} + \mathbf{V}\,\mathbf{X} = 0 \]

This is a set of coupled linear differential equations. Because both $\mathbf{T}$ and $\mathbf{V}$ are symmetric matrices, the system is well suited for a normal-mode analysis.

Concept of Normal Coordinates

The system is linear, and therefore we expect the motion to be harmonic. Accordingly, we assume a trial solution of the form

\[ \mathbf{X} = \mathbf{A}\,e^{i\omega t} \]

where $\mathbf{A}$ is the amplitude vector and $\omega$ is the angular frequency.

Differentiating twice with respect to time gives

\[ \ddot{\mathbf{X}} = -\omega^2 \mathbf{A}\,e^{i\omega t} \]

Substituting these expressions into the matrix equation of motion,

\[ -\omega^2 \mathbf{T}\mathbf{A}e^{i\omega t} + \mathbf{V}\mathbf{A}e^{i\omega t} = 0 \]

Canceling the common exponential factor,

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A} = 0 \]

For a non-trivial solution ($\mathbf{A} \neq 0$), the determinant of the coefficient matrix must vanish. This gives the secular determinant.

\[ \left|\mathbf{V} - \omega^2 \mathbf{T}\right| = 0 \]

Solving the Secular Determinant

For the linear triatomic molecule, the potential and kinetic energy matrices are

\[ \mathbf{V} = \begin{bmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{bmatrix}, \qquad \mathbf{T} = \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix} \]

Hence, the secular determinant becomes

\[ \left| \begin{matrix} k-\omega^2 m & -k & 0 \\ -k & 2k-\omega^2 m & -k \\ 0 & -k & k-\omega^2 m \end{matrix} \right| = 0 \]

Evaluating the determinant,

\[ (k-\omega^2 m)\Big[(2k-\omega^2 m)(k-\omega^2 m)-k^2\Big] - k^2(k-\omega^2 m) = 0 \]

Simplifying,

\[ (k-\omega^2 m)\Big[\omega^4 m^2 - \omega^2 k(2m+m)\Big] = 0 \]

Normal Mode Frequencies

From the secular equation, the allowed values of $\omega$ are obtained.

The first factor gives

\[ \omega^2 = 0 \]

This corresponds to the translational mode, in which all three atoms move together with no restoring force.

The remaining factors give the non-zero normal mode frequencies:

\[ \omega_2 = \sqrt{\frac{k}{m}} \]

\[ \omega_3 = \sqrt{\frac{k(2m+m)}{m m}} = \sqrt{\frac{k}{m}\left(1+\frac{2m}{m}\right)} \]

Thus, the linear triatomic molecule has three normal modes:

One zero-frequency translational mode
Two vibrational normal modes with distinct frequencies

Each normal mode corresponds to an independent collective motion of the atoms, even though the original coordinates were coupled. This is the power of the normal-coordinate transformation.

Collecting the Results

From the secular equation, the three normal mode frequencies of the linear triatomic molecule are obtained as:

\[ \omega_1 = 0 \]

This corresponds to the translational mode, in which all atoms move together without any restoring force.

\[ \omega_2 = \sqrt{\frac{k}{m}} \]

This mode represents symmetric stretching of the molecule.

\[ \omega_3 = \sqrt{\frac{k}{m}\left(1+\frac{2m}{m}\right)} \]

This corresponds to the asymmetric stretching mode.

Importance of the Amplitude Matrix $ \mathbf{A} $

Recall the assumed harmonic solution:

\[ \mathbf{X} = \mathbf{A} e^{i\omega t} \]

Here, the matrix (or vector) $\mathbf{A}$ contains the amplitudes of motion of the individual atoms. It plays a crucial role in defining the nature of the normal modes.

We may also write the transformation

\[ \mathbf{X} = \mathbf{A}\mathbf{q} \]

where $\mathbf{q}$ represents the normal coordinates. This transformation is useful because it converts coupled equations into independent ones.

