Phonon Dispersion in 1D Diatomic Chains

Recommended resources

The freely available resources this lesson will direct you to.

noc19-ph02 Lecture 49-Displacement of the atoms for the acoustic and optical Phononswww.youtube.com
Diatomic Lattice vibrations_ Concepts of Accoustic branch & Optical branch_lec onSolid State Physicswww.youtube.com
7 Lattice Vibrationswww3.itp.tu-berlin.de
Introduction to the theory of lattice dynamicswww.neutron-sciences.org

Hello! Welcome to the first lesson of our new module on lattice dynamics.

In the previous module, we treated electron scattering as a phenomenological process characterized by a relaxation time, $τ$ . We acknowledged that a major cause of this scattering in real crystals is the interaction of electrons with the vibrations of the crystal lattice. To build a proper quantum theory of this interaction, we first need to understand the nature of these lattice vibrations themselves.

Today, we will begin this journey by quantizing these collective vibrations, which will lead us to the concept of phonons—the quanta of lattice vibration. Our goal is to set up the dynamical matrix for a 1D diatomic chain and solve for the acoustic and optical phonon dispersion relations. This is the simplest model that captures the essential new physics that arises when there is more than one atom in the crystal's basis. It will introduce us to the fundamental distinction between acoustic and optical modes of vibration.

1. The Diatomic Chain: Setting up the Model

Let's imagine an infinite one-dimensional crystal with a two-atom basis. The atoms have masses $M_{1}$ and $M_{2}$ and are arranged in an alternating chain. We'll assume they are connected by identical springs with a force constant $f$ , representing the harmonic potential that holds the crystal together. The length of a unit cell (containing one atom of each type) is $a$ .

One-Dimensional Diatomic Chain Model — A one-dimensional diatomic chain with alternating masses $M_1$ (blue) and $M_2$ (red). The lattice constant is $a$, and we denote the displacement of the atoms from their equilibrium positions in the $n$-th unit cell as $s_{1,n}$ and $s_{2,n}$.

Our goal is to find the collective modes of vibration for this system. We start, as is often the case in mechanics, by writing down Newton's second law for each atom in the $n$ -th unit cell. The force on each atom is determined by the stretching of the two springs connecting it to its nearest neighbors.

For mass $M_{1}$ in cell $n$ :

M_{1} s ¨_{1, n} = - f (s_{1, n} - s_{2, n}) - f (s_{1, n} - s_{2, n - 1}) = f (s_{2, n} + s_{2, n - 1} - 2 s_{1, n})

For mass $M_{2}$ in cell $n$ :

M_{2} s ¨_{2, n} = - f (s_{2, n} - s_{1, n}) - f (s_{2, n} - s_{1, n + 1}) = f (s_{1, n} + s_{1, n + 1} - 2 s_{2, n})

We now have an infinite set of coupled differential equations. To solve this, we exploit the translational symmetry of the lattice.

2. The Dynamical Matrix and Dispersion Relation

Since the lattice is periodic, we expect the solutions to be plane waves. We make the following ansatz for the displacements, where $ϵ_{1}$ and $ϵ_{2}$ are the complex amplitudes for each atom in the basis:

s_{1, n} (t) = ϵ_{1} e^{i (q an - ω t)}

s_{2, n} (t) = ϵ_{2} e^{i (q an - ω t)}

Here, $q$ is the wavevector, and $ω$ is the angular frequency. Substituting this ansatz into our equations of motion and simplifying (e.g., $s_{1, n + 1} = s_{1, n} e^{i q a}$ ), we transform the differential equations into a set of algebraic equations for the amplitudes $ϵ_{1}$ and $ϵ_{2}$ :

- ω^{2} M_{1} ϵ_{1} = f (ϵ_{2} + ϵ_{2} e^{- i q a} - 2 ϵ_{1})

- ω^{2} M_{2} ϵ_{2} = f (ϵ_{1} + ϵ_{1} e^{i q a} - 2 ϵ_{2})

