Brewster H.D. Solid State Physics
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252
Broken Translational Invariance
The
Hamiltonian
of
this system,
H =
fd
3
r
(~p(i7)2
+
~(f.l
+
A)
(V.
ii)2
-1
~ii
. V
2
ii
)
can be rewritten in terms
of
creation and annihilation operators,
1
[~-
-
.ce=-
-,-]
ak,s
=
J2ii
"POOk,s
Es
.
uk
+
'~
OOk,s
Es'
Uk
t 1
[~-
-
·fE_..:.]
a =
--
POOk
E •
u_k
- I
--
E •
u_k
k,s
J2ii
,s s
OOk
s s
as
H
=
"hOOk
s
(a
k
t
ak
s
+.!.)
~
,
,s,
2
k,s
Inverting the above definitions, we can express the displacement ii(r) in
terms
of
the creation and annihilation operators:
ii(r)
=
"~Es
(a-
+a
t
_
)i~·r
~
2 V s
k,s
-k
s
k,s
P
OOk
'
S =
1,
2, 3 corresponds to the longitudinal and two transverse polarizations.
Acting with
ii
k either annihilates a phonon
of
momentum k or creates a phonon
of
momentum
k.
The allowed k values are determined by the boundary conditions in a
finite system.
For
periodic boundary conditions in a cubic system
of
size
V =
L3,
the allowed k
's
are 2n (n}, n
2
,
n
3
).
Hence, the
-k-space
volume per
L
allowed k is
(2n)3N.
Hence,
we
can take the inn finite -volume limit by
making the replacement:
DEBYE
THEORY
LJ(k)=~
LJ(k)(l1k)3
k
(11k)
k
=_V_
fd
3
k J(k)
(2n)3
Since a solid can be modelled as a collection
of
independent oscillators,
we canobtain the energy
in
thermal equilibrium using the Planck distribution
function:
Br.oken Translational Invariance
253
E -
VI
f d
3
k hros(k)
- s
.ls.z.
(2n)3
e~1!(J)s(k)_1
where s =
1,
2, 3 are the three polarizations
of
the phonons and the integral
is over the Brillouin zone.
This can be rewritten in terms
of
the phonon density
of
states, g(m) as:
where
r
hro
E = V drog(ro)
~1!(J)
e
-1
d
3
k
g(ro) = I f
--38(ro-ros(k»
s
.B.z.
(2n)
The total number
of
states is given by:
r r
d3k
drog(ro) =
droI
f
--38(ro-ros(k»
s
Js.z.
(2n)
d
3
k
=
31.z
(2n)3
=3
N
ions
V
The total number
of
normal modes
is
equal to the total number
of
ion
degrees
of
freedom.
For a continuum elastic medium, there are two transverse modes with
velocity
v
t
and one longitudinal mode with velocity
vI.
In
the limit that the
lattice spacing
is
very small, a
~
0,
we expect this theory to be valid.
In
this
limit, the Brillouin zone
is
all
of
momentum space, so
d
3
k
gCEM(ro)=
f--3
(28(ro-v
t
k»
+
8(ro-vlk»
(2n)
1
(2
1)
2
=--
3+3
ro
2n2 V
t
VI
In
a crystalline solid, this will be a reasonable approximation to
gem)
for
kBT«
hv
la where the only phonons present will be at low energies, far from
the Brillouin zone boundary. At high temperatures, there will
be
thermally
excited phonons near the Brillouin zone boundary, where the spectrum
is
definitely not linear, so we cannot use the continuum approximation.
In
particular, this g(m) does not have a finite integral, which violates the condition
that the
integral, should
be
the total number
of
degrees
of
freedom.
