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1

October 14th

, 2010

Class 11

Crystal Vibrations: Monoatomic basis

Consider a family of planes in the crystal and assume one of these planes is moved out of its

equilibrium position. Assuming all the planes are connected to each other by elastic interactions,

this action will produce a force in all other planes. Assuming a simple harmonic model, a modelwhere a force is a linear function of the displacement, the force on plane s is proportional to the

change in the distance between this and another plane locatedp planes from it (remember that for

a spring F=-k(x-xo)) andE= k(x-xo)2). Considering only consecutive planes,p=1

ssssssss

s uuuCuuCuuCdt

udMF 211112

2

where C is the elastic constant. Since in the equation above the second derivative of us isproportional to us, the function itself is a sine, a cosine or a complex exponential. It is easier to

elaborate on this if we think that the above is the force on 1 atom in the plane rather than on theentire plane.

Assuming an exponential function of time as a solution in time:

tiexpaut,au ss where us was used for both the space-time function and the spacefunction of the displacement to follow the book convention.

By using the above definition

ssss uuuCuM 211

2 where now, u is the oscillation amplitude.

This is a difference equation in the displacement and has traveling waves solutions of the form

Kasiuaus 1,0exp1,0

so KasiKasiKasiuCKasiuM exp21exp1expexp2 What gives, after some simple algebra and canceling Kasiuexp at both ends of the equal

KaCiKaiKaCM cos122expexp2 or using that

2

1sin2cos1

2

2

4 22 Kasin

M

C what leads to the dispersion relation

2

4 Kasin

M

CK

We can sample the limits of this equation. For the long wavelength limit (small K),

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2

22

11cos KaKa and 22 cos12 Ka

M

CaK

M

C

or KM

Ca2

On the other end, the wave is stationary for K=/a. These are the limits of the first Brillouin

zone.

for Kmax=/a 2/min=/a or min=2a and

ss uisexpuaa

isexpuu 1

Thus, consecutive planes move in opposite phase.

All the physics of the problem is already in the first Brillouin zone, second and others are

redundant and do not produce new information.

To understand this, notice that us are actually the amplitude of the oscillation at each atom, thus

any wave for which the amplitude at each atomic position is NOT a new solution since there arenot atoms to see intermediate amplitudes. Notice in Figure 5 (page 93) that the two waves predict

the same amplitude at each point.

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October 19th

, 2010

Class 12

Crystal Vibrations: Monoatomic basis

We found that the dispersion relation for monoatomic basis was

24 22 KasinMC what leads to the dispersion relation

24 KasinMCK

That for the long wavelength limit (small K) leads to,

22

11cos KaKa and 22 cos12 Ka

M

CaK

M

C

or KM

Ca2

and for K=/a leads to stationary waves (the derivative is zero at those points). These are thelimits of the first Brillouin zone.

for Kmax=/a 2/min=/a or min=2a and

ss uisexpuaa

isexpuu 1

Thus, consecutive planes move in opposite phase.

All the physics of the problem is already in the first Brillouin zone, second and others are

redundant and do not produce new information.

To understand this, notice that us are actually the amplitude of the oscillation at each atom, thus

any wave for which the amplitude at each atomic position is the same and NOT a new solution

since there are no atoms to see intermediate amplitudes. Notice in Figure 5 (page 93) that the twowaves predict the same amplitude at each point.

Notice that the ratio of the amplitude of the atomic displacement between consecutive planes is

iKaexpsKaiexpu

Kasiexpu

u

u

s

s

11

All independent values of this ratio are obtained for Ka between -and . Given any value ofK

outside the mentioned [-,] interval, there exist another wavevector K=K-2n/a (with n andinteger) that it is in the interval. For this wavevector Kin the first Brillouin zone

aiKaa

niiKaa

a

nKiiKa

u

u

s

s 'exp2

expexp2

'expexp1

Thus the relative amplitude is exactly the same for Kas it is for a K in the first Brillouin zone.

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Group Velocity

When a packet of waves with different wave lengths (and frequency) travel together, it is not thespeed of each of the components what matters, but the velocity of the group (imagine a pulse

sent through space). In non-dispersive mediums, frequency and wavevector are proportional and,

as we will see, phase velocity (the one of individual components) is equal to group velocity (like

light in vacuum), all the components travel at the same speed. But in general that is not the case,each component travels at a different speed, and the actual package travels at a speed giving by:

dK

dvg

To see this, consider two waves with a slightly different frequency and wavevector (and just forillustration purpose, the same amplitude too). The composed waves will be:

x'Kt'iKxti ueue

This equation can be re-arranged to give

x

Ktcosue

xKti

222

Where and K are the average frequency and wave vector.

That is a high frequency wave with a low frequency envelope. The velocity at which the

package moves is the velocity of the envelope, thus

Kvg

thus the vg indeed give the speed of the package.

