The linear term vanishes because the expansion is about equilibrium. The
harmonic approximation keeps the quadratic term and discards everything
beyond it, so that
\begin{equation}
U \propto u^{2}, \qquad F = -\frac{\partial U}{\partial u} \propto u ,
\end{equation}
which is Hooke's law. This is legitimate when the displacements are small —
that is, at low temperature. Real amplitudes at room temperature are a few
percent of the interatomic spacing, so the approximation is good but not exact.
What the harmonic approximation throws away
A purely harmonic crystal would have:
no thermal expansion — the potential is symmetric about
\(R_{0}\), so raising the temperature increases the amplitude but not the mean
separation;
no thermal resistance — normal modes of a harmonic system do
not scatter off one another, so heat would propagate ballistically and the
thermal conductivity would be infinite;
temperature-independent elastic constants, and no phonon
lifetime.
Two further preliminaries. Each atom in three dimensions has three degrees of
freedom, so it behaves as a three-dimensional harmonic oscillator. And the
vibrations are coupled: heating one end of a rod raises the amplitude
there, and because each atom is tied to its neighbours by springs the
disturbance travels. In a metal the free electrons carry most of the heat and it
travels fast; in an insulator only the lattice can carry it, so it travels more
slowly — but it does travel, which is the observable consequence of the
coupling.
The one-dimensional monatomic chain
Take a chain of identical atoms of mass \(m\), spaced \(a\) apart, each joined to its
nearest neighbours by springs of force constant \(c\) (the figure above).
Let \(u_{n}\) be the displacement of the \(n\)-th atom from its equilibrium position.
The one-dimensional monatomic chain: identical
masses \(m\) at spacing \(a\), joined by springs of force constant \(c\). Only
nearest-neighbour interactions are retained. The displacements \(u_{n}\) are
measured along the chain (longitudinal), and are drawn exaggerated.
The force on atom \(n\) comes from the two springs attached to it. The spring on
the right is stretched by \(u_{n+1}-u_{n}\) and the one on the left by
\(u_{n}-u_{n-1}\), so
This is one equation for every atom in the chain — \(10^{23}\) coupled ordinary
differential equations. The trick that decouples them is to look for solutions in
which every atom executes the same motion, delayed by a phase that advances
uniformly along the chain:
\begin{equation}
u_{n} = u_{0}\,e^{i\left(\omega t - kna\right)} ,
\end{equation}
a travelling wave of amplitude \(u_{0}\), angular frequency \(\omega\) and wavevector
\(k\). The position of atom \(n\) enters only through \(na\), so
Substituting into the relation above, the common factor \(u_{n}\) cancels — which
is the whole point of the ansatz — and leaves a single algebraic equation:
This is the dispersion relation of the monatomic chain. Note \(\omega_{m}\),
the maximum frequency, reached at \(k = \pi/a\).
The dispersion relation
\(\omega = \omega_{m}\left|\sin(ka/2)\right|\) for the monatomic chain. It is
periodic in \(k\) with period \(2\pi/a\), so all the physical content lies in the
first Brillouin zone \(-\pi/a \le k \le \pi/a\) (shaded); the repeated zones
(grey) carry no new information. At small \(k\) the curve is linear,
\(\omega \simeq v_{s}k\), and the chain behaves as a continuous elastic medium.
At the zone boundary the curve is flat, the group velocity vanishes, and the
mode is a standing wave.
The long-wavelength limit: sound
For \(ka \ll 1\) — long wavelength, small \(k\), low frequency — we may write
\(\sin(ka/2) \simeq ka/2\), so
Group and phase velocities are equal and independent of \(k\): the chain is
non-dispersive, behaving exactly like a continuous elastic medium in which
sound travels at speed \(v_{s}\). Indeed, treating the chain as a rod of linear
density \(\rho = m/a\) and elastic modulus \(s = ca\),
When \(\lambda \gg a\) the wave cannot resolve the individual atoms; it sees only
a continuum with a certain stiffness and a certain density. Every crystal
therefore has a branch whose frequency goes to zero linearly as \(k\to0\), and that
branch is sound. This is the acoustic branch, and its slope at the
origin is the velocity of sound.
The zone-boundary limit: standing waves
At the other extreme, take \(k\) towards \(\pi/a\). The group velocity
Meanwhile the phase velocity \(v_{p} = \omega/k\) is still finite, so \(v_{p} \neq
v_{g}\): the chain has become strongly dispersive.
Physically, a wave of wavelength \(2a\) makes adjacent atoms move in exact
antiphase. Such a disturbance cannot propagate — it is Bragg-reflected by the
lattice, exactly as an X-ray at the Bragg condition of Chapter is
— and the forward and backward waves superpose into a standing wave.
Energy is not transported, which is what \(v_{g} = 0\) means.
\(v_{g}=0\) does not mean \(\omega=0\)
At the zone boundary the frequency is at its maximum,
\(\omega = \omega_{m}\), while the group velocity is zero. Students routinely
confuse “the wave does not travel” with “the atoms do not move”. The atoms
are moving as fast as they ever do; it is the energy that stands still.