A simple approximation, due
to
Debye,
is
to replace the Brillouin zone
254
Broken Translational Invariance
by a sphere
of
radius
kD
and assume that the spectrum
is
linear up to
koo
In
other words, Debye assumed that:
{
3
of
--
1 ro<roD
gD(ro)
= 2n
2
v
3
o ifro >
roD
Here, we have assumed, for simplicity, that vI = v
t
and we have written
roD
=
vkDo
roD
is chosen so that
l.eo
With this choice,
3
N~ns
= r drog(ro)
£
OD
3 2
=
dro~ro
2n v
1
roD
= (6n
2
v
3
N
ions
IV)3
r
D 3 2
hro
E = V
dro-
2
-
3
ro
pliOl
2n
v e
-1
= V 3(kBT)4
rPliOlD
dx~
2n
2
v
2
h
3
.b
eX
-1
In the low temperature limit,
~hroD
---7
00
, we can take the upper limit
of
the integral to
00
and:
k
4 3
E~V3(
BT)
i
dx_
X
_
2n
2
v
3
h3.b
eX
-1
The specific heat is:
C
v
~T3[V
12k~
i
dx~l
2n2v2h3.b
eX
-I
The
T3
contribution to the specific heat
of
a solid is often the most
important contribution to the measured specific heat.
For
T
---7
00,
E
~
V 3(kBT)4
rPliOlD
dxx
2
2rc
2
v
2
h
3
.b
ro
2
=V+,kBT
2rc
v
= 3NionskBT
Broken Translational Invariance
255
so
C
v
"'"
3N
ions
kBT
The high-temperature specific heat is
just
ksl2 times the number
of
degrees offreedom, as in classical statistical mechanics.
At
high-temperature,
we were guaranteed the right result since the density
of
states was normalized
to give the correct total number
of
degrees
of
freedom.
At
low-temperature,
we obtain a qualitatively correct result since the spectrum is linear.
To obtain the exact result, we need to allow for longitudinal and transverse
velocities
which
depend
on
the
direction,
v(k),
vl(k)
,
since
rotational
invariance is not present.
Debye's
formula interpolates between these well-understood limits.
We can define
aD
by
kBa
D
=
hOOD'
For lead, which is soft, aD"'" 88K,
while for diamond, which is hard,
aD"'"
1280K.
MORE
REALISTIC
PHONON
SPECTRA:
OPTICAL
PHONONS,
VAN
Although
Debye's
theory is reasonable, it clearly oversimplifies certain
aspects
of
the physics. For instance, consider a crystal with a two-site basis.
Half
of
the phonon modes will be optical modes. A crude approximation for
the optical modes is an Einstein spectrum:
N
g£(oo)
=~
8(00-00£)
2
In
such a case, the energy will be:
E = V 3( k BT)
4
fl3l1ro
max
dx
~
+ V N
ions
noo
£
2n
2
v
3
n
3
.b
eX
-I
2 e
PllroE
_I
with
OO
max
chosen so that
3
3 N
ions
_
OO
max
-2--
2n
2
v
3
Another feature missed by
Debye's
approximation is the existence
of
singularities in the phonon density
of
states. Consider the spectrum
of
the linear
chain:
1
oo(k) =
2(~y
ISin
~I
The minimum
of
this spectrum is at k =
O.
Here, the density
of
states is
well described by Debye theory which, for a
ID
chain predicts
g(oo)
- const:.
The maximum is at k
= n/a. Near the maximum, Debye theory breaks down;
the density
of
states is singular:
2 1
g(
(0)
= -
-;=====
na
'00
2
-oi
"
max
256
Broken Translational Invariance
In
3D, the singularity will be milder, but still present. Consider a cubic
lattice.
The spectrum can be expanded about a maximum as:
co(k)=cornax
-uxCk~nax
-k
x
)2
-uy(k~ax
_ky)2
-u=(k::
ax
_k=)2
Then (6 maxima;
112
of
each ellipsoid
is
in
the B.Z.)
G(co)
==
1
max
dcog(co)
Differentiating:
=
6.!.~
(vol.
of
ellipsoid)
2
(21t)
3
=3~~1t
(COrnax
-co)2
(21t)3
3
(u=,uy,uJ
1
3V
(CO
rnax
_(0)2
g«(O)=-
1
41t2
(u=,u
y
,u=)2
In
2D and 3D, there can also be saddle points, where V kco(k) =
0,
but
the eigenvalues
of
the second derivative matrix have different signs. At a
saddle point, the phonon spectrum again has a square root singularity. van
Hove proved that every 3D phonon spectrum has at least one maximum and
two saddle points (one with one negative eigenvalue, one with two negative
eigenvalues).
