the band theory of solids · the band theory of solids w1.1 introduction this is a special section...

© Michael de Podesta, 2002

CHAPTER W1

The band theory of solids

Explanation

This chapter is a supplement to the book:

Understanding the properties of matter

by Michael de Podesta.

Please note:• The copyright of this chapter resides with Michael de Podesta.• The text and figures may be used freely for educational pur-

poses, but their source must be acknowledged.• The figures for this chapter are in colour, but should reproduce

satisfactorily when reproduced with a grey-scale printer.• For more details see www.physicsofmatter.com

© Michael de Podesta, 2002 1

CHAPTER W1

The band theory of solids

W1.1 Introduction

This is a special section of Understanding the Properties of Matter. It is special because it is an almostentirely theoretical look at solids. In the rest of the book I have focussed on experimental properties, andthen developed theories only to the extent that they are needed to understand the experimental data. Mymotivation in making this additional section available is that it addresses a question that is central to un-derstanding of what ‘really happens’ inside solids. This question can be simply summarised as ‘How dowe describe the electronic states within solids?’

The challengeConsider the solid form of two elements: argon and potassium. We know from §6.2 and §7.3, that argonis described as a molecular solid. However, when we describe potassium, we would naturally discountthe theory for argon as irrelevant, and use the theory for a free electron gas described in §6.5. This mayseem obvious: after all, one is a metal and the other an insulator. Well the essential question is: ‘Why isit obvious?’ Argon and potassium atoms differ by only a single electron in their outer shells. In describ-ing the internal electronic structure of their atoms, we use the same ‘language’ to describe both argonand potassium. What we would like is a unified picture, which will describe the nature of electronicstates in all solids. From this we should be able to deduce that argon is an insulator and potassium is ametal. This is the challenge.

The problemThe problem with this approach is that the electronic states in solids are not simple. Some states are likethe corresponding states on the isolated atoms, and some appear to be like free-electron states i.e. planewaves. Most states, at least most of the valence states, in solids are in a state somewhere between thesetwo extremes, and the challenge is to find a way to describe such states. The solution to this problem hasbeen ‘known’ for half a century or more, but it is not at all clear to be who it is who has been doing the‘knowing’. I say this, because while physical descriptions of the nature of electronic states are extremelyrare, but dense mathematical tomes are plentiful. In this exposition, It is my intention to keep themathematical complexities to an absolute minimum. However, the level of mathematical ability required(particularly with regard to the tight-binding theory) is definitely higher than for the topics in the maintext. Some familiarity with the apparatus of quantum mechanics is also assumed.

We shall proceed as follows:§W1.2 and §W1.3 k-space, the reciprocal lattice and free electron theory of metals. These sections

repeat some of the material Chapters 6 and 7 of the main text. The purpose of this repetition isto clearly establish the context in which the following discussion takes place.

§W1.4 Nearly-free electron theory of metals and insulators. In the free electron theory, all materi-als with mobile valence electrons are metals. In this section we will see that the introductionof a weak periodic potential allows us to understand how such materials can behave as metals,semi-metals, semiconductors, or insulators.

§W1.5 Tight-binding theory. In this section, we introduce a theory which does not ignore the factthat solids are made of atoms. The result is theory of electronic states in solids which is almostrealistic.

THE QUEST FOR AN UNDERSTANDING OF MATTER


W1.1 k-space

General descriptionAs we saw in §6.5, k-space is a geometrical con-struction used for counting quantum states. Asidefrom its use in counting states, the use of k-vectorsprovides an extremely compact way of describingquantum states. Just by specifying a single point,k = (kx, ky, kz) we describe a function Acos(k . r)which takes up the whole of the volume of a metal

You should be familiar with the above descriptionif you have read §6.5. There are two points aboutk-space that I would like to draw to your attention.

The first concerns the quantisation of k-spacearising from the finite extent of the crystal. Wewill revisit this topic in §W1.1.2 and use amathematical trick known as Born-von Karmenboundary conditions to try to justify the resultssummarised in Equation 6.59

km

Lk

m

Lk

m

Lxx

xy

y

yz

z

z= = =2 2 2π π π

, , (6.59*)

The second concerns the extent of k-space. Itmight seem that the mesh of allowed states would

simply extend, uninterrupted, forever. This isnearly true: k-space does extended indefinitely,but it is not a completely barren landscape. Thereare certain special values of k that reflect the pe-riodicity of the crystalline potential energy. Theseform a pattern known as the reciprocal lattice, andare discussed in §W1.1.3

Let us look at each of these two points in turn.

W1.1.2 Quantisation

Born-von Karmen boundary conditionsLet us think back to the solutions of the particle ina box problem that we considered in §2.3. Therewe required the wave functions to be zero at theedge of the box. This was because we imaginedthe edge of the box to represent walls of somekind. For a metal, we considered the edge of thebox to be the surface of the metal. Now, however,we consider a more complex situation: we imaginea volume which is representative of the situationwithin the box as a whole and we imagine it tosurrounded by identical volumes.

Figure W1.1 Illustration of two approaches to the particle in a box problem. (a) The first approach uses the idea of anisolated box representing the entire crystal. The second approach (b) uses the idea of a box of representative materialsurrounded by identical copies of itself. Thus the wave functions need not go to zero but must have (i) the same valueand (ii) the same slope at either side of the box. Thus one could imagine placing the boxes side by side and the wavewould be identical in each box The edges of the actual crystal are imagined to be far enough away that they do notsignificantly affect the particles deep inside the box. (c) The same idea applied in two dimensions

(a) (b) (c)

Ene

rgy

Ene

rgy

L L

k-space


The differences between this situation and thestanding wave situation are illustrated in FigureW1.1. The reader may consider this change to be‘trickery’, and indeed it is. It is a mathematicaltrick, but it will allow us to see more clearly thephysics of what is happening in the metal. First ofall, we expect any small region of a metal crystalto be typical of the crystal as a whole. Further-more, we would not expect its properties to de-pend at all upon the state of the surface of thecrystal. The motivation for using this trick is thatwe are able to use travelling wave solutions to theparticle in a box problem.

