phonon energy

5/20/2018 phonon energy

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Chapter 6

Lattice vibrations - Phonons

6.1 Phonons in mono-atomic crystals

How do we treat the motion of the ions within a crystal structure? Or in other words, how will the motion of

a bound ion influence its immediate neighbours and indeed beyond? This is most easily visualised in a simple

cubic lattice along a direction of high symmetry - for example the [100] direction where we shall regard the

displacement of lattice planes :

n 1n

n+1

n+2

un+3

n 1 n n +1 n +2 n +3

Figure 6.1: Schematic illustration of displacements of planes along one spatial direction. In the two orthogonal

directions translational invariance is assumed.

The above schematic can easily be understood as a longitudinal wave, but clearly transverse excitations can also

be envisaged, as shown below.

K

Figure 6.2: Schematic illustration of transverse displacements.

How can we calculate the dynamics? Assume elastic forces are linear and given by Hookes law - F = kx,

where the vector x is the difference of the displacements. The force resulting from the displacementsun, um is

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CHAPTER 6. LATTICE VIBRATIONS - PHONONS 27

then given by:

Fnm = Cnm(un um) (6.1)

For simplicity we shall only consider nearest neighbours, so we find the total force acting upon an atom within

a plane:Fn = C(un+1 un) +C(un1 un) (6.2)

n 1 n n+1

a a

Figure 6.3: The displacements for a periodic configuration of lattice planes is shown.

and C is assumed linear. Newtons second law then yields,

Md2un

dt2 =C[un+1 un1 2un] (6.3)

the equation of motion for a plane (M is the mass of a atom in the plane). Assume all displacements have the

same frequency of dependence (normal modes) - exp (it):

M 2un = C[un+1+un1 2un] (6.4)

This differential equation can be solved by assuming a wavelike solution with wave vector kand with an additional

phase factor which scales linearly with the plane position, i.e.

un = u exp(inka) (6.5)

Here we are considering a simple cubic lattice with lattice constant a. This assumption simply means that a

fixed phase relationship exists between any two neighbouring planes. Thus we find:

2Mu exp(inka) =C u [exp(i(n+ 1)ka) + exp (i(n 1)ka) 2exp(inka)] (6.6)

Which may be simplified by using the exponential definition of the cosine - exp (ix) + exp(ix) = 2 cos(x),

2M = 2C[1 cos(ka)] (6.7)

2 = 4CM sink a2

2. (6.8)

Where we used the trigonometric relation cos (2) = cos2 sin2 = 1 2sin2 2sin = 1 cos(2). This

is thedispersion relationfor an elastic wave of wave vector k and of frequency .


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Figure 6.4: Illustration of the dispersion relationEquation 6.8,the maxima lies at (4C/M)1/2 but has been normalised

in the above schematic./a k /a

What range of k makes physical sense? The wave is only defined for a particular lattice plane located at nka.

The phase difference between two plans can be written as

un+1um

= u exp(i(n+ 1)ka)

u exp(inka) = exp (ika) (6.9)

as expected.

A sensible range for the argument (ratio is periodic, 2i) is merely ka and hence /a k /a; i.e.

we restrict ourselves to solutions within the first Brillouin zone of our reciprocal space, defined by the Wigner-

Seitz cell of our reciprocal lattice.

-2a2a -aa 0 2aaa

Figure 6.5: The above two waves clearly differ by a extra wavelength between adjacent planes.

The above two oscillations show the displacements for the lattice planes (separated by a) - no additional infor-

mation is to be gained by considering the smaller (or larger k). The wave vectork outside of the first Brillouin

zone can always be transformed back into the first Brillouin zone by the addition of an appropriate reciprocal

lattice vector.k =k+ K= exp (inka) = exp(inka)exp(in Ka)

=1

1 (6.10)

1see chapter 3 - this is simply the definition of the reciprocal lattice considered earlier.


