---
title: Anharmonicity, Thermal Expansion, and Heat Conduction
module: Lattice Dynamics
moduleNumber: 4
lessonNumber: 4
order: 404
summary: >
  A perfectly harmonic crystal neither expands when heated nor resists heat flow.
  Both effects come from the cubic and higher terms the harmonic approximation
  discards. This lesson derives thermal expansion from an asymmetric interatomic
  potential, treats phonon-phonon scattering as the decay channel these terms
  open, shows why Umklapp processes are what make lattice thermal conductivity
  finite, and traces the temperature dependence of the conductivity and the phonon
  mean free path.
topics: [Lattice Dynamics]
draft: false
sources:
  - book: Ashcroft & Mermin
    ref: "Ch. 25 — Anharmonic Effects in Crystals"
  - book: Kittel
    ref: "Ch. 5 — Phonons II: Thermal Properties"
  - book: Hook & Hall
    ref: "Ch. 2 — Phonons"
---

The [harmonic crystal](/condensed-matter/lattice-dynamics/phonon-dispersion) is
an idealization with two unphysical consequences: it does not expand when heated,
and it conducts heat without resistance. Both failures trace to the same
omission. Truncating the interatomic potential at second order makes the normal
modes exact, independent, and eternal — phonons never scatter off one another, and
the mean atomic position never shifts. Restoring the cubic and quartic terms
couples the modes and displaces the potential minimum. This lesson shows how
those anharmonic terms produce thermal expansion, phonon–phonon scattering, and a
finite, temperature-dependent thermal conductivity.

## Why the harmonic crystal fails

Two statements about a purely harmonic solid must be established before repairing
them.

- **No thermal expansion.** In a harmonic potential $U = c\,x^2$ the average
  displacement $\langle x\rangle$ is zero at every temperature, because the
  parabola is symmetric about its minimum: positive and negative excursions are
  equally likely and equally weighted. Heating raises the amplitude but not the
  mean position, so the crystal does not expand.
- **Infinite thermal conductivity.** The harmonic normal modes are exact
  eigenstates. A phonon wavepacket, once launched, propagates forever at its
  group velocity without decaying into other modes. A temperature gradient would
  drive an unimpeded phonon flux, giving infinite conductivity. Any real
  resistance requires a mechanism that scatters phonons, and the harmonic
  Hamiltonian provides none.

The cure for both is the same: keep the next terms in the expansion of $U$.

## The anharmonic potential and thermal expansion

Expand the interaction of a pair of atoms about the equilibrium separation,
writing $x$ for the displacement from it:

$$
U(x) = c\,x^2 - g\,x^3 - f\,x^4 + \cdots
$$

The cubic term $-g\,x^3$ (with $g>0$) makes the well asymmetric: it is softer on
the side of increasing separation (large positive $x$) and stiffer on the side of
compression. An atom oscillating in this well spends more time at larger
separations, so its mean position drifts outward as the amplitude grows. That
drift is thermal expansion.

$$
% caption: An asymmetric interatomic potential. The harmonic parabola (light,
% symmetric) is compared with the true well (accent), softer on the
% large-separation side. As temperature rises the thermal average of the position
% shifts outward along the dashed midline, producing expansion.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black!70] (-2.4,0) -- (3.4,0) node[below right] {$x$};
  \draw[->, black!70] (0,-0.3) -- (0,3.2) node[left] {$U(x)$};
  % symmetric parabola (light)
  \draw[black, thick, densely dotted, domain=-1.7:1.7, samples=80, variable=\x]
    plot ({\x},{0.85*\x*\x});
  % anharmonic well (accent): softer for x>0
  \draw[acc, very thick, domain=-1.55:2.9, samples=140, variable=\x]
    plot ({\x},{0.85*\x*\x - 0.14*\x*\x*\x});
  % thermal levels (horizontal chords) with rightward-shifting midpoints
  \foreach \y/\xl/\xr in {0.6/-0.82/0.93, 1.4/-1.22/1.55, 2.2/-1.5/2.15}{
    \draw[black] (\xl,\y) -- (\xr,\y);
    \pgfmathsetmacro{\xm}{(\xl+\xr)/2}
    \fill[black] (\xm,\y) circle (1.6pt);
  }
  % midline through the shifting centers
  \draw[acc, densely dashed] (0.05,0.6) -- (0.32,2.2);
  \node[acc, anchor=west] at (0.35,2.35) {mean position};
  \node[black, anchor=east] at (-1.0,2.4) {harmonic};
\end{tikzpicture}
$$

The mean displacement follows from a classical Boltzmann average. For small
anharmonicity the cubic and quartic terms are corrections, and expanding the
Boltzmann factor to first order in $g$ and $f$ gives

$$
\langle x\rangle = \frac{\int x\,e^{-U(x)/k_B T}\,\d x}
{\int e^{-U(x)/k_B T}\,\d x}
= \frac{3g}{4c^2}\,k_B T.
$$

The mean position grows linearly with temperature, so the linear thermal-
expansion coefficient $\alpha = (1/x_0)\,\d\langle x\rangle/\d T = 3g k_B/(4c^2
x_0)$ is a constant at high temperature, set entirely by the cubic coefficient
$g$. A symmetric potential ($g=0$) gives $\langle x\rangle = 0$ and no expansion,
confirming that thermal expansion is an anharmonic effect. The full quantum
treatment replaces $k_B T$ by the mode energy and reproduces the same link to the
heat capacity through the [Grüneisen
parameter](/condensed-matter/lattice-dynamics/debye-einstein-heat-capacity).