To see why this might be true, draw the spectrum
in
the full k-space,
repeating the Brillouin zone. Imagine drawing lines connecting the minima
of
the spectrum to the nearest neighboring minima (i.e. from each copy
of
the
. B.Z. to its neighbors). Imagine doing the same with the maxima. These lines
intersect; at these intersections, we expect saddle points.
LATTICE
STRUCTURES
Thus far, we have focussed on general properties
of
the vibrational physics
of
crystalline solids. Real crystals come in a variety
of
different lattice
structures, to which we now turn our attention.
BRAVAIS
LATTICES
Bravais lattices are the underlying structure
of
a crystal. A 3D Bravais
lattice
is
defined by the set
of
vectors R
{R
I R =
nlQl
+
n2
Q
2
+
n3
Q
3;ni
E
Z}
Broken Translational Invariance 257
where the vectors a
i
are the basis vectors
ofthe
Bravais lattice. (Do not confuse
with a lattice with a basis.) Every point
of
a Bravais lattice is equivalent to
every other point. In
an
elemental crystal, it is possible that the elemental ions
are located
at
the vertices
of
a Bravais lattice. In general, a crystal structure
will be a Bravais lattice with a basis.
The
symmetry group
of
a Bravais lattice is the group
of
translations
through the lattice vectors together with some discrete rotation group about
(any) one
of
the lattice points. In the problem set you will show that this rotation
group can only have 2-fold, 3-fold, 4-fold, and 6-fold rotation axes.
There are 5 different types
of
Bravais lattice in 2D: square, rectangular,
hexagonal, oblique, and body-centered rectangular. There are
14
different types
of
Bravais lattices in 3D.
We
will content ourselves with listing the Bravais
lattices and discussing some important examples.
Bravais lattices can be grouped according to their symmetries. All but
one can be obtained by deforming the cubic lattices to lower the symmetry.
• Cubic symmetry: cubic, FCC, BCC
• Tetragonal: stretched in one direction, a x a x
c;
tetragonal, centered
tetragonal
• Orthorhombic: sides
of
3 different lengths a x b x c,
at
right angles
to
each other; orthorhombic, base-centered, face-centered, body-
centered.
• Monoclinic:
One
face
is a
parallelogram,
the
other
two
are
rectangular; monoclinic, centered monoclinic.
• Triclinic: All faces are parallelograms.
• Trigonal: Each face is an a x a rhombus.
• Hexagonal:
2D
hexagonal lattices
of
side a, stacked directly above
one another, with spacing
c.
Examples:
• Simple cubic lattice: a
i
=
ax,
.
• Body-centered cubic (BCC) lattice: points
of
a cubic lattice, together
with the centers
of
the cubes
==
interpenetrating cubic lattices opset
by I =2 the body-diagonal.
•
a2
=
aX2
,
a3
=
~
(Xl
+
X2
+
X3
)
Examples: Ba, Li, Na, Fe, K, TI
Face-centered cubic (FCC) lattice: points
of
a cubic lattice, together
with the centers
of
the sides
of
the cubes,
==
interpenetrating cubic
lattices opset by 1 =2 a face-diagonal.
_
a(h
h)
_
a(h
h)
al
="2
X2
+ x3 ,
a2
="2
Xl +
x3
,
_
a(h
h)
a3
=-
Xl
+x2
2
Examples: AI, Au, Cu, Pb, Pt, Ca, Ce, Ar.
258
Broken Translational Invariance
• Hexagonal Lattice: Parallel planes
of
triangular lattices.
