1
Condensed M
atte
r Physic
sC
ondensed M
atte
r Physic
sJ. E
llis (1
0 L
ectu
res)
J. E
llis (1
0 L
ectu
res)
Perio
dic S
yste
ms: O
verv
iew o
f crystal stru
ctures, th
e recipro
cal lattice.
Ph
on
on
s: Ph
on
ons as n
orm
al mo
des –
classical and
qu
antu
m p
icture. 1
D
mo
nato
mic ch
ain, 1
D d
iatom
ic chain
, exam
ples o
f ph
ono
ns in
3D
. Deb
ye
theo
ry o
f heat cap
acity, th
ermal co
nd
uctiv
ity o
f insu
lators.
Elec
tron
s in so
lids:
Free electro
n m
od
el: Ferm
i-Dirac statistics, co
ncep
t of F
ermi lev
el,
electron
ic con
tribu
tion
to h
eat capacity
. Bu
lk m
od
ulu
s of a n
early free
electron
metal. E
lectrical and
therm
al con
du
ctivity
. Wied
eman
n-F
ranz law
.
Hall effect.
Nearly
free electron
mo
del: D
erivatio
n o
f ban
d stru
cture b
y co
nsid
ering
effect of p
eriod
ic lattice on
1-D
free electron
mo
del. B
loch
’s theo
rem
.
Co
ncep
t of effectiv
e mass. T
he d
ifference b
etween
con
du
ctors,
sem
icon
du
ctors an
d in
sulato
rs exp
lained
by co
nsid
ering
the b
and
gap
in
2D
. Ho
le and
electron
con
du
ction
.
Do
pin
g o
f semico
nd
ucto
rs, pan
d n
typ
es, pn
jun
ction
s –d
iod
es, LE
Ds
and
solar cells.
Bo
ok
s
In g
eneral th
e cou
rse follo
ws th
e treatmen
t in S
olid
Sta
te P
hysics, J.R
.
Ho
ok
and
H.E
. Hall (2
nd
editio
n, W
iley, 1
99
1).
Intro
du
ction
to S
olid
Sta
te P
hysics, C
harles K
ittel(8
th ed
ition
, Wiley
,
20
05
) is hig
hly
recom
men
ded
. (need
no
t be th
e latest editio
n)
An
oth
er bo
ok
, gen
erally av
ailable in
Co
llege lib
raries and
may
usefu
lly b
e
con
sulted
is Th
e So
lid S
tate, R
osen
berg
H M
(3rd
edn
OU
P 1
98
8)
Web
pa
ge h
ttp://w
ww
-sp.p
hy.ca
m.ac.u
k/~
je10
2/
2
•C
ourse deals with crystalline m
aterials –can be
extended later to amorphous m
aterials.•
Crystalline structure characterised by set of lattice
points –each in equivalent environm
ent, but not necessarily at the position of an atom
.•
Each lattice point w
ill have associated with it one or
more atom
s -the ‘basis’. e
.g.N
aCl
•(M
athematically, the lattice w
ould be represented by an array of delta functions, and the crystal described by a convolution of the lattice w
ith a function that described e
.g.the electron density associated w
ith the basis.)
Condensed M
atte
r Physic
s:
Condensed M
atte
r Physic
s:
Perio
dic
Stru
ctu
res
Perio
dic
Stru
ctu
res
Structure
LatticeB
asis
=*
3
•Lattice described by a unit cell –
which m
ay have one lattice point per unit cell (a ‘prim
itive’unit cell) or m
ore than one (‘non-primitive). e
.g. for cubic:
•P
rimitive C
ubic
•F
ace Centred C
ubic (fcc)
•B
ody Centred C
ubic(b
cc)
•H
ow m
any lattice points per unit cell? Either count
those at corners and face centres with w
eight 1/8 and ½
respectively, or move w
hole cell so that no lattice points are on the sides/corners, and count lattice points inside the cell.
Condensed M
atte
r Physic
s:
Condensed M
atte
r Physic
s:
Unit C
ells
Unit C
ells
Prim
itive unit cell, 1 lattice point per unit cell
Non prim
itive unit cell, 4 lattice points per unit cell
Non prim
itive, 2 lattice points per unit cell
4
Bra
vais
Bra
vais
Lattic
es
Lattic
es
P=
primitive
I= body centred
F=
Face centred
on all faces
A,B
,C = centred
on a single face
Need to
remem
ber the P
,I, and F form
s of the cubic unit cells
In 3D there are
14 different lattices –
know
as ‘Bravais’
lattices.
5
Dire
ctio
ns
Dire
ctio
ns
•U
nit cells characterised by the 3 ‘lattice vectors’(a
,b,and c) that define their edges.
e.g. for a face centred cubic (fcc) lattice
•D
irections given in terms of basis vectors –
a direction:
would be w
ritenas [u,v,w
].
•In a cubic lattice are all related by sym
metry. T
hey are together denoted by .
cb
ar
wv
u+
+=
Non prim
itive
Lattice Vectors
a
b
c
Prim
itive
a bc
Unit C
ell
]1
00
[],
01
0[],
00
1[],
00
1[
],0
10
[],
10
0[
__
_
10
0
6
•T
he notation describing a set of uniformly spaced
planes within a crystal is defined as follow
s:•
Assum
e one of the planes passes through the origin
•Look at w
here the next plane cuts the three axes that are defined by the three lattice vectors.
•If the plane cuts the three axes at a/h
, b/k, c/l , then
the set of planes is described by the Miller indices
(h,k,l), and {h,k,l} indicates all planes related to (h,k,l) by sym
metry.
•If h,k, or lis zero it indicates that the plane is parallel to the respective axis.
e.g. (show
ing only the plane next to one that contains the origin)
Pla
nes
Pla
nes
x
y
z
x
y
z
a a/ /l l
b b/ /l l
c c/ /l l
a a/ /l l
b b/ /2 2
(120)(111)
7
Fourie
r Tra
nsfo
rms a
nd T
he
Fourie
r Tra
nsfo
rms a
nd T
he
Recip
rocal L
attic
eR
ecip
rocal L
attic
e
• •1D
periodic functions1D
periodic functions•
A 1D
periodic function, f(x)=f(a+
x), can be represented as a F
ourier series:
The w
ave vectors used, kh
are a uniformly separated set of points
in 1D w
ave vector (k) space.
• •T
o illustrate how a 3D
Fourier series is built up consider the
To illustrate how
a 3D F
ourier series is built up consider the orthorhom
bic case (orthorhom
bic case (a a≠ ≠
b b≠ ≠
c c, 90, 90
° °betw
een axes)betw
een axes)•
In 2D, the coefficients C
hvary w
ith y:
•B
ut the function is periodic in y, so represent Ch (y) as a F
ourier series:
•In 3D
the Ch
k vary with z:
•A
nd again since the function is periodic in z, Ch
k(z) can be
written as a F
ourier series:
•H
ence the 3D series is:
•k
vectors, (kh ,k
k ,kl ), needed for the F
ourier transform form
a lattice in 3D
reciprocal space known as the R
EC
IPR
OC
AL
LAT
TIC
E.
ah
ke
Cx
fh
h
xik
hh
π2:
wh
ere)
(=
=∑ ∞−∞=
()
∑ ∞−∞=
=h
xik
hh
ey
Cy
xf
),
(
bk
ke
Cy
Ck
k
yik
hk
hk
π2:
wh
ere)
(=
=∑ ∞−∞=
∑ ∞
−∞=
++
=l
kh
zk
yk
xk
i
hkl
lk
he
Cz
yx
f,
,
)(
),
,(
()
∑ ∞
−∞=
+=
kh
yk
xik
hk
kh
ez
Cz
yx
f,
),
,(
cl
ke
Cz
Cl
l
zik
hkl
hk
lπ2
:w
here
)(
==∑ ∞−∞=
8
The R
ecip
rocal L
attic
e: th
e
The R
ecip
rocal L
attic
e: th
e
Genera
l Case
Genera
l Case
•F
or a periodic function in 3D w
ith a lattice described by lattice vectors a, b
and c, all the wavevectors you need in
3D k
space for a 3D F
ourier transform representation are:
(Alw
ays use primitive unit cells.)
•T
he set of Gvectors given by all possible integer values of
h,k, and lis known as the re
cip
rocal la
ttice. T
he Gvectors
are know as re
cip
roca
l lattic
e v
ecto
rs.