The mathematical effect of this change in repre-sentation concerns the boundary conditions thatthe wave function Ψ must satisfy. Originally werequired that the wave functions be zero at theedges of the box i.e.

Ψ Ψ( / ) ( / )− = − =L Lx x2 2 0 (W1.1)

where Lx is the length of the box in the x-directionNow we require that the wave functions be identi-cal in neighbouring boxes i.e.

Ψ Ψ( ) ( )x x L= + x (W1.2)

If the solutions are of the form sin( )k xx orcos( )k xx , this is equivalent to requiring that:

cos( ) cos( [ ])k x k x Lx x x= + (W1.3a)

or

sin( ) sin( [ ])k x k x Lx x x= + (W1.3b)

The requirements given by these Equations W1.3are known as Born-von Karmen boundary condi-tion. Their effect is to allow only discrete valuesof k as solutions. Let us follow one of EquationsW1.3 through explicitly to find these conditionsfor kx . The boundary conditions require that:

cos( ) cos( [ ])

cos( )

k x k x L

k x k Lx x x

x x x

= += + . (W1.4)

Using the trigonometric identity:

cos cos cos sin sin( + ) = – A B A B A B (W1.5)

Equation W1.4 can be rewritten as:

cos( ) cos( )

cos( ) cos( )cos( ) sin( )sin( )

k x k x k L

k x k x k L k x k Lx x x x

x x x x x x x

= += −

(W1.6)

Clearly, this equation is not true in general! It isonly true when:• cos( )k Lx x = 1, and• sin( )sin( )k x k Lx x x is zero. Since x can vary,

this will only be true when sin( )k Lx x = 0 .Now cos( )k Lx = 1 when :

k Lx x = ± ± …0 2 4, , ,π π . (W1.7)

and these values of k Lx x also cause sin( )k Lx x = 0 .So, the Born–von Karmen boundary conditionsare satisfied when Equation W1.7 is true i.e.when:

kL Lx

x x= ± ± …0

2 4, , ,

π π (W1.8)

or in general when:

km

Lxx

x= 2 π (W1.9)

where mx = ± ± …0 1 2, , , Similar equations apply inthe y and z directions and, as we stated in §6.5 wefind that for a three dimensional box::

km

Lk

m

Lk

m

Lxx

xy

y

yz

z

z= = =2 2 2π π π

, , (6.59*)

Note for pedantsNotice that the solution where mx = my = mz = 0 isnot an allowed solution, even though it is permit-ted by the boundary conditions. It is not allowedbecause (since p kx x= h and λ πx x= 2 / k ) it cor-responds to a solution with zero momentum andinfinite wavelength. This is only an allowed solu-tion in free space, and not inside any finite con-tainer. This does not make much difference exceptin cases where the number of states is very smalllike the Option A – Option B counting exercisethat we carried out for an eight atom metal.

‘Volume’ of k-space per quantum stateAs illustrated in Figure 6.19, each allowed k- state

UNDERSTANDING THE PROPERTIES OF MATTER: WEB CHAPTER 1:THE BAND STRUCTURE OF SOLIDS


can be considered to ‘occupy’ a small cuboid of‘volume’:

∆ ∆ ∆ = × ×

=

k k kL L L

V

x y zx y z

2 2 2

8 3

π π π

π(W1.10)

where V is the total volume of the crystal. I haveused the word ‘volume’ in quotation marks aboveEquation W1.10 because I am referring to ‘vol-ume’ on a k-space graph. As you can see fromEquation W1.10, the ‘volume’ on the k-spacegraph actually has the dimensions of inverse vol-ume i.e. m–3 rather than m3. The final point weneed to make is that we need to take account ofthe internal spin of electrons. This allows twoelectrons with opposite spin to occupy each k-state. Thus, the volume of k-space required foreach electron is given by:

∆ =k4 3πV

(W1.11)

Equation W1.11 tells us that the larger the volumeof the crystal V, the more closely spaced are thequantum states. Thus if we imagine doubling thephysical size of crystal from say 1 mm3 to 2 mm3,then the ‘volume’ of k-space per allowed quantumstate decreases from 1.24 × 108 m – 3 to6.2 × 107 m–3. At first sight you might be puzzledas to how this change occurs: how does k-space‘know’ about the volume of the crystal. In fact, ithighlights clearly the nature of the k-space: it isnot a real place: it is a concept used to help uscount quantum states. So the change in the densityof allowed states in k-space is not a real physicalchange, but is simply the way in which we accountfor the fact that larger crystals obviously containmore particles and so must have more quantumstates to accommodate them.

W1.1.3 The reciprocal lattice

IntroductionThe reciprocal lattice, is, like k-space itself, a con-cept. It is an abstract idea. If you persist in your

studies of solids, I am sure you will find it a usefulconcept, but it is quite possible that initially atleast, you will not see the point of the reciprocallattice. All I can say is this: you will not be alone!