## Phonon–phonon scattering

The cubic term does more than shift the minimum: it couples the normal modes. A
term proportional to $x^3$, written in normal coordinates, is a product of three
mode amplitudes, so it connects three phonons. To lowest order it allows a phonon
to decay into two, or two phonons to merge into one — a **three-phonon process**.
These are the collisions that give phonons a finite lifetime and a mean free path.

Every three-phonon process conserves energy and crystal momentum:

$$
\omega_1 = \omega_2 + \omega_3,
\qquad
\vec k_1 = \vec k_2 + \vec k_3 + \vec G,
$$

for the decay of phonon $1$ into $2$ and $3$ (merging is the reverse). The
reciprocal-lattice vector $\vec G$ distinguishes the two kinds of collision, and
the distinction controls whether the process resists heat flow at all.

> **Definition (Normal and Umklapp processes).** A three-phonon process is a
> **normal** ($N$) process when $\vec G = 0$: the total crystal momentum
> $\vec k_1 + \vec k_2 + \cdots$ is exactly conserved. It is an **Umklapp** ($U$)
> process when $\vec G \ne 0$: the summed wavevector leaves the first Brillouin
> zone and is folded back, reversing the net direction of crystal-momentum flow.
> Only Umklapp processes degrade a phonon heat current, because normal processes
> conserve the total momentum that carries the heat.

$$
% caption: Two three-phonon collisions. In the normal process the outgoing
% wavevectors sum, within the zone (dashed), to the incoming one, so the momentum
% flow is unchanged. In the Umklapp process the sum exceeds the zone and a
% reciprocal-lattice vector G returns it, flipping the direction of the heat-
% carrying momentum.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % Normal
  \begin{scope}
    \draw[black, dashed] (-1.5,-1.5) rectangle (1.5,1.5);
    \draw[acc, very thick, ->] (0,0) -- (0.9,0.6);
    \node[acc, anchor=south east] at (0.85,0.6) {$k_2$};
    \draw[acc, very thick, ->] (0.9,0.6) -- (1.25,-0.15);
    \node[acc, anchor=west] at (1.2,0.15) {$k_3$};
    \draw[black, thick, ->] (0,0) -- (1.25,-0.15);
    \node[black, anchor=north] at (0.9,-0.25) {$k_1$};
    \node[anchor=south] at (0,-2.1) {normal};
  \end{scope}
  % Umklapp
  \begin{scope}[xshift=6.0cm]
    \draw[black, dashed] (-1.5,-1.5) rectangle (1.5,1.5);
    \draw[acc, very thick, ->] (0,0) -- (1.1,0.8);
    \node[acc, anchor=south east] at (1.0,0.8) {$k_2$};
    \draw[acc, very thick, ->] (1.1,0.8) -- (2.0,1.5);
    \node[acc, anchor=west] at (1.7,1.25) {$k_3$};
    \draw[black, thick, ->] (2.0,1.5) -- (0.15,0.5);
    \node[black, anchor=south] at (1.2,1.15) {$G$};
    \draw[acc, thick, ->] (0,0) -- (0.15,0.5);
    \node[acc, anchor=east] at (0.1,0.35) {$k_1$};
    \node[anchor=south] at (0.3,-2.1) {Umklapp};
  \end{scope}
\end{tikzpicture}
$$

An Umklapp process requires phonons with wavevectors large enough that their sum
reaches the zone boundary — of order $k_D/2$. Exciting such phonons costs an
energy of order $k_B\Theta_D/2$, so the number of Umklapp-active phonons carries a
Boltzmann factor $e^{-\Theta_D/bT}$ (with $b$ of order $2$) that becomes tiny at
low temperature. This freeze-out of Umklapp scattering is the key to the
temperature dependence of the conductivity.

## Thermal conductivity

Treat the phonons as a gas carrying heat down a temperature gradient. Kinetic
theory gives the lattice thermal conductivity

$$
\kappa = \frac{1}{3}\,C\,v_s\,\ell,
$$

with $C$ the phonon heat capacity per unit volume, $v_s$ the mean phonon speed,
and $\ell = v_s\tau$ the phonon mean free path set by the scattering time $\tau$.
The temperature dependence of $\kappa$ is the competition between $C(T)$ and
$\ell(T)$, and it produces a characteristic peak.

