Lecture 7
Crystal Vibrations
… and Neutron Scattering
7.1 Vibrations of crystals with monatomic basis
7.2 Two atoms per primitive basis
7.3 Quantization of elastic waves
7.4 Phonon momentum
7.5 Inelastic neutron scattering for phonons
References:
1) Kittel, Chapter 4
2) Marder, Chapter 13
3) Ashcroft, Chapters 22, 24
4) Burns, Chapters 12
5) Ziman, Chapter 2
6) Ibach, Chapter 6
Lecture 7
1
Longitudinal and transverse waves
Planes of atoms as displaces for longitudinal and transverse waves
Lecture 7
2
1
Continuous Elastic Solid
We can describe a propagating vibration of amplitude u along a rod of material
with Young’s modulus Y and density ρ with the wave equation:
∂ 2 u Y ∂ 2u
=
∂t 2
ρ ∂x 2
for wave propagation along the x-direction
By comparison to the general form of the 1-D wave equation:
∂ 2u
∂ 2u
= v2 2
2
∂t
∂x
ω = 2πf = 2π
group velocity
v=
we find that
v
λ
So the wave speed is independent of
wavelength for an elastic medium!
ρ
ω (k ) is called the
dispersion relation of
the solid, and here it is
linear (no dispersion!)
ω
= kv
dω
dk
vg =
Y
k
Lecture 7
3
7.1 Vibrations of Crystals with Monatomic Basis
By contrast to a continuous solid, a real solid is not uniform on an atomic scale,
and thus it will exhibit dispersion. Consider a 1-D chain of atoms:
M
a
In equilibrium:
s −1
s+ p
s +1
s
Longitudinal wave:
u s −1
Fs =
For atom s
(for plane s)
∑ c (u
p
p
s+ p
− us )
us
u s +1
us + p
p = atom label
p = ± 1 nearest neighbors
p = ± 2 next nearest neighbors
cp = force constant for atom p
Lecture 7
4
2
The equation of motion of the plane s
Elastic response of the crystal is a linear function of the forces (elastic energy is
a quadratic function of the relative displacement)
The total force on plane s comes from planes s ± 1
FS = C (uS+1 – uS) + C (uS - 1 – uS)
C – the force constant between nearest-neighbor planes (different for
transverse and longitudinal waves)
Equation of motion
of the plane s:
M
d 2u S
= C (u S +1 + u S −1 − 2u S )
dt 2
Solutions with all displacements having time dependence exp(-iωt). Then
d 2u S
= −ω 2u S
dt 2
− Mω 2u S = C (u S +1 + u S −1 − 2u S ) This is a difference equation in
the displacements u
u S ±1 =Lecture
ue isKa7 e ± iKa
5
Equation of Motion for 1D Monatomic Lattice
Applying Newton’s second law: Fs = M
u s = uei ( kxs −ωt )
For the expected harmonic
traveling waves, we can write
Thus:
∂ 2us
= ∑ c p (u s + p − u s )
∂t 2
p
xs = sa position of atom s
(
Mu ( −iω ) 2 e i ( ksa −ωt ) = ∑ c p ue i ( k ( s + p ) a −ωt ) − ue i ( ksa −ωt )
)
p
Or:
(
)
− Mω 2 e i ( ksa −ωt ) = e i ( ksa −ωt ) ∑ c p e ikpa − 1
p
So:
(
= ∑ c (e
)
− Mω 2 = ∑ c p e ikpa − 1
p
− Mω 2
p >0
p
ikpa
Now since c-p = cp by symmetry,
)
+ e − ikpa − 2 = ∑ 2c p (cos( kpa ) − 1)
Lecture 7
p >0
6
3
Dispersion relation of the monatomic 1D lattice
The result is:
ω2 =
2
M
∑c
p >0
p
(1 − cos(kpa )) =
Often it is reasonable to make the
nearest-neighbor approximation (p = 1):
ω2 ≅
4
M
∑c
p >0
p
sin 2 ( 12 kpa)
4c1
sin 2 ( 12 ka)
M
ω
The result is periodic in k
and the only unique
solutions that are physically
meaningful correspond to
values in the range: − π ≤ k ≤ π
a
a
4c1
M
k
−
2π
a
Lecture 7
−
π
0
a
π
a
2π
a
7
Theory vs. Experiment
In a 3-D atomic lattice we expect to
observe 3 different branches of
the dispersion relation, since there
are two mutually perpendicular
transverse wave patterns in
addition to the longitudinal pattern
we have considered
Along different directions in the
reciprocal lattice the shape of the
dispersion relation is different
Note the resemblance to the simple
1-D result we found
Lecture 7
8
4
Counting Modes and Finding N(ω
ω)
v
A vibrational mode is a vibration of a given wave vector k (and thus λ),
frequency ω , and energy E = hω . How many modes are found in the
v
v
v
interval between (ω , E , k ) and (ω + dω , E + dE , k + dk ) ?