The ‘metal as a box’ approachFigure 6.16 illustrates the rationale that we chosefor describing a metal as a box. We know that in ametal, the potential energy V( )r will vary in acomplex way from place to place. Importantly, weassumed that V( )r was periodic so that:

V V( ) ( )r r R= + (W1.12)

where R is a vector which is some multiple of thecrystalline period. We noted that since the wavefunctions Ψ( , )k r did not have this period, then wewould expect that in each unit cell of the crystal,the wave function would have a different value ofpotential energy. Thus, we replaced the ‘corruga-tions’ on the ‘bottom of the box’ with a ‘flat bot-tom’.

The reciprocal lattice is a set of points K whichreflects the periodicity of the crystalline potentiali.e. K is the set of values of k which describewaves which have exactly the same value at everylattice site in the crystal.

Now three-dimensional geometry is hard, and I donot want to tax you too much. However, you mustnow face up to the realities of our three-dimensional world. I will try to make life easierfor you by discussing mainly (but not exclusively)just one crystal structure, simple cubic. This crys-tal structure is illustrated in Figure 6.6 and also inFigure W1.2.

In Figure W1.2 I have also illustrated a couple ofthe waves which have same periodicity as the lat-tice. For example, if the lattice has periodicity a,the wave with k = π( / , , )2 0 0a clearly has thesame value at every point in the lattice. You canverify this by noting that:

cos cos( )2 2π = π +x

a

x ma

a(W1.12)

To see this we need merely to separate out theterms:

k-space


cos cos2 2 2π

= π + π

x

a

x

a

ma

a(W1.13)

cos cos2 2

2π

= π + π

x

a

x

am (W1.14)

and then note that all trigonometric functions re-peat after any integer number of 2π increments.We thus dignify the vector k = π( / , , )2 0 0a bysaying it belongs to the reciprocal lattice of thesimple cubic crystal structure. This point, and sev-eral other reciprocal lattice vectors, are plotted inFigure W1.2

Reciprocal lattices in generalYou may have noticed that we started with acrystal with simple cubic structure with latticeconstant a, and we ended up with reciprocal latticewhich was also simple cubic with a lattice con-stant 2π / a . In fact, every type of repeating lattice

has a corresponding reciprocal lattice. The recip-rocal lattice is the set of all the wave vectors Kthat are special to a particular lattice. By ‘special’we mean that the wave has exactly the same phasein every unit cell of the crystal. However, mostreciprocal lattices are more difficult to visualisethan the simple cubic lattices that we are dealingwith in this section.

It is also interesting that we can plot any kind ofwaves in k-space. So, for example, we can plotpoints representing phonons (§7.6) or points rep-resenting photons of different wavelengths such aslight, or X-rays or points representing the wave-lengths of electron wave functions. We will seethat the relationship between the wave vectors ofthese various waves and particles, and the recipro-cal lattice will be very important for determiningthe properties of crystals.

Figure W1.2 (a) A simple cubic lattice with lattice constant a (b) to (f) Various waves belonging to the reciprocal latticei.e. waves which have the same value at every lattice site within the crystal. (g) A segment of the reciprocal latticeshowing the points which represent the waves illustrated in (b) to (f).

(a) (b) k = (2π/a, 0 , 0) (c) k = (0, 0, 2π/a) (d) k = (2 × 2π/a, 0 , 0)

a

(e) k = (2π/a, 2π/a , 0) (f) k = (2π/a, 2π/a , 2π/a ) (g) Reciprocal lattice

ky

2πa

kz

b

c

d

e

f

kx



W1.1.4 The Brillouin zone

So now we know two features about k-space.• We know that when it comes to finding al-

lowed solutions of the particle in a box prob-lem, only certain allowed values of( , , )k k kx y z are allowed.

• And we also know that there are special setsof values of k which are given the symbol K.,These tell us which wave vectors are specialto the crystal lattice that plays host to ourelectrons.

Let me now introduce a new feature: the Brillouinzone. This is the region of k-space around eachreciprocal lattice point which is nearer to that re-ciprocal point than it is to any other. Sounds com-plicated? Not really. Although the Brillouin zonesof three-dimensional crystals in are indeed quitedifficult to visualise, in two dimensions the idea isquite simple.

Brillouin zones in 2- dimensionsA two-dimensional Brillioun zone is the area of alattice which is nearer to a particular reciprocallattice point than to any other reciprocal latticepoint. To find this area we proceed as follows:• Start at the reciprocal lattice point and draw

the K vectors to nearby reciprocal latticepoints.

• Bisect these K vectors with lines.

• The area enclosed by all the bisectors is thearea nearest to the original point. This iscalled the first Brillouin zone, or often just theBrillouin zone.

This procedure is illustrated in FigureW1.3

Brillouin zones in 3-dimensionsA three-dimensional the procedure is similar. TheBrillioun zone is the volume which is nearer to aparticular reciprocal lattice point than to any otherreciprocal lattice point. To find this volume weproceed as follows:• Start at the reciprocal lattice point and draw

the K vectors to nearby reciprocal latticepoints.

• Bisect these K vectors with planes perpen-dicular to the K vectors.

• The volume enclosed by all the bisectingplanes is the volume nearest to the originalpoint. This is called the first Brillouin zone, oroften just the Brillouin zone.

