---
title: The Harmonic Crystal and Phonon Dispersion
module: Lattice Dynamics
moduleNumber: 4
lessonNumber: 1
order: 401
summary: >
  Atoms in a crystal vibrate about their equilibrium sites, and expanding the
  potential to second order turns the whole lattice into a set of coupled
  harmonic oscillators. This lesson sets up the harmonic approximation and the
  dynamical matrix, solves the monatomic linear chain for its dispersion
  omega(k) = 2 sqrt(K/M) times the absolute sine of ka over two, explains why
  wavevectors outside the first Brillouin zone are redundant, and extends the
  chain to two atoms per cell to produce acoustic and optical branches with a
  frequency gap.
topics: [Lattice Dynamics]
draft: false
sources:
  - book: Ashcroft & Mermin
    ref: "Ch. 22 — Classical Theory of the Harmonic Crystal"
  - book: Kittel
    ref: "Ch. 4 — Phonons I: Crystal Vibrations"
  - book: Simon
    ref: "Ch. 8–9 — Vibrations of a One-Dimensional Monatomic and Diatomic Chain"
---

A crystal is not a rigid array of fixed points. Each atom sits in a potential
well formed by its neighbors and oscillates about its equilibrium site, and
because the wells are coupled the oscillations propagate as waves. Those waves
carry the crystal's sound, most of its heat capacity, and the scattering that
limits electrical conduction. This lesson treats them classically: expand the
interatomic potential to second order, reduce the equations of motion to an
eigenvalue problem, and solve the one-dimensional chains that already contain
every qualitative feature of the three-dimensional result.

## The harmonic approximation

Label the equilibrium positions of the ions by the Bravais lattice vectors $\vec
R$ and let $\vec u(\vec R)$ be the displacement of the ion nominally at $\vec R$.
The potential energy of the crystal is some function $U$ of all the instantaneous
positions $\vec R + \vec u(\vec R)$. Expand it about the equilibrium
configuration in powers of the displacements:

$$
U = U_{\text{eq}} + \sum_{\vec R,\,i} \frac{\partial U}{\partial u_i(\vec R)}\,
u_i(\vec R) + \frac{1}{2}\sum_{\substack{\vec R,\,i\\ \vec R',\,j}}
\frac{\partial^2 U}{\partial u_i(\vec R)\,\partial u_j(\vec R')}\,
u_i(\vec R)\,u_j(\vec R') + \cdots
$$

The constant $U_{\text{eq}}$ is the cohesive energy and sets the zero. The linear
term vanishes: equilibrium is defined by every first derivative of $U$ being
zero, so no net force acts when all displacements vanish. The first surviving
term is quadratic. Truncating there is the **harmonic approximation**, and it
defines the force constants

$$
\Phi_{ij}(\vec R,\vec R') = \frac{\partial^2 U}{\partial u_i(\vec R)\,
\partial u_j(\vec R')}\bigg|_{\text{eq}},
$$

a matrix of second derivatives evaluated at equilibrium. Because the crystal is
periodic, $\Phi_{ij}(\vec R,\vec R')$ depends only on the separation $\vec R -
\vec R'$, not on $\vec R$ and $\vec R'$ separately.

> **Definition (Harmonic approximation).** The **harmonic approximation** retains
> the interatomic potential only to second order in the ionic displacements. The
> crystal then behaves as a system of masses connected by ideal (Hooke's-law)
> springs, with force constants $\Phi_{ij}(\vec R - \vec R')$ set by the curvature
> of the potential at equilibrium. Cubic and higher terms — the **anharmonic**
> corrections — are treated separately and are responsible for thermal expansion
> and finite thermal conductivity.

The force on the ion at $\vec R$ is $-\partial U/\partial u_i(\vec R)$, giving
Newton's equations

$$
M\,\ddot u_i(\vec R) = -\sum_{\vec R',\,j} \Phi_{ij}(\vec R - \vec R')\,
u_j(\vec R').
$$

These are coupled linear differential equations, one for each Cartesian
component of each ion. Their solution is the program of lattice dynamics.