- **High temperature ($T \gg \Theta_D$).** The heat capacity saturates at the
  Dulong–Petit value, so $C$ is constant. The number of phonons available for
  Umklapp scattering grows linearly with $T$, so the scattering rate $\propto T$
  and $\ell \propto 1/T$. Hence $\kappa \propto 1/T$: hotter crystals conduct heat
  worse, a signature that Umklapp phonon–phonon scattering dominates.
- **Low temperature ($T \ll \Theta_D$).** Umklapp processes freeze out
  exponentially, $\ell \propto e^{\Theta_D/bT}$, so the mean free path grows
  enormous and is eventually cut off by the sample boundaries or by defects,
  becoming constant. The heat capacity then controls $\kappa$, and with $C \propto
  T^3$ the conductivity rises as $\kappa \propto T^3$.

Between the rising $T^3$ regime and the falling $1/T$ regime the conductivity
reaches a maximum, typically at $T \sim \Theta_D/20$. The peak height and
position depend on sample size and purity, because those set where boundary
scattering takes over.

$$
% caption: Lattice thermal conductivity versus temperature (log-log schematic).
% At low T the conductivity rises as T cubed, tracking the heat capacity while
% the mean free path is limited by boundaries. At high T it falls as one over T,
% limited by Umklapp phonon-phonon scattering. The two regimes meet at a peak.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black!70] (0,0) -- (6.2,0) node[below right] {$T$};
  \draw[->, black!70] (0,0) -- (0,3.4) node[left] {conductivity};
  % rising T^3 branch then falling 1/T branch, peaked
  \draw[acc, very thick, domain=0.35:2.0, samples=80, variable=\t]
    plot ({\t},{0.42*\t*\t*\t});
  \draw[acc, very thick, domain=2.0:5.9, samples=120, variable=\t]
    plot ({\t},{3.36*2.0/\t - 0.14*\t + 0.28});
  % annotate branches
  \node[acc, anchor=west] at (0.55,1.35) {$T^3$};
  \node[acc, anchor=west] at (3.4,1.6) {$\frac{1}{T}$};
  \draw[black, dashed] (2.0,0) -- (2.0,3.35);
  \node[black, anchor=south] at (2.0,3.35) {peak};
\end{tikzpicture}
$$

## The mean free path and its limits

The mean free path is set by whichever scattering mechanism is shortest, through
a sum of rates (Matthiessen's rule):

$$
\frac{1}{\ell} = \frac{1}{\ell_U} + \frac{1}{\ell_{\text{def}}}
+ \frac{1}{\ell_{\text{bdy}}}.
$$

Each term dominates a different temperature range:

- **Umklapp scattering** $\ell_U \propto e^{\Theta_D/bT}$: dominant at high
  temperature, where it gives the $1/T$ falloff. It grows without bound as $T\to
  0$ and stops limiting $\ell$ at low temperature.
- **Point-defect and isotope scattering** $\ell_{\text{def}}$: roughly
  temperature-independent, set by the concentration of impurities and isotopic
  disorder. Rayleigh-like ($\propto \omega^{-4}$) scattering makes it weak for
  long-wavelength phonons, so it caps $\ell$ at intermediate temperatures.
- **Boundary scattering** $\ell_{\text{bdy}} \approx d$, the sample dimension:
  temperature-independent and the ultimate limit at the lowest temperatures, where
  the wavelength-long phonons scatter only off the crystal surfaces. In this
  regime $\kappa \propto C \propto T^3$ with a coefficient proportional to the
  sample size, one of the few transport quantities that depends on how large the
  crystal is.

$$
% caption: Phonon mean free path versus temperature (log-log schematic). The
% Umklapp-limited path (accent) diverges exponentially as T falls; the
% defect-limited and boundary-limited paths (light, dashed) are flat. The
% observed mean free path follows the lowest curve at each temperature.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black!70] (0,0) -- (6.2,0) node[below right] {$T$};
  \draw[->, black!70] (0,0) -- (0,3.4) node[left] {mean free path};
  % Umklapp: falling with T (diverges at low T) -> draw as decreasing curve
  \draw[acc, very thick, domain=0.6:5.9, samples=120, variable=\t]
    plot ({\t},{3.1/(\t*\t) + 0.15});
  \node[acc, anchor=west] at (1.4,1.75) {Umklapp};
  % boundary limit (flat, high)
  \draw[black, densely dashed] (0.3,2.6) -- (2.2,2.6);
  \node[black, anchor=south] at (1.1,2.62) {boundary};
  % defect limit (flat, lower)
  \draw[black, densely dashed] (2.0,1.35) -- (5.9,1.35);
  \node[black, anchor=south] at (4.3,1.37) {defect};
\end{tikzpicture}
$$

The two anharmonic phenomena share one cause. The cubic term in the potential
both shifts the equilibrium separation, producing thermal expansion, and couples
the phonon modes, producing the scattering that makes thermal conductivity
finite. A crystal with no anharmonicity would be a perfect, non-expanding,
infinite-conductivity idealization that no material realizes. With lattice
dynamics complete, the next module turns from the ions to the conduction
electrons and the [free-electron Fermi
gas](/condensed-matter/free-electron-fermi-gas/free-electron-gas-and-conduction).