The result of applying this procedure to a threedimensional simple cubic lattice is illustrated inFigureW1.4

The ‘volume ‘of the Brillouin zoneFor a crystal with exactly N lattice sites, the Bril-louin zone contains exactly N allowed values of k.This is important. Suppose we are using k-space tovisualise the occupancy of states of a particle in abox problem i.e. the free electron gas. We recall

Figure W1.3 Illustration of the way in which a Brillioun zone is constructed.

vector to nearby reciprocallattice vector

bisector of vector

area of k-space nearerto origin than the chosen

vector

Brillioun zone

Figure W1.3 Illustration of the way in which a Brillioun zone is constructed.

vector to nearby reciprocallattice vector

bisector of vector

area of k-space nearerto origin than the chosen

vector

Brillioun zone

k-space


(§6.5) that each allowed value of k can ‘accom-modate’ two quantum states with opposite valuesof intrinsic electron spin. This was summarised nEquation W1.11 where we saw that the 'volume ofk-space per allowed quantum state was:

∆ =k4 3πV

(*W1.11)

Thus, if there is one atom at each lattice site (N)which donates one electron to the electron gas(N × 1), then the total ‘volume’ of occupied statesis:

N NV

× ×∆ = ×14 3

kπ (W1.15)

Since volume of the crystal is just Na3 we find thatthe ‘volume’ of occupied states in k-space is:

NNa a

× =4 43

3

3

3π π (W1.16)

Notice that this does not depend at all on the sizeof the crystal, but does depend on its lattice con-stant a.

The Brillioun zone of a simple cubic lattice is acube of side 2¹/a and so has volume given by:

2 83 3

3π

= π

a a(W1.17)

By comparing Equations W1.16 and W1.17 wesee that exactly half the quantum states in theBrillioun zone (i.e. half the volume of the Bril-lioun zone) would be occupied by electrons from amonovalent metal. Similarly, all the quantumstates in the Brillioun zone would be occupied byelectrons from a divalent metal.

This result, that exactly half of the quantum statesin the Brillioun zone would be occupied by elec-trons in a monovalent metal is not restricted tosimple cubic lattices but applies to all lattices:

The significance of the Brillouin zoneWhy have I introduced the idea of a Brillouinzone? What is the significance of the Brillouinzone boundaries? After 20 years of studying thissubject I can find no obvious significance, inthemselves. However, I find that time and timeagain interesting physics occurs in a crystal whenany kind of wave takes a value near a Brillouinzone boundary. For each different type of wave(X-rays, electrons, phonons) the physics is slightlydifferent, but it is always interesting

Figure W1.4 Illustration of the 3 dimensional Brillioun zones of (a) the simple cubic (sc), (b) the face centred cubic(fcc) direct lattice and (c) the body-centred cubic (bc) direct lattice. .(a) The sc direct lattice has a reciprocal latticewhich also forms a sc pattern with a lattice constant of side 2π/a and the first Brillioun zone forms a cube of side 2π/a .(b) The fcc direct lattice has a reciprocal lattice which forms a bcc pattern with a cube of side 4π/a and the first Brilliounzone forms a 14-sided shape which can be described as a truncated octahedron. (c) The bcc direct has a reciprocallattice which forms an fcc pattern with a cube of side 4π/a and the first Brillioun zone forms a 12 sided shape known asa regular rhombic dodecahedron. Notice that for all of the figures drawn:

• The origin of k-space lies at the centre of the shape depicted.

• The region inside the shape depicted is closer to the origin than it is to any other reciprocal lattice point.

4πa

4πa

4πa



W1.2 Free electron theory of metals

W1.2.1 General descriptionWe have already presented the basic ideas of thefree electron theory of a metal in §6.5. Here wewill remind ourselves of the relationships betweenthe k vector that describes each electron quantumstate and (a) the momentum and (b) the energy, ofthe particle occupying that state. However as wemake our calculations, we are now in a position tokeep in mind the ‘landscape’ in which these elec-tron states belong: the reciprocal lattice that wedescribed in §1.1. It will turn out to be importantto understand the relationship between:• the wave vectors describing the electron

quantum states, and• the wave vectors describing the reciprocal

lattice,If the electronic quantum states were one-dimensional, then we could relate the wave vectorto the wavelength of the wave by:

λ = π2

k(W1.18)

In three dimensions, things are a little more com-

plicated. Recall that the ki ’s are components of avector k that points from the origin to the pointk = ( , , )k k kx y z . Now the wavelength of the waverepresented by k = ( , , )k k kx y z is given by:

λ = π

= π

+ +

2

22 2 2

k

k k kx y z

(W1.19)

Equation 6.63 allows us to use the de Broglie rela-tion (Equation 2.42) to infer the momentum andenergy of the wave represented by ( , , )k k kx y z . Themomentum may be written as:

ph h

p k k k

= = =

= + +

πλ 2

2 2 2

k

kh

h x y z

(W1.20)

Notice that the quantum states with small amountsof momentum have small values of k . Thus in k-

Figure W1.5 (a) In three-dimensions the occupied states form a sphere in k-space known as the Fermi sphere which isshown shaded in the figure. If we imagine slicing through the Fermi sphere in plane where kz = 0 we would find a situa-tion represented in (b). Each small circle represents an allowed travelling wave solution to the Schrödinger Equation.The filled circles represent occupied quantum states and the unfilled circles represent empty quantum states. At abso-lute zero, only the lowest energy states (low k = long wavelength = low energy) are occupied. For any macroscopicpiece of metal the quantum states would be much more densely packed than in the figure.(a) (b)

kx

kz

ky

+kx

+ky

–kx

–ky

FREE ELECTRON THEORY OF METALS


space they will be represented by points near theorigin. The energy, which is just the kinetic energyof a free particle, may be written as:

E m vp

m

Em

Em

k k k

= =

=

= + +[ ]

1

2 2

2

2

22

2 2

2

ee

e

ex2

y2

z2

h

h

k (W1.21)

Notice again that the low-energy quantum stateshave small values of k . Thus in k-space they willbe represented by points near the origin. Impor-tantly, the last Equation in W1.21 is the equationof a sphere. Thus, all quantum states with energy Ewill be represented by points in k-space which lieon the surface of a sphere with radius:

k = 22

m Eeh

(W1.22)

The density of states function.The density of states function g(E) is the functionwhich answers the question ‘How many quantumstates are there with energy between E andE + dE ’. We derived an expression for this in§6.5.2 where we found:

g EV m E

V mE

( ) =π

=π

2

2

2 3

2 3

e3

e3

h

h

(*6.70 & W1.23)

Notice that g E( ) depends only a group of funda-mental constants, and increases as the square rootof the energy.