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wherea is a lattice vector. The solution is a travelling wave everywhere except for k= /a. There the phase

factor between adjacent planes is exp (ika) = exp(i) = 1, therefore the motion of neighbouring planes is in

exact antiphase and we have a standing wave solution.

6.1.1 Velocity of solutions

Thephase velocityis, as always, defined as

vp =

k (6.11)

but we are more interested in the motion of the wave packet, i.e. in the group velocity

vg =

k = gradk((k) (6.12)

We now calculate the group velocity from Equation 6.8:

vg =

4C

M

12 a

2cos

ka

2

0< k

a

vg = Ca2

M 1

2

cos ka2 (6.13)

g,max

/adirection of motionreversed

zero, standing

wave solution

Figure 6.6: Illustration of the group velocity calculated in Equation 6.13. Near extrema the group velocity is roughly

speaking independent of the wave vectork.

For small wave vectorsk we arrive at the long wavelength limit:

2 = 2CM [1 cos(ka)] 2CM1 1 12 k2a2 +...

2C

M

1

2k2a2

=

CM

12 ka (6.14)

The velocity of sound is independent of frequency, this is obvious in the sense that when k goes to zero then ka

goes to zero, which means our crystal looks like a continuous medium.

How many modes are there? Every atom within the plane has essentially three degrees of freedom. There

are three modes of vibration - 2 transverse modes plus one longitudinal mode, and almost certainly they have

different C !

6.2 Phonons in diatomic crystals

So far we have considered a simple cubic structure (a lattice point is one atom). What happens when our

primitive unit cell contains more than one atom? If there are p atoms and N unit cells, then there are 3pN


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degrees of freedom. 3N degrees of freedom are contained in the modes we have just discussed and (3p 3)N in

some other modes. What is the nature of these additional modes?

Consider, for example, a diatomic chain of atoms with masses m1 and m2. Clearly associated with each lattice

plane there is an atom of mass m1 andan atom with mass m2.

1 1m2 m2 2

s 1 s + 1 s + 1

a

Figure 6.7: Representation of a chain of atoms consisting of two different types of atoms with masses m1 and m2.

However the displacements of the atoms from their equilibrium positions are, of course, considered separately.

Thus we find two coupled differential equations for the dynamics of the two atomic species:

m1d2nsdt2

= C[vs+vs1 2us] (6.15)

m2d2vsdt2

= C[us+1+us 2vs]

where we assume, as before, that the wavemotion may be written as:

us = u exp(iska)exp(it) (6.16)

vs = v exp(iska)exp(it). (6.17)

Inserting this into equationEquation 6.15we find:

2m1u = Cv [1 + exp (ika)] 2Cu (6.18)

2m2v = Cu [exp (ika) + 1] 2Cv. (6.19)

We find the solutions by considering the following determinant (and requiring it to be 0):

2C m12 C(1 + exp (ika))

C(1 + exp (ika)) 2C m22

= 0 (6.20)

4C2 2C(m1+m2)2 +m1m24 C2(1 + exp (ika))(1 + exp (ika)) = 0 (6.21)

The last term can be written as 2 + 2 cos(ka):

m1m24 2C(m1+m2)

2 + 2C2(1 cos(ka)) = 0 (6.22)

This is a quadratic equation in 2. So we can solve this equation for 2 and then take the square root for

= (k). Consider initially the limiting behaviour for k 0. As before we can expand cos (ka) 1 1/2k2a2:

4m1m2 2C(m1+m2)2 +C2k2a2 = 0 (6.23)

2 = 2C(m1+m2)

4C2(m1m2)

2 4m1m2C2k2a2

1

2

2m1m2 k (6.24)

One possible solution (at k = 0) is:

2 = 2C

m1+m2

m1m2

= 2C

1

m1+

1

m2

optical branch (6.25)


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The other possible root yields

2 = 1

2k2a2

C

(m1+m2) acoustic branch. (6.26)

and is also found for the mono-atomic chain.