## The monatomic linear chain

Strip the problem to its essentials: a line of identical ions of mass $M$,
spacing $a$, connected by springs of stiffness $K$ that couple only nearest
neighbors. Let $u_n$ be the longitudinal displacement of the $n$-th ion from its
site $na$.

$$
% caption: A one-dimensional monatomic chain: identical masses M at spacing a,
% joined by nearest-neighbor springs of stiffness K. The displacement of ion n
% is u_n, measured from its equilibrium site.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % springs (zigzag drawn as short segments) between sites
  \foreach \i in {0,1,2,3,4}{
    \pgfmathsetmacro{\xa}{1.5*\i}
    \pgfmathsetmacro{\xb}{1.5*\i+1.5}
    \draw[black, thick]
      (\xa+0.22,0) -- (\xa+0.42,0.16) -- (\xa+0.62,-0.16)
      -- (\xa+0.82,0.16) -- (\xa+1.02,-0.16) -- (\xa+1.22,0.16) -- (\xb-0.22,0);
  }
  % masses
  \foreach \i/\lab in {1/{n-2},2/{n-1},3/{n},4/{n+1},5/{n+2}}{
    \pgfmathsetmacro{\x}{1.5*\i}
    \fill[black] (\x,0) circle (5pt);
  }
  % index labels below
  \node[anchor=north] at (1.5,-0.35) {n-2};
  \node[anchor=north] at (3.0,-0.35) {n-1};
  \node[anchor=north] at (4.5,-0.35) {n};
  \node[anchor=north] at (6.0,-0.35) {n+1};
  \node[anchor=north] at (7.5,-0.35) {n+2};
  % spacing a
  \draw[black, <->] (4.5,0.75) -- (6.0,0.75) node[midway, above] {$a$};
  % displacement arrow on ion n
  \draw[acc, thick, ->] (4.5,-0.75) -- (5.0,-0.75) node[right] {$u_n$};
\end{tikzpicture}
$$

The stretch of the spring to the right of ion $n$ is $u_{n+1} - u_n$; the stretch
of the spring to the left is $u_n - u_{n-1}$. The net force is the difference,
so the equation of motion is

$$
M\,\ddot u_n = K(u_{n+1} - u_n) - K(u_n - u_{n-1}) = K(u_{n+1} + u_{n-1} - 2u_n).
$$

Look for travelling-wave solutions $u_n = A\,e^{i(kna - \omega t)}$: a plane wave
sampled at the lattice sites, with wavevector $k$ and angular frequency $\omega$.
Substituting and cancelling the common factor $A\,e^{i(kna-\omega t)}$,

$$
-M\omega^2 = K\left(e^{ika} + e^{-ika} - 2\right) = 2K(\cos ka - 1)
= -4K\sin^2\!\frac{ka}{2}.
$$

Solving for $\omega$ gives the dispersion relation of the monatomic chain:

$$
\omega(k) = 2\sqrt{\frac{K}{M}}\,\left|\sin\frac{ka}{2}\right|.
$$

Frequency is not proportional to wavevector except near $k=0$. The chain is a
dispersive medium: waves of different wavelength travel at different phase
speeds.

$$
% caption: Dispersion of the monatomic chain over the first Brillouin zone. The
% curve rises from a linear (sound-like) region near k = 0 to a maximum of
% 2 sqrt(K/M) at the zone boundary k = pi/a, where the group velocity vanishes.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % axes
  \draw[->, black!70] (-3.4,0) -- (3.6,0) node[below right] {$k$};
  \draw[->, black!70] (0,0) -- (0,3.0) node[left] {frequency};
  % dispersion 2*sin(|x|/2) scaled; x=ka runs -pi..pi -> here map k axis units
  % use omega = 2.5*|sin(pi*k/(2*3))| with k in [-3,3] representing [-pi,pi]
  \draw[acc, very thick, domain=-3:3, samples=160, variable=\x]
    plot ({\x},{2.5*abs(sin(deg(pi*\x/6)))});
  % zone boundaries
  \draw[black, dashed] (3,0) -- (3,2.5);
  \draw[black, dashed] (-3,0) -- (-3,2.5);
  \node[anchor=north] at (3,-0.05) {zone edge};
  \node[anchor=north] at (-3,-0.05) {zone edge};
  \node[anchor=north east] at (-0.05,0) {$0$};
  % max marker
  \draw[black, dashed] (0,2.5) -- (3,2.5);
  \node[anchor=east] at (-0.05,2.5) {max};
  % linear tangent near origin
  \draw[black, densely dotted] (0,0) -- (1.15,2.5);
  \node[black, anchor=west] at (0.7,1.9) {slope $=$ sound speed};
\end{tikzpicture}
$$

Two limits read off the curve. Near $k=0$, $\sin(ka/2) \approx ka/2$ and

$$
\omega \approx a\sqrt{\frac{K}{M}}\;\lvert k\rvert = v_s\,\lvert k\rvert,
\qquad v_s = a\sqrt{\frac{K}{M}},
$$

a linear (nondispersive) branch: long-wavelength vibrations are ordinary sound,
travelling at the speed $v_s$. At the zone boundary $k = \pi/a$ the frequency
reaches its maximum $\omega_{\max} = 2\sqrt{K/M}$, and the group velocity
$\d\omega/\d k = a\sqrt{K/M}\,\cos(ka/2)$ falls to zero: the boundary mode is a
standing wave in which alternate ions move in opposite directions and no energy
propagates.