Eigenvalue Formulation

From the equation

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A} = 0 \]

we can rewrite it as

\[ \mathbf{V}\mathbf{A} = \omega^2 \mathbf{T}\mathbf{A} \]

or, equivalently,

\[ \mathbf{V}\mathbf{A} = \lambda \mathbf{A} \]

This is an eigenvalue problem, where:

$\mathbf{V}$ represents the potential energy coupling
$\lambda = \omega^2$ is the eigenvalue
$\mathbf{A}$ is the eigenvector

Each eigenvector corresponds to a specific mode shape, and each eigenvalue gives the square of the corresponding normal-mode frequency.

Physical Meaning of Eigenvectors

An eigenvector represents a collective displacement pattern of the atoms. Once the system is excited in the direction of an eigenvector, it continues to vibrate in that same pattern indefinitely.

One normal mode does not influence another — this is the concept of independent frequencies.

The physical significance of the amplitude matrix $\mathbf{A}$ is that:

$\mathbf{A}$ defines the shape of the vibration
$e^{i\omega t}$ defines the time evolution

Thus, shape and time are cleanly separated in normal-mode analysis.

If the system is initially set into motion along one of its eigenvectors, the resulting motion is perfectly simple and remains confined to that mode.

Mode Shapes and Normal Coordinates

Starting from

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A}e^{i\omega t} = 0 \]

or equivalently,

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A}\mathbf{q} = 0, \qquad \mathbf{q} \neq 0 \]

we see that the condition for non-trivial motion is

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A} = 0 \]

The columns of $\mathbf{A}$ therefore give the mode shapes, and the corresponding coordinates $\mathbf{q}$ are the normal coordinates.

In normal coordinates, the original coupled system is completely decoupled, and each mode behaves like an independent simple harmonic oscillator.

Explicit Determination of Mode Shapes

Having obtained the normal-mode frequencies, we now determine the corresponding eigenvectors (mode shapes) by solving

\[ (\mathbf{V} - \omega^2 \mathbf{T})\mathbf{A} = 0 \]

Each value of $\omega$ defines a separate case.

Case I: $\omega_1 = 0$ (Translational Mode)

For $\omega = 0$, the eigenvalue equation reduces to

\[ \mathbf{V}\mathbf{A}_1 = 0 \]

Using the potential energy matrix,

\[ k \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} a_{11} \\ a_{12} \\ a_{13} \end{bmatrix} = 0 \]

This leads to the following set of linear equations:

\[ \begin{aligned} a_{11} - a_{12} &= 0 \\ - a_{11} + 2a_{12} - a_{13} &= 0 \\ a_{12} - a_{13} &= 0 \end{aligned} \]

From these equations, we immediately obtain

\[ a_{11} = a_{12} = a_{13} \]

Since only the relative amplitudes matter, we may choose

\[ \mathbf{A}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

This corresponds to pure translation of the molecule, where all atoms move together with equal displacement.

Case II: $\omega_2^2 = \dfrac{k}{m}$ (Symmetric Stretching)

For the second normal mode, the eigenvalue equation becomes

\[ (\mathbf{V} - \omega_2^2 \mathbf{T})\mathbf{A}_2 = 0 \]

Substituting $\omega_2^2 = k/m$,

\[ \left[ \begin{bmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{bmatrix} - \frac{k}{m} \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix} \right] \begin{bmatrix} a_{21} \\ a_{22} \\ a_{23} \end{bmatrix} = 0 \]

This simplifies to

\[ \begin{bmatrix} 0 & -k & 0 \\ -k & 2k-\frac{k m}{m} & -k \\ 0 & -k & 0 \end{bmatrix} \begin{bmatrix} a_{21} \\ a_{22} \\ a_{23} \end{bmatrix} = 0 \]

The resulting equations give

\[ a_{22} = 0, \qquad a_{21} = -a_{23} \]

Choosing a convenient normalization,

\[ \mathbf{A}_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \]

This represents a symmetric stretching mode, in which the outer atoms move symmetrically in opposite directions, while the central atom remains stationary.

Thus, by solving the eigenvalue equation explicitly, we obtain not only the frequencies but also the corresponding mode shapes that describe the physical motion of the molecule.