If for a particular medium, the frequency and the wavevector are proportional, then group and

phase velocity are the same, and equal to the speed of each component. In this case the mediumis non-dispersive and all the components move at the same speed, thus that package does not

change shape. If instead different components travel at different speeds, the package will spread,

or disperse as it moves on (thus the relationship between and K is known as the dispersion

relationship)

For a continuum of frequencies,dK

dvg

In the case of phonons, using the dispersion relation obtained above.

02

cos

02cos

2

2

KaKa

M

Ca

Ka

Ka

M

Ca

dK

dvg

vg is discontinuous for K=0 where it goes from -1

to 1 in going from negative values ofKto positive values ofK. Which is not surprising, it onlymeans that the wave now moves in the opposite direction.

In the limits of the Brillouin zone, Ka=and vg=0, consistent with standing waves.

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As we saw before, for the long wavelength limit (small K),

KM

Ca2

Thus the group velocity is equal to /K. This can be understood by realizing that in the long

wave length limit K~0, >>a and the interaction with the lattice is minimal leading to a non-

dispersive behavior.

Force constant from the interaction with non-consecutive planesIf we consider now the effect of a planep places from the reference plane s then

aKpsiexpuau ps

from where the dispersion condition can be easily shown to be

pKacosCM p

122 and considering all the planes

0

21

2

p

p pKacosCM

Where Cp, the force constant associated with the interaction between plane s andp is given by:

(See derivations provided)

dKpKaMa

C

a

a

p

/

/

2cos

2

More than one type of atoms per primitive basisWhen the basis consist on one atom, there are three independent vibration modes, 1 longitudinal

(the atoms in the plane oscillate in a direction perpendicular to the plane) and 2 transversal where

the atoms oscillates parallel to the plane (the two modes come from the two dimensions of the

plane). When the basis have more than one type of atom, then each type of atom give rise tothree independent vibration modes, thus in a crystal with a p-atoms basis, there are 3p

independent modes. Out of the 3p modes, 3 and called acoustical modes and 3p-3 are called

optical modes for a reason that will be evident soon. Also both optical and acoustical modeshave transversal and longitudinal in the rate of 2-to-1.

In what follows we will consider a cubic crystal with a two atoms base with alternate planes ofatoms 1 (with massM1) and atoms 2 (with massM2).

ssss uvvC

dt

udMAtom 2:1 12

2

1 sss

s vuuCdt

vdMAtom 2:2 12

2

2

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Solutions to the above equations can be proposed to be of the form

sKatiuu s exp and sKativvs exp Where a is the distance between plane of atoms of the same kind

Using these solutions in the equation above,

uiKavvCuM 2exp21 viKauuCvM 2exp22 What leads to a set of two homogeneous equations with two unknowns,

021

012

2

1

2

1

MCueC

eCuMC

iKa

iKa

This set of equations has non-trivial solution only if the following determinant is zero

22

2

1

2exp1

exp12

MCiKaC

iKaCMC

what leads to

0122 22214

21 KacosCMMCMM

with solutions

21

2

21

2

21

2

212

2

1842

MM

KacosCMMMMCMMC

For the limit Ka

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5

2121

21

21

21

22

2121

2 1122

4

2

4

44

MMC

MM

MMC

MM

MMC

KaCMMMMC

To the first order in Kaand

2

2121

21

22

21

2

22

4

4

KaMM

C

MM

MMC

KaCMM

By using these solutions in the set of equations for u and v and solving for them, it can be seen

that for the first of the solutions in the K~0 limit,1

2

M

M

v

u , thus the consecutive planes vibrate

in opposite direction keeping the center of mass fixed. If the two atoms are ions of opposite sign,this mode can be excited by an electromagnetic field, thus this mode is called optical mode. The

other solution for the K~0 leads to u=v thus the planes move in the same direction as it is forlong wavelengths sound waves, thus this mode is known as acoustical mode.

For the other limit, Ka= cos(Ka)=-1 and thus the solutions are :

21

2121

21

2

21

2

21

2

212

2

22

2

1642

MM

MMCMMC

MM

CMMMMCMMC

Or2

2 2

M

C and

1

2 2

M

C

This leads to a solution where each type of atom vibrates independently from the other. One

solution correspond to one type of atom vibrating while the other is at rest and viceversa

There is a gap between optical and acoustical mode at the boundary of the first Brillouin zonewhere frequencies between (2C/M2)

2 and (2C/M1)2 are not allowed.

Figure 8 shows both modes. Also in Figure 8, it can be seen that a 3-D crystal, longitudinal and

transversal vibrations split in frequency.

In a crystal with ap atom base there are 3 acoustical modes (1 longitudinal and 2 transversal) and3p-3 optical modes.

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Quantization of Elastic Waves

By describing vibration in the lattice by a quantum harmonic oscillator model, it is observed thatthe lattice energy is quantized, in other words, the energy of a lattice vibrating with a single

frequency is:

21n

That means that no just any value of energy is allowed but the ones in that equation. For n=0 the

lowest energy state is obtained and it is known as zero point energy. Other values ofn representexited states of the crystal vibration where n photons populate the state. Phonons are not

subjected to the Pauli Exclusion Principle (as photons are not) thus any number of phonons can

occupy a given state, however you add phonons one by one, a fraction of a phonon would not do.

