Appunti in inglese di Solde Physics State

V

q rec

= N

Where V is the volume of the elementary cell of the reciprocal lattice. We call V the volume of

rec dir

the elementary cell of the direct lattice. When we studied the reciprocal lattice, we saw that this

relation holds: ( ) 3

2π

V =

rec V

dir q

If in the volume V there are N vectors q we can express in this way the volume in the reciprocal

rec

space available to each of the vectors q: ( ) ( )

3 3

2π

2π

V 1

q rec

= = =

N N V V

dir

Where V = NV is the volume of the sample. As we can see, each vector q occupies a portion of the

dir

reciprocal lattice inversely proportional to the volume of the crystal, in analogy with the 1D case

where for each q there is at disposal a length inversely proportional to the total length L. In the 3D

3

case the density of the vectors q in the reciprocal space, that is the inverse of q, is V/(2) .

4.6 Density of states

The vibrational energy of a crystal is the sum of the energies of all the oscillators that make it up:

 

1



+ + ω( , )

 

E = E n(

q

, s ) q s

o  

2

q ,s

E is the energy corresponding to the equilibrium position of the atoms, and it is constant. The second

o

term expresses the total energy of a set of oscillators each characterized by the two indices: s (the

mode of oscillation) and q. Each oscillator contributes to the total energy with n quanta each one of



ω

them of energy . The frequency and the wave vector q are not independent but are linked by

the dispersion relations.

If in the expression already seen above we neglect constant terms, and we omit the index s for

simplicity, we can write: - 55 -



= ω( ,s)

E n(

q

,s) q

q

Each possible oscillator, characterized by a particular value of q, contributes to the total energy with

ω

a multiple n of the corresponding quantum of energy . Let us think about the structure of this

formula by distinguishing different aspects. .

- First of all, let us consider a single wave vector oscillator q and frequency Its contribution

,

to the total energy depends on which in turn is a function of q, because its energy is a

ω

multiple of .

- Always considering a single oscillator, its contribution to the total energy is n times this

amount. One must therefore find the way to know n, and this is the task of statistical physics.



We can already anticipate that n depends only on and is the same for all oscillators having



the same frequency. This explains why is the most physically significant variable for energy

considerations. 

- The contribution to the total energy of the oscillators with frequency is therefore equal to

ω

the average energy of each of them (i.e. n ) times the number of oscillators of this type

present in the crystal. It is therefore useful to have a function that expresses the number of



oscillators for each frequency or better, a function g() such that g()d represents the



number of oscillators in the interval d centered on a particular vaule of the frequency. The

function g() is called density of states. 

For the moment we know how to calculate, or measure, the dispersion relations (q). Now we are

s

interested in moving from these to the density of states g(). Let us see how to proceed with reference

to the simple case of the dispersion relation valid for a linear chain of atoms represented in the figure.



Let us consider an interval d centered around a value of of the frequency. We are interested in

1

how many oscillators there are within this range. To do this it is better to consider the corresponding

interval dq and the reason for this is that in the reciprocal space the possible oscillators constitute a

1

discretized set with constant density and this obviously makes them easy to count. The number of

oscillators within dq (and therefore within d) is equal to the product of dq for the density of the

1 1

possible values of q which, as we know, is equal to L/2. If we refer to a sample with a unit length

(L = 1) the density is simply 1/2. In this case then the number of oscillators is:

1 dq

Number of oscillators within d = g()d = 1

2π

The number of oscillators within the same frequency range depends on the value of the frequency

around which it is centered. Consider in fact the case in which d is centered around the frequency

 . The corresponding range dq is wider, which means that the number of oscillators contained in it

2 2

is higher.

In general, the number of oscillators g()d is: 1

= dq

g(ω)dω 2π

- 56 -





2 d

d



1 dq

dq 2

1 q

q q

1 2

From which: 1 dq

=

g(ω) dω

2π  .

In the expression appears the derivative of q with respect to and the result is a function of To

(q)

calculate this function we have to start from the dispersion relation and derive its inverse

function q(). Alternatively, the density of states g() can be expressed directly in terms of the

derivative of the dispersion relationship:

1 dq 1 1 1 1 1 1

= = = =

g(ω)  

dω 2π 2π ω'(q) 2π ω' q(ω)

dω

2π dq 

In this case the result obtained is a function of q which, however, we know to be a function of and

this allows us to express the density of states as a function of frequency.