We can derive an expression for the Fermi energyusing the density of states function. To do this wenote that at the absolute zero of temperature, all ofthe quantum states below the Fermi energy will beoccupied, and all the states above the Fermi en-ergy will be empty. So, we can write that:

N g E EE

E E

= ∫=

=( )d

F

0(W1.24)

Substituting for g E( ) from Equation W1.23 andtaking the constants outside the integral we find:

NV m

E E

V mE E

NV m

E

E

E E

E

E E

=π

∫

=π

∫

=π

=

=

=

=

2 30

2 30

2 3

2

2

2 2

3

12

12

e3

e3

e3

F

d

d

F

F

32

h

h

h

(W1.25)

Rearranging this as an expression for the Fermienergy:

3

2 2

2 3n

mE

π =h

e3 F

32

En

mFe2= π( )h2 23

2

23

(W1.27)

which agrees with Equation 6.73.

Figure W1.6 Showing g(E) occupied up to the Fermienergy.

En

erg

y

Density of states g(E)

Fermi Energy

Occupied states

Empty states



Potassium: a simple metalIn §6.5 we worked out the free electron estimatesfor the Fermi parameters of the metal copper. Herewe compare those estimates with those for potas-sium. Potassium is a group 1 metal which shouldsatisfy well our idea of a material which ‘ionises’easily in the solid, leaving a gas of electrons rela-tively free to move through the metal. We expecteach atom to ‘donate’ one electron to the electrongas.

As in §6.5, we estimate the number density ofelectrons from the mass density of the elementalmetal. First we divide the mass density by themass of an atom to get the number density of at-oms. We then multiply by the valence to arrive atn. From Table 7.2 we see that the density of potas-sium is 862 kg m–3, and the mass of an atom is39.1 × 1.66 × 10–27 kg. Assuming potassium has avalence of one, then our estimate for n is:

n = ×× ×

= ×

−

−

1 862

39 1 1 66 10

1 33 10

27

28 3

. .

. electrons m(W1.27)

which we can compare with nC u = 8.47 × 1028

electrons m–3 for copper. It is interesting to notethat copper has an electron density some six timesgreater than potassium. As discussed in §7.2, thisis due to the effect of additional covalent bondingbetween copper atoms by the d-electrons.

Similarly we can estimate the Fermi wave vector,Fermi energy, density of states and the density ofstates at the Fermi energy as we did for copper.The results are summarised in Table W1.1

A metal in a lattice.We can now compare the periodicities encounteredin the lattice with those occurring in the free elec-tron gas. We will use the fact that potassium crys-tallises in the bcc structure with a conventionalunit cell of side a = 0 52. nm to compare the sizeof the Fermi sphere with the size of the Brillouinzone. In particular, for reasons which will becomeclear in the next section, we will find out whetherthe Fermi sphere fits entirely within the Brillouinzone.

The reciprocal lattice of a bcc crystalWhat does the reciprocal lattice of a bcc crystallook like? The reciprocal lattice of bcc of conven-tional unit cell side a is fcc with conventional unitcell of side 4¹/a. Numerically, this amounts to≈12.6/a or (for a = 0.532 nm) the lattice constantamounts to 2 36 1010. × m–1. The reciprocal lattice isillustrated in Figures W1.4 and W1.7.

Figure W1.7 gives an impression of the Brillouinzone with the Fermi sphere illustrated inside thezone. Although three-dimensional geometry ishard, we can make a relatively simple calculationif we confine our attention to a cross-section of theFermi sphere in the kz = 0 plane.

First we need to calculate the distance from theorigin to the point on the Brillouin zone boundarywhich most closely approaches the origin. This isat the midpoint of the line joining the origin to the(1,1,0) reciprocal lattice point. From the scale ofthe diagram we can see that this distance (OA)from the origin to the nearest reciprocal latticepoint is:

21

2

41 67 1010 1× × π = × −

a. m (W1.28)

and so the distance (OB) from the origin to theBrillouin zone boundary at its closest point of ap-proach is:

Table W1.1 Free electron parameters for potassium andcopper. Notice that in the free electron approximation, thedensity of states function g(E) is the same for all metals.

Potassium Copper

kF 7 33 109 1. × −m 13 6 109 1. × − mEF 2 05. eV 7 06. eV

g E( )1 062 1056

1 3

. ×− −

E

states J mg E( )F 6 08 1046

1 3

. ×− −states J m

1 13 1047

1 3

. ×− −states J m

FREE ELECTRON THEORY OF METALS


1

22

1

2

40 835 1010 1× × π

= × −

a. m (W1.29)

We now need to compare this distance with theradius of the Fermi sphere, kF.. At the risk of bor-ing you, let us do this calculation yet again. Recallthat the Fermi sphere must ‘accommodate’ N elec-trons, each of which occupies a ‘volume’ givenEquation W1.11.