Rearranging these into a matrix equation gives us an eigenvalue problem:

(2 f - M_{1} ω^{2} - f (1 + e^{i q a}) - f (1 + e^{- i q a}) 2 f - M_{2} ω^{2}) (ϵ_{1} ϵ_{2}) = (00)

For a non-trivial solution to exist (i.e., for $ϵ_{1}, ϵ_{2}$ not both zero), the determinant of the matrix must be zero. This condition, known as the secular equation, will give us the allowed frequencies $ω$ for each wavevector $q$ .

The following reading provides a clear and complete derivation of this process. It formally constructs the dynamical matrix and solves the resulting eigenvalue problem. Given your preference for formal derivations, following this step-by-step will be valuable.

7 Lattice Vibrations

These lecture notes, 'Lattice Vibrations', formally derive the dispersion relation for the 1D diatomic chain using the dynamical matrix formalism.

Please read subsection 7.2.5.2, 'Linear Chain with Two Atomic Species' (pages 191-193). Start from the equations of motion (Eq. 7.45) and follow the derivation through to the final dispersion relations for the two branches (Eq. 7.58). The text also shows how to set up the 2x2 dynamical matrix (Eq. 7.56), which is central to this topic.

After solving the secular equation, you find a quadratic equation for $ω^{2}$ , which yields two solutions, $ω_{+}^{2}$ and $ω_{-}^{2}$ . This means that for each wavevector $q$ , there are two possible vibrational frequencies, leading to two distinct branches in the dispersion relation:

ω_{\pm}^{2} = f (\frac{1}{M _{1}} + \frac{1}{M _{2}}) \pm f (\frac{1}{M _{1}} + \frac{1}{M _{2}})^{2} - \frac{4 sin ^{2} ( q a /2 )}{M _{1} M _{2}}

These two branches, $ω_{+} (q)$ and $ω_{-} (q)$ , are known as the optical branch and the acoustic branch, respectively.

3. Physical Interpretation: Acoustic and Optical Modes

The mathematical solution gives us two branches, but what is the physical difference between them? To find out, we need to examine the eigenvectors—that is, the ratio of the displacement amplitudes $ϵ_{1} / ϵ_{2}$ —for each branch in different limits of $q$ .

Case 1: The Zone Center (q → 0)

Let's expand our solution for small $q$ , where $sin (q a /2) \approx q a /2$ .

Acoustic Branch ( $ω_{-}$ ): In the limit $q \to 0$ , the frequency goes to zero linearly:
$ω_{-} (q \to 0) \approx \frac{2 f}{M _{1} + M _{2}} \frac{a}{2} ∣ q ∣ = v_{s} ∣ q ∣$
Here, $v_{s}$ is the speed of sound in the material. If you solve for the eigenvector in this limit, you'll find that $ϵ_{1} \approx ϵ_{2}$ . This means the two atoms in the unit cell move together, in phase, with the same amplitude. This collective, long-wavelength motion is precisely what a sound wave is.
Optical Branch ( $ω_{+}$ ): In contrast, the optical branch approaches a finite, maximum frequency at $q = 0$ :
$ω_{+} (q \to 0) = 2 f (\frac{1}{M _{1}} + \frac{1}{M _{2}})$
The eigenvector in this case shows that the atoms move against each other, out of phase, such that the center of mass of the unit cell remains stationary: $M_{1} ϵ_{1} \approx - M_{2} ϵ_{2}$ . If the two atoms carry opposite charges (as in an ionic crystal like NaCl), this motion creates an oscillating electric dipole. This dipole can be excited by, and couple strongly to, electromagnetic radiation (i.e., light), which is why this is called the "optical" mode.

The video below provides an excellent and detailed walkthrough of the physical motion of the atoms for each mode.

noc19-ph02 Lecture 49-Displacement of the atoms for the acoustic and optical Phonons

To get a clear physical picture of the atomic motion in these modes, watch the following video. It rigorously derives the displacement ratios for both branches at the key points of the Brillouin zone.