# modes
v
dN = N (ω )dω = N ( E )dE = N (k )d 3k
We will first find N(k) by examining allowed values of k. Then we will be able
to calculate N(ω)
First step: simplify problem by using periodic boundary conditions for the
linear chain of atoms:
We assume atoms s and
s+N have the same
displacement—the lattice
has periodic behavior,
where N is very large
s+N-1
L = Na
s
s+1
x = sa
x = (s+N)a
Lecture 7
9
s+2
Step one: finding N(k)
Since atoms s and s+N have the same displacement, we can write:
us = us + N
ue i ( ksa −ωt ) = ue i ( k ( s + N ) a −ωt )
This sets a condition on
allowed k values:
So the separation between
allowed solutions (k values) is:
Thus, in 1-D:
kNa = 2πn →
∆k =
k=
1 = e ikNa
2πn
Na
2π
2π
∆n =
Na
Na
n = 1, 2, 3, ...
independent of k, so
the density of
modes in k-space is
uniform
# of modes
1
Na
L
=
=
=
interval of k − space ∆k 2π 2π
Lecture 7
10
5
Next step: finding N(ω
ω)
Now for a 3-D lattice we can apply periodic boundary
conditions to a sample of N1 x N2 x N3 atoms:
N3c
# of modes
N a N 2b N 3c
V
= 1
= 3 = N (k )
volume of k − space 2π 2π 2π 8π
Now we know from before that
we can write the differential # dN
of modes as:
We carry out the integration
in k-space by using a
“volume” element made up
of a constant ω surface with
thickness dk:
N2b
N1a
v V
v
= N (ω )dω = N (k )d 3 k = 3 d 3 k
8π
v
d 3 k = ( surface area) dk =
[∫ dS ]dk
ω
Lecture 7
11
Finding N(ω
ω)
Rewriting the differential number
of modes in an interval:
We get the result:
N (ω ) =
dN = N (ω )dω =
V
dSω dk
8π 3 ∫
V
dk
V
1
dSω
= 3 ∫ dSω ∂ω
3 ∫
8π
dω 8π
∂k
A very similar result holds for N(E) using constant energy surfaces for the
density of electron states in a periodic lattice!
This equation gives the prescription for calculating the density of modes
N(ω) if we know the dispersion relation ω(k).
Lecture 7
12
6
7.2 Two Atoms per Primitive Basis
Consider a linear diatomic chain of atoms (1-D model for a crystal like NaCl):
a
In equilibrium:
M1
M1 < M2
M2
M1
M2
Applying Newton’s second law and the nearest-neighbor approximation to this
system gives a dispersion relation with two “branches”:
2

 M + M 2   2  M1 + M 2 
4c12
 ± c1 
 −
ω = c1  1
sin 2 ( 12 ka)

 M 1M 2    M 1M 2  M 1M 2
1/ 2
2
ω-(k)
ω 0 as k 0
acoustic modes
ω+(k)
ω ωmax as k 0
optical modes
(M1 and M2 move in phase)
(M1 and M2 move out of phase)
Lecture 7
13
Optical and acoustical branches
Two branches may be presented as follows:
M1 < M2
(2a/M1)1/2
gap in allowed frequencies
(2a/M2)1/2
If there are p atoms in the primitive cell, there are 3p branches to the
dispersion relation: 3 acoustical branches and 3p-3 optical branches
Lecture 7
14
7
Optical and acoustical branches
Lecture 7
15
7.3 Quantization of Elastic Waves
The energy of a lattice vibrations is quantized
The quantum of energy is called a phonon (analogy with the photon of the
electromagnetic wave)
Energy content of a vibrational mode of frequency ω is an integral number of
energy quanta hω . We call these quanta “phonons”.