4

3

4 3

π = × πk N

VF3 (W1.30)

Remembering that the total volume of the crystalis just N times the volume of a unit cell containingone atom we write:

V Na=

3

2(W1.31)

Substituting this result into Equation W1.30

4

3

4

2

83

3

3

3π = × π = πk N

Na aF

3 (W1.32)

and solving for kF we find:

ka a

ka

ka

F3

F

F

= ππ = π

π

= π

π

= π

3

4

8 3

4

2

3

4

2

0 6202

3

3

3

13

.

(W1.33)

Evaluating this for a = 0.532 nm we find:

kF = 0.73 × 10–10 m–1 (W1.34)

Comparing Equation W1.34 with W1.29 we seethat the Fermi sphere extends 0.73/0.84 ≈ 87% ofthe way to the Brillouin zone boundary. So, basedsolely on an estimate of electron density, the freeelectron theory predicts that the Fermi sphere liesentirely within the Brillouin zone. This quite asimple prediction, and as we mention in just amoment, an amazingly accurate one.

Is this correct?In the next section (§W1.5), on the nearly-freeelectron model of a metal, we shall see that theprediction that the Fermi surface of potassium is asphere is true because it does not come near theBrillioun zone boundary. In §W1.5 on the tightbinding model we will see that it is possible tohave Fermi surfaces which are quite different inshape from a sphere. Astonishingly, in my opinionat least, one can determine the Fermi surface of

Figure W1.7 Illustration of the relationship between the size of the Fermi sphere and Brillioun zone for potassium, amonovalent metal with a bcc crystal structure. (a) Shows the fcc reciprocal lattice of the bcc direct lattice showing theBrillioun zone from Figure W1.4 and the kz = 0 plane. The origin is shown as a white dot in this picture and the otherreciprocal lattice points are shown as black dots. (b) The first Brillioun zone 'chopped' in kz = 0 plane showing the Fermisphere. (c) Detailed view of the kz = 0 plane showing the relative sizes of the Fermi sphere and Brillioun zone.

(a) (b) (c)

4πa

4πa

4πa

4πa

0.7071

kFO

A

B



many metals experimentally. The best establishedtechnique for achieving this is by means of the deHaas-van Alphen effect. This effect gives rise totiny oscillations in the magnetisation of very puremetals at very low temperatures (<≈ 1 K) and invery high magnetic fields (>≈ 3 T). From a de-tailed examination of these oscillations, in par-ticular their periodicity as the magnetic field isvaried, the Fermi surface may be reconstructed.

Using techniques such as this, the Fermi surface ofpotassium is found to be in excellent agreementwith this simple theory. The diameter of the Fermisphere is found to agree with the simple estimateswe made to within 1% and the ‘sphere’ deviatesfrom being a perfect sphere by less than 1 part in103. To put this in perspective, it is a more perfectsphere than a typical snooker (pool) ball.

So consider this: the free electron theory (which isbased solely on an estimate of electron density andhas no free parameters, correctly predicts the elec-tronic structure of at least one real metal to withinone percent! This result is even more astoundingthan the predictions for the bulk modulus of potas-sium that we calculated in §7.3 where the agree-ment was only at the 10% level.

However, there is a ‘fudge’ here. It is that we hadto decide arbitrarily what the valence was. Thusthe theory does not tell us what we asked at thestart of the chapter, namely, it does not tell us thatargon (one electron and one proton less than potas-sium) will not be a metal. So the free electron the-ory is sometimes surprisingly good, but there aresome questions for which it is not able to provideanswers.


W1.4 The nearly-free electron theory of metals and insulators

W1.4.1 IntroductionIn the last section we considered the case of elec-trons which were free to move within the metaland so the only contribution to their energy wastheir kinetic energy:

In this section we will consider how the electronsare affected by the crystalline potential arisingfrom the ion cores. Essentially, this is an attempt tolook at the effect of the ‘corrugations’ at the bot-tom of the box in Figure W1.8. As long as the cor-rugations are not too large, we will find that theenergy varies with k in a way which is similar tothat predicted by the free-electron model. Thissupports the use of the free electron model that weconsidered in the last section where we ignored thepotential energy completely! However, we willfind that when k is close to the boundary of theBrillouin zone then the potential energy contribu-tion becomes significant. We will also find that weare able to take a first step towards understandinghow to represent materials that are not metals.

W1.4.2 The total energyThe band structure of a material is (essentially), adescription of the way the total energy of elec-tronic states varies as a function of their quantumnumber. k = ( , , )k k kx y z . In other words, the bandstructure of a material is a statement of the totalenergy relationship E( )k , which is in turn the sumof the kinetic and potential energy contributions.

Kinetic energyThe kinetic energy contribution is related to theaverage curvature of the wave function which inone dimension is given by:

− ∂∂

h2 2

22m xx

eΨ( ) (W1.35)

In the free electron approximation the potentialenergy contribution is assumed to be zero. Thewave functions are plane waves with the generalform A k xcos x and the total energy is just the ki-netic energy:

E kk

m( )x

x2

e= −h2

2(W1.36)

Potential energyThe potential energy contribution V( )k is causedby the electrostatic (Coulomb) interactions be-tween the electron and all the other electricallycharged objects in its vicinity i.e. the ions, and theother electrons. In the nearly-free electron modelwe acknowledge that a potential energy contribu-tion exists, but we assume that it is (a) small com-pared to the kinetic energy contribution and (b) hasthe same periodicity as the crystal.