The quantization of the energy has the consequence to produce quantization on the amplitude of

the oscillation.

To see this consider a vibration described by

tcosKxcosuu o

As for a harmonic oscillator, the temporal average of the energy is half kinetic and half potential.

Lets consider the kinetic energy density

2

2

1

t

uK , where is the mass density

The volume integral of the kinetic energy density gives (notice that the integral of the cos2 isLx/2

while the integral iny andz givesLy andLz), thus the volume integral is V/2

V2uo2sin

2t

with a time average of8

1 V2uo

2 thus (since =1/2). Making this average kinetic

energy equal to of the total energy

81 V

2uo

2= (n+)

we obtain that

uo2= 4(n+)/V= 4(n+)/M

Thus the vibration amplitude is related to the occupancy n of the mode.

The solution of the motions equation for atoms in the crystal gives 2 so, what is the sign of?

Conventionally is assumed positive. For unstable structures, 2 is negative and is

imaginary.

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Phonon Momentum

A phonon of wavevector Kcan interact with photons, neutrons, electrons, and other particles asif they had a momentum K. Looking at this in a different way, if a photon (or other quantum

particle) collides with the crystal, without losing energy, the selection rule for the incoming

particle wave vector imposes that

k=k+G

No energy is transferred to the crystal but the crystal as a whole recoils with a momentum -G.

If inelastic scattering occurs, some energy is indeed lost by the incoming particle to the crystal

that will excite phonon modes, besides the fact that the transferred energy should be at least ,

the energy of a phonon of frequency, there is a generalized selection rule that can be written as

k+K=k+G

whereKis the phonons wave vector, thus the wavevector of the scattered particle is conditioned

to the wavevector of the created phonon such that the above equation is satisfied. In the sameway, if a phonon is annihilated in the interaction, the equivalent selection rule is:

k=K+k+G

Neutrons are often used to determine the dispersion relations for the phonon. After scattering,

the energy and direction of scattered neutrons is recorded thus k and kare determined.

The kinetic energy of the neutrons is p2/2Mn and the momentum is p= k, so conservation of

energy imposes

nn M'k

Mk

22

2222

Where is the energy of the phonon created (+) or absorbed (-) from

where is obtained.

By knowing the crystal lattice (G) and from the formula kK=k+G, Kis determined, where G

is chosen such thatKis in the first Brillouin zone.

The sign applies to phonon created or absorbed.

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October 21th

, 2010

Class 13

Phonon Heat Capacity

The heat capacity of a material is the ratio between the heat (energy) provided to the material andthe change in its temperature. In other words, the amount of heat necessary to increase the

temperature ofsaid material in 1 degree. A good analogy is a liquid container, the heat is the

water poured into the container, the temperature is the liquid level on the container, the heatcapacity is a relationship between the two, how much water you need to add so the level raisee 1

unit. A good analogy for a large heat capacity material in comparison to a small heat capacity is

a wide and a narrow container, in a narrow container (small heat capacity), a small amount ofwater will be enough to raise the level significantly, while for a wide container (large heat

capacity), more water is needed to produce the same water level.

When heat is added to a material, the energy can be used for many things, one is to excite

phonon modes and that is what we will be concerned about here. The heat capacity can be

defined at a constant volume or a constant pressure, while experimentally heat capacity is often

measured in a constant pressure condition, the heat capacity at a constant volume is morefundamental (since it only involves changes in internal energy as opposed to the case at constant

pressure where some of the energy is used to change the lattice volume involving mechanical

work). The two heat capacities CV and Cp are related such that Cp-CV=f(T) thus knowing one,the other can be obtained.

V

VT

UC

Where U is the energy. The different contributions to the heat capacity can be

obtained by taking the derivative with respect to temperature of the corresponding energy.

Particularly, the contribution to the heat capacity of the phonon modes, is known as the lattice

heat capacity Clat. The total phonon energy (ignoring the zero point energy) is:

K p

p,Kp,Klat nU

is the average occupancy of each of the phonon modes at thermal equilibrium, K is the

wave number related with through the dispersion relation, andp in the polarization (acoustical,

optical, transversal, longitudinal)

The average occupancy of a given mode is given by the Plank distribution. (See book for

the steps leading to this equation)

1

1

Tkexp

n

B

Combining the two

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2

K p

B

p,K

p,K

lat

Tkexp

U

1

At this point it is important to remember that the Ks are discrete but too close to each other and

can be considered a continuum, but they form a continuum and so an integral is more appropriate

than a summation. If the number of modes in the frequency range+dwith polarizationp is

given by Dp()d, then

p

B

plat

Tkexp

DdU

1

Where of course the dispersion relation allows us to integrate in instead ofK. Thus

p

x

x

pBlat

lat

e

exDdk

T

UC

2

2

1

WhereTk

xB

Now the heat capacity can be obtained if the density of modes is known. Coming up next, we

revise some models to determine the density of states and with that to solve the integral andobtain the density of states.