•A
periodic function f(r) can then be expressed as the 3D
Fourier series.
•S
ince the dot product of a lattice vector (ua+
vb
+w
c) with a
reciprocal lattice vector Gh
kl is 2π(u
h+vk+
wl) –
an integer m
ultiple of 2π-
if you move by a lattice vector the phase of
the exponentials remains unchanged giving the sam
e value for f(r) and the correct periodicity in real space.
'recip
rocal
' hen
ce :
0,
0,
2:
and
2,
2,
2
:W
here
),
,in
teger
(
etc ππ
π
lk
hl
kh
hkl
=⋅
=⋅
Α=
⋅×
⋅ ×=
×⋅ ×
=×
⋅ ×=
++
=
cA
ba
A
cb
a
ba
Cc
ba
ac
Bc
ba
cb
A
CB
AG
π
∑ ∞
−∞=
⋅=
,,
,
)(
lk
h
i
hkl
hkl
eC
fr
Gr
9
•O
rthorhombic: a
≠b≠
c, 90°betw
een axes, a, b, c
form a
right handed set.•
Reciprocal lattice vectors:
•V
iew structure dow
n ‘c’axis:
•If the angles betw
een the a,b, and caxes are not 90°
then a
axis in real space will not necessarily be parallel to the A
axis in reciprocal space. (See hexagonal exam
ple later.)
The R
ecip
rocal L
attic
e. A
n
The R
ecip
rocal L
attic
e. A
n
Orth
orh
om
bic
Exam
ple
Orth
orh
om
bic
Exam
ple
a b
A=
2π/a
B=
2π/b
cC
bB
aa
cb
a
cb
A
))
))
c π
b π
aa
bc
bc
π
2
,2
,2
22
==
==
×⋅ ×
=π
π
Reciprocal
Space Lattice
Real S
pace Lattice
10
The R
ecip
rocal L
attic
e a
nd
The R
ecip
rocal L
attic
e a
nd
Mille
r Index P
lanes
Mille
r Index P
lanes
•T
he first plane (after the plane going through the origin) w
ith a Miller index (h, k, l ) goes through the points:
•T
he normal to this plane is parallel to the cross product of
two vectors in this plane, and hence to G
hkl :
•F
or a plane wave, w
avevectorG
hkl , the difference in phase
between a point on the plane that goes through the origin,
and a point in the plane shown in the diagram
above is:
(The phase difference betw
een two points separation r
is k.r. G
hkl is perpendicular to the planes and so any vector, r
joining a pair of points, one in each plane, will do.)
•T
hus the set of planes with M
iller indices (h,k
,l) are perpendicular G
hkl , and the phase of a w
ave, wavevector
Gh
kl changes by 2π
between one plane and the next. T
he set of planes have the sam
e spacing, therefore, as w
avefrontsof the w
ave with w
avevectorG
hkl
lk
hc
ba
,,
()(
)(
)
()
hkl
lk
hhkl
lk
hhkl
lh
kh
G
CB
Ac
ba
ba
ac
cb
ca
ba
∝
++
×⋅
=
×+
×+
×=
−×
−1
()
π2=
⋅+
+=
⋅h
lk
hh
hkl
aC
BA
aG
x
y
z
a a/ /h h
b b/ /k k
c c/ /l lG
hkl
11
Re
cip
roca
l La
ttice
an
d M
iller In
de
x
Re
cip
roca
l La
ttice
an
d M
iller In
de
x
Pla
ne
s: O
rtho
rho
mb
ic E
xa
mp
leP
lan
es: O
rtho
rho
mb
ic E
xa
mp
le
• •E
xamples of M
iller indices and G vectors
Exam
ples of Miller indices and G
vectors
a
Reciprocal
Space
(01•) planes
b
A
B
Real S
pace
G01•
G000
3rd
index undefined as w
e are looking in 2D
a
Reciprocal
Space
Real S
pace
b
A
B
(02•) planes
G02•
12
Re
cip
roca
l La
ttice
: Orth
orh
om
bic
R
ecip
roca
l La
ttice
: Orth
orh
om
bic
an
d H
exa
go
na
l Exa
mp
les
an
d H
exa
go
na
l Exa
mp
les
a
Reciprocal
Space
Real S
pace
b
AB
(12•) planesG12•
Note:[1] G
vector perpendicular to planes, and of length inversely proportional to the plane spacing.
[2] If the lines (=planes in 3D
) drawn in the figure w
ere ‘wave
crests’then the wavevector
of that wave w
ould be the associated G
vector
Reciprocal
Space
Real S
pace
a
b
(10•) planesA
B
G10•
G000
13
X X- -ra
y a
nd N
eutro
n D
iffractio
nra
y a
nd N
eutro
n D
iffractio
n
•T
he diffraction of x-rays and neutrons from a solid is used
to study structure.
•T
he phase of a wave changes by k.r
over distance r
•T
he condition for diffraction from a crystal relates to the
scatte
ring w
ave v
ecto
rk
s , which is the difference:
ks
= k
f -k
i between the w
avevectors of the outgoing (kf )
and incoming (k
i ) beams.
•If the scattering w
avevectoris equal to a reciprocal lattice
vector then since the product of a lattice vector with a
reciprocal lattice vector is an integer multiple of 2π, all
equivalent points within the crystal (e
.g.all identically
located atoms) w
ill scatter in phase and give a strong outgoing beam
. i.e.the diffraction condition is:
ks
= G
hkl
and:k
f = k
i + G
hkl
incoming w
ave w
avevectork
i
outgoing wave
wavevector
kf
r
phase difference (k
f –k
i ).r=k
s .rS
cattering objectsE
xtra phase ki .r
Extra phase k
f .r
14
X X- -ra
y a
nd N
eutro
n D
iffractio
n:
ray a
nd N
eutro
n D
iffractio
n:
Energ
y C
onserv
atio
nE
nerg
y C
onserv
atio
n
•C
onservation of energy requires that the incoming and
outgoing wavevectors
(once the scattering particle is free of the crystal) m
ust be of equal magnitude.
•C
ondition for diffraction neatly represented by Ew
ald’ssphere construction: both k
i and kf m
ust lie on the surface of a sphere, and be separated by a G
vector.
•It is clear that it is quite possible that for a particular incident condition there is no diffraction from
a crystal –the E
wald
construction is quite specific on ki and k
f .
•D
iffraction from pow
ders overcomes this by having a large
number of crystals in different orientations.
ki k
f
G
Ew
ald’sC
onstruction
15
Stro
ng S
catte
ring o
f Waves in
S
trong S
catte
ring o
f Waves in
Cry
sta
lsC
rysta
ls
•N
eutrons (provided the sample is typically thinner than 1cm
) andx-rays are scattered at m
ost once as they pass through a crystal.
•If you try to send a beam
of electrons through a crystal it is very strongly scattered –
the mean free path depends on energy, but
takes a minim
um at 50-100eV
of about 6Åin a typical m
etal.
•If you im
agine starting a beam of electrons inside a crystal w
ith a particular w
ave vector k, it will quickly be scattered into a set of
waves travelling w
ith wavevectors
k+
Ghkl
.
•T
here will then be m
ore scattering, but now, since diffraction
simply adds a G
vector to the intialwave vector, it w
ill be from
one of these new set of w
aves to another –indeed som
e may be
scattered back into the original wave w
ith wave vector k.
•A
fter a while a sort of equilibrium
is reached with the rate of
scattering out of a particular set of waves equalling the rate of
scattering into it. Once this has happened, no further effect ofthe
scattering can be seen, and this explains why despite the large
scattering cross sections, as we shall see later, electrons can
behave as if they move through a crystal unim
peded.
•W
e shall see later that because electrons moving through a
crystal with a certain w
ave vector (k) can in fact have some of
their ‘probability amplitude’in w
hole set of associated waves
(wavevectors
k+
Ghkl ), one m
ay have to allow for this by m
aking a correction to the ‘effective m
ass’that they seem to have.
16
Condensed M
atte
r Physic
s:
Condensed M
atte
r Physic
s:
Phonons
Phonons
• •A
ims:
Aim
s:•
Lattice vibrations ‘normal m
odes’/’phonons’•
Establish concepts by considering m
odes of a 1-dim
ensional, harmonic chain, both m
onatomic and
diatomic.
•E
xamples of phonons in a 3D
lattice.