Introducing the potential energy in this way is ob-viously not very realistic. However, even at thislevel of approximation it introduces an importantphysical feature which is also true of more com-plex theories. This is that the energy of a planewave is no longer determined solely by the wavevector k. It can also depend on the phase of thewave function. In this case it means that the func-tion Ψ = sinkx can have a different energy than thefunction Ψ = coskx. This is a feature of all morecomplex theories and the nearly-free electron the-ory is the simplest theory which includes this fea-ture.

Figure W1.8 Illustration of the potential energy in thenearly-free electron model.

The ‘free electron’ or ‘particle in a box’ approximation to the real potential

The expected potential energy for an electron

in a crystal

The potential energy in the

nearly free electron model

States described by the nearly-free electron model have energies in

this range



Total energyThe total energy is the sum of the kinetic energyand the potential energy. However, we need totake account of the fact that the potential energydepends not just on k, but on the phase of the wavefunction. Thus we shall need to list separately theway in which total energy of sine-based wavefunctions depends on k, and then way in whichtotal energy of cosine-based wave functions de-pends on k. Using the labels A and B to indicatesine- and cosine-based wave functions respectivelywe write:

E KE V

E KE VA A

B B

( ) ( ) ( )

( ) ( ) ( )

k k k

k k k

= += + (W1.37)

W1.4.2 A ‘simple’ calculationLet us now follow our introductory discussion bymaking some estimates of the effect of introducinga periodic potential into the free-electron ‘box’.

Considering only one-dimension, we will considerthe effect a periodic potential of the form:

V x Vx

a( ) cos= π

o

2 (W1.38)

We will evaluate this for electron wave functionsof the form:

ΨAx

x= 2

Lk xsin (W1.39)

ΨBx

x= 2

Lk xcos (W1.40)

where Lx is the size of the ‘box’ in the x-direction.We will work out the total energy in four stages:• First, we will demonstrate that Equations

W1.39 and W1.40 are normalised.• Second, we will calculation the kinetic energy• Third, we will calculation the potential energy• Finally, we will add the potential and kinetic

energies and then describe a more accurateway of calculating the total energy

Stage 1: NormalisationFirstly, let us notice that these wave functions arenormalised. This means that if we evaluate theprobability of finding the electron anywhere theanswer is unity. This reflects the fact that the elec-tron must be some somewhere. Mathematically wewrite:

Ψ ΨA*

Aall space

∫ = 1 (W1.41)

We now substitute using W1.39 to find:

IL

k xL

k x xL

Ax

xx

x dx= ×∫ 2 2

0sin sin (W1.42)

Taking the constants outside the integral yields:

IL L

k x x

Lk x x

L

L

Ax x

x

xx

d

d

x

x

= ∫

= ∫

2 2

2

2

0

2

0

sin

sin

(W1.43)

If we had started with ΨB we should have found

IL

k x xL

Bx

x dx= ∫2 2

0cos (W1.44)

To proceed with either integral we need to sim-plify the kernel of the integral using the trigono-metric identities:

cos cos2 1

21 2θ θ= +[ ] (W1.45)

Figure W1.9 Illustration of the kinetic energy parabolain two dimensions.

kx

ky

E(kx, ky)

E

NEARLY-FREE ELECTRON THEORY OF METALS AND INSULATORS


sin cos2 1

21 2θ θ= −[ ] (W1.46)

We can now treat the two wave functions A and Bat the same time by writing:

IL

k x xL

A/Bx

x dx= ±[ ]∫2 1

21 2

0cos (W1.47)

where wave function A takes the lower option (–)of the ± sign and wave function B takes the upperoption (+). The integral can then be separated intotwo integrals as follows:

IL

k x x

Lx k x x

L

L L

A/Bx

x

xx

d

d d

x

x x

= ±[ ]∫

= ∫ ± [ ]∫

2 1

21 2

12

0

0 0

cos

cos

(W1.48)

Now the first integral is straightforward and thesecond integral is conveniently very small. We cansee that this is so because the integral itself evalu-ates to somewhere between 1 and –1 as for anyoscillating function, but the pre-factor 1/Lx makesthe integral extremely small for any reasonablylarge crystal. We thus write:

IL

x k x x

Lx

LL

I

L L

L

A/Bx

x

x

xx

A/B

d d

d

x x

x

= ∫ ± [ ]∫

= ∫ ±

= ±[ ]=

12

10

10

1

0 0

0

cos

(W1.49)

So we conclude that these are indeed normalisedwave functions.

Stage 2: Kinetic energyWe estimate the kinetic energy using the kineticenergy operator:

− ∇ = − + +

h h22

2 2

2

2

2

2

22 2m m x y z

∂∂

∂∂

∂∂

(W1.50)

which can be written in one-dimension as:

− h2 2

22m x

∂∂ (W1.51)

The kinetic energy is found by evaluating the ex-pectation value of the kinetic energy operator :

KEm x

xL

A A Adx= −

∫ ∗Ψ Ψh2 2

20 2

∂∂ (W1.52)

KEL

k xm x L

k x xL

Ax

xx

x dx= −

∫ 2

2

22 2

20

sin sinh ∂

∂(W1.53)

KEm L

k xx

k x xL

Ax

x x dx= −

∫h2 2

202

2sin sin

∂∂

(W1.54)

Differentiating in two stages we find:

KEm L

k xx

k k x xL

Ax

x x x dx= −

∫h2

02

2sin cos

∂∂

(W1.55)

and then:

KEm L

k x k k x xL

Ax

x x2

x dx= − −[ ]∫h2

02

2sin sin

(W1.56)

which simplifies to:

KEk

m Lk x x

L

Ax2

xx d

x= ∫h22

02

2sin (W1.57)

Similarly we find for ΨB that:

KEk

m Lk x x

L

Bx2

xx d

x= ∫h22

02

2cos (W1.58)

and so in general we can write:

KEk

m Lk x x

L

A/Bx2

xx d

x= ±[ ]∫h2

02

2 1

21 2cos

(W1.59)

As we saw in Equations W1.47 to W1.49, the inte-gral evaluates to Lx/2 and so we find:



KEk

mA/Bx2

= h2

2(W1.60)

which does not depend on the phase of the wavefunction (sine or cosine). Now let us consider thepotential energy terms.

