QUEEN MARY, UNIVERSITY OF LONDON
SCHOOL OF PHYSICS AND ASTRONOMY
Condensed Matter A
!
Exercise Set 9
!
!
Friday, 10 March, 2017
“do
200
pag
184 Lattice dynamics
The following questions refer to the phonon dispersion curves for the face-centred cubic material neon:
(0,0,ξ)
(ξ,ξ,0)
T(z)
1.5
Frequency (THz)
L
Fig. 8.7 Acoustic mode dispersion curves for
neon (ccp structure) measured by inelastic
neutron scattering. L and T denote longitudinal and transverse modes respectively.
For the [001] and [111] directions the two
transverse modes are degenerate (as indicated
by the label 2T). Along the [110] direction, the
(x,y) and (z) labels indicate the directions of
the atomic motions. (Data taken from Endoh
et al., Phys. Rev. B 11, 1681, 1975.)
(ξ,ξ,ξ)
L
1.0
2T
T(x,y)
0.5
0.0
0
1
0
Reduced wave vector (ξ)
0.5
Useful information: The lattice parameter of neon is a = 4.43 Å and the mass of a neon atom is 3.351 × 10−26 kg.
[110] are symmetrically equivalent points in reciprocal space (see below, and
1. (a) What is the speed Problem
of sound
for longitudinal waves travelling in the (1, 0, 0) direction?
4.5).
(b)
!
!
There are a number of overall features that should be noted at the outset.
The transverse acoustic modes for wave vectors along [100] and [111]
Solution: The graph looks roughly linear up to a point of about k = 0.4a∗ , ωare= 0.9 THz. So
degenerate as a result of the 4-fold and 3-fold rotational symmetry along these
directions respectively. On the other hand,12the [110] direction contains only
× 10 modes have different frequencies.
a 2-fold rotation axis,
so the two 0.9
transverse
c=
= 160 ms−1 .
The dispersion curves along
the
[100]
and ×
[111]
directions closely follow the
0.4 × 2π/4.43
10−10
sinusoidal relationship developed for the simple monatomic chain with nearestneighbour interactions. The more complex shapes of the dispersion curves along
[110] arise from second-neighbour interactions between planes of atoms which
arise
even when
are onlyspring
interactions
between
atoms.
Hence or otherwise,
estimate
thethere
effective
constant
Jeffnearest-neighbour
between neighbouring
planes perpendicular to
This will be discussed in more detail below.
(1, 0, 0), and hence the
value
of
the
spring
constant
J
between
nearest
neighbour
Ne
atoms.
Another point that should be noted is that the [100] and [110] points in
reciprocal space are symmetrically equivalent (see Problem 4.5), and this
means
the dispersion
along
the two
converge to to
common
Solution: There
are that
many
ways to curves
do this.
One
waydirections
is as suggested
use the value of the speed of
frequencies at this point. In particular, it is necessary for two of the modes along
sound from the [110]
previous
part.degenerate at the Brillouin zone boundary. The reason why it
to become
p
is the longitudinal and one of the transverse modes that become degenerate
is
From our 1D monoatomic
chain model, the speed of sound is c = a1D J1D /m.
When we apply this to
explored in Problem 8.1.
the present model,The
J1Ddata
is the
effective
spring
constant
between
planes
J
and
a
in Fig. 8.7, as in other plots of dispersion curves that will
effbe given1D is the distance between
1
later
in
this
chapter,
were
obtained
by
inelastic
neutron
scattering
experiments.
planes, or 2 a. So
The basic theory of the method of inelastic neutron scattering will be discussed
in Chapter 10,
for the moment
we can outline
−26
2 the basic idea. Neutrons are
2 but 3.351
4mc
× 10
× (160)
−2
−19), and the
by quanta
of lattice
vibrations
(to be discussed
Chapter
Jscattered
=
=
= 17.2inmNm
= 1.07 meVÅ .
eff
2 wave vector
2 be obtained by measuring the
frequencyaand
of any×vibration
( 21 × 4.43
10−10 )can
change in energy and wave vector of the scattered neutron beam. The change
!