What happens at the zone boundary at ka= ?

m1m24 2C(m1+m2)

2 + 4C2 = 0 (6.27)

4 2C

1

m1+

1

m2

2 +

4C2

m1m2= 0 (6.28)

2 = 2Cm1 or 2 = 2Cm2 (6.29)

Ifm1 is larger than m2 we can conclude 21 <

22 :

optical branch

acoustic branch

/a

C 1

m1+

1

m2

2

2C

m2

2

2C

m1

2

k

Figure 6.8: Dispersion relations for the diatomic chain.

In long wavelength limit (k 0) the acoustic and optical mode differ with regard to the direction of motion of

the atomic species as shown below:

Figure 6.9: For the acoustic mode, neighbouring atoms of differing species are displaced in the same direction. For the

optical mode the converse is true.

For the optical mode the behaviour can be mathematically derived by substituting 2 into Equation 6.18 or


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Equation 6.19and solving foruv =

m2m1

. (6.30)

For optical modes the centre of mass is fixed and the atoms oscillate against each other.

For 2C/m1 < 2 < 2C/m2 there are no solutions for the problem, k must be complex and therefore decay-

ing in space.

6.3 Phonon momentum

A lattice vibration is characterised by its frequency and its wave vectork. What is the physical meaning ofk?

Is the wave vector k related to the momentum ? No it is not, because the motion we are describing with k is a

motion of the relative coordinates. The physical momentum is given by the following relation:

p = M d

dt

us with us= u exp(iska) (6.31)

= Mdu

dt sexp(iska) (6.32)

For N atoms the sum extends from s = 0 to s = N 1.

s

xs =1 xN

1 x =M

du

dt

1 exp(iNka)

1 exp(ika) (6.33)

The allowed values of k2 are k = 2m/Na or k = m/N 2/a, where 2/Na is the spacing in k space as

before.

Mdu

dt

1 exp(i2m)

1 exp(i2m/N) = 0 (6.34)

We therefore conclude that phonons carry no real physical momentum but they do behave (in their interactions

with other particles) as if they carried a definite momentum k. This quantity is known as thecrystal momentum.

For the elastic scattering of photons we havek =k+ K (6.35)

wherek is the initial state,k the final state and |k| = |k| and Kis a reciprocal lattice vector. We see that the

elastic scattering of a photon implicitly yields momentum conservation only if the crystal absorbs the momentum

K (recoil of the crystal).

General principal: The total wave vector is conserved in a periodic lattice but we can always add an ad-

ditional reciprocal lattice vector such that kph lies in the first Brillouin zone:

k + kph = k+ K (6.36)

For example the above equation describes the inelastic scattering of a photon with the creation of a phonon

with crystal momentum kph. This is called phonon emission.

The process of phonon absorption may be written as,

k =k+ K+ kph. (6.37)

The measurement of phonon dispersion relations(k) require the simultaneous determination of phonon frequency

and its momentumkph, which can, by an appropriate choice K(reciprocal lattice vector), always be placed into

the first Brillouin zone.

As before, we write for phonon emission/absorption (+/):

k+ K=k kph (6.38)

2This is an applications of the BvK boundary conditions as considered earlier for a system with length L = Na: k= 2/L

mk= 2m/L with L = Na.


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In this case we need to determine the vectorskand k, therefore we require not just the directions ofkand k but

also thedifferingmagnitudes ofkand k. An apparatus which performs this analysis is known as a spectrometer.

The above discussion can be immediately extended to the case of inelastic neutron scattering. In this case

energy conservation then yields the frequency of the phonon:

2k2

2mn = 2k2

2mn (6.39)

where we have assumed that the energy of the phonon with frequency can be described, as for photons, by the

simple relation

= . (6.40)

Why this is so is considered in the next section?

6.4 Phonon modes and the Planck distribution function

The photons are the quanta of energy associated with a mode of frequency f = /2. Each photon has the

energy. The energy in a particular mode is given by (n+1/2), with /2 the zero point energy of the mode.What are the modes? In electromagnetism they can be regarded as the modes compatible with the boundary

condition (E= 0) at the surface of some cavity (This picture leads to Plancks law for black-body radiation).

