5.2 Atomic Case
87
ω(q) =
2k
m
(1 − cos qa)
= 2
k
m
sin
qa
2
.
(5.6)
The amplitude r
0 is, in principle, also a function of q, but has no meaning in this
simplest case. Owing to the cyclic boundary condition, Nqa must be an integer multiple of 2π, i.e., Nqa = 2nπ (n ∈ Z). Besides, spatial arrangements of displacements
corresponding to q and q +
2π
a
are the same because of
r
0 exp
i
−ωt + la
q +
2π
a
= r
0 exp[i(−ωt + laq)].
(5.7)
Thus, we can set n max − n min = N − 1. Considering that the total number of motional
degrees of freedom equals N , we restrict the range of n as −N /2 < n ≤ N /2. The
range of q is
−
π
a
< q ≤
π
a
.
(5.8)
This range of q is an example of the so-called first Brillouin zone, the region consisting
of q, of which the nearest lattice point is the origin (|q| = 0) in the q-space.
Since the equation of motion (Eq. 5.2) is linear, a general solution of r l (t) is given
by an arbitrary superposition of r l (q). Namely,
r l (t) =
q
w(q) exp[i(−ω(q)t + laq)],
(5.9)
where w(q) is a complex weight determined by the initial condition (according to
the spirit of deterministic mechanics). Each vibrational mode represents a single
motional degree of freedom. Thus, the so-called equipartition law (of energy) determines the magnitude (absolute value) of the weight w(q). The equipartition law (per
motional degree of freedom) holds for atoms because the numbers of atoms and
independent modes coincide with each other.
A standard procedure of quantization reveals that ω and q are energy and
momentum of a phonon, a quasi-particle of lattice vibrations. Thus, Eq. 5.6 is
regarded as a relation between energy and momentum. Such a relation between
energy and momentum is called a dispersion relation. The main issue of lattice
dynamics is to find dispersion relations for lattice vibration. Reminding the dispersion relation for a free particle, (energy) = (momentum)
2
/2(mass), we assume that
a quantity corresponding to the mass of a quasi-particle is given by (∂
2
ω/∂q
2
)
−1 .
It is interesting to see that this effective mass is dependent on momentum.
1
1 Because of ω ∝ q for small q from Eq. 5.6, this definition encounters difficulty. The mass of
acoustic phonon at q = 0 is instead regarded as zero after the Dirac equation.
87
ω(q) =
2k
m
(1 − cos qa)
= 2
k
m
sin
qa
2
.
(5.6)
The amplitude r
0 is, in principle, also a function of q, but has no meaning in this
simplest case. Owing to the cyclic boundary condition, Nqa must be an integer multiple of 2π, i.e., Nqa = 2nπ (n ∈ Z). Besides, spatial arrangements of displacements
corresponding to q and q +
2π
a
are the same because of
r
0 exp
i
−ωt + la
q +
2π
a
= r
0 exp[i(−ωt + laq)].
(5.7)
Thus, we can set n max − n min = N − 1. Considering that the total number of motional
degrees of freedom equals N , we restrict the range of n as −N /2 < n ≤ N /2. The
range of q is
−
π
a
< q ≤
π
a
.
(5.8)
This range of q is an example of the so-called first Brillouin zone, the region consisting
of q, of which the nearest lattice point is the origin (|q| = 0) in the q-space.
Since the equation of motion (Eq. 5.2) is linear, a general solution of r l (t) is given
by an arbitrary superposition of r l (q). Namely,
r l (t) =
q
w(q) exp[i(−ω(q)t + laq)],
(5.9)
where w(q) is a complex weight determined by the initial condition (according to
the spirit of deterministic mechanics). Each vibrational mode represents a single
motional degree of freedom. Thus, the so-called equipartition law (of energy) determines the magnitude (absolute value) of the weight w(q). The equipartition law (per
motional degree of freedom) holds for atoms because the numbers of atoms and
independent modes coincide with each other.
A standard procedure of quantization reveals that ω and q are energy and
momentum of a phonon, a quasi-particle of lattice vibrations. Thus, Eq. 5.6 is
regarded as a relation between energy and momentum. Such a relation between
energy and momentum is called a dispersion relation. The main issue of lattice
dynamics is to find dispersion relations for lattice vibration. Reminding the dispersion relation for a free particle, (energy) = (momentum)
2
/2(mass), we assume that
a quantity corresponding to the mass of a quasi-particle is given by (∂
2
ω/∂q
2
)
−1 .
It is interesting to see that this effective mass is dependent on momentum.
1
1 Because of ω ∝ q for small q from Eq. 5.6, this definition encounters difficulty. The mass of
acoustic phonon at q = 0 is instead regarded as zero after the Dirac equation.