But we know from class that Jeff = 2J. So
−2
J = 8.59 mNm−1 = 0.536 meVÅ .
An alternative way, which results in a slightly higher answer, would be to compare the frequency at the
p
Brillouin zone boundary to its ideal value 4Jeff /m.
(c) Why are the curves identical at the points (0, 0, 1) and (1, 1, 0), and why does a longitudinal mode at (0, 0, 1)
match a transverse mode at (1, 1, 0)? Sketch the displacements of atoms for a vibration at one of these k values
to help your explanation.
Solution: Using our usual projection onto the xy plane:
1
2
1
2
1
2
y
1
2
x
This is the longitudinal mode with wavevector (1, 1, 0) (check you can see this). But note that all the
atoms in the plane of the paper are moving together, and so are all the atoms at z = 12 . So it is also a
transverse mode with wavevector (0, 0, 1).
Another way of looking at this is as a consequence of the periodicity of the dispersion curves in reciprocal
space. The point (1, 1, 0) is a vector (0, 0, 1) away from the reciprocal lattice point at (1, 1, 1). (In fact, it
is therefore closer to this point than to Γ, so lies outside the Brillouin zone.)
(d) Why are there only two branches in the (0, 0, ξ ) and (ξ , ξ , ξ ) directions, but three in the (ξ , ξ , 0)?
Solution: In the (0, 0, ξ ) and (ξ , ξ , ξ ) directions, the two transverse modes are degenerate – by symmetry
there can be no difference between them. Thus these two branches fall on top of one another.
In the (ξ , ξ , 0) direction, however, the two transverse modes are different: moving in the (x − y) direction
is distinct from moving in the z direction. Thus all three branches are visible.
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2. (a) Draw a plan of the unit cell of neon, and indicate the planes of atoms that move together for wavevectors
parallel to (1, 1, 0).
Solution:
1
2
1
2
1
2
y
1
2
x
(b) Use your diagram to show that, even if we consider only interactions between nearest-neighbour atoms (as
usual), in this direction we will still need to consider interactions between second-nearest-neighbour planes.
Hence explain why the (ξ , ξ , 0) branch of the phonon dispersion curve doesn’t match our result for a 1D
monoatomic chain.
Solution: As the springs in the diagram above show, each atom has four nearest neighbours in the nearest
plane in each direction, and one nearest neighbour in the second-nearest plane. Thus we would need to
extend our monoatomic chain model to consider second-nearest neighbours in order to reproduce these
data.
(c) Looking closely at the experimental curve, it is specifically the longitudinal mode that doesn’t match – the
transverse modes seem to match rather well. Can you explain this too?
Solution: The “spring” that complicates things points directly along the wavevector, perpendicular to
the planes of atoms that move together. Thus, using Jeff = ∑(x2 /r2 )J, we see that this spring contributes
J to the longitudinal mode and nothing at all to the transverse modes. These therefore behave according
to our monoatomic model.
√
(d) Show that the ratio between the maximum frequencies of the two transverse branches is approximately 2.
(Hint: Set up a coordinate system (x0 , y0 , z) in which x0 is parallel to k, so that the two transverse branches
correspond to motion along y0 and z.)
Solution: In the (x0 , y0 , z) coordinate system the four nearest neighbours in the neighbouring plane are at
√
√
( 14 2, ± 14 2, ± 12 ), and transverse motion in the xy plane corresponds to motion in the y0 direction. Thus
Jeff,T(xy) = ∑
y2
J = 4×
r2
while in the z direction
2
16
2
2
16 + 16
+ 41
=J
1
z2
4
= 2J
J
=
4
×
2
1
2
r2
+
16
16 + 4
√
As the maximum frequency is proportional to Jeff , this gives the correct ratio, as well as matching our
answers from considering the equivalent vibrations for k in the (0, 0, 1) direction.
Jeff,T(z) = ∑
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