Case III: $\omega_3$ (Asymmetric Stretching Mode)

For the third normal mode, the angular frequency is

\[ \omega_3 = \sqrt{\frac{k}{m}\left(1+\frac{2m}{m}\right)} \]

so that

\[ \omega_3^2 = \frac{k}{m}\left(1+\frac{2m}{m}\right) \]

The eigenvalue equation is

\[ (\mathbf{V} - \omega_3^2 \mathbf{T})\mathbf{A}_3 = 0 \]

Substitution into the Eigenvalue Equation

Substituting the explicit forms of $\mathbf{V}$ and $\mathbf{T}$,

\[ \left[ k \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} - \frac{k}{m}\left(1+\frac{2m}{m}\right) \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix} \right] \begin{bmatrix} a_{31} \\ a_{32} \\ a_{33} \end{bmatrix} = 0 \]

After simplification, this leads to the following set of linear equations:

\[ \begin{aligned} -\frac{2m}{m}k\,a_{31} - k a_{32} &= 0 \\ - k a_{31} + \frac{k m}{m} a_{32} - k a_{33} &= 0 \\ - k a_{32} - \frac{2m}{m}k\,a_{33} &= 0 \end{aligned} \]

Solving for the Eigenvector

From the first and third equations, we obtain

\[ a_{31} = -\frac{m}{2m}\,a_{32}, \qquad a_{33} = -\frac{m}{2m}\,a_{32} \]

Choosing a convenient normalization $a_{32} = 1$, the eigenvector becomes

\[ \mathbf{A}_3 = \begin{bmatrix} -\dfrac{m}{2m} \\ 1 \\ -\dfrac{m}{2m} \end{bmatrix} \]

This represents the asymmetric stretching mode, in which the central atom moves opposite to the outer atoms, with unequal amplitudes.

Summary of Normal Modes

The linear triatomic molecule has three independent normal modes:

Translational mode
$\displaystyle \omega_1 = 0,\quad \mathbf{A}_1 = \begin{bmatrix}1\\1\\1\end{bmatrix}$
Symmetric stretching mode
$\displaystyle \omega_2 = \sqrt{\frac{k}{m}},\quad \mathbf{A}_2 = \begin{bmatrix}1\\0\\-1\end{bmatrix}$
Asymmetric stretching mode
$\displaystyle \omega_3 = \sqrt{\frac{k}{m}\left(1+\frac{2m}{m}\right)},\quad \mathbf{A}_3 = \begin{bmatrix} -\dfrac{m}{2m}\\ 1\\ -\dfrac{m}{2m} \end{bmatrix}$

Each normal mode represents an independent collective vibration of the molecule. Once expressed in normal coordinates, the system behaves as a set of uncoupled simple harmonic oscillators.

This completes the normal-mode analysis of a linear triatomic molecule using matrix methods and linear algebra.

The Normal Coordinate Transformation

We now collect the eigenvectors obtained for the three normal modes into a single matrix, called the amplitude matrix $\mathbf{A}$.

\[ \mathbf{A} = \begin{bmatrix} \mathbf{A}_1 & \mathbf{A}_2 & \mathbf{A}_3 \end{bmatrix} \]

Using the eigenvectors obtained earlier, the matrix $\mathbf{A}$ may be written as

\[ \mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & -1 \\ -\dfrac{m}{2m} & 1 & -\dfrac{m}{2m} \end{bmatrix} \]

This matrix contains, column by column, the displacement patterns corresponding to each normal mode.

Transformation to Normal Coordinates

We define the transformation from the physical coordinates $\mathbf{X}$ to the normal coordinates $\mathbf{q}$ as

\[ \mathbf{X} = \mathbf{A}\mathbf{q} \]

In component form,

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & -1 \\ -\dfrac{m}{2m} & 1 & -\dfrac{m}{2m} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix} \]

This gives the explicit relations:

\[ \begin{aligned} x_1 &= q_1 + q_2 + q_3 \\ x_2 &= q_1 - q_3 \\ x_3 &= -\frac{m}{2m}q_1 + q_2 - \frac{m}{2m}q_3 \end{aligned} \]

Thus, each physical displacement is expressed as a linear combination of the normal coordinates.

Alternative Choice of Eigenvector Normalisation

Since eigenvectors are defined only up to a multiplicative constant, we may choose a convenient normalization. For example, choosing

\[ a_{31} = 1, \quad a_{32} = -\frac{2m}{m}, \quad a_{33} = 1 \]

gives an alternative form of the third eigenvector:

\[ \mathbf{A}_3 = \begin{bmatrix} 1 \\ -\dfrac{2m}{m} \\ 1 \end{bmatrix} \]

Such rescaling does not change the physics of the normal mode.