•D
ebye theory of heat capacity•
3k
B /atom at ‘high’tem
peratures.
•α
T3
at low tem
peratures.•
What are high/low
temperatures,
concept of Debye tem
perature.
•T
hermal C
onductivity of insulating crystals•
αT
3at low
temperatures.
•α
T-1
at high temperatures.
•S
trong effect of defects andspecim
en dimensions at low
tem
peratures.
Th
erm
al C
on
du
ctiv
ity o
f Ge
v.
Te
mp
era
ture
0.1 1 10
100
110
1001000
Tem
pera
ture
/K
Thermal
Conductivity/W/cmK
17
Ato
mic
Motio
n in
a L
attic
eA
tom
ic M
otio
n in
a L
attic
e
•In a solid, the m
otion of every atom is coupled to that of its
neighbours –so cannot describe m
otion atom by atom
–use ‘norm
al mode’approach instead.
•T
he motion of a ‘harm
onic’system (objects connected by
‘Hook’s law
’springs), can be described as a sum of
independent‘normal m
odes’in which the coordinates all
oscillate at same frequency
and maintain fixed ratios
to each other.
•M
otion of atoms m
ust be described quantum
mechanically, but w
e will use the results that:
•T
he displacement patterns of the classical norm
al modes are
the same as the ratios of the coordinates in the quantum
m
echanical ones.
•T
he energy of the quantum m
echanical modes is expressed
in terms of the frequency (ω
) of the classical mode:
•In a solid these quantised norm
al modes are called
phonons.
+=
2 1n
Eωh
18
Norm
al M
odes
Norm
al M
odes - -
Cla
ssic
al
Cla
ssic
al
Vie
w: 2
Coord
inate
Exam
ple
Vie
w: 2
Coord
inate
Exam
ple
•M
ode 1:M
ode 2:
•G
eneral solution.
()
()2
22
11
1
21
2 1
cos
1 1
2 1co
s1 1
2 1
1 1
2 1
1 1
2 1
εω
εω
+
−+
+
=
−
+
=
tA
tA
uu
x x
k
mm
x1
x2
kk
()
23
12
21
1,
,:
righ
tto
leftten
sion
s,S
prin
gkx
Fx
xk
Fkx
F−
=−
==
()
()
()
()
()
()
()
()
.
so
0
2,
m
od
eIn
.
so
0
1
,
mo
de
In
.am
plitu
des
t in
dep
end
en
are
,
and
3
,w
here
cos
,co
s
:so
lutio
ns
m
otio
n,
o
f
equ
ation
st
ind
epen
den
with
sco
ord
inate
n
orm
al
the
are
21
an
d2
1
33
:(2
)(1
)
Su
btactin
g
:(2
)(1
)
Ad
din
g
(2)
2
(1)
2:
mo
tion
of
Eq
uatio
ns
21
12
12
21
21
22
22
11
11
21
22
11
22
21
21
11
21
21
21
23
2
21
12
1
xx
ux
xu
AA
mk
mk
tA
ut
Au
xx
ux
xu
ku
um
xx
kx
xm
ku
um
xx
kx
xm
kx
kx
FF
xm
kx
kx
FF
xm
−=
==
=
==
+=
+=
−=
+=
−=
⇒−
−=
−−
−=
⇒+
−=
++
−=
−=
+−
=−
=
ωω
εω
εω
&&&&
&&
&&&&
&& && &&
1/√2 for normalisation
1/√2 for normalisation
19
Norm
al M
odes
Norm
al M
odes - -
Quantu
m
Quantu
m
Vie
w: 2
Coord
inate
Exam
ple
Vie
w: 2
Coord
inate
Exam
ple
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
()
+=
+=
=+
=
+
∂
∂−
+∂
∂−
=
=
+∂
∂−
+∂
∂−
+=
+−
+=
=+
∂
∂−
∂
∂−
2 1
and
2 1
:an
d
app
liesso
lutio
n
S
HO
stan
dard
th
eW
here
32 1
2
1
2 1
2
1
:an
d,
can
write
w
ean
d
,
,3
2 1,
2,
2 1,
2
:s
coo
rdin
ate
,in
separates
equ
ation
s
er'S
chro
edin
g.
.
32 1
2 1,
as
written
b
ecan
,
ho
wev
er
2 1
2 1
2 1,
:w
here
,,
,,
2
,
2
:eq
uatio
n s
er'S
chro
edin
g
22
21
11
21
22
2
22
2
22
22
22
11
2
12
1
11
22
11
22
11
21
21
21
2
22
2
21
22
21
2
12
1
21
22
21
2
2
2
12
12
1
2
2
2
12
2
12
1
21
21
21
2
2
21
22
2
1
21
22
nE
nE
EE
Eu
kuu
u
mu
uku
u
u
mu
uu
uu
uu
E
uu
kuu
uu
mu
uku
u
uu
m
uu
ei
kuku
uu
Vx
xV
kxx
xk
kxx
xV
xx
Ex
xx
xV
x
xx
mx
xx
m
ωω
ψψ
ψ
ψψ
ψ
ψψ
ψ
ψ
ψψ
ψψ
ψψ
ψψ
hh
h
h
hh
hh
•T
he frequencies and amplitude ratios are the sam
e as for the classical case, but the energy is quantized
Each term
has only one variable but their sum
is constant, so eachm
ust be constant giving two
Independent equations
Each term
has only one variable but their sum
is constant, so eachm
ust be constant giving two
Independent equations
20
Lattic
e v
ibra
tions
Lattic
e v
ibra
tions
• •1 1- -D
harmonic chain
D harm
onic chain•
Take identical m
asses, m, separation a
connected by springs (spring constant, α
):
•T
his is a model lim
ited to “nearest-neighbour”interactions.
Equation of m
otion for the nth
atom is:
•W
e have N coupled equations (for N
atoms).
•T
ake cyclic boundary conditions –N
+1
thatom
equivalent to first (w
ill be discussed later).
•A
ll masses equivalent –
so the normal m
ode solutions m
ust reflect this symm
etry and all have the same
amplitude ( u
0 ) and phase relation to their neighbours, i.e.
()
()
{}
()n
nn
n
nn
nn
n
uu
uu
m
uu
uu
um
21
1
11
−+
=
−−
−=
−+
−+
α α
&& &&
()
()δ
δi
uu
iu
un
nn
n−
==
−+
exp
,
exp
11
21
Lattic
e V
ibra
tions: F
requency
Lattic
e V
ibra
tions: F
requency
of M
odes
of M
odes
• •Look for norm
al mode solutions
Look for normal m
ode solutions
•E
ach coordinate has time dependence:
•S
ubstitute into equation of motion:
•P
hase, δ, only has unique meaning for a range of
2π: makes m
ost sense to consider ωas a function
of δover the range –π
to π, giving:
()t
iω−
exp
()
()
()
()
()
=
=−
=
−−
+=
−−
−
2sin
4
2sin
4co
s2
2
exp
2e
eex
p
22 2
δα
δω
δα
δα
ω
ωα
ωω
δδ
m
m
ti
ut
iu
mn
ii
n
Ph
aseδ
-ππ
22
Lattic
e V
ibra
tions: N
atu
re o
f Lattic
e V
ibra
tions: N
atu
re o
f
Modes
Modes
•C
an write the am
plitude of the nth
atom as:`
•C
an write the phase difference betw
een successive atoms
in terms of a w
avevector, conventionally written as q
for phonons:
•N
ow it is clear that the m
odes are waves travelling along
the chain of atoms:
•T
he dispersion relationfor these w
aves is:
•S
ince the phase, δ, only has unique meaning for the range
– π/2 to π/2, qonly has a unique value over the range:
•E
nergy stored in mode is
i.e. a ground state of energy ħω
/2
plus: nphonons each of energy ħ
ω.
•M
omentum
of a phononturns out to be ħ
q.
•V
elocity = ∂ω
/∂q
(If you can see the wave m
ove you must
have formed a w
avepacket, so velocity is group velocity)
qa
=δ(
)
=
2sin
4q
a
mq
αω
{}
()t
ni
uu
nω
δ−
=ex
p0
aq
a
ππ
≤≤
− {}
()
{}
()t
qx
iu
tq
na
iu
un
ωω
−=
−=
exp
exp
00
+=
2 1n
Eωh
na
is the distance xalong the chain
na
is the distance xalong the chain
23
The M
eanin
g o
f phonon
The M
eanin
g o
f phonon
wavevecto
rw
avevecto
rq.
q.