For the lattice we argue a mode is a lattice vibration of any frequency . How can n and the energy in the

mode be seen from a classical point of view to be related to each other? For a standing wave solution we have:

u= u0cos (ka)cos(t) (6.41)

For the energy of a harmonic oscillator we know that the energy oscillates between potential and kinetic energy:

< 12

p.E. >=< 12

k.E > (6.42)

Thus the time average of the kinetic energy is 1/2 the total energy. The kinetic energy can be written as

1/2

ut

2, with the mass density. We now consider a crystal with volume V and integrating over the volume

of the crystal and averaging over space1

4V 2u0sin

2 (t) (6.43)

and now averaging over time yields:18V

2u20= 12

n+ 12

(6.44)

This equation is the connection between n the number of quanta in the mode and the classical amplitude.

Now we wish to consider the thermal properties of these phonon modes. The energies are as for a simple

harmonic oscillator:

n =

n+

1

2

= s for simplicity (6.45)

Where n is the quantum number for the harmonics oscillator, but here we talk of the number of phonons (pho-

tons) in the mode. This picture has a physical basis in the sense that the motion of a particle in a potential is

localised aboutx0 - 1/2 m2(x x0)

2 whereas we are dealing with an excitation over the whole crystal.

To proceed further we need to consider the thermal average occupancy of these phonon modes.

Z is the so-called partition function, which we require to normalise the thermal occupation probabilities and is

equal to the sum over all Boltzmann factors.

Z=

n=0

exp(n/kBT) =

xn geometric series for x


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= 1

1 exp(/kBT) (6.47)

Hence the probability of finding n phonons in the mode is given by:

P(n) = exp(n/kBT)

Z (6.48)

What is the thermal average occupancy < n > ?

< n >=

n=0

nP(n) =Z1

n exp(n/kBT) (6.49)

Setting /kBT =x we have

n exp(nx) =

d

dx

exp(nx) =

1

1 exp(x) =

exp(x)

[1 exp(x)]2 (6.50)

< n >= exp(x)

1 exp(x). (6.51)

or using our physical notation we have

< n >= 1exp (/kBT)1

Planck distribution function. (6.52)

The Planck distribution function is valid for photons, phonons and other comparable bosons. But how do we

calculate the total energy in the lattice vibrations of the crystal? This can be done as follows:

< >= < s >=

exp(/kBT) 1 (6.53)

gives the energy associated with the thermal average occupancy of a single mode. We now have to sum over all

allowed frequencies i.e. over all allowed wave vectors kph and indeed polarisations:

U=

p

kph

k,pexp(k,p) 1

(6.54)

Clearly this sum depends critically upon the dispersion of the crystal. It can be most readily evaluated by

introducing (as in the free electron model) the concept of the phononic density of states Dp() whereby:

Dp()d= the number of phonon modes of polarisation p between and +d

U =

p

d Dp()

exp(/kBT)

(6.55)

But what is the density of states in practice? Let us consider initially a 1D chain of atoms where the position ofthe first and the last atom is assumed fixed.

0 1 2 3 4 5 6

L

a

7

lenght L

separation a

N+1 particles

L=Na

Figure 6.10: Illustration of the phononic density of states.


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Clearly we may write a standing wave solution for the amplitudes of the oscillations.

un = u0exp (ik,pt)sin(nkpha) (6.56)

In order for the solutions to vanish at n = 0 and n = N, i.e. at the ends of the string, we can only allow the

wave vectorkph to have the following values:

k= L

,2L

, 3L

... (N 1)L

(6.57)

0 is clearly physically not allowed because in this case un would be zero for all n, as is the case for N/Lbecause

sin(n N/La) = sin(n) = 0 n. The number of allowed k-modes is equal to the number of atoms (lattice

places) which are free to move.

Note the difference between the fixed boundary conditions and the Born- von Karman boundary conditions.