What Happens in This “Magic” Transformation?

The original equations of motion were

\[ \mathbf{T}\ddot{\mathbf{X}} + \mathbf{V}\mathbf{X} = 0 \]

Substituting $\mathbf{X} = \mathbf{A}\mathbf{q}$,

\[ \mathbf{T}\mathbf{A}\ddot{\mathbf{q}} + \mathbf{V}\mathbf{A}\mathbf{q} = 0 \]

Multiplying from the left by $\mathbf{A}^T$,

\[ \mathbf{A}^T\mathbf{T}\mathbf{A}\,\ddot{\mathbf{q}} + \mathbf{A}^T\mathbf{V}\mathbf{A}\,\mathbf{q} = 0 \]

Because the eigenvectors are chosen appropriately, we obtain

\[ \mathbf{A}^T\mathbf{T}\mathbf{A} = \mathbf{I}, \qquad \mathbf{A}^T\mathbf{V}\mathbf{A} = \mathrm{diag}(\omega_1^2,\omega_2^2,\omega_3^2) \]

Decoupled Equations of Motion

The equations of motion in normal coordinates become

\[ \ddot{q}_i + \omega_i^2 q_i = 0 \]

Thus, the original problem of three coupled particles has been transformed into a problem of three independent simple harmonic oscillators.

This is the central result of normal-mode analysis: linear algebra converts complex coupled motion into simple independent motions.

What Are Normal Coordinates?

Normal coordinates are not artificial coordinates invented for mathematics. They are the coordinates that the physical system itself prefers.

In normal coordinates, each coordinate describes an independent motion, and the complicated coupled dynamics in physical space disappears naturally.

Physical Visualization of Normal Modes

We now visualize each normal mode physically for a linear triatomic molecule (such as O=C=O).

Mode 1: $\omega_1 = 0$ — Translational Mode

\[ \mathbf{A}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

In this mode, all atoms move together in the same direction.

There is no stretching of any spring, and therefore

No restoring force
No vibration

This is pure translation of the molecule as a whole, not a vibrational mode.

Because no spring is stretched, the frequency is zero.

Mode 2: $\omega_2 = \sqrt{\dfrac{k}{m}}$ — Symmetric Stretching

\[ \mathbf{A}_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \]

In this mode:

The two outer atoms move symmetrically in opposite directions
The middle atom remains stationary

The middle atom does not move because the forces acting on it from the two springs cancel by symmetry.

The mass of the central atom does not appear in the frequency because it does not participate in this motion.

Mode 3: $\omega_3 = \sqrt{\dfrac{k}{m}\left(1+\dfrac{2m}{m}\right)}$ — Asymmetric Stretching

\[ \mathbf{A}_3 = \begin{bmatrix} 1 \\ -\dfrac{2m}{m} \\ 1 \end{bmatrix} \]

In this mode:

The two outer atoms move in the same direction
The central atom moves in the opposite direction
Both springs stretch simultaneously

Because all three atoms participate, the frequency depends on both the spring constant and the masses.

This is a true vibrational mode involving collective motion of all atoms.

Energy in Normal Coordinates

In normal coordinates, the kinetic and potential energies take the simple forms:

\[ T = \frac{1}{2}\sum_i \dot{q}_i^2, \qquad V = \frac{1}{2}\sum_i \omega_i^2 q_i^2 \]

Thus, each normal coordinate behaves like an independent simple harmonic oscillator.

This is why normal coordinates are so powerful: they reveal the true independent motions hidden inside a coupled system.

Normal Modes and IR / Raman Spectroscopy

Normal modes form the natural bridge between molecular vibrations and vibrational spectroscopy. Each normal mode may or may not be observable in infrared (IR) or Raman spectroscopy, depending on how it interacts with electromagnetic radiation.

Concept of Phonons

When a normal mode of vibration is quantized, its energy is exchanged in discrete packets called phonons.

A phonon is simply a quantized normal mode — the vibrational analogue of a photon.

