Download Lattice Vibrations

Lattice Vibrations
Chris J. Pickard
500
400
-1
ω (cm )
300
200
L
100
K W
0
W
L
Γ
X
X
W
K
The Breakdown of the Static Lattice Model
• The free electron model was refined • The classical static lattice can only be
by introducing a crystalline external valid for T=0K
potential
• It is even wrong for T=0K:
∆x∆p ≥ h̄ ⇒ Zero point motion
• This allows much progress, but is not
the full story
• This is a particular problem for
insulators: unless kB T > Eg there
• Ions are not infinitely massive, nor are no degrees of freedom to account
held in place by infinitely strong forces for their many properties
Equilibrium Properties
cv
saturated
– the electronic degrees of freedom
alone cannot explain experiment
• Density and cohesive energy
– zero point motion is important for
solid neon and argon, and dominant
for solid helium (a quantum solid )
cubic (lattice)
linear (electronic)
T
• Specific heat of a metal:
• Thermal expansion of insulators
– electronic contribution negligible
for kbT < Eg
Transport Properties
• Conductivity
– no vibrations, no superconductors
– in a perfect metallic crystal there
are no collisions and perfect • Thermal conductivity of insulators
conduction
– the electronic degrees of freedom
– lattice vibrations provide the
are not sufficient
scattering mechanisms
• Superconductivity
– interaction between two electrons
via lattice vibrations
• Transmission of sound
– sound waves are carried
vibrations of the lattice
by
Interaction with Radiation
• Reflectivity of ionic crystals
– Sharp maximum in infrared, far
below h̄ω = Eg
– E-field applies opposite forces on ±
ions
• Inelastic scattering of light
– small frequency shifts (Brillouin or
Raman scattering)
– understood via lattice vibrations
• X-ray scattering
– thermal vibrations and zero point
motion diminish the intensity of the
peaks
– there is a background in directions
not satisfying the Bragg condition
• Neutron scattering
– momentum transfer with the lattice
is discrete, and provides a probe of
the lattice vibrations
A Classical Theory of the Harmonic Crystal
• A general treatment of the deviation
of ions from their equilibrium positions
is intractable, so proceed in stages:
Bravais lattice remains as an average
of the instantaneous configurations
1. Treat small deviations classically • Denote the position of an atom whose
2. Proceed to a quantum theory
mean position is R by r(R):
3. Examine implications of larger r(R) = R + u(R)
movements
• To treat the small deviations, we • The dynamics of the lattice is
governed by the classical Hamiltonian:
assume each ion stays in the vicinity P
P(R)2
of its equilibrium position R, and the
R 2M + U
The Harmonic Approximation
• 3D Taylor expand the potential energy
around the equilibrium configuration:
f (r + a) = f (r) + a · ∇f (r)+
1
2
3
(a
·
∇)
f
(r)
+
O(a
)
2
U
• At equilibrium the net force is zero,
and the potential energy is given by:
U = U eq + U harm
r
U
harm
=
1
2
P
RR0 ,µν
• The general form for U harm is:
uµ(R)Dµν (R − R0)uν (R0)
The Adiabatic Approximation
• The quantities D in the harmonic
expansion are in general very difficult
to calculate
configuration ⇒ the wavefunctions
change as the ions move
• Make the adiabatic approximation by
• In ionic crystals the difficulties are the separating the typical timescales of
long ranged coulomb interactions
the motion of the electrons and ions
• In covalent/metallic crystals the – the electrons are in their
groundstate for any configuration
difficulty comes from the fact that the
contribution to the total energy of the
valence electrons depends on the ionic • D is still difficult to calculate
Normal Modes of a 1D Bravais Lattice
K
M
K
M
K
M
K
M
K
M
K
M
K
M
K
M
Masses M and springs K
M
• Consider ions of mass M separated by
distance a
• For simplicity,
assume
neighbour interactions only
u([N+1]a) = u(a) ; u(0) = u(Na)
Born-von Karman BCs
nearest
• In the Harmonic approximation, this
is equivalent to masses connected by
springs of strength K:
harm
M ü(na) = − ∂U
, U harm =
∂u(na)
P
1
2
K
[u(na)
−
u((n
+
1)a)]
n
2
Normal Modes of a 1D Bravais Lattice
ω( k)
• Seek solutions of the form:
u(na, t) ∝ ei(kna−ωt)
2Κ
Μ
• The PBCs ⇒ eikN a = 1 ⇒ k =
with n integer, N solutions and
− πa ≤ k < πa
2π n
a N
• Substitution into
qthe dynamical eqn.