(as considered for free electron motion)

Fixed: Mode spacing L ; onlyk > allowed

B.v.K.: mode spacing 2L

; but the allowed values for k are 0 ,2L

,4L

...

Butthe total number of modes is the same for B.v.K. boundary conditions and for fixed boundary conditions.

For N atoms and their lattice vibrations the total number of modes is 3N (= 2 transverse + 1 longitudinal

polarisation).

What we now have to calculate is the phononic density of states D(k) or,

D(k)dk= D()d D() = D(k)dk

d

=

L

dk

d

. (6.58)

For a one dimensional system the above formulation is valid, but in a real crystal we expect lattice vibrations in

alldirections in real space. Hence we need to consider a three dimensional picture of lattice vibrations.

In order to calculate the density of states in 3D we consider, as before for free electrons a box of volume V =L3

and periodic Born - von Karmann boundary conditions.

exp(ikr) = exp(ikxx+ikyy+ikzz) = exp(ikx(x+L)) +iky(y+L) +ikz(z+L) (6.59)

allowed values for kx, ky, kz = 0,2

L,

4

L; ...

2n

L n N (6.60)

The volume of a single state in k-space is (2/L)3. The number of states within radius k can be calculated as

follows:

= 4

3k3/

2

L

3 N=V

1

62k3 (6.61)

for each polarisation, i.e. for two transverse and for one longitudinal polarisation.3

Thus for each polarisationwe may be write:

D() = V 1

22k2

k

(6.62)

Where the last term has to be found from the concrete form of the dispersion.

6.4.1 The Debye approximation

At low enough temperatures we can write = vsk, i.e. only low energy acoustic phonons will be excited and

hence we can safely approximate our (k) through a linear, constant speed of sound.

k =vs and D() =

V 2

22v2s . (6.63)

3To account for the spin degeneracy for free electrons, we multiply by 2 - we cannot do something similar here for the various

polarisations are not necessarily degenerate - L(k)=T(k).


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How many modes does the acoustic branch contain? As before we expect 3N degrees of freedom per atom. For

one atom per primitive cell and for one polarisation we therefore require N modes:

N=V 1

62

3Dv3s

(6.64)

where we have defined the so-called Debye-frequencyand Debye temprature

kD = D

v =

62 NV1/3

. (6.65)

Only k-vectors smaller than kD are allowed! We now can calculate the total energy:

Utot =

p

kph

k,pexp(/kBT) 1

(6.66)

U =

p

dD()< n()>

= p

p0

V 222v3

exp(/KBT) 1 (6.67)

For simplicity we assumevs,L= vs,T =v and therefore summing over polarisations simply yields a factor of 3:

U= 3V

22v3

D0

3

exp(/kBT) 1d (6.68)

We now substitute /kBT =x d = kBT /dx and xD = D/kBT = /Twhere is the so-calledDebye

temperature,

= kB = vkB

62N

v

1/3(6.69)

The substitution yields:

U= 3V k4BT

4

22v33

D0

dx x3

ex 1= 9N kBT

T

3 D0

x3

ex 1 dx (6.70)

Note that this integral is not independent of temperature since the temperature is implicitly contained in the

upper limit of the integral. Now we may derive the specific heat from CV = U

T . It follows fromEquation 6.70

CV = 9N kB

T

3 xD0

dx x4ex

(ex 1)2 (6.71)

At very low temperatures xD = /kBT :

0

dx x3

ex 1=

0

dx x3

s=1

exp(sx) = 61

1

s4 =

4

15 (6.72)

Utot= 34N kBT /5

3 (6.73)

CV = 124N kB(T)

3 /5 234NkB

T

3(6.74)

We thus conclude that we have found the source of the T3 term in our empirical heat capacity.

CV =aT+bT3 (6.75)


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This law is found for metals at low temperatures; empirically godd agreement is found for temperatures T = N

exp(/kBT) 1 (6.77)

CV = U

T =N kB

kBT

2exp (/kBT)

exp (/kBT)1 (6.78)

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