Why \(k\) is restricted to the first Brillouin zone
The relation above is periodic in \(k\) with period \(2\pi/a\). Replacing
\(k \to k + 2\pi/a\) leaves \(\omega\) unchanged, and more importantly leaves the
motion unchanged:
\begin{equation}
e^{-i\left(k+2\pi/a\right)na} = e^{-ikna}\,e^{-2\pi i n} = e^{-ikna},
\end{equation}
since \(n\) is an integer. Two wavevectors differing by a reciprocal lattice vector
\(G = 2\pi/a\) describe identical sets of atomic displacements. They are not
two waves; they are one wave with two labels.
This is a genuine difference from a continuous medium, and it arises because a
lattice samples the wave only at discrete points. There is no physical content in
the shorter wavelength — nothing exists between the atoms to be displaced. The
convention is therefore to keep only
\begin{equation}
\boxed{\;-\frac{\pi}{a} < k \le \frac{\pi}{a}\;}
\end{equation}
which is the first Brillouin zone of Chapter, now arrived
at from a completely different direction.
Why wavevectors outside the first Brillouin
zone are redundant. The blue wave (\(k = \pi/4a\)) and the red wave
(\(k' = k - 2\pi/a\), outside the zone) prescribe identical displacements at
every lattice site. Since only the atoms exist, the two are physically the
same mode. This is the lattice version of aliasing in signal sampling.
Worked example 4.1. Reading a dispersion relation
The dispersion relation of a one-dimensional monatomic crystal of lattice spacing
\(a\) is \(\omega = A\left|\sin(ka/2)\right|\), where \(A\) is a constant. Find (i) the
group velocity at the zone boundary and (ii) the force constant, given that the
atoms have mass \(M\).
(i) \(v_{g} = \mathrm{d}\omega/\mathrm{d}k = (Aa/2)\cos(ka/2)\), which is
\(\boxed{0}\) at \(k = \pi/a\). The frequency is maximum there, so its derivative
vanishes — no calculation is really needed.
(ii) Comparing with the relation above, \(A = \sqrt{4c/M}\), so
\[
c = \frac{MA^{2}}{4} .
\]
Worked example 4.2. Phase and group velocity in the long-wavelength limit
For a monatomic chain with
\(\omega(k) = \sqrt{\dfrac{2C}{M}\left[1-\cos(ka)\right]}\), find the ratio of
phase to group velocity when \(\lambda \gg a\).
For small \(ka\), \(\cos(ka) \simeq 1 - (ka)^{2}/2\), so
Then \(v_{p} = \omega/k = a\sqrt{C/M}\) and
\(v_{g} = \mathrm{d}\omega/\mathrm{d}k = a\sqrt{C/M}\), so
\[
\frac{v_{p}}{v_{g}} = 1 .
\]
The ratio is unity for any branch that is linear through the origin; a
value different from \(1\) is the signature of dispersion.
Worked example 4.3. A three-dimensional dispersion relation
The phonon dispersion of a cubic solid is
\(\omega^{2}(\vec{k}) = \omega_{0}^{2}\left(3 - \cos k_{x}a - \cos k_{y}a
- \cos k_{z}a\right)\). Find the sound velocity.
For long wavelengths all three \(k_{i}a\) are small, and
\(\cos k_{i}a \simeq 1 - k_{i}^{2}a^{2}/2\). Each bracket contributes
\(k_{i}^{2}a^{2}/2\), so
Beware the near-miss alternatives: \(2\omega_{0}\sin^{2}(ka/2)\) has the wrong
dimensions of curvature at the origin (it is quadratic, not linear, in \(k\), so it
would give zero sound velocity), and \(2\omega_{0}\left[1-\cos(ka/2)\right]\) fails
the same test.
The Diatomic Chain: Acoustic and Optical Branches
Put two different atoms in each primitive cell and something qualitatively new
appears. The single branch of the section on the Harmonic Approximation and the Monatomic Chain splits into two, and the upper
one does not go to zero at \(k=0\): there is a whole family of vibrations whose
frequency stays finite however long the wavelength. These are the optical
modes, and they are responsible for the infrared absorption of ionic crystals,
for Raman scattering, and for a large fraction of the questions asked about this
chapter.
Setting up
The diatomic chain. Two different masses
alternate along the line with nearest-neighbour separation \(a\), so the
primitive cell has length \(2a\) and contains one atom of each kind. All springs
are identical. Compare the figure above, where the nearest-neighbour
distance and the lattice constant were the same thing.
Which length is called \(a\)?
This is the single most common source of wrong answers in this section, because
two conventions are in circulation and they give different-looking formulae for
the same physics.
This book: \(a\) = nearest-neighbour
Alternative: \(a\) = lattice constant
primitive cell
\(2a\)
\(a\)
dispersion
\(\sin^{2}(ka)\)
\(\sin^{2}(ka/2)\)
zone boundary
\(k = \pi/2a\)
\(k = \pi/a\)
sound velocity
\(a\sqrt{\dfrac{2c}{m_{1}+m_{2}}}\)
\(a\sqrt{\dfrac{c}{2\left(m_{1}+m_{2}\right)}}\)
The two are related by \(a_{\text{alt}} = 2a\), and substituting that into either
column reproduces the other. Read the figure or the phrase “lattice
parameter” in the question before substituting, and check your answer against
the zone-boundary position quoted there. Every result below is stated in the
left-hand convention.