Density of States in One DimensionThe values ofK that are possible in a given lattice would be a continuum unless appropriate

boundary conditions are applied. One common type is the periodic boundary condition, in one

dimension that simply means that the first and last point in the chain are entirely equivalent (thusstarting the sequence again). A condition like that implies

u(sa)= u(sa+L)

By imposing this condition to the phonon wave function in a line ofN+1 sites, such that site 1

and siteN+1 are equivalent by the boundary conditions (Nindependent sites)

uoei(Ksa-(K)t)= uoe

i(K(sa+L)-(K)t)=uoei(Ksa-(K)t) eiKL

What implies that eiKL=1 or KL=2n or

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L

N,,

L,

L,

L,K

6420 Notice that for N/L (where L/N=a) we have e

iwhere +

and -are actually the same angle and thus the same solution, so there are exactly Nindependent

solutions. For n >Nthe same solutions are obtained.

LN

LLLK ,,6,4,2,0

Thus for a chain ofN+1 atoms with periodic boundary conditions, there are exactly Nstates, 1per each atom moving independently in the chain.

Also notice that K=n/L actually refers to a wave with the same frequency only propagating inopposite directions.

The distance between two points in the K-space is 2/L. The number of states per unit of length

in the K space is then L/2and thus the number of states of absolute value in the range [0,K]is

then given by

KL

NK2

2

(where the 2 account for the two waves with the same Kpropagating in opposite direction n/a).

The density of states, number of states per unit of length can be defined as dN/dKforNlargeenough that the states can be consider a continuum. So

dKd

dLd

d

dKLd

d

dK

dK

dNdNdD KK

2

2

The dependence of the density of states in terms of has been forced, but now the density of

states is expressed in terms of the group velocity which can be obtained if the dispersion relationis known.

Density of States in Three Dimensions

By applying periodic boundary conditions in the same way that we did for 1-D, the number of

states inside a sphere of radius Kis:

3

4

2

33KL

NK

whereL

3=V

Thus the density of states is then given by:

dK

d

dVKd

d

dK

dK

dNd

d

dNdD KK

2

2

2

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The problem is now reduced to finding the group velocity (or the dispersion relation) for a

particular crystal and to apply the equations above.

Debye Model for the Density of States

In this model, the velocity of sound is taken as a constant for each polarization type, thus

Kp and 32

2

2

2

1

2p

V

dKd

VKD

Now we have all the ingredients and we can calculate the lattice energy (under the Debye

approximation).

p TkppTk

plat

d

B

d

B e

Vd

e

DdU

0 32

2

0

12

1

Where D is the Debye frequency and correspond to that frequency associated with themaximum value ofK, KD. Now, if we assume that the velocity is independent of the polarization

mode and we consider three polarization modes (3 acoustical, 1 per dimension)

dd

B

x

x

B

Tklat

e

xdx

TVk

e

dV

U0

3

332

44

0

3

3212

3

12

3

Where the following variable substitutions has been made

from where

Here it is convenient to introduce another definition, the Debye temperature

what leads to

The Debye temperature is then the temperature at which xD=1 or the temperature at which themaximum phonon energy is equal to kBT.

Now using that

3

33

3

4

2

LNK to obtain an expression for .

WhereNis the total number of occupied states

With these definitions, now

1519

12

343

0

33

0

3

32

TTNk

e

xd

TTNk

e

dV

U Bx

xBTk

lat

dd

B

What leads to

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5

3

3

378.233

TT

UC latV

This is known as the Debye T3 approximation. See figure 9, page 115 for the heat capacity of

Argon, where this law is in very good agreement with the experiment. For Argon, the Debye

temperature is 92K. The agreement with this law worsens as T approaches the Debyetemperature, the law is most approximately met for Twithin 1-2% of the Debye temperature.

Thus the Debye approximation is good for very low temperature (compare to the Debyetemperature).

Einstein Model of the Density of States

This model considersNoscillators all vibrating at the same frequency o and thus

D()=N(-o)

1

3

1

Tk

o

p Tkolat

B

oB

e

N

e

NdU

22

1

3

Tk

o

B

oBV

B

o

eTk

NkC

For large T, CV~3NkB, the Dulong and Petit Value.

In summary, the Einstein model accounts for the high temperature limit while the Debye model

accounts for the low temperature regime.

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October 25th

, 2010

Class 14

Phonon Heat Capacity

V

V

T

UC

Where U is the energy. The different contributions to the heat capacity can be

obtained by taking the derivative with respect to temperature of the corresponding energy.Particularly, the contribution to the heat capacity of the phonon modes, is known as the lattice

heat capacity Clat.

p

B

plat

Tkexp

DdU

1

Density of States in One Dimension

We found that,

KL

NK2

2

Thus,

dK

d

dLd

d

dKLd

d

dK

dK

dNdNdD KK

22

Density of States in Three Dimensions

By applying periodic boundary conditions in the same way that we did for 1-D, the number of

states inside a sphere of radius Kis:

3

4

2

33KL

NK

whereL

3=V

Thus the density of states is then given by:

dK

d

dVKd

d

dK

dK

dNd

d

dNdD KK

2

2

2

The problem is now reduced to finding the group velocity (or the dispersion relation) for a

particular crystal and to apply the equations above.