• •T
he T
he wavevector
wavevector
q qgives the phase shift betw
een gives the phase shift betw
een successive unit cells.successive unit cells.•
qis defined on the range
where G
=2π
/ais the sm
allest reciprocal lattice vector.•
qhas no m
eaning between lattice points, so is
equivalent to q+
G.
•R
emem
ber: a wave vector that is a reciprocal lattice
vector gives a phase shift of 2nπ
between tw
o points separated by a lattice vector.
•F
ree space is uniform, so a phase shift along a w
ave given by φ
=kr
works for any r. In a crystal, space is not
uniform -
equivalent points are separated by a lattice vector, and φ
=kr
only has meaning if r
is a lattice vector.
n
-1.2
-0.7
-0.2
0.3
0.8
01
23
45
67
8
q
q+
G
Amplitude
2
2
i.e.
Gq
G
aq
a≤
≤−
≤≤
−π
π
Phase shift betw
eenlattice points ism
eaningless
Phase shift betw
eenlattice points ism
eaningless
24
Phonon d
ispers
ion
Phonon d
ispers
ion
• •D
ispersion curvesD
ispersion curves�
ωversus q
gives the wave dispersion
• •K
ey pointsK
ey points•
The periodicity in q
(reciprocal space) is a consequence of the periodicity of the lattice in real space. T
hus the phonon at some
wavevector, say, q
1is the sam
e as that at q
1 +n
G, for all integers n, w
here G=
2π/a
(a reciprocal lattice vector).
•In the long w
avelength limit (q
→0) w
e expect the “atom
ic character”of the chain to be
unimportant.
25
Lim
iting b
ehavio
ur
Lim
iting b
ehavio
ur
• •Long w
avelength limit
Long wavelength lim
it•
dispersion formula
•leads to the continuumresult (see IB
waves course)
•T
hese are conventional sound-waves.
• •S
hort wavelength lim
itS
hort wavelength lim
it•
“Atom
ic character”is evident as the w
avelength approaches atom
ic dimensions q
→π
/a. λ=
2a
is the shortest, possible w
avelength.•
Here w
e have a standing wave ∂ω
/∂q
=0
ρω
αω
Y
qa
m
aq
q
=→
→
;0
=2
sin4
qa
m αω
()
22
sinq
aq
a≈
Continuum
resultY
-Y
oung’s modulus
ρ -density
Continuum
resultY
-Y
oung’s modulus
ρ -density
q→
0q→
0
mα
ω4
max
=
26
Mom
entu
m o
f a P
honon:
Mom
entu
m o
f a P
honon: ħ ħ
q q
•N
eed to extend our concept of mom
entum to som
ething that w
orks for phonons –a so called ‘crystal m
omentum
’.
•If, for exam
ple, a neutron hits a crystal and creates a phonon, we
want a definition of phonon m
omentum
such that mom
entum w
ill be conserved in the scattering/phonon creation process.
•F
or a static lattice we sim
ply have diffraction, and to get a large scattered intensity all the scatterers
have to scatter in phase, i.e. (k
f -k
i ).r=
2nπ
(ris a lattice vector: separation of identical atom
s) and for this to be true k
f -k
i=
G.
•If the lattice is now
distorted by a phonon, the way each atom
scatters w
ill be modified by an extra phase term
, q.r, so, if the
scattered amplitudes are all to add up, the scattering w
avevectorw
ill have to give an extra phase difference between lattice
positions of q.r. i.e
:
(kf -
ki ).r
= 2
nπ+
q.r
and kf -
ki =
G +
q .
•T
his means that on scattering the crystal changes m
omentum
by ħ
(G +
q). ħ
G is the m
omentum
transfer due to diffraction from
the lattice causing the whole crystal to recoil, and so it is sensible
to define the mom
entum of the phonon as ħ
q–
after scattering either you have created a phonon m
omentum
-ħq
or you have annihilated one of m
omentum
ħq.
•B
ut you say, you can’t just define mom
entum anyhow
you like –surely it is som
ething that exists and we have to m
easure it. Not
at all. Mom
entum and energy entered physics as constructions
created to make the m
aths of doing physics easy. Consider
potential energy –to w
hat measurable ‘real’quantity can you add
an arbitrary offset and everything is still ok? Why is energy
conserved ?–because w
e carefully define all forms of energy so
that it is, at least that is how the idea started.
27
Mo
me
ntu
m o
f a P
ho
no
n:
Mo
me
ntu
m o
f a P
ho
no
n: ħ ħ
q,b
ut
q,b
ut
is it
is it
rea
so
na
ble
? (n
on
exa
min
ab
le)
rea
so
na
ble
? (n
on
exa
min
ab
le)
•T
he problem is that if you really do have a infinite uniform
wave in
an infinite lattice, there are as many atom
s going forwards as
backwards and it carries no m
omentum
.
•T
he total (classical) mom
entum is carried som
ehow by all the
atoms in the crystal –
what w
e are trying to do is divide it notionally betw
een mom
entum carried by the phonon and that
associated with m
otion of the centre of mass –
so its complicated.
•H
owever, if you m
ake a wavepacketout of a sm
all spread of w
avevectors, then if you give the wavepacketan energy ħ
ωand
add up the mom
entum associated w
ith all the vibrating particles they don’t quite cancel out, but do indeed give ħ
q.
•E
ffectively, after the neutron scatters the whole crystal starts
to m
ove, carrying mom
entum ħ
G, and inside the crystal is a
wavepacketof vibrations travelling through the crystal that carries
a net extra mom
entum of ħ
q.
•If you are considering scattering of a neutron, you are not considering an infinite crystal, so one can reconcile norm
al m
omentum
with crystal m
omentum
.
•T
o understand the infinite case (non quantum m
echanically) –you have to take the lim
it of the wavepacketgoing to infinite
length –w
hich is approaching infinitiyin a different w
ay from
saying that we have uniform
oscillations throughout the crystal and let the crystal size go to infinity, so you get a different result for the m
omentum
when you go to the lim
it in a different way.
•If an inifinte
lattice has to supply ħG
or ħq
of mom
entum it
does not change the state of motion of the lattice (i.e
.the lattice does not start to m
ove) because it has infinite mass.
28
1 1st
stB
rillouin
Zone
Brillo
uin
Zone
• •P
eriodicity:P
eriodicity:A
ll the physically distinguishable modes
All the physically distinguishable m
odes lie w
ithin a single span of lie w
ithin a single span of 2 2π π/ /a a. .
• •F
irst Brillouin Z
one (BZ
)F
irst Brillouin Z
one (BZ
)•
we chose the range of q
to liew
ithin |q|<
π/a. T
his is the 1stB
Z.
•N
umber of m
odes must equal the num
ber of atoms, N
, in the chain and for finite N
the allowed q
values are discrete, separation 2π/N
a (see ‘w
aves in
a b
ox’
later).
• •T
o Sum
marise:
To S
umm
arise: E
ach mode (at particular q) is a quantised, sim
ple-harm
onic oscillator, E=
ħω(n
+1
/2). P
honons have particle character –
bosons: each mode can have
any number of phonons in it w
ith:E
nergy=ħω
, Mom
.= ħq, V
elocity = ∂ω
/∂q.
The unique m
odes lie within the first B
.Z..
1stB
rillouin zone(shaded)
1stB
rillouin zone(shaded)
29
Measure
ment o
f Phonons
Measure
ment o
f Phonons
• •B
asic principle:B
asic principle:•
Need a probe w
ith a mom
entum and energy
comparable to that of the phonons e.g. therm
al energy neutrons for bulk, and H
e atoms at surfaces.
X-rays can have correct w
avelength, but the energy is so high it is hard to resolve the sm
all changes induced by phonon interactions. (A
t λ=1Å
, energy is 12.4keV
–typical phonon energies are up to 40m
eV)
•P
article hits the lattice and creates/annihilates phonons.
•Illum
inate sample w
ith a monochrom
atic beam –
incident wavevector
ki
•E
nergy analyse scattered signal –peaks in signal
correspond to single phonon creation/annihilation occurring at a particular k
f .