K
|sin(ka/2)|
gives: ω(k) = 2 M
−π/a
0
π/a
k
• The group and phase velocities differ
substantially at the zone boundaries
Normal Modes of a 1D Bravais Lattice with a Basis
ω( k)
O
2(K+G)
Μ
2K
Μ
2G
Μ
A
−π/a
0
π/a
k
• The analysis can be repeated
• The are 2 solutions for each k ⇒ 2N
solutions in total:
√
K+G
1
2
ω = M ± M K 2 + G2 + 2KGcos ka
• There are acoustic and optical modes
O
A
Normal Modes of a 3D Bravais Lattice
ω( k)
• The
matrix is D(k) =
P dynamical
−ik·R
, where D(R − R0)
R D(R)e
is the second derivative of U with
respect to the displacement of ions at
R and R0 at eqbm..
L
• The solution of the dynamical
equation is given by the eigenequation M ω 2e = D(k)e, where e
is the polarization vector
T2
T1
0
k
• 3N solutions for each ion in the basis
Normal Modes of a Real Crystal
• The dynamical matrix can be built up
from first principles calculations
500
400
• Can use a supercell approach to study
certain high symmetry k-vectors
-1
ω (cm )
300
• For arbitary k use linear response
theory – a perturbation theory
200
L
100
K W
0
W
L
Γ
X
X
W
Dispersion curves for Silicon
K
Connections with the theory of Elasticity
• The classical theory of elasticity slowly over the atomic length scale
ignores the microscopic atomic
structure
• Using the symmetries
of D:
P
Uharm = 14 RR0
• The continuum theory of elasticity can [u(R0) − u(R)]D(R − R0)[u(R0) − u(R
be derived from the theory of lattice
vibrations
• Slowly varying displacements ⇒
u(R0) =
• Consider displacements that vary u(R) + (R0 − R) · ∇u(r)|r=R
³
´³
´
P
P
∂u
(
R
)
∂u
(
R
)
µ
1
1
ν
harm
Eσµτ ν Eσµτ ν = − 2 R Rσ Dµν (R)Rτ
U
= 2 R,µνστ
∂xσ
∂xτ
A Quantum Theory of the Harmonic Crystal
• In a Quantum theory the system can
be in a set of discrete stationary states
independent oscillators ⇒ the 3N
classical normal modes
• These stationary states are the •
eigenstates
of
the
harmonic
P
P (R)2
harm
= R 2M +
Hamiltonian:
H
P
1
0
0
RR0 u(R)D(R − R )u(R )
2
•
• The result is: an N -ion harmonic
crystal is equivalent to 3N
The energy in each mode is discrete,
and the
is:
P total energy
E = ks(nks + 12 )h̄ωs(k)
The integer nks is the excitation
number of the normal mode in branch
s at wave vector k
Normal Modes or Phonons
• So far we have described the state in • Instead of saying that the mode k,s is
in the nks excited state we say there
terms of the excitation number nks
are nks phonons of type s with wave
• This is clumsy if describing processes vector k
involving the exchange of energy
– between normal modes, or other • Photons ⇒ of the correct frequency
systems (electrons, neutrons or X- are visible light
Phonons ⇒ of the correct frequency
rays)
are sound
• As for the QM theory of the EM
field we use an equivalent corpuscular • Don’t forget phonons/normal modes
are equivalent
description
Classical Specific Heat: Dulong-Petit
• The thermal energy density is given •
by averaging over all configurations
weighted by e−βE with β = k 1T
B
R
−βH
H
dΓe
•
u = V1 R dΓe−βH
R
1 ∂
= − V ∂β ln dΓe−βH
The thermal energy density is:
u = ueq + 3nkB T, (n = N/V )
The specific heat is independent of T :
cv = 3nkB
• This is not observed experimentally
– only approximately at high
• By
a change of variables:
temperature where the harmonic
R making
eq
dΓe−βH = e−βU β −3N × constant approximation is bad anyway
The Quantum Mechanical Lattice Specific Heat
• The QM
thermal energy
density is:
P
P
u = V1 i Eie−βEi / i e−βEi
P −βE
1 ∂
i
= − V ∂β ln i e
• The energy
density is :
P
u = V1 ks h̄ωs(k)[ns(k) + 12 ]
• The mean excitation number ns(k) =
• The sum is over the stationary states eβh̄ωs1(k)−1 is the Bose-Einstein
with energy:
distribution function
P
1
i
Ei = ks(nks + 2 )h̄ωs(k),
niks = 0, 1, 2, . . .