Debye Model for the Density of StatesIn this model, valid for small values of T, the velocity of sound is taken as a constant for eachpolarization type, thus

Kp and 32

2 p

VD

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2

15

43

TTNkU Blat where

What leads to

33

378.233

TT

UC latV

This is known as the Debye T3

approximation. The agreement with this law worsens as Tapproaches the Debye temperature, the law is most approximately met for Twithin 1-2% of the

Debye temperature. Thus the Debye approximation is good for very low temperature (compare

to the Debye temperature).

Einstein Model of the Density of States

This model considersNoscillators all vibrating at the same frequency o and thus

D()=N(-o)

1

3

1

Tk

o

p Tkolat

B

oB

e

N

e

NdU

222

1

3

Tk

o

B

oBV

B

o

eTk

NkC

For large T, CV~3NkB, the Dulong and Petit Value.

In summary, the Einstein model accounts for the high temperature limit while the Debye modelaccounts for the low temperature regime.

Anharmonic Crystal InteractionsThe harmonic approximation is just that, an approximation and many of the disagreement

between the theoretical predictions (within the harmonic approximation) and the experiment can

be tracked down to the limitations of this approach. For instance, the thermal expansionpredicted by the harmonic model is zero. To see this, the average displacement is

dxe

dxxex

Tkcx

Tkcx

B

B

2

2

Where the Boltzmann distribution was used to weight the displacements according to theirthermodynamical probability. However, the integral on the top is the integral of an odd

functions (a function for which f(x)=- f(-x)) integrated is a symmetrical interval, and thus it iszero. The denominator is the intergral of a non-zero, positive defined function what warranties

that the denominator does not go to zero. So the average displacement is zero at ALL

temperatures and the material is NOT predicted to expand with temperature.

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If anharmonic terms are added to the energy (terms of power larger than 2 of the displacement),

this prediction changes. For instance

432 fxgxcxxU then

dxe

dxxex

Tk

fxgxcx

Tkfxgxcx

B

B

432

432

If we assume the cubic and quartic terms are very small compared to kBT, we can expand in

series to get

2/32/5

2/1

54

4

32432

Tkc

gdx

Tk

fx

Tk

gxxedxxe B

BB

Tk

cx

Tk

fxgxcx

BB

and

2/1

2/3

2/5

2/12/143

4

31

2432

Tkc

Tkc

fTk

cdx

Tk

fx

Tk

gxedxe BBB

BB

Tk

cx

Tk

fxgxcx

BB

Leading to

Tkc

gx B2

4

3

Thermal Conductivity

Without anharmonicity there is no interaction between phonons, without this interaction thermalconductivity would be infinite as shown below.

The steady state flow of heat is given by

dx

dTKju

This diffusion process requires a tortuous transport as opposed to ballistic, otherwise the flux

would only depend on the difference in temperature between the ends of the sample and not on

the gradient, (like electrons accelerated by an electric field, their kinetic energy change only

depends on the voltage difference and no on how far they move). So this behavior requires heatcarriers to often collide and scatter.

Using kinetic theory we can find a model for K.

Consider a gas of particles (like phonons) with n particles per unit of volume moving with

average velocity in the x direction of xv , in equilibrium, half will move in the positive

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direction and half in the negative direction. Thus, the flux of particle is xp vnj2

1 where

there is a flux of equal magnitude on each direction.

Ifc is the heat capacity per particle, each particle will lose an amount of energy ofcT when it

moves from a place at temperature T+T to a place at temperature T. If lx is the distancebetween collisions, then

xx vdx

dTl

dx

dTT where is the average time between collision

So, the flux of energyjU

dx

dTCvl

dx

dTcnv

dx

dTcvnTcjj xpU

3

1

3

1 22

Where nc=C(heat capacity per unit volume), and v=l

So CvlK3

1

Thus, if there is no interaction with the lattice, l is infinite and the thermal conductivity is infinite.

The interaction occurs because of the third order

anharmonic term presence of one phonon causes

the modulation in space and time of the latticeelastic constant, a second phonon sees that strain

and it is scattered producing then a third phonon in

a three-phonon process.

K1+ K

2=K

3Picture from wikipedia

This process, although it explains phonon collision and finite thermal conductivity, it is still not

enough to account for phonon local thermal equilibrium as the total momentum of the phonongas remains unchanged (thus one end of a metal cannot get in equilibrium with a temperature

source) any momentum lost by one phonon goes to another. An interaction with the lattice as a

whole is necessary, thus the process must be of the form

K1+ K2 =K3+G

These processes are known as Umklapp processes. However this is not a forced phenomena,

remember that all physically meaningful values for the phonon wave number K must be in thefirst brillouin zone, it is possible that the K1+ K2 leads to a phonon with wave number K3 outside

the zone, but that is not a physically meaningful state, thus a G vector must be used to bring thephonon back into the first Brillouin zone. Looking at this issues from a different point of view,

when two colliding phonons that have a wave number so large that the sum will lead to a

momentum outside the first Brillouin zone, the process is not possible as a phonon with such amomentum cannot exist and the lattice as a whole must absorb some momentum, thus the total

momentum of the phonon gas changes and the phonons can thermalize. At high T, the