• •U
se of conservation laws
Use of conservation law
s•
Energy of phonon (+
= creation, -
= annihilation):
•C
rystal mom
entum conservation for phonon creation:
•C
rystal mom
entum conservation for phonon
annihilation:
() 2
22
2f
ik
km
−=h
hω
Gq
kk
++
=f
i
fi
kG
qk
=+
+
30
Relative Intensity
En
ergy
Tran
sfer/meV
Measure
ment o
f Phonons II
Measure
ment o
f Phonons II
•T
o measure energy of probe can use tim
e of flight techniques –
e.g. helium atom
scattering (HA
S):
•T
ime flight of individual atom
s through apparatus –to
determine energy transfer on scattering.
Phys. R
ev. B4
8, 4917, (1993)P
hys. Rev. B
48, 4917, (1993)
Elastic peak
Single phonon
creation peaks
HA
S T
oFdata for C
u(100) surface
Rotating D
isk C
hopper
31
Dia
tom
ic la
ttice
Dia
tom
ic la
ttice
• •T
echnically a lattice with a basis
Technically a lattice w
ith a basis
•proceeding as before. E
quations of motion are:
Trial solutions:
substituting gives
homogeneous equations require determ
inant to be zero giving a quadratic equation for ω
2.
mA
mAm
Bm
B
()
()1
22
22
12
21
21
22
2 2
++
+
−+
−+
=
−+
=
nn
nn
B
nn
nn
A
uu
uu
m
uu
uu
m
α α
&& &&
()
{}
()
()
{}t
qa
ni
Uu
tn
qa
iU
u
n n
ω
ω
−+
=
−=
+1
2ex
p
2ex
p
21
2
12
()
()
()
()
02
cos
2
0co
s2
2
2
2
1
21
2
=−
+
=+
−
Um
Uqa
Uqa
Um
B
A
αω
α
αα
ω
()
[
()
{}
]2
12
2
2
sin4
qa
mm
mm
mm
mm
BA
BA
BA
BA
−+
±+
=α
ωT
wo solutions
for eachq
Tw
o solutionsfor each
q
32
Acoustic
and O
ptic
modes
Acoustic
and O
ptic
modes
• •S
olutionsS
olutions•
q→
0:•
Optic m
ode (higher frequency)
•A
coustic mode (low
er frequency)
()
µ αα
ω2
2=
+=
BA
BAm
m
mm
()
[
()
()
()
]
qm
m
a
qa
mm
mm
mm
mm
mm
BA
BA
BA
BA
BA
BA
+≈
+−
+
−+
≈
2
21
2
2
2
2
41
αω
αω
Effective
mass
µE
ffectivem
assµ
Periodic: all
distinguishablem
odes lie in|q
|<π/2
a
Periodic: all
distinguishablem
odes lie in|q
|<π/2
a
ω=√(2α
/mB )
ω=√(2α
/mB )
ω=√(2α
/mA )
ω=√(2α
/mA )
33
Dis
pla
cem
ent p
atte
rns
Dis
pla
cem
ent p
atte
rns
•D
isplacements show
n as transverse to ease visualisation.
• •A
coustic modes:
Acoustic m
odes: Neighbouring atom
s in phase
• •O
ptical modes:
Optical m
odes: Neighbouring atom
s out of phase
• •Z
oneZ
one- -boundary m
odesboundary m
odes•
q=
π/2
a; λ
=2π
/q=
4a
(standing waves)
•H
igher energy mode –
only light atoms m
ove
•Low
er energy mode –
only heavier atoms m
ove
34
Orig
in o
fO
rigin
of
optic
and a
coustic
bra
nches
optic
and a
coustic
bra
nches
• •E
ffect of periodicityE
ffect of periodicity•
The m
odes of the diatomic chain can be seen
to arise from those of a m
onatomic chain.
Diagram
matically:
Monatom
ic chain,period a
Monatom
ic chain,period a
period in qis π
/a
for diatomic chain
period in qis π
/a
for diatomic chain
Acoustic and
optical modes
Acoustic and
optical modes
Energy of optical andacoustic m
odessplit if alternatingm
asses different
Energy of optical andacoustic m
odessplit if alternatingm
asses different
Modes w
ith q out-side new
BZ
period ‘backfolded’into new
BZ
by adding ±
G= π
/a
Modes w
ith q out-side new
BZ
period ‘backfolded’into new
BZ
by adding ±
G=π
/a
35
Dia
tom
ic c
hain
:D
iato
mic
chain
:
sum
mary
sum
mary
• •A
coustic modes:
Acoustic m
odes:•
correspond to sound-waves in the long-
wavelength lim
it. Hence the nam
e.ω→
0 as q→
0
• •O
ptical modes:
Optical m
odes:•
In the long-wavelength lim
it, optical modes
interact strongly with electrom
agnetic radiation in polar crystals. H
ence the name.
•S
trong optical absorption is observed. (P
hotons annihilated, phonons created.)ω→
finitevalue as q
→0
•O
ptical modes arise from
folding back the dispersion curve as the lattice periodicity is doubled (halved in q-space).
• •Z
one boundary:Z
one boundary:•
All m
odes are standing waves at the zone
boundary, ∂ω/∂
q=
0: a necessary consequence of the lattice periodicity.
•In a diatom
ic chain, the frequency-gap betw
een the acoustic and optical branches depends on the m
ass difference. In the limit of
identical masses the gap tends to zero.
36
Phonons in
3P
honons in
3- -D
cry
sta
ls:
D c
rysta
ls:
Monato
mic
lattic
eM
onato
mic
lattic
e
• •E
xample: N
eon, an E
xample: N
eon, an f.c.cf.c.c. solid:
. solid: •
Inelastic neutron scattering results in different crystallographic directions
•M
any features are explained by our 1-D m
odel:•
Dispersion is sinusoidal (nearest neighbour.
interactions).•
All m
odes are acoustic (monatom
ic system)
Phys. R
ev. B1
1, 1681, (1975)P
hys. Rev. B
11, 1681, (1975)
(00)ξ
()ξ
ξξ
()
ξξ0
37
Neon:
Neon:
a m
onato
mic
, a m
onato
mic
, f.c.c
f.c.c
. solid
. solid
• •N
otes: (continued)N
otes: (continued)•
There are tw
o distinct types of mode:
•Longitudinal (L), w
ith displacements parallel to
the propagation direction,•
These generally have higher energy
•T
ransverse (T), w
ith displacements
perpendicular to the propagation direction•
These generally have low
er energy•
They are often degenerate in high
symm
etry directions (not along ( ξξ0))
•M
inor point (demonstrating that real system
s are subtle and interesting, but also com
plicated):•
L mode along (ξξ0) has 2 F
ourier components,
suggesting next-n.n. interactions (see Q 3, sheet
1). In fact there are only n.n. interactions•
The effect is due to the
fccstructure. N
earest-neighbour interactions from
atom
, A(in plane I) join to
atom C
(in plane II) and to atom
B(in plane III) thus
linking nearest-and
next-nearest-planes.I II III
A
BC
C
(110)
38
Phonons in
3P
honons in
3- -D
cry
sta
ls:
D c
rysta
ls:
Dia
tom
ic la
ttice
Dia
tom
ic la
ttice
• •E
xample:
Exam
ple: NaC
lN
aCl, ,has sodium
chloride structure!
•T
wo interpenetrating f.c.c. lattices
• •M
ain points:M
ain points:•
The 1-D
model gives several insights, as
before. There are:
•O
ptical and acoustic modes (labels O
and A);
•Longitudinal and transverse m
odes (L and T).
•D
ispersion along (ξξξ) is simplest and m
ost like our 1-D
model
•(ξξξ) planes contain, alternately, N
a atoms and
Clatom
s (other directions have Na and C
lmixed)
Phys. R
ev. 17
8
1496, (1969)P
hys. Rev. 1
78
1496, (1969)
39
NaC
lN
aC
lphonons
phonons
• •N
otes, continuedN
otes, continued… …
•N
ote the energy scale. The highest energy
optical modes are ~8 T
Hz (i.e. approxim
ately 30 m
eV). H
igher phonon energies than in N
eon. The strong, polar bonds in the alkali
halides are stronger and stiffer than the weak,
van-der-Waals bonding in N
eon.•
Minor point:•
Modes w
ith same sym
metry cannot cross,
hence the avoided crossing between acoustic
and optical modes in (00ξ) and (ξξ0) directions.
