Phonons
April 21, 2022
1
, One-dimensional Monatomic Chain
Figure 1: A one-dimensional monatomic chain.
Phonons are defined to be quantized lattice vibrations in a crystal. To understand how phonons work,
we first consider a one-dimensional atomic lattice comprised of chains of an atom of mass m that are
periodically arranged in space with lattice constant a. We visualize the atoms in this one-dimensional
lattice to be harmonic oscillators, connected to their nearest neighbors with a ”spring” with spring
constant κ. If we consider lattice sites un−1 , un , and un+1 , the classical equation of motion for the
system is given by
d2 un
m = κ(un+1 − un ) + κ(un−1 − un ) = κ(un+1 + un−1 − 2un ) (1)
dt2
Since the lattice is periodic, the solution to this differential equation can be written as
ukn (t) = Ak ei(kna−ωt) (2)
where k is the wave vector and ω is the angular frequency. Substituting equation (2) back into
equation (1), we get
−mω 2 Ak ei(kna−ωt) = κAk ei[k(n+1)a−ωt] + ei[k(n−1)a−ωt] − 2ei(kna−ωt)
−mω 2 Ak ei(kna−ωt) = κAk ei(kna−ωt) (eika + e−ika − 2)
−mω 2 = κ(eika + e−ika − 2) (3)
−mω 2 = κ[2(cos(ka)) − 2]
−mω 2 = 2κ[cos(ka) − 1]
We use the trigonometric identities
x
1 − cos(x) = 2sin2 (4)
2
x
1 + cos(x) = 2cos2 (5)
2
to get
ka
− mω 2 = −4κsin2 (6)
2
Thus, we get an expression for the dispersion relation given by
r
κ ka
ω(k) = 2 sin (7)
m 2
We note the following Taylor series expansion:
2
April 21, 2022
1
, One-dimensional Monatomic Chain
Figure 1: A one-dimensional monatomic chain.
Phonons are defined to be quantized lattice vibrations in a crystal. To understand how phonons work,
we first consider a one-dimensional atomic lattice comprised of chains of an atom of mass m that are
periodically arranged in space with lattice constant a. We visualize the atoms in this one-dimensional
lattice to be harmonic oscillators, connected to their nearest neighbors with a ”spring” with spring
constant κ. If we consider lattice sites un−1 , un , and un+1 , the classical equation of motion for the
system is given by
d2 un
m = κ(un+1 − un ) + κ(un−1 − un ) = κ(un+1 + un−1 − 2un ) (1)
dt2
Since the lattice is periodic, the solution to this differential equation can be written as
ukn (t) = Ak ei(kna−ωt) (2)
where k is the wave vector and ω is the angular frequency. Substituting equation (2) back into
equation (1), we get
−mω 2 Ak ei(kna−ωt) = κAk ei[k(n+1)a−ωt] + ei[k(n−1)a−ωt] − 2ei(kna−ωt)
−mω 2 Ak ei(kna−ωt) = κAk ei(kna−ωt) (eika + e−ika − 2)
−mω 2 = κ(eika + e−ika − 2) (3)
−mω 2 = κ[2(cos(ka)) − 2]
−mω 2 = 2κ[cos(ka) − 1]
We use the trigonometric identities
x
1 − cos(x) = 2sin2 (4)
2
x
1 + cos(x) = 2cos2 (5)
2
to get
ka
− mω 2 = −4κsin2 (6)
2
Thus, we get an expression for the dispersion relation given by
r
κ ka
ω(k) = 2 sin (7)
m 2
We note the following Taylor series expansion:
2