## Brillouin-zone periodicity

The dispersion depends on $k$ only through $\sin(ka/2)$, which is periodic in $k$
with period $2\pi/a$. Shifting $k \to k + 2\pi/a$ leaves $\omega$ unchanged. More
than that, it leaves the physical motion unchanged. The displacement pattern is
determined by the phase $kna$ evaluated at integer $n$, and

$$
e^{i(k + 2\pi/a)na} = e^{ikna}\,e^{i2\pi n} = e^{ikna}.
$$

A wavevector outside the first Brillouin zone $(-\pi/a, \pi/a]$ describes exactly
the same lattice motion as its image inside the zone. There is no physical
content in $k$ beyond the zone: sampling a wave only at discrete sites cannot
resolve wavelengths shorter than $2a$. This aliasing is the lattice analogue of
the Nyquist limit.

$$
% caption: Aliasing on the lattice. A short-wavelength wave (light curve) and a
% long-wavelength wave (accent curve) pass through identical displacements at
% every lattice site, so they represent the same motion; only the wave with
% wavevector inside the first zone is retained.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % baseline
  \draw[black] (0,0) -- (8,0);
  % long-wavelength wave (in zone)
  \draw[acc, very thick, domain=0:8, samples=120, variable=\x]
    plot ({\x},{0.9*sin(deg(pi*\x/4))});
  % short-wavelength alias (out of zone): agrees at integer sites x=0,1,..8
  \draw[black, densely dotted, domain=0:8, samples=200, variable=\x]
    plot ({\x},{0.9*sin(deg(pi*\x/4 + 2*pi*\x))});
  % lattice sites
  \foreach \i in {0,1,2,3,4,5,6,7,8}{
    \fill[black] (\i,{0.9*sin(deg(pi*\i/4))}) circle (2.2pt);
  }
  \node[acc, anchor=south] at (2,1.0) {in-zone wave};
  \node[black, anchor=north] at (5.5,-0.95) {aliased short wave};
\end{tikzpicture}
$$

Counting confirms that one zone holds every mode. Impose Born–von Kármán periodic
boundary conditions on a ring of $N$ ions: $u_{n+N} = u_n$ forces $e^{ikNa} = 1$,
so $k = 2\pi m/(Na)$ for integer $m$. Exactly $N$ of these values fall in the
first zone, one allowed $k$ per ion. A chain of $N$ ions has $N$ degrees of
freedom and $N$ normal modes, and the first Brillouin zone accounts for all of
them.

## The diatomic chain

Real crystals often have more than one atom per primitive cell, and the extra
atoms produce qualitatively new modes. Take the simplest case: a chain with two
different masses $M_1$ and $M_2$ alternating along the line, joined by identical
springs $K$, with lattice constant $a$ (so the two-atom cell repeats every $a$).
Let $u_n$ be the displacement of the mass-$M_1$ atom in cell $n$ and $v_n$ that
of the mass-$M_2$ atom.

$$
% caption: A diatomic chain: two masses M1 and M2 alternate along the line,
% joined by identical springs K. The primitive cell of length a holds one of
% each; u_n and v_n are the two displacements in cell n.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % springs between successive atoms at half-spacing 0.75
  \foreach \i in {0,1,2,3,4,5}{
    \pgfmathsetmacro{\xa}{0.9*\i}
    \draw[black, thick]
      (\xa+0.16,0) -- (\xa+0.30,0.13) -- (\xa+0.45,-0.13)
      -- (\xa+0.60,0.13) -- (\xa+0.74,0);
  }
  % atoms: large (M1) and small (M2) alternating
  \foreach \i in {0,2,4}{
    \pgfmathsetmacro{\x}{0.9*\i}
    \fill[black] (\x,0) circle (5.5pt);
  }
  \foreach \i in {1,3,5}{
    \pgfmathsetmacro{\x}{0.9*\i}
    \fill[black] (\x,0) circle (3.5pt);
  }
  \node[anchor=north] at (0,-0.35) {$M_1$};
  \node[anchor=north] at (0.9,-0.35) {$M_2$};
  % cell bracket length a spans one M1 and one M2
  \draw[black, <->] (0,0.7) -- (1.8,0.7) node[midway, above] {$a$};
  \node[anchor=south] at (3.6,0.2) {$u_n$};
  \node[black, anchor=south] at (4.5,0.2) {$v_n$};
\end{tikzpicture}
$$