•Ignore the detail for present purposes
40
Conserv
atio
n L
aw
s a
nd
Conserv
atio
n L
aw
s a
nd
Sym
metry
Sym
metry
• •LagrangianLagrangian
Mechanics
Mechanics
•N
ewton 2 norm
ally considered in Cartesian coordinates: ‘F
=m
a’
•C
an generalise to non-Cartesian coordinates, but now
write
equations of motion in term
s of derivatives of the La
gra
ng
ian:
L=
K.E
. –P
.E. . F
ield of ‘analytical dynamics’based on this idea.
• •C
onservation Laws
Conservation Law
s•
A key result of analytical dynam
ics is Noether’s
theorem –
for every sym
metry in the Lagrangian
(i.e. in the system), there is an
associated conservation law.
e.g
. it turns out that:•
If the system’s behaviour is independent of the tim
eyou set it going,
energyis conserved.
•If the system
’s behaviour is independent of where it is in space
, m
omentum
is conserved. •
If the system’s behaviour is independent of the its angular orientation,
angular mom
entumis conserved.
•In a crystal –
space is no longer uniformbut has a new
symm
etry –
its periodic, so the law of conservation of m
omentum
is replaced by a new
law –
the conservation of ‘crystal mom
entum’
in which m
omentum
is conserved to within a factor of ħ
G.
E.g
.
•D
iffractio
n: w
avevecto
rallo
wed to
change b
y fa
cto
rs o
f G
•P
honon c
reatio
n:
Gq
kk
++
=f
i
Gk
k+
=i
f
41
A m
ath
em
atic
al a
sid
e
A m
ath
em
atic
al a
sid
e
(for in
tere
st
(for in
tere
st - -
non e
xam
inable
).non e
xam
inable
).
•L=
K.E
.-P.E
(e.g. S.H
.O. )
•E
quations of motion:
Euler-Lagrange equations:
SH
O:
•C
onjugate mom
entum: (S
HO
: )
•If L
is independent of qi :
•E
nergy conservation? Under m
any circumstances the
Ham
iltonian H(defined as ) is the
energy, and since , if Ldoes not have explicit
time dependence , H
and hence energy is conserved.
0=
∂ ∂−
∂ ∂
ii
q L
q L
dt d
&
22
2 1kx
xm
L−
=&
()
kx
xm
kx
xm
dt d
−=
⇒
=+
&& &0
i
iq L
p&
∂ ∂=(
)0
=∂ ∂
−i
iq L
pd
t d
ith tim
e.not v
ary w
does
and
0
hen
ce
and
0i
i
i
pdt
dp
q L
==
∂ ∂
t L
dt
dH
∂ ∂−
=
∑=
−∂ ∂
=f
Nii
iL
q Lq
H1
&&
xm
q L
i
&&
=∂ ∂
Noether’s
theoremN
oether’stheorem
42
The U
se o
f Conserv
atio
n L
aw
sT
he U
se o
f Conserv
atio
n L
aw
s
• •W
hat do conservation laws tell you?
What do conservation law
s tell you?
•C
onservation laws te
ll you w
hat is
allo
wed to
happe
n–
it is not possible to have an outcome of an event that
violates a valid conservation law.
•U
nle
ss c
onserv
atio
n la
ws p
erm
it on
ly o
ne o
utc
om
e,
they d
o n
ot te
ll you w
hat w
ill actu
ally
ha
ppe
n, n
or h
ow
fast it w
ill happ
en.
e.g
.conservation of crystal mom
entum inside a
periodic solid tells you what possible outgoing
mom
entaa diffracted particle m
ay have, ( ) but they do not tell you how
intense the outgoing beam
s will be.
Gk
k+
=i
f
43
Th
erm
al P
rop
ertie
s o
f Insu
latin
g
Th
erm
al P
rop
ertie
s o
f Insu
latin
g
Cry
sta
ls: H
ea
t Ca
pa
city
Cry
sta
ls: H
ea
t Ca
pa
city
• •T
hermal energy is stored in the phonons
Therm
al energy is stored in the phonons•
Need to know
how m
uch energy is stored in each m
ode.•
Need to know
how m
any phonon modes there
are.•
Need to sum
the thermal energy over all
modes
•H
eat capacity is then the derivative of the therm
al energy.
• •E
nergy stored in a phonon E
nergy stored in a phonon ‘ ‘normal m
odenorm
al mode
’ ’•
Each m
ode has an energy E=ħω
(n+ ½
) where
n is the number of phonons in the m
ode.•
The factor of ½
is the ‘zero point’energy –it
cannot be removed. S
ince thermal energies
are taken to be zero in the ground state, it will
be ignored in this treatment.
•It the solid is in therm
al contact with som
e fixed tem
perature ‘reservoir’then the probability of the m
ode having n phonons relative to the chance of it having none is given by a B
oltzmann factor: P
n = exp(-n ħω
/kB T
)
44
Energ
y/n
orm
al m
ode,
Energ
y/n
orm
al m
ode,
contin
ued
contin
ued
•C
alculate average energy stored in a particular normal
mode (i th
) by averaging over all possible values of n(0 to
∞).
[]
1)
exp
()
exp
(1
)ex
p(
:is
mo
de
p
articular
a
in
sto
reden
ergy
averag
e
Hen
ce
)ex
p(
1
)ex
p(
)ex
p(
)
exp
(
:r
den
om
inato
o
f
is
Nu
merato
r
)ex
p(
1
1)
exp
(
:)
exp
(
ratio
series
geo
metric
a
isr
Den
om
inato
1
wh
ere
)ex
p(
)ex
p(
)ex
p(
)ex
p(
20
0 0
0
0
0
0
−=
−−
−=
−−
−=
−∂ ∂
−=
−
∂ ∂−
−−
=−
−
=−
−=
−
−=
∑∑ ∑
∑
∑
∑
∑
∞=
∞= ∞=
∞=
∞=∞=
∞=
Tk
E
nn
n
n
Tk
β
n
nn
Tk
n
Tk
nn
E
B
i
i
i ii
i
i
i
n
i
n
ii
in
i
i
B
n
i
n
ii
nB
i
nB
ii
i
ωω
βω
βω
ω
βω
βω
ωβ
ωβ
βω
ω
β
βω
βω
βω
βω
βω
ω
ω
ωω
h
h
h hh
h hh
hh
h
hh
h
h
hh
h
hh
Planck’s form
ulafor a singleoscillator
Planck’s form
ulafor a singleoscillator
45
Heat C
apacity
at H
igh
Heat C
apacity
at H
igh
Tem
pera
ture
sT
em
pera
ture
s
•Low
temperature (k
B T<<ħω
) limit of energy:
•H
igh temperature lim
it of energy (1>>ħω
/ kB T
)
• •H
ow m
any phonon modes?
How
many phonon m
odes?•
If a crystal contains N atom
s, you need 3N
coordinates to describe position of all N atom
s and so there w
ill be 3N norm
al modes.
• •T
hermal behaviour of w
hole crystal at high T
hermal behaviour of w
hole crystal at high tem
peratures:tem
peratures:•
Since each m
ode stores kB T
of energy at ‘high’tem
peratures, and there are 3N
modes, then the total
energy stored at high temperatures is 3
Nk
B Tand the
heat capacity for kB T
>>ħω
ι is 3N
kB . (B
asis of Dulong
and Petit’s
law, 1819 –
heat capactiy/atomic w
eight constant.)
T sm
allfo
r )
exp
(
1)
exp
(T
k
Tk
EB
ii
B
i
i
i
ωω
ωω
hh
h
h−
≈−
=
Tk
Tk
Tk
EB
B
i
i
B
i
i
i=
−+
≈−
=1
11
)ex
p(
ωω
ωω
h
h
h
h
46
De
bye
Th
eo
ry: T
he
Aim
De
bye
Th
eo
ry: T
he
Aim
• •T
hermal behaviour of w
hole crystal at intermediate
Therm
al behaviour of whole crystal at interm
ediate and low
temperatures.
and low tem
peratures.•
At ‘non-high’
temperatures, E
i depends on ħωi so w
e need a w
ay of summ
ing the contribution all the modes:
•T
he first step is to convert the sum to an integral:
where g
(ω)=
dN
/dωis the ‘density of states and g(ω
)δωgives the num
ber of phonon states δN
with energies
lying between ω
and ω+δω
.•
The actual g
(ω)
is complicated –
the Debye theory of
heat capacity works by producing a sim
plified model for
g(ω
)so that the integral for E
total can be performed.