While a photon is a quantized unit of electromagnetic energy, a phonon is a
quantized unit of vibrational (elastic) energy.
Lecture 7
16
8
7.4 Phonon Momentum
v
Associated with each mode of frequency ωv is a wavevector k , which leads
to the definition of a “crystal momentum”: hk
Crystal momentum is analogous to but not equivalent to linear momentum. No
net mass transport occurs in a propagating lattice vibration, so a phonon does
not carry physical momentum
But phonons interacting with each other or with electrons or photons obey a
conservation law similar to the conservation of linear momentum for interacting
particles
Lecture 7
17
Conservation Laws
Lattice vibrations (phonons) of many different
frequencies can interact in a solid. In all
interactions involving phonons, energy must be
conserved and crystal momentum must be
conserved to within a reciprocal lattice vector:
hω1 + hω2 = hω3
r
v
v
v
hk1 + hk 2 = hk3 + hG
v
ω1 k1
v
ω3 k 3
Schematically:
v
ω2 k 2
Compare this to the special case of elastic
scattering of x-rays with a crystal lattice:
v v r
k′ = k +G
Photon wave vectors
Lecture 7
Just a special case
of the general
conservation law!
18
9
Back to Brillouin Zones
The 1st BZ is the region in reciprocal space containing all information about the
lattice vibrations of the solid
v
Only the
v k values in the 1st BZ correspond to unique vibrational vmodes.
Any k outside this zone is mathematically equivalent to a value k1 inside the
1st BZ
This is expressed in terms of a general translation vector of the reciprocal lattice:
ω
r
G
4c1
M
v v r
k = k1 + G
k
−
4π
a
−
3π
a
−
2π
a
−
π
a
0
v
k1
π
a
2π
a
v
k
3π
a
4π
a
Lecture 7
19
7.5 Neutron scattering measurements
What is a neutron scattering measurement?
- neutron source sends neutron to sample
- some neutrons scatter from sample
- scattered neutrons are detected
2
Conservation of energy:
2
h2k f
h 2 ki
=
± ∆E
2M
2M
When a phonon of wavelength |K| is created by the inelastic scattering of a
photon or neutron, the wavevector selection rule:
r
r r r
k f + K = ki + G creation of a phonon
r
r r r
Lecture 7on of a phonon
k f = ki + K + G annihilati
20
10
Why neutrons?
Wavelength: λ =
9.044
E
- At 10 meV, λ=2.86Å ⇒ similar length scales as structures of interest
Energy:
- thermal sources: 5-100meV
- cold sources: 1-10meV
- spallation sources: thermal and epithermal neutrons (>100meV)
can cover range of typical excitation energies in solids and liquids!
Lecture 7
http://www.ncnr.nist.gov/summerschool/ss05/Vajklecture.pdf
21
Energy and Length Scale
Lecture 7
22
http://www.ncnr.nist.gov/index.html
11
Effective Cross Section
Cross Section, σ: an effective area which represents probability
that a neutron will interact with a nucleus
σ varies from element to element and even isotope to isotope
Typical σ ~ 10-24 cm2 for a single nucleus
One unit of cross section is a 1 barn= 10-24 cm2
… as in “it can’t hit the size of the barn”
Total nuclear cross section
for several isotopes
http://www.ncnr.nist.gov/index.html
Lecture 7
23
Neutron Scattering
Number of scattered neutrons is proportional to scattering function, S (G, ω)
Scattering:
elastic;
quasielastic;
Inelastic
Neutrons are sensitive to
components of motion
parallel to the momentum
transfer Q
Angular width of the scattered
neutron beam gives information
on the lifetime of phonons
Lecture 7
24
12
Phonon dispersion of bcc-Hf
LA-Phonon
TA-Phonon
Trampenau et al. (1991)
Lecture 7
25
Inelastic neutron scattering data for KCuF3
measured
Bella Lake, D. Alan Tennant, Chris D. Frost and Stephen E.
Nagler Nature Materials 4, 329 - 334 (2005) Lecture 7
26
13
2D: Inelastic Scattering on the Surfaces
Thermal Energy Helium Atom Scattering
Lecture 7
27
Time-of-Flight Spectra and Dispersion Curves
Time-of-flight spectrum for He atoms scattering from an LiF(001) surface along the [100] azimuth. The sharp
Lecture 7
28
peaks are due to single surface phonon interactions (From Brusdeylins et al, 1980)
14