Each $M_1$ atom is pulled by the $M_2$ atoms on either side, and vice versa:

$$
\begin{aligned}
M_1\,\ddot u_n &= K(v_n + v_{n-1} - 2u_n),\\
M_2\,\ddot v_n &= K(u_{n+1} + u_n - 2v_n).
\end{aligned}
$$

Try coupled travelling waves $u_n = A\,e^{i(kna-\omega t)}$ and $v_n = B\,
e^{i(kna-\omega t)}$ with different amplitudes $A$ and $B$. Substituting gives a
$2\times 2$ homogeneous linear system for $A$ and $B$:

$$
\begin{aligned}
(2K - M_1\omega^2)\,A - K(1 + e^{-ika})\,B &= 0,\\
-K(1 + e^{ika})\,A + (2K - M_2\omega^2)\,B &= 0.
\end{aligned}
$$

A nontrivial solution requires the determinant to vanish, giving the
characteristic equation

$$
M_1 M_2\,\omega^4 - 2K(M_1 + M_2)\,\omega^2 + 2K^2(1 - \cos ka) = 0.
$$

This is quadratic in $\omega^2$, so at each $k$ there are two frequencies:

$$
\omega^2_\pm(k) = K\!\left(\frac{1}{M_1}+\frac{1}{M_2}\right)
\pm K\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2
- \frac{4\sin^2(ka/2)}{M_1 M_2}}.
$$

The two roots are the two branches of the dispersion.

> **Definition (Acoustic and optical branches).** In a lattice with $p$ atoms per
> primitive cell, the dispersion relation has $p$ branches in one dimension (and
> $3p$ in three). The branch whose frequency vanishes linearly as $k\to 0$ is an
> **acoustic branch**: neighboring atoms move in phase and long-wavelength modes
> are ordinary sound. Each remaining branch is an **optical branch**: atoms within
> a cell move out of phase, the frequency stays finite as $k\to 0$, and in an
> ionic crystal these modes carry an oscillating dipole that couples to light.

## Acoustic and optical branches

The two roots separate cleanly in the long-wavelength limit. For small $k$,
$\sin^2(ka/2)\approx (ka/2)^2$, and expanding the square root gives

$$
\omega_-^2 \approx \frac{K/2}{M_1+M_2}\,(ka)^2,
\qquad
\omega_+^2 \approx 2K\!\left(\frac{1}{M_1}+\frac{1}{M_2}\right).
$$

The lower root $\omega_-$ is the **acoustic branch**: it starts from zero with
slope $v_s = a\sqrt{K/2(M_1+M_2)}$, and the two atoms move together as a rigid
unit — the amplitude ratio $B/A \to 1$. The upper root $\omega_+$ is the
**optical branch**: it approaches a finite frequency at $k=0$ where the two atoms
move exactly out of phase against each other, keeping the center of mass fixed
($M_1 A + M_2 B = 0$). At the zone boundary $k = \pi/a$ the two branches reach

$$
\omega_-(\pi/a) = \sqrt{\frac{2K}{M_{\text{heavy}}}},\qquad
\omega_+(\pi/a) = \sqrt{\frac{2K}{M_{\text{light}}}},
$$

leaving a **frequency gap** between $\sqrt{2K/M_{\text{heavy}}}$ and
$\sqrt{2K/M_{\text{light}}}$ in which no travelling wave exists. Waves at those
frequencies are evanescent and decay into the crystal; the gap is the mechanical
analogue of the electronic band gap.

$$
% caption: The two branches of the diatomic chain over the first Brillouin zone.
% The acoustic branch rises from zero; the optical branch runs across the top.
% Between the zone-boundary values a frequency gap opens, forbidden to
% travelling waves. Insets show the in-phase and out-of-phase cell motions.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black!70] (0,0) -- (3.7,0) node[below right] {$k$};
  \draw[->, black!70] (0,0) -- (0,3.4) node[left] {frequency};
  % k axis: 0..3 represents 0..pi/a
  % acoustic: rises to ~1.4 at boundary (heavy mass)
  \draw[very thick, domain=0:3, samples=120, variable=\x]
    plot ({\x},{1.4*abs(sin(deg(pi*\x/6)))});
  % optical: starts at ~2.9 (k=0), dips to ~2.1 at boundary (light mass)
  \draw[very thick, densely dashed, domain=0:3, samples=120, variable=\x]
    plot ({\x},{sqrt(8.41 - 2.5*sin(deg(pi*\x/6))*sin(deg(pi*\x/6)))});
  % gap band
  \draw[acc, dashed] (0,1.4) -- (3,1.4);
  \draw[acc, dashed] (0,2.1) -- (3,2.1);
  \node[acc, anchor=west] at (3.05,1.75) {gap};
  % boundary line
  \draw[black, dashed] (3,0) -- (3,2.9);
  \node[anchor=north] at (3,-0.05) {zone edge};
  % labels
  \node[anchor=west] at (1.15,0.55) {acoustic};
  \node[anchor=south] at (1.5,2.75) {optical};
\end{tikzpicture}
$$