Watch the segment from 5:35 to 20:16. The presenter analyzes the eigenvectors (the ratio of displacement amplitudes, V/U in his notation) for the optical and acoustic modes at the zone center (k=0) and zone boundary (k=pi/a). This analysis is crucial for understanding the physical difference between the modes.

Case 2: The Zone Boundary (q = ±π/a)

At the edge of the first Brillouin zone, where $q = \pm π / a$ , we have $sin^{2} (q a /2) = 1$ . The frequencies become:
$ω_{+}^{2} (π / a) = \frac{2 f}{M _{2}} and ω_{-}^{2} (π / a) = \frac{2 f}{M _{1}} (assuming M_{1} > M_{2})$
At this point, the group velocity $v_{g} = d ω / d q$ is zero for both branches, meaning these are standing waves. The motion is also distinct:

On the acoustic branch, only the heavier atoms ( $M_{1}$ ) are moving, while the lighter atoms are stationary.
On the optical branch, only the lighter atoms ( $M_{2}$ ) are moving, while the heavier atoms are stationary.

Test your understanding!

What happens to the two dispersion branches in the limit that the two masses become equal, i.e., $M_{1} = M_{2} = M$ ? How does your result relate to the dispersion of a monoatomic chain?

Show answer

If you set $M_{1} = M_{2} = M$ , the gap between the acoustic and optical branches at the zone boundary $q = π / a$ closes, because $ω_{+} (π / a) = 2 f / M$ and $ω_{-} (π / a) = 2 f / M$ . The dispersion curve becomes that of a monoatomic chain, but with a lattice constant of $a /2$ . The Brillouin zone is now twice as large, extending from $- π / (a /2)$ to $π / (a /2)$ . The optical branch essentially "folds down" to become the upper half of the single acoustic branch of the monoatomic chain in this larger Brillouin zone.

4. The Phonon Dispersion Curve

Plotting the two solutions $ω_{+} (q)$ and $ω_{-} (q)$ across the first Brillouin zone gives the complete phonon dispersion relation for our model system.

Phonon Dispersion Relation for a 1D Diatomic Chain — The dispersion relation for a 1D diatomic chain. The lower curve is the acoustic branch, which is linear near q=0. The upper curve is the optical branch, which has a finite frequency at q=0. A frequency gap exists between the two branches.

This plot encapsulates all the physics we have discussed. It shows the two distinct branches, the linear behavior of the acoustic mode near $q = 0$ , the "band gap" in allowed vibrational frequencies, and the flattening of the bands at the Brillouin zone boundary where the group velocity vanishes.

Conclusion

Today we took a crucial first step in understanding lattice dynamics. By applying classical mechanics to a simple 1D diatomic chain, we derived its vibrational spectrum and uncovered a fundamental feature of crystals with a multi-atom basis.

Key Takeaways:

A lattice with a basis of $p$ atoms has $3 p$ phonon branches in 3D (or $p$ branches in 1D). For our diatomic chain, $p = 2$ , giving us two branches.
The acoustic branch has a dispersion that goes to zero as $q \to 0$ . It corresponds to the in-phase motion of atoms within a unit cell, analogous to sound waves.
The optical branch has a finite frequency at $q = 0$ . It corresponds to the out-of-phase motion of atoms within the unit cell. In ionic crystals, these modes can be excited by light.
The entire problem can be elegantly formulated as an eigenvalue problem, $ω^{2} e = D (q) e$ , where $D (q)$ is the dynamical matrix.

Preview of Next Lesson:
We will generalize this framework from our 1D toy model to a realistic 3D crystal. This will involve expanding the dynamical matrix to a 3x3 block for each atom in the basis, and we will see how this naturally leads to the distinction between longitudinal and transverse phonon modes.