Label the light atoms (\(m_{1}\)) with even indices and the heavy atoms (\(m_{2}\))
with odd ones, as in the figure above. Each atom is still pulled by its two
neighbours, so
Because the two sublattices are physically different, they need
independent amplitudes:
\begin{equation}
u_{2n} = A\,e^{i\left(\omega t - 2kna\right)},
\qquad
u_{2n+1} = B\,e^{i\left(\omega t - \left(2n+1\right)ka\right)} .
\end{equation}
Two amplitudes, not one
Writing a single amplitude for both sublattices is the commonest set-up error
here, and it destroys the problem: the optical branch exists precisely
because \(A\) and \(B\) can differ in magnitude and in sign. The monatomic
ansatz the relation above had one amplitude because all atoms were identical.
Substituting the relation above into the relations above and
cancelling the common exponentials gives two coupled algebraic equations:
\begin{align}
-m_{1}\omega^{2}A &= 2c\left(B\cos ka - A\right), \\
-m_{2}\omega^{2}B &= 2c\left(A\cos ka - B\right),
\end{align}
Linear in \(k\): sound again, exactly as for the monatomic chain. Note that only
the total mass per cell enters — at long wavelength the wave cannot
resolve the two sublattices.
Optical branch, \(k\to0\)
Here the square root tends to \(\left(1/m_{1}+1/m_{2}\right)\), so
where \(\mu = m_{1}m_{2}/(m_{1}+m_{2})\) is the reduced mass. This is a
finite constant: the optical branch starts at a non-zero frequency and
is nearly flat near \(k=0\).
Zone boundary, \(k = \pi/2a\)
Here \(\sin^{2}(ka) = 1\), and the discriminant collapses neatly:
At the zone boundary each branch depends on one mass only. The optical
branch is governed by the lighter atom, the acoustic branch by the heavier one.
This is the direct consequence of the eigenvectors, examined below.
The two branches of the diatomic
chain, drawn for \(m_{2} = 3m_{1}\). The acoustic branch (red) rises linearly
from the origin and saturates at \(\sqrt{2c/m_{2}}\); the optical branch (blue)
starts at \(\sqrt{2c/\mu}\) and falls to \(\sqrt{2c/m_{1}}\). Between
\(\sqrt{2c/m_{2}}\) and \(\sqrt{2c/m_{1}}\) lies a band of frequencies that the
chain cannot propagate at all. Setting \(m_{1}=m_{2}\) closes the gap and
recovers the figure above folded into the smaller zone.
The frequency gap
Between \(\omega_{-}(\pi/2a) = \sqrt{2c/m_{2}}\) and
\(\omega_{+}(\pi/2a) = \sqrt{2c/m_{1}}\) no mode exists. A wave driven at a
frequency in this window cannot propagate; it is attenuated exponentially. The
gap width is
The gap is not proportional to \(\sqrt{m_{2}/m_{1}}\)
It is the ratio of the two zone-boundary frequencies that equals
\(\sqrt{m_{2}/m_{1}}\). The gap itself is the difference in
the relation above, which is not proportional to that ratio — it also depends on
the absolute masses. The two statements agree only on the qualitative trend.
That trend is worth stating carefully, because it is asked directly. Increasing
\(m_{2}\) lowers \(\omega_{-}\) and widens the gap; increasing \(m_{1}\) lowers
\(\omega_{+}\) and narrows it. When \(m_{1} = m_{2}\) the gap closes
completely — as it must, since the chain is then monatomic and the “two”
branches are just the single monatomic branch drawn in a zone half as wide. The
optical branch is the part of the monatomic curve between \(\pi/2a\) and \(\pi/a\),
folded back.
which is small when \(m_{2} \gg m_{1}\): heavy-plus-light crystals have nearly
flat, nearly dispersionless optical branches.
What the atoms actually do
The frequencies come from the determinant of the relation above; the
motions come from its eigenvectors. Substituting each root back into the
first row gives the amplitude ratio \(B/A\).
Atomic motions in the three cases
that examinations ask about. Vertical guides mark the equilibrium sites, arrows
show the instantaneous displacements. In the long-wavelength acoustic mode
(a) both sublattices move together, which is why only \(m_{1}+m_{2}\) appears in
the sound velocity. In the long-wavelength optical mode (b) they move in
opposition about a fixed centre of mass, which is why the reduced mass \(\mu\)
appears. At the zone boundary (c) one sublattice is completely at rest.
The four eigenvectors worth memorising
\(k\to0\)
\(k = \pi/2a\)
Frequency there
Acoustic
\(A=B\): in phase, cell translates
\(A=0\): light atoms at rest
\(\sqrt{2c/m_{2}}\)
Optical
\(m_{1}A+m_{2}B=0\): antiphase
\(B=0\): heavy atoms at rest
\(\sqrt{2c/m_{1}}\)
Mnemonic: at the zone boundary, each branch belongs to the mass that is
still moving
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