$$
% caption: Cell motions at k near zero. In the acoustic mode the two atoms of a
% cell move in the same direction (in phase); in the optical mode they move in
% opposite directions (out of phase), the lighter atom swinging farther.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % acoustic row
  \node[anchor=east] at (-0.2,0.9) {acoustic};
  \foreach \i in {0,1,2,3}{
    \pgfmathsetmacro{\x}{1.3*\i}
    \fill[black] (\x,0.9) circle (5pt);
    \fill[black] (\x+0.65,0.9) circle (3pt);
    \draw[black, thick, ->] (\x,0.9) -- (\x+0.35,0.9);
    \draw[black, thick, ->] (\x+0.65,0.9) -- (\x+0.95,0.9);
  }
  % optical row
  \node[anchor=east] at (-0.2,-0.3) {optical};
  \foreach \i in {0,1,2,3}{
    \pgfmathsetmacro{\x}{1.3*\i}
    \fill[black] (\x,-0.3) circle (5pt);
    \fill[black] (\x+0.65,-0.3) circle (3pt);
    \draw[black, thick, ->] (\x,-0.3) -- (\x+0.22,-0.3);
    \draw[black, thick, ->] (\x+0.65,-0.3) -- (\x+0.30,-0.3);
  }
\end{tikzpicture}
$$

The optical name comes from ionic crystals such as NaCl. There the two atoms
carry opposite charges, so the out-of-phase optical motion at $k\approx 0$
produces an oscillating electric dipole. That dipole couples strongly to
infrared light at the optical frequency, giving the reststrahlen (residual-ray)
absorption band that reflects a narrow range of infrared wavelengths almost
perfectly.

## The three-dimensional generalization

A real crystal replaces the scalar displacement $u_n$ with a vector $\vec u(\vec
R)$ and the scalar force constant with the tensor $\Phi_{ij}(\vec R - \vec R')$.
The same plane-wave substitution $\vec u(\vec R) = \vec\epsilon\,e^{i(\vec k \cdot
\vec R - \omega t)}$ reduces the equations of motion to an eigenvalue problem for
the **dynamical matrix**

$$
D_{ij}(\vec k) = \frac{1}{M}\sum_{\vec R} \Phi_{ij}(\vec R)\,e^{-i\vec k \cdot
\vec R},
\qquad
\omega^2(\vec k)\,\vec\epsilon = D(\vec k)\,\vec\epsilon.
$$

$D(\vec k)$ is a $3p \times 3p$ Hermitian matrix for a crystal with $p$ atoms per
cell. Its eigenvalues $\omega_s^2(\vec k)$ are the squared frequencies of the
$3p$ branches; its eigenvectors $\vec\epsilon_s$ are the **polarization vectors**
giving the direction each atom moves. The branch structure generalizes directly:

- **Three acoustic branches**, one for each polarization (one longitudinal, two
  transverse), all with $\omega \to 0$ linearly as $\vec k \to 0$. These are the
  three sound waves of the elastic continuum.
- **$3(p-1)$ optical branches**, finite frequency at $\vec k = 0$, in which atoms
  within a cell oscillate against one another.

A crystal with $p=1$ (a monatomic Bravais lattice such as copper) has only the
three acoustic branches and no optical modes. A crystal with $p=2$ (silicon,
NaCl, diamond) has three acoustic and three optical branches. The dynamical
matrix, its symmetry, and the eigenvectors organize the full three-dimensional
problem, but the counting and the acoustic-versus-optical distinction are already
visible in the one-dimensional chains.

So far the treatment is entirely classical: the modes are continuous oscillations
of arbitrary amplitude. The [quantization of these normal
modes](/condensed-matter/lattice-dynamics/phonons-quantization-and-dos) turns
each into a ladder of energy levels and gives the phonon, the quantized unit of
lattice vibration that carries crystal momentum and sets the thermal properties
of the solid.