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