• The specific
heat is given by:
P
P −βE
−βh̄ωs (k)/2
h̄ωs k
e
∂
1
i = Π
e
=
c
k
s
v
−βh̄ω
(
k
)
βh̄ω
i
ks ∂T e
s
s (k) −1
V
1−e
The High-Temperature Lattice Specific Heat
• When kB T À h̄ωs(k) all the normal
modes are highly excited
cv =
1
V
P
∂
ks ∂T kB T
=
3N
V kB
• Writing βh̄ωs(k) = x, then x is small • The next term is constant in T
and we can expand:
1
1
x
x2
3
• We might try to correct the Dulong=
[1
−
+
+
O(x
)]
x
e −1
x
2
12
Petit law, but anharmonic terms
• Keeping just the leading term we are likely to dominate where the
regain the Dulong-Petit law:
expansion holds – or the crystal melts!
ω( k)
L
T2
T1
0
k
The Low-Temperature Lattice Specific Heat
• In the limit of a large crystal integrate
over the P
1st RBrillouin zone:
h̄ωs (k)
∂
dk
cv = ∂T
3
βh̄ω
i
s (k) −1
(2π) e
1. Ignore the optical modes
2. Use the dispersion relationship:
ωs(k) = cs(k̂)k
3. Integrate over all k
• Modes with h̄ωs(k) À kB T will not
contribute – but the acoustic branches
will at long enough wavelengths for • Making the substitution βh̄cs(k̂)k =
x we obtain, and c as the average
any T
speed of sound:
• Make some approximations:
cv =
∂
∂T
const ×
(kB T )4
(h̄c)3
∝ T3
Intermediate Temperature: The Debye and Einstein
Models
• The T 3 relation only remains valid
while the thermal energy is small
compared to the energy of phonons
with a non-linear dispersion (much •
lower than room temperature)
acoustic modes, all with ω = ck, and
integrate up to kD
Einstein: optical modes represented
by modes of ωE
• The Debye and Einstein models • The Debye temperature divides the
approximate the dispersion relations
quantum and classical statistical
regimes:
√
3
• Debye: all branches modelled by 3 kB ΘD = h̄ωD = h̄ckD = h̄c 6π 2n
Measuring Phonon Dispersion Relations
• Normal mode dispersion relations phonon
ωs(k) can be extracted from
experiments in which lattice vibrations
• The same applies or X-rays or visible
exchange energy with an external
light
probe
• Energy lost (or gained) by a neutron • Neutrons carry more momentum than
⇒ emission (or absorption) of a photons in the energy range of interest
Neutron Scattering by a Crystal
• Neutrons only interact strongly with ∆nks = n0ks − nks
the atomic nuclei, and so will pass
through a crystal, possibly with a • The
conservation
of
crystal
2
changed E = p /2Mn and p
momentum:
P
p0 − p = ks h̄k∆nks + K
• Conservation laws allow the extraction
of information from the scattering
• This is the same crystal momentum as
for the Bloch states – important for
theories of electron-phonon scattering
• The conservation
of energy:
P
E 0 = E − ks h̄ωs(k)∆nks,
• Different numbers of phonons can be
involved in a scattering event
Zero Phonon Scattering
• The final state is identical to the initial
state
changes by h̄K: q0 = q + K
• These are just the von Laue conditions
• Energy conservation implies that the
energy of the neutron is unchanged
• We
can
extract
the
same
(elastically scattered): q 0 = q
crystallographic information of the
• Crystal momentum conservation static lattice as from X-ray diffraction
implies that the neutron’s momentum experiments
One Phonon Scattering
• The situation where one phonon is • For the absorption case:
p02
p2
p 0 −p
absorbed or emitted conveys the most
2Mn = 2Mn + h̄ωs ( h̄ )
information
• In an experiment we control p and E
• The conservation laws imply:
E 0 = E ± h̄ωs(k)
p0 = p ± h̄k + h̄K
• We can choose a direction in which to
measure, and record the energy E 0 to
map out the dispersion curves ωs(k)
• The additive K can be ignored • Multi-phonon
scattering
because ωs(k + K) = ωs(k)
produce a background
events