These formulas are important because they relate the density of states to the dispersion relations. We

will come back to this subject later to talk about some of the density of states properties that can be

deduced from this formula.

In the 3D case we proceed in the same way. What we need to know is the number of oscillators

(q) (q)

present in the portion of space between the surface and the surface + d, which we can

formally express in this way. ω( )+dω

q d

q



=

g(ω)dω 3

8π

ω( )

q

If different modes of oscillations are present, then the corresponding density of the states must be

calculated for each of them; the overall density will be the sum of all the individual ones.

- 57 -

In the previous expression dq represents the volume of an element of the reciprocal space that,

multiplied by the density of the vectors q in the reciprocal space in three dimensions which is equal

d

1 q

to , gives precisely the number of vectors q present in the element dq.

3

3 8π

8π dSdq

To proceed, we express the volume of element dq as the product of an element dS of the

⊥

 dq

surface where is constant for an element perpendicular to the surface:

⊥

Moving in the direction perpendicular to the surface, the frequency varies and we can write:

= grad ω dq

dω ⊥

q

In this way the volume of the element dq assumes the following expression:

dSdω

dSdq =

⊥ ω

grad q

and then, by integrating over the entire surface: 1 dS



=

g(ω) 3

ω 8π

grad

S q

4.7 Van Hove singularities

The previously defined density of states g() is used to express the number of states available for

each particular frequency value. In order to express it correctly, it is necessary to search in the

reciprocal space for all the states characterized by the same frequency, whatever the value of their

wave vector q.

The functional relationship between density of states and frequency depends on the dimensionality

of the system. If we consider only acoustic excitations, where the dispersion relation is linear, we can

easily demonstrate that g() is constant in 1D systems, linear in the case of 2D systems and quadratic

in the case of 3D systems. The reality is different, however, since the dispersion relations of the

acoustic modes are linear only for small values of q and this is reflected in the form of the g() which

is never a regular curve as in the simplified models but presents singularities known as Van Hove

singularities, named after the Belgian physicist who first dealt with them in the 1950s. To understand

the origin of these singularities it is worth considering the case of a linear chain of equal atoms whose

dispersion relations is, as is known, the sine functions:

K 1

ω(q) = 2 sin qa

M 2

Previously we have obtained the relation that binds the density of the states g() to the dispersion

(q)

relations and therefore we can calculate it for the case of the linear chain:

- 58 -

1 1 1 1

= =

g(ω) ω'(q) 2π cosqa

2π

This formula expresses the density of states as a function of q. However, we have seen that by

,

inverting the dispersion relation and expressing q as a function of we obtain the desired formula

that expresses the density of states as a function of frequency. The graph of the density obtained is



the one shown in the figure. For small values of the function is approximately constant, as it should

be in the case of a linear dispersion relationship in 1D. For larger values, however, the trend changes

until it reaches, at the edge of the zone, a singularity. This is a simple example of a Van Hove

singularity. g() 



max

The situation in 3D is more complex but even in this situation in the densities of states singularities

are still present. They exist in different forms and occur at particular values of the wave vector q

called critical points.

4.8 A few remind of statistical physics

At this point we are able to "count" the different types of oscillators present in a crystal. What we

must now learn to do is to understand how the globally available energy is distributed among them.

This aspect is the subject of statistical physics of which we will summarize the essential concepts

useful for the study of specific heats.

A crystal is a thermodynamic system in thermal balance with the surrounding environment. To study

its properties, therefore, the so called formalism of the canonical ensemble is used. The energy of the

system can fluctuate and its value can assume any of the eigenvalues of the energy of the Hamiltonian

describing the system itself, which we call E . The statistical nature of the problem manifests itself in

n

the fact that we are not able to say, moment by moment, what is the value of the energy, we can only

say what is the probability w that the energy is equal to E . The corresponding statistical distribution

n n

is the Gibbs distribution: E n

– kT

w =Ae

n - 59 -

This law applies both in classical and quantum mechanics. The important point to note is that the

energies E are not single particle energy levels, they are the possible values of the energy of the

n

system intended as a wh