Selection Rules: IR vs Raman

A vibrational mode is IR active if it produces a change in the dipole moment of the molecule:

\[ \mu = q\,d \]

Thus, IR spectroscopy “sees” changes in charge separation during vibration.

A vibrational mode is Raman active if it produces a change in the polarizability of the molecule:

\[ \alpha \propto \text{electron cloud deformation} \]

Raman spectroscopy is sensitive to how easily the electron cloud is distorted.

IR and Raman Activity of Normal Modes (Linear CO₂)

Mode 1: Translation ($\omega = 0$)

In translational motion, the entire molecule moves as a whole.

No vibration
No spring stretching
No change in dipole moment
No change in polarizability

This mode is IR inactive and Raman inactive.

Mode 2: Symmetric Stretching ($\omega_2 = \sqrt{k/m}$)

In symmetric stretching:

The two outer atoms move symmetrically
The central atom remains fixed
Charge distribution remains symmetric

Since the dipole moment does not change,

Symmetric stretching is IR inactive.

However, the vibration causes a large symmetric deformation of the electron cloud.

Symmetric stretching is Raman active.

Mode 3: Asymmetric Stretching

In asymmetric stretching:

The symmetry of the molecule is broken
The dipole moment changes with time

Asymmetric stretching is IR active.

Because the electron cloud distortion is less symmetric, this mode is comparatively weaker in Raman scattering.

Bending Modes (Out-of-Plane and In-Plane)

In linear molecules like CO₂, bending vibrations occur in planes perpendicular to the molecular axis.

In-plane bending
Out-of-plane bending

These bending modes change the dipole moment and are therefore IR active.

Important Non-Intuitive Points

A common misconception is:

“Symmetric stretching sounds strong, so it should be IR active.”

This is incorrect.

Another misconception is:

“Any vibration should absorb IR radiation.”

The correct physical understanding is:

IR spectroscopy does not care about vibration strength
IR spectroscopy cares only about change in dipole moment

Why Symmetric Stretching Is Raman Active

During symmetric stretching, the polarizability of the molecule changes significantly because the electron cloud expands and contracts symmetrically.

Raman spectroscopy detects exactly this:

Change in electron cloud size
Change in electron cloud deformation

Thus, symmetric stretching is strongly Raman active even though it is IR inactive.

This completes the conceptual connection between normal modes, phonons, and vibrational spectroscopy.

Conclusion: Normal Modes and Spectroscopic Activity

We summarize the IR and Raman activity of the normal modes of a linear triatomic molecule as follows:

Mode	IR Activity	Raman Activity
Translation	✗	✗
Symmetric Stretching	✗	✓
Asymmetric Stretching	✓	✗
Bending	✓	✗

Translational motion involves no change in relative atomic positions. Therefore, it produces neither a time-dependent dipole moment nor a change in polarizability.

Hence, translation is neither IR active nor Raman active.

From Normal Modes to Phonons

For a linear triatomic molecule, we have obtained three independent normal modes with frequencies $\omega_1, \omega_2,$ and $\omega_3$.

In classical mechanics, these modes describe independent oscillations. However, when we move from classical mechanics (CM) to quantum mechanics (QM), each normal mode becomes quantized.

Energy Quantization of Normal Modes

For each normal mode, the allowed energy levels are

\[ E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n = 0, 1, 2, \dots \]

Thus, the vibrational energy is quantized.

A phonon is one quantum of vibrational energy associated with a normal mode.

Physical Meaning of Phonons

Each normal mode can be populated by an integer number of phonons.

The three normal modes combine to produce the actual atomic motion that we observe in a molecule.

A refined and precise statement is:

The actual atomic motion observed in a molecule is a superposition of normal modes, with each normal mode being populated by an integer number of phonons.

Final Physical Picture

Each normal mode oscillates independently with its own frequency:

$\omega_1$: translational mode
$\omega_2$: symmetric stretching mode
$\omega_3$: asymmetric stretching mode

The combined motion of atoms is obtained by superposing all these oscillations.

A phonon is the quantum of a normal mode of vibration, and real atomic motion is obtained by superposing all populated normal modes.

This completes the discussion of normal modes, their physical interpretation, their spectroscopic signatures, and their quantized description in terms of phonons.