-
~
-
a(~
r;;~)
-
~
al
= axl'
a2
="2 XI + ,,
3x
2'
a3
=
cx3·
Bravais lattices can be broken up into unit cells such that all
of
space can
be recovered by translating a unit cell through all possible lattice vectors. A
primitive unit cell is a unit cell
of
minimal volume. There are many possible
choices
of
primitive unit cells. Given a basis, ai'
a2'
a3'
a simple choice
of
unit cell is the region:
{f!i==Xlal +x2
a
2 +x3a3;xi E[O,ll}
The volume
of
this primitive unit cell and, thus, any primitive unit cell
is:
al
.
a2
x
a3
An alternate, symmetrical choice is the Wigner-Seitz cell: the set
of
all
points which are closer to the origin than to any other point
of
the lattice.
Examples:
Wigner-Seitz for square=square, hexagonal=hexagon (not parallelogram),
oblique
= distorted hexagon, BCC=octohedron with each vertex cut
01t
to give
an extra square face (A
+ M).
Reciprocal Lattices
If
al,a2,a3' span a Bravais lattice, then
~
=
21t
_ a
2
_x
a
3
_
al
.
a2
x
a3
b =
21t
a3
x
al
2 - _ _
al
·a2 xa3
~
=21t _ a
l
_xa
2
_
al
.
a2
x
a3
span the reciprocal lattice, which is also a bravais lattice.
The reciprocal
of
the reciprocal lattice is the set
of
all vectors f satisfYing
e
iG
.;:
= 1 for any recprocal lattice vector
G,
i.e. it is the original lattice.
As we discussed above, a simple cubic lattice spanned by
aXl>
aX2'
aX3
has the simple cubic reciprocal lattice spanned by:
21t
~
21t
~
21t
~
-XI'
-x2,
-x3
a a a
An FCC lattice spanned by:
a(~
~)
a(~
~
-
x2
+
x3
, -
xI
+
x3),
2 2
Broken Translational Invariance
259
has a BCC reciprocal lattice spanned by:
41t
1 A A A )
41t
1
(A
A
A)
41t
1.
(A
A
A)
--~+~-~,
--~+~-~,
--~+~-~
a2
a2
a2
Conversely, a BCC lattice has an FCC reciprocal lattice.
The Wigner-Seitz primitive unit cell
ofthe
reciprocal lattice is the1t first
Brillouin zone. In the problem set, you will show that the Brillouin zone has
volume
(21t)3
Iv
if
the volume
of
the unit cell
of
the original lattice is
v.
The
first Brillouin zone is enclosed in
the planes which are the perpendicular
bisectors
of
the reciprocal lattice vectors. These planes are called Bragg planes
for reasons which wili become clear below.
Bravais
Lattices
with a
Basis
Most crystalline solids are not Bravais lattices: not every ionic site
is
equivalent to every other. In a compound this is necessarily true; even in
elemental crystals it is often the case that there are inequivalent sites in the
crystal structure. These crystal structures are lattices with a basis. The
classification
of
such structures is discused in Ashcroft and Mermin. Again,
we will content ourselves with discussing some important examples.
• Honeycomb Lattice (2D): A triangular lattice with a two-site basis.
The triangular lattice is spanned by:
iiI
=~
(.J3
Xl
+
3X2)'
ii2
=
a.J3
Xl
• The two-site basis is:
Example: Graphite
• Diamond Lattice: FCC lattice with a two-site basis: The two-site
basis is:
•
•
0,
a(A A
A)
-
Xl
+x2 +x3
4
Example: Diamond, Si, Ge
Hexagonal Close-Packed (HCP): Hexagonal lattice with a two-site
basis:
aA
a A c
A
0,
iXI
+
2.J3
x2
+i
X3
Examples: Be,Mg,Zn, ...
Sodium Chloride: Cubic lattice with
Na
and
CI
at alternate sites
==
FCC lattice with a two-site basis:
0,
a("
"
A)
-
Xl
+x2 +x3
2
Examples: NaCl, NaF,
KCI
.260
Broken Translational Invariance
- BRAGG SCATTERING
One way
of
experimentally probing a condensed matter system involves
scattering a photon or neutron
on:
the system and studying the energy and
angular dependence
of
the resulting cross-section. Crystal structure experiments
have usually been done with X-rays.