∑=
−=
Ni
B
i
i
tota
l
Tk
E1
1)
exp
(ω
ωh
h
()
∫ ∞
−=
01
)ex
p(
ωω
ωω
dg
Tk
E
B
tota
lh
h
47
Bo
un
da
ry C
on
ditio
ns a
nd
Mo
de
ls
Bo
un
da
ry C
on
ditio
ns a
nd
Mo
de
ls
for
for g
(g
(ω ω): P
erm
itted
k v
alu
es.
): Pe
rmitte
d k
va
lue
s.
• •R
eflecting B.C
.R
eflecting B.C
.•
Reflecting boundary conditions give standing w
ave states.•
At boundary m
ay have node (photons, electrons) or antinode(phonons).
• •C
yclic boundary conditionsC
yclic boundary conditions•
N+
1th
atom equivalent to 1
st.•
Travelling w
ave solutions.•
1D you can w
rap into a circle, but cyclic b.c.harderto justify in
2 and 3D.
•C
an consider repeating your block on N atom
s with identical
units to fill infinite space and requiring all blocks to have identical atom
ic displacement patterns. O
k classically, but hard to get the quantum
mechanics correct.
• •‘ ‘InfiniteInfinite
’ ’extent:extent:
•A
s soon as you use the modes derived you m
ake a w
avepacketof some sort w
ith zero amplitude at infinity, w
hich fits any b.c. at infinity, but this m
ethod does not give density of states.
A
nk
n A
n n
π
λ
2= =
A
Reflectin
g B
.C.
Cyclic B
.C.
A nk
n A
n n
π
λ
= =2
n=
1
n=
2
n=
3
n=
1
n=
2
n=
3
48
Debye M
odel:
Debye M
odel: g
(g(ω ω
) for
) for ‘ ‘W
aves
Waves
in a
Box
in a
Box’ ’
•F
or small values of k
(long wavelengths) phonons look
like sound waves –
with a linear dispersion relation
ω=
vs k
where v
sis a m
ean speed of sound (see discussion later).
•T
he Debye m
odel assumes this is true for all
wavelengths –
not just long ones –i.e. it ignores the
structure in g( ω) due to the atom
ic nature of the m
aterial.
•g(ω
) is calculated by assuming that the crystal is a
rectangular box of side lengths A,B
,C. W
e use reflecting b.c., though cyclic b.c. give sam
e results
•In each dim
ension the there must be a w
hole number of
half wavelengths across the box so as to fit the
boundary conditions, i.e.in each direction A
=nλ/2
and k=
nπ/A. T
he total wavevector
of the phonon must be:
•V
olume per state in k
space is π3/(A
BC
) i.e. π3/V
where
V is the volum
e of the box.
•(k
not qis used in this derivation because the idea of w
aves in a box applies to m
any problems in physics –
including black body radiation and the free electron m
odel of a solid, and by convention kis used in these
derivation.)
=C n
B n
A nz
yx
ππ
πk
49
g(
g(ω ω
) for
) for ‘ ‘W
ave
s in
a B
ox
Wa
ve
s in
a B
ox’ ’II II
•F
or cyclic BC
, states 2x as far apart –vol/state = 8π
3/Vbut
require full shell, not just +ve
octant, –net result sam
e g(k).
•C
an show (W
igner) results independent of shape of box.
()
()
()
()
()
()
()
bo
x)
th
eo
f
vo
lum
e th
eis
(
2 3
and
1
so
wav
eso
un
d
aF
or
,
NS
ince
23
.8
43
32
2
22
32
AB
CV
v
Vg
vd
dk
vk
dd
kk
gg
gk
kg
AB
Ck
kg
AB
Ck
kk
kg
N
ss
s
=
==
=
==
=
=⇒
⋅=
=
πω
ωω
ω
ωω
δωω
δδ
ππ
δπ
δδ
All states in the shell
have same |k|
All states in the shell
have same |k|
States uniform
lydistributed in
k-space
States uniform
lydistributed in
k-space
1 state “occupies avolum
e”(π
3/AB
C)
1 state “occupies avolum
e”(π
3/AB
C)
Volum
e of shell V
olume of shell
Vol. of one state
Vol. of one state
3 Polarisations/k state
3 Polarisations/k state
•F
irst, work out g(k) from
no. of states, δ
N, that have a
wavevector
of magnitude
between k
and k+ δ
k.These
states lie within the positive
octant of a spherical shell of radius k
and thickness δk. (k
+ve
for standing waves.)
•F
or each phonon mode
there are two transverse and
one longitudinal polarisation, i.e. 3 m
odes per point in kspace.
50
Inte
rnal E
nerg
y in
Debye
Inte
rnal E
nerg
y in
Debye
Model
Model
•H
eat capacity follows from
differentiating the internal energy (as usual).
•F
or the present we w
ill ignore the zero point motion.
•N
eed to make sure you integrate over the correct num
ber of m
odes –use the fact that if there are N
atoms in the
crystal (volume V
) then there are 3N
modes. D
ebye suggested sim
ply stopping the integral (at the ‘Debye
frequency’, ωD ) once 3
Nm
odes have been covered, i.e.
•Internal energy
•H
ence:
(We can now
see that the appropriate mean velocity is
where v
Land v
Tare the longitudinal and
transverse sound wave velocities)
()
VN
vv
Vd
v
Vd
gN
sD
s
D
s
DD
/6
22 3
33
23
32
3
0
32
2
0
πω
π ωω
πω
ωω
ωω
=⇒
==
=∫
∫
()
()
∫−
=D
dg
kTU
ωω
ωω ω
01
exph
h
No. of phononsin dω
at ω
Energy per phonon(P
lanck formula)
()
()
∫
∫
−=
−=
D
D
dkT
v
V
dkT
v
VU
s
s
ω
ω
ωω
ωπ
ωω
ωπ
ω0
3
32
03
2
2
1ex
p
1
2 3
1ex
p
1
2 3
h
h
hh
+=
33
3
21
3 11
TL
sv
vv
51
Heat C
apacity
With
in th
e
Heat C
apacity
With
in th
e
Debye M
odel
Debye M
odel
•D
ifferential Uto get heat capacity C
:
()
()
()
[]
()
()
()
[]
[]
by
giv
en
is
re
temperatu
Deb
ye
th
ew
here
19
1ex
p exp
6
19
1ex
p
exp
2
3
1ex
p
1
2
3
T
02
4
3
02
4
3
32
3
02
23
32
0
3
32
DB
DD
x
x
D
B
Tk
BB
B
B
B
s
B
B
B
s
Bs
k
dx
e
ex
TN
k
Tk
dT
k
Tk
Tk
Tk
Nv
VN
k
dT
k
Tk
kT
v
V
T UC
dT
kv
VU
D
B
D
D
D
θω
θ
θ
ω
ω
ωω
π
ωω
ωω
ωπ
ωω
ωπ
θ
ω
ω
ω
=
−
=
−
=
−=
∂ ∂=
−=
∫
∫
∫
∫
h
h
h
hh
h
h
hh
h
h
h
h
C/3NkB
T/θ
D
31
Dω
52
Debye T
em
pera
ture
s
Debye T
em
pera
ture
s
•D
ebye frequency controlled by mass of atom
s and stiffness of lattice.
•A
s you go down a group in the periodic table (e.g. Li to
Cs or C
to Pb) the m
ass increases and the atoms
become bigger and m
ore deformable –so the rigidity
goes down.
•T
ransition metals tend to lie betw
een 200-600K –
so D
ulongand P
etit’slaw
works roughly at room
tem
perature.
•N
ote very high value for carbon –a light atom
and a very rigid lattice.
3838
5656
9191
158158
344344
θ θD D /K /K
Cs
Cs
Rb
Rb
K KN
aN
aLiLi
Elem
entE
lement
105105
200200
374374
645645
22302230
θ θD D /K /K
Pb
Pb
Sn
Sn
Ge
Ge
Si
Si
C CE
lement
Elem
ent
θ θD D /K /K
Elem
entE
lement
437437
343343
450450
445445
470470
410410
630630
380380
420420
360360
Zn
Zn
Cu
Cu
Ni
Ni
Co
Co
Fe
Fe
Mn
Mn
Cr
Cr
V VT
iT
iS
cS
c
53
Heat C
apacity
at
Heat C
apacity
at
Low
Tem
pera
ture
Low
Tem
pera
ture
• •C
heck high temperature behaviour:
Check high tem
perature behaviour:
• •Lim
iting behaviour as Lim
iting behaviour as T T→ →
0 0. .•
At low
temperature the higher frequency m
odes are not excited. T
hus contributions to the integral for large ω
(>ω
D ) can be ignored and ωD
replaced by ∞
.