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5

population of high momentum states is very large and Umklapp processes dominate. In this

regime the number of phonon is proportional to T (See Einstein model) and the number ofcollision is proportional to the population, the mean free path is proportional to 1/T, thus thermal

conductivity is proportional to 1/T. At low T, most phonons have low enough momentum that it

is very unlikely that in a collision a phonon is created with momentum outside the first Brillouin

zone and thus heat conduction is dominated by normal processes.Free Electron TheoryMany of the properties of metals can be understood if we consider the electrons in the metal as

particles that are free to move in a box. This model it is clearly an approximation since itassumes two completely wrong hypothesis, 1) electrons do not interact with each other, 2)

electron do not interact with the lattice. Both are totally wrong even in classical terms since

electrons and the lattice are charged, electrons are negative and the lattice bears a net positivecharge. Surprisingly, this theory explain acceptably well some of the properties of metals as we

will see below.

Consider alkali metals, Li, Na, K, Rb, Cs. All of them have the structure of an inert gas

plus 1 electron on a s level. For these metals it is not hard to share that electron and in treatingthese system theoretically, it is assumed each atom contribute with 1 electron to the electron gas.

All those electrons now navigate freely around the entire metal.

Classical theory of metals, produced long before quantum mechanics (Drudes model)surprisingly predict fairly well conduction properties, ohms law, thermal conduction and the

relationship between them (although the proportionality constant was a little bit off compared tothe experiment). However it overestimates by about 2 orders of magnitudes the electronic

contribution of the heat capacity. How can this be possible? The answer is cancellation of

errors, the heat capacity is proportional to the number of electrons that are able to absorbthermal energy, while conduction properties is proportional to that and to the average kinetic

energy of those electrons. The classical theory (as we will see later) overestimate the number of

electrons that can actually absorb thermal energy in about 2 orders of magnitude (thus the heat

capacity is overestimated) but it underestimates, by about the same factor, the energy of thoseelectrons and thus, by coincidence, conductive properties are very well predicted.

Electrons in a one-dimensional infinite boxThe closest problem to the one in hand is that of electrons in a box, a typical problem in quantum

mechanics.

Any particle is characterized by a kinetic energy and a potential energy. Classically that means

that the total energy of the system is given by:

Vm

p 2

2

H WhereHis known as the Hamiltonian or energy operator,pis the particles

momentum, m is its mass and Vis the potential acting on the particle.

In quantum mechanics,p can be represented by the operatorp=i d/dx where the action of this

operatorp is to take the derivative of the function and then multiply that by the constant i.

p2 means applying this operator twice, thus

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6

nn

n

nH

nV

xm

2

22

2

The above is known as the Schrdinger equation and all that there is to know about the electron

can be obtained from the equation above (except that this equation does not have an exact

solution for most potential functions)

For the electron in a box case, the potential energy is defined as V=0 inside, V=infinite outsideWhat lead to the equation:

Exm

2

22

2

With boundary conditions that force the wave function to be zero at z=0 and z=L (where L is the

box width)

The condition for that the wave function needs to be zero for z=0 is satisfied proposing asolution of the form:

kxAn sin The condition that the wave function must be zero at L is satisfied if k=n/L or,

,...3,2,1sin

n

L

xnAn

=> k=2/n

Thus the solutions are quantized and the allowed energy values are given by

2222

22

L

n

mm

kn

Normalization of the wave function

LA

LAdz

L

znAdzdz

n

n

L

n

LL

2

2sin1*

2

0

22

0

2

0

Energy difference between two consecutive levels increases as n increases

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7

222

2

8fi nn

mL

hhE

Transition between levels leads to absorption or emission of a photon with energy equal to the

difference between levels.

The above solutions correspond to a single electron in an infinite well. If we consider now thatthe metal haveNelectrons (assuming 1 per atom), and invoke the Pauli exclusion principle, then

we can only accommodate 2 electrons per energy level and thus only N/2 levels will be filled

with electrons. The energy level with the maximum energy that holds electrons is known as theFermi energy and it is given by

222

2

222

L

N

mL

n

m

f

f

Class activity 5.

Effect of TemperatureThe picture described above, where energy levels are being filled one by one (with two electronseach) until we run out of electrons, is only valid at zero Kelvin. As the temperature increases,

some electrons with energies of the order ofkBTaround the Fermi energy can be promoted to

higher energy levels and thus some levels above the Fermi level are occupied while some levelsbelow the Fermi level are empty. The Fermi-Dirac distribution accounts for that effect:

1

1

Tkexp

f

B

where is the energy and is the chemical potential (we will define it in a little bit). Notice thatas T tends to zero, the equation above tends to two different values, 0 or 1, depending on the

relative values ofand, for the exponential

goes to infinity. As the temperature increase, this function decreases smoothly as increases.