Let
us
first examine this problem intuitively and then in a more systematic
fashion. Consider, first, elastic scattering
of
X-rays. Think
of
the X-rays as
photons which can take different paths through the crystal. Consider the case
in which
k is the wave vector
of
an incoming photon and
k'
is
the wave
vector
of
an outgoing photon. Let us, furthermore, assume that the photon
only scatters
off
one
of
the atoms in the crystal (the probability
of
multiple
scattering is very low). This atom can be
anyone
of
the atoms in the crystal.
These different scattring events will interfere constructively
ifthe
path lengths
differ by an integer number
of
wavelengths. The extra path length for a
scattering
off
an atom at
R,
as compared to an atom at the origin
is:
R . k + ( - R .
k')
= R .
(k
.
k')
If
this is an integral multiple of2n: for all lattice vectors
R,
then scattering
interferes constructively. By definition, this implies that
k _
k'
must be a
reciprocal lattice vector. For elastic scatering,
Ikl
=
Ik'i
' so this implies that
there is a reciprocal lattic vector
G
of
magnitude
I G I =
2/k/
sine
where e is the angle between the incoming and outgoing X-rays.
To rederive this result more formally, let
us
assume that our crystal is in
thermal equilibrium at inverse temperature
~
and that photons interact with
our crystal via the Hamiltonian
H'. Suppose that photons
of
momentum
ki'
and energy
{O/
are scattered by our system. The differential cross-section for
the photons to be scattered into a solid angle
dO.
centered about k j and into
the energy range between
(OJ±
d{O
is:
d
2
(j
=
Ik
j
l(k
j
;mI
H
'lk
i
;n)1
2
e-
13En
8(ro+E
n
-Em)
dQdro
k/,
m,n
where
{O
= {O/-
{Ojand
nand
m label the initial and final states
of
our crystal.
Let
q
-k
j
-k/.
Let
us
assume that the interactions between the photon and the ions
in
our system
is
of
the form:
H'
=
I.U(x
-
(R
+
u(R»)
R
Then
Broken Translational Invariance
261
(kj;mIH'lki;n)
=
fd
3
x
~
e,q·;
\
ml~U(X
-(R +
U(R)))ln)
~
H
~~e-,q(R+.(R»H
=
~
I [e-
iq
.
R
(mle-,q·u(R)ln)
]
V(q)
R
[
1"
-iq'R] ( I I ) -
= V
L:
e
m,e-iq'u(O)
n
U(q)
Let us consider,1t first, the case
of
elastic scattering, in which the state
of
the crystal does not change. Then
In)
=
1m),
IkA
=
Ikjl
==
k,
Iql
2k
sin
~
, and:
2
(kf;nIH'lk,;n)
~
[~
~e
-,qR]
(.le-".(O)ln)
U(q)
Let us focus on the sum over the lattice:
~
Ie-
iq
.
R
= I
°q,G
R R.L.V.G
The sum is 1
if
q is a reciprocal lattice vector and vanishes otherwise.
The scattering cross-section is given by:
d~ro
=
Ie-~En
l(nl
e
-
i
q,u(O)ln)1
2
IV(q)f
I
°q,G
n R.L.V.G
In other words, the scattering cross-section is peaked when the photon
is
- - -
scattered through a reciprocal lattice vector k j =
k;
+ G . For elastic scattering,
this requires
(
-)2
(-
-)2
k;
= k,
+G
or,
(6)2 =
-2k,
·6
This is called the Bragg condition. It is satis1t ed when the endpoint
of
k
is
on a Bragg plane. When it
is
satisfied, Bragg scattering occurs.
When there
is
structure within the unit cell, as
in
a lattice with a basis,
the formula is slightly more complicated. We can replace the photon-ion
interaction
by:
H'=
LLUb(x-(R+h
+u(R+h)))
R b