•N
ote similarity to heat capacity of a vacuum
–photons in black body radiation.
()
()
()B
D
B
x
x
D
x
x
D
B
Nk
dx
xT
Nk
C
xx
e
eT
dx
e
ex
TN
kC
D D
39
:an
d1
11
1
1fo
r
19
T
0
2
3
22
2
T
02
4
3
=
≈
=−
+≈
−>>
−
=
∫ ∫
θ θ
θ
θ
θ
Integral = 4π
4/15
Integral = 4π
4/15
Debye, T
3Law
Debye, T
3Law
()
3
34
02
4
3
5
12
19T
C
TN
kC
dx
e
ex
TN
kC
D
B
x
x
D
B
∝
=
−
=
∫∞
θπ
θ
54
Measure
d D
ensity
of S
tate
sM
easure
d D
ensity
of S
tate
s
• •E
xample: A
luminium
E
xample: A
luminium
(shows com
mon features)
(shows com
mon features)
•M
easured density of states compared w
ith D
ebye approximation.
•B
oth measured and D
ebye density of states are sim
ilar at low ω
, as expected (ω ∝
q).•
Debye frequency chosen to give sam
e total num
ber of modes (i.e. equal area under both
curves)•
Largest deviations where phonon m
odes approach zone boundary.
•M
easured curve is complex because the 3-D
zone has a relatively com
plicated shape, and the transverse and longitudinal m
odes have different dispersions (as w
e have seen earlier).
55
Therm
al C
onductiv
ityT
herm
al C
onductiv
ity
• •P
honons and thermal conductivity
Phonons and therm
al conductivity•
Phonons are travelling w
aves that carry energy and can therefore conduct heat.
•K
inetic theory gives the thermal conductivity
•E
xcess temperature of phonons crossing plane
•E
xcess energy in each phonon mode
•l= m
ean free path –phonons assum
ed to therm
aliseat each collision.
θco
sl
dz
dT
zd
z
dT
T−
=∆
=∆
θco
sl
dz
dT
cT
cp
hp
h−
=∆
l
θ
z
∆θ
zl
= -
cos
heat capacity ofa phonon m
ode
heat capacity ofa phonon m
ode
56
Definition of therm
al conductivityD
efinition of thermal conductivity
()
[][
] []
()
()
3
3 1
cos
cos
2 1
cos
sin2 1
cos
cos
2sin
1
1
2
00
2
00
cn
lc
dz
dT
dz
dT
cn
lc
H
dc
dz
dT
nl
cH
dc
cf
cd
dz
dT
nl
cH
dz
dT
lc
cd
dc
cn
fH
ph
ph ph
ph
ph
=
−=
−=
+=
−=
−=
∫
∫∫
∫∫
−
∞
∞
κ
κ
θθ
θθ
θ
θθ
θθ
π
π
Therm
al C
onductiv
ity II
Therm
al C
onductiv
ity II
•N
umber density of phonon m
odes, n
•H
eat flux across plane
•T
hermal conductivityl
cC
3 1=
κM
ean free pathM
ean free path
Average speed
Average speed
Heat capacity per unit vol
Heat capacity per unit vol
θdθ
2 sin
dπ
θθ
()d
cc
fn
number w
ithspeed c
to c+d
c
number w
ithspeed c
to c+d
c
fraction with
angles θto θ
+dθ
fraction with
angles θto θ
+dθ
2sin
4sin
2θ
θπ
θθ
πd
d=
c
speed normal to plane
speed normal to plane
net heat per mode
net heat per mode
57
Mean F
ree P
ath
for P
honons.
Mean F
ree P
ath
for P
honons.
•M
ean free path –lim
ited by scattering processes
•W
ith many scattering processes add scattering rates:
Thus, the shortest m
ean free path dominates.
•“G
eometric”
scattering:•
Sam
ple boundaries (only significant for purest samples
at low tem
peratures).•
Impurities/grain boundaries: lindependent of T.
•P
honon-phonon scattering:•
True norm
al modes do not interact w
ith each other. •
How
ever, in an anharmonic
lattice, phonons can scatter. A
s a phonon extends/compresses the bonds it changes
the spring constants and one phonon can diffract from
the grating of variable elastic properties produced by another phonon.
LL
++
==>
++
=2
12
11
11
ll
ll
cl
cl
c
V(r)
Sp
ring
Con
st.
r/r0
Lennard-Jones 6-12 potential
Mean phonon am
plitude at 20K
in Ne is 1%
of m
ean nearest neighbour distance and gives significant changes in the spring constant.
58
Te
mp
era
ture
De
pe
nd
en
ce
of
Te
mp
era
ture
De
pe
nd
en
ce
of
Th
erm
al C
on
du
ctiv
ity o
f Insu
lato
rsT
he
rma
l Co
nd
uctiv
ity o
f Insu
lato
rs
•In insulators there are no contributions from
free electrons. •
In pure crystalline form the conductivity can be very high
(larger than metals e
.g. at 70K diam
ond has κ=
12000W/m
/K, at 300K
Copper has κ=
380W/m
/K)
•N
on-crystalline systems have m
uch lower conductivity l~
local order e
.g.at 300K
for glass has l~ 3Å, and rubber
has l~ 10-20Å
.•
Therm
al conductivity shows strong tem
perature dependence.
•Low
temperatures:
-few
phonons, geom
etric scattering dom
inates, lconstant.-C
and hence κ α T
3.•
High tem
peratures:-C
constant (3N
kB )
-no of phonons α
T, so
l α 1/T
and κα
1/T
Th
erm
al C
on
du
ctiv
ity o
f Ge
v.
Te
mp
era
ture
0.1 1 10
100
110
1001000
Tem
pera
ture
/K
Thermal
Conductivity/W/cmK
αT
3
α1/T
lc
C3 1
=κ
•Interm
ediate temperatures
–expect conductivity to
be below 1/T
and T3
asymptotes, (heat capacity is
dropping and there is both phonon and geometric
scattering) but actually get a large rise for pure, crystalline sam
ples.
59
Therm
al C
ond
uctiv
ity a
t Inte
rmed
iate
T
herm
al C
ond
uctiv
ity a
t Inte
rmed
iate
Te
mp
era
ture
s:P
hono
nT
em
pera
ture
s:P
hono
n- -P
ho
no
n
Pho
no
n
Scatte
ring a
nd
Scatte
ring a
nd U
mkla
pp
Um
kla
pp
Pro
cesses.
Pro
cesses.
•A
typical phonon-phonon collision process is coalescence:
•H
owever, if the resulting phonon has a m
omentum
that is sim
ply the addition of the two colliding phonons –
the collision is very ineffective as far as therm
al conductivity goes –
you still have the same energy going in roughly the
same direction.
•T
o really change the direction energy is flowing –
the sum of
the mom
entaof the incom
ing phonons must be outside the
first Brillouin
zone –so that you get an ‘um
klapp’(german
for ‘fold around’) i.e.
the new phonon m
omentum
has a G
vector subtracted from the sum
of the old ones, which
changes significantly the direction of travel.
q1
q2
q3
q1
q2
q3
G•
As the tem
perature drops few
er phonons have enough energy to have enough m
omentum
to give a resulting phonon that is produced by an um
klappprocess and so the
mean free path and hence κ
rise dramatically
Um
kapp
Process
60
Um
kla
pp
Um
kla
pp
Pro
cesses
Pro
cesses – –
a 1
D
a 1
D
exam
ple
exam
ple
•D
irection of travel determined by group velocity dω
/dq
•T
o change direction you need to have enough mom
entum
in the incoming phonons, q
1 +q
2so that q
3is outside the
first Brillouin
zone, so that its group velocity has a direction very different from
q1
and q2 .
•B
y convention if q3
is outside the first B.Z
. a G vector is
subtracted to show it in the first B
.Z.-
hence the ‘umklapp’.
q3
= q1
+ q
2 L
T
q3 -G
q2
q1
Direction of travel
(group velocity)
G/2
-G/2
1stB
rillouinZ
one
Phonons w
ith enough qfor
umklapp
poorly excited at interm
ediate temperatures