The Fermi-Dirac distribution is a probability distribution; it indicates the probability of a given

energy level to be occupied. At 0K, the probability of occupation of energy levels below the

chemical potential is 1 (all are occupied) while the probability of occupation of energy levelsabove the chemical potential is zero (all empty). The chemical potential is an empirical

parameter defined as the energy where the probability of occupation is exactly . For very low

temperature kBT

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1

October 28th

, 2010

Class 15

Electrons in a three-dimensional infinite box

For the case of electrons in a box, the finite size box may be not the best way to go considering

the macroscopic dimensions of a metal that can be regarded, without much approximation, asinfinite in size. For a case like that, periodic boundary conditions may be more appropriate.

Thus for the three dimensional cases, periodic boundary conditions will be used.

By following the same mathematical procedure than the one we followed to determine the

number of vibrational modes in a crystal, we found that the solutions for an electron in a three

dimensional box with periodic boundary conditions are:

nm

22

2

nzyxm

2

2

2

2

2

22

2

Since rikrk exp

xikikdx

dxx exp

and

222

2

exp xxx kxikkdx

d

yikikdy

dyy exp

and 22

2

2

exp yyy kyikkdy

d

zikikdz

dzz exp

and

222

2

exp zzz kzikkdz

d

Thus,

n

zyxm

2

2

2

2

2

22

2

becomes

nzyx kkkm

2222

2

or

2222

2zyxn kkk

m

22

2k

mn

where 2222 zyx kkkk Our relationship between wavevector and wavelength is such that,

x

xx

L

nk

2 ,

y

y

yL

nk

2 ,

z

zz

L

nk

2 .

The quantization comes from the boundary conditions zyxnn LLL ,,0,0,0

Using the quantum mechanical definition ofp

kkk kip

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2

So the linear momentum of an electron with a wavevectork is kand the speed of such an

electron is v=k/m

The energy at the Fermi level (now in 3D called Fermi surface), is given by:

22

2FF k

m

where kFis the wavevector at the Fermi surface.

In the same way we did with the vibrational modes, we can count the number of electronic states

and since the solutions are the same, we can see that in the unit of volume there are3

2

Lelectronic states in thek space. Thus in a sphere of volume 3

3

4

Fk (volume of the Fermi

sphere) there are 32

3

3

334

22 Ff k

Vk

L

states (counting both spin states). If in the same

volume there areNatoms and each contributes with 1 electron, then that number above must be

equal toNfrom where

31

23

V

NkF

32

22 3

2

V

N

mF

and

31

23

V

N

mm

kFF

N/Vis the electron density (equal to the atomic density assuming each atom contributes with 1

electron).

From the above equation it is useful to define the Fermi temperature TF as the temperature for

which kBTF=F. This temperature has nothing to do with the thermodynamical temperature,

TF~104 is the temperature Tg an electron gas need to be for its thermal energy be kBTg=F.

Density of States

The equation3

222 3

2

V

N

m

, is valid for any value of energy not only for the Fermi energy,

in that caseN represents the number of electrons up to that energy. So inverting the equation we

obtain

23

22

2

3

mVN

whereN is the number of one electron states (with or without electrons) considering each energy

level as 2 states (one for spin up and the other for spin down) with energy between 0 and for

any larger or smaller than the Fermi energy.

so the density of states is

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3

2

32

32

32

2

23

222

123

22

NmVmV

d

dND

Heat Capacity

As said before, only a few electrons near the Fermi level contribute to the heat capacity,electrons with energy much smaller that the Fermi level cannot take advantage of the thermalenergy because energy levels above them are occupied and the Pauli principle does not allow

them to also occupy those energy levels.

Classically, the energy of an electron gas is 3/2kBT per electron, thus each electron contributes to

the heat capacity by a quantity 3/2kB ( per degree of freedom). In the Drudes model (no Pauli),

all theNelectrons contribute to the heat capacity and thus the electronic contribution to the heat

capacity was predicted to be 3/2NkB, or the specific heat capacity 3/2 N/V kB where N/V is theelectron concentration.

Considering the Pauli principle, electrons with energy within kBTof the Fermi energy can absorbenergy, thus only a fraction T/TF can actually contribute to the heat capacity and then

qualitatively the total thermal kinetic energy of electrons is

Uel(NT/TF)kBT

leading to a heat capacity of

CelNkB(T/TF)

which is not the exact value but retains the T dependence found in the experiment. At room

temperature, T/TF~0.01 for TF~3x10

4

K which is in the order of a typical value for the Fermitemperature

For a more quantitative calculation consider the change in energy of the metal when the

temperature increases to a certain temperature Tis given by

dDdDTfUF

00

,

Multiplying the identity

dDdDTfNF

00

,

by F

dDdDTfF

FF

00

, what leads to 0,00

dDTfdDFF

F

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4

Adding this zero to the equation above and reorganizing we get

F

F

F

F

dDdDTfdDTf

dDdDTfU

FFF

FF

00

00

,,

,

or

F

F

dDTfdDTfU FF

0

,1,

The first term is the product of the energy gained by one electron when it goes from the Fermi

level to the energy level , times the number of states at the energy times the probability thatthe state has been occupied integrated in all the level above the Fermi level. This is equal to the

total amount of energy invested to populate energy levels about the Fermi level. The second

term is the product of the energy gained by one electron when it goes from the energy level below the Fermi level to the Fermi level times the number of states at the energy times theprobability that the state has indeed been vacated integrated in all the level below the Fermi level,

that is the total amount of energy invested to de-populate energy levels below the Fermi level.

So now the heat capacity can be obtained simply by taken the derivative of the last equation with

respect to T. Noticing that onlyf() depends on T

dDdT

TdfdD

dT

TdfdD

dT

Tdf

dT

dUC FFFel

F

F

00

,,,

Since we are mostly interested in temperatures around room temperature, then T/TF= kBT/F~0.01where the density of states can be approximated by the density of states at the Fermi level and

the chemical potential can be considered independent of temperature and ~F

Then

ddT

TdfDC FFel

0

,

1exp

1,

Tk

Tf

B

F

22

1exp

exp,

Tk

Tk

TkdT

Tdf

B

F

B

F

B

F

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5

dx

e

exTDkd

Tk

Tk

TkDkC

Tk

x

x

FB

B

F

B

F

B

FFBel

B

F

2

22

0

2

2

11exp

exp

x=(-F)/kBT

In order to integrate, we can replace the lower limit with -, which is approximately true for

kB

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6

2) The interaction between electrons and phonons what distorts the lattice that in turninteract with other electrons increasing their effective mass

3) Electron-electron interaction what also increases the effective massElectrical Conductivity

The momentum of a charge is given by mv=k, thus its equation of motion in an electric fieldcan be written as:

qEdt

dk

dt

dmF

v

(where q=-e for electrons)

What leads to

qEtk the same for ALL electrons, thus the entire Fermi sphere is moving!

What is consistent with a free charge in an electric field.

In a real system, electrons are not really free, they interact with the ions that are hopping in place

(phonon) and they interact with imperfections in the crystal. A more appropriate model consists

in assuming that these electrons are free to accelerate just during a time t=after which they juststop and start over. The overall effect is that these electrons find a steady state at a higher speedthan before the electric field is applied (sort of like a terminal velocity). So the shift in the

position of the Fermi sphere is qE/ while the speed increases an amount v=k/m= qE/m.

At equilibrium, the average speed of the electron gas is zero, thus vis the drift velocity.

Consider n electrons per unit of volume moving at a speed v, the number of electrons that will

cross a surface of unit area perpendicular to the direction of motion in a time to is equal to the

number of electrons in a cylinder of unit cross section area and length l=vto, thus, in the unit of

time, the total charge crossing a unit are is:

j=nqv=nq2E/m

jis the current density and the above is known as the Ohms law where

= nq2/m

is the conductivity and the inverse

= m/nq2 is the resistivity.

Another important property of charge carriers is their mobility which is defined as q=v/EwhereEis the electric field. Carriers with high mobility will pick up speed at small values of electricfield thus leading to larger currents. With this definition

j=nqv=nq qE and

=nq q orq=q/mThe two main sources for resistivity are the electron-phonon interaction (interaction with the ions

that are hopping in their place) and the interaction with impurities and crystal defects. The

former is more relevant at high temperature while the later is relevant at all temperature. If the

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electric field is then removed, collisions will make the electrons go back to their unbiased speed.

The rate of these collisions is independent from each other, thusf=fL+fi

wherefis the total number of collisions per unit of time fL is the number of collisions with the

lattice (charge-phonon collisions) per unit of time, and fi is the number of collisions withimpurities per unit of times

thus the characteristic time can be obtained as:

iL

111

The equation above then indicates that the total number of collisions per unit of time is the sum

of the total number of collisions with the lattice (phonons) per unit of time plus the total number

of collisions with impurities per unit of time) or multiplying by m/ne2 =L+i

Hall EffectIf a magnetic field is applied perpendicular to a current, the electrons will suffer a force

perpendicular to both, the drift velocity and B, of a magnitudeF=qvxB

This force produces an accumulation of charge and an electric field that generates anelectrical force that opposes to the magnetic force. In equilibrium

-eEH=evB or EH=-vB

By defining the Hall coefficientRHas

EH=-RHJB

vB= RHJB or RH=v/J

(notice that RH can be measured by measuring the electric field perpendicular to the current and

it is a property of the material).

UsingJ=nqv RH=1/nq (where q is negative for electrons)

For Li, monovalent, the measured Hall coefficient at room temperature is -1.7x10-10 m3 /C and

the calculated value -1.4x10-10 m3 /C. For Al (trivalent) both values are ~ -0.3x10-10 m3 /C. But

for Zinc, the calculated value is -0.5x10-10

m3

/Cwhile the measured one is +0.3x10

-10m

3 /C. The

reason is that the carriers are positive. Although

electrons are always the ones moving, some

material, especially divalent, behave as if the

carriers were positive. Well talk about that later.