Stage 3: Potential energyWe estimate the potential energy of these wavefunctions in this cosine potential using the poten-tial energy operator V(r)

PE V xL

A A Adx= [ ]∫ ∗Ψ Ψ( )r

0(W1.61)

which in one dimensions reduces to:

PE V x xL

A A Adx= [ ]∫ ∗Ψ Ψ( )

0(W1.62)

PE Vx

ax

L

A A o Adx= π

∫ ∗Ψ Ψcos

2

0(W1.63)

Substituting for ΨA we find:

PEL

k x Vx

a Lk x x

L

Ax

x ox

x dx= π

∫ 2 2 2

0sin cos sin

(W1.64)

PEV

Lk x

x

ak x x

V

Lk x

x

ax

L

L

Ao

xx x

o

xx

d

d

x

x

= π

∫

= π

∫

2 2

2 2

0

2

0

sin cos sin

sin cos

(W1.65)

Again we use the trick of writing sin2 in terms ofthe cosine of twice the angle to find:

PEV

Lk x

x

ax

L

Ao

xx d

x= −[ ] π

∫2 1

21 2

2

0cos cos

(W1.66)

If we had started with wave function B we wouldhave found

PEV

Lk x

x

ax

L

Bo

xx d

x= +[ ] π

∫2 1

21 2

2

0cos cos

(W1.67)

and so in general we can write:

PEV

Lk x

x

ax

L

A/Bo

xx d

x= ±[ ] π

∫2 1

21 2

2

0cos cos

(W1.68)

Where ΨΑ corresponds to the lower option of the ±sign and ΨΒ corresponds to the upper option. Mul-tiplying out this bracket yields:

PEV

L

x

ax

V

Lk x

x

ax

L

L

A/Bo

x

o

xx

d

d

x

x

= π

∫

± π

∫

2 2

22

2

0

0

cos

cos cos

(W1.69)which amounts to:

PEV

Lk x

x

ax

L

A/Bo

xx d

x= ± π

∫0

22

2

0cos cos

(W1.70)

Notice that the first integral in Equation W1.69 iszero because we are integrating over a functionwith period a, and Lx (the length of the crystal) isan integer number of unit cell spacings in length.

Now we arrive at an interesting integral. If kx takesalmost any value then the integral is zero for thesame reason we just discussed. However, if kx

takes on the value ±π / a , then the integral is notzero and we find:

PEV

L

x

ax

L

A/Bo

xd

x= ± π

∫2 22

0cos (W1.71)

Which can be written in terms of the same trigo-nometric identity we discussed in Equation W1.45

PEV

L

x

ax

L

A/Bo

xd

x= ± + π

∫2 1

21

4

0cos

(W1.72)

PEV

Lx

V

L

x

ax

L

L

A/Bo

x

o

x

d

d

x

x

= ± [ ]∫

± π

∫

1

2 1

2

4

0

0cos

(W1.73)



Once again the second term is zero because it is aperiodic function integrated over an integer num-ber of periods and we find:

PEV

LL

V

A/Bo

xx

o

= ± [ ]= ±

(W1.74)

or explicitly:

PE V

PE VA o

B o

= −= + (W1.75)

Stage 4: Total energy: a more complete calcula-tionThe results of our nearly-free electron calculationcan be summarised as follows:

E k

k

mif k

a

k

mV if k

a

A x

x2

ex

x2

eo x

(2

2

) =≠ ± π

− = ± π

h

h

2

2(W1.76)

E k

k

mif k

a

k

mV if k

a

B x

x2

ex

x2

eo x

(2

2

) =≠ ± π

+ = ± π

h

h

2

2(W1.77)

These equations indicate that energy is the same asfor the free electron model, except when the wavevector is exactly kx = ±¹/a. The physical differencebetween the two cases A and B, and the signifi-cance of the special case when kx = ±¹/a, is illus-trated in Figure W1.10 and described in the legendto the figure.

The result encapsulated in Equations W1.76 andW1.77 is interesting, but in fact is only approxi-mately correct. It has been arrived using what iscalled first-order perturbation theory. A more ac-curate calculation (using, for example, second-order perturbation theory) would take account ofthe fact that in response to the potential V(x) thewave functions change slightly from simple sineand cosine waves. Before we go on to see whatsecond-order perturbation theory tells us, it is im-portant to note again the key result of first ordertheory: it tells us that ‘something happens’ at

kx = ±¹/a. Those with a keen memory will recallthat this exactly at the Brillioun zone boundary.

Second-order resultsThe results of the second-order calculation are insome ways only slightly different from the first-order calculation, but they will change the way wethink about electrons in solids profoundly. Thesecond-order calculation takes account of the factthat wave functions are composed of linear super-positions of sine and cosine wave functions. Formost values of kx the sine and cosine functions aredegenerate (i.e. have the same energy), just as wecalculated in first order theory. Because of the de-generacy, the actual wave functions are composedof equal amounts of sine and cosine functions.However, near the Brillioun zone boundary, thesine and cosine functions are no longer degenerate(as revealed by the first-order calculation) and theactual wave functions are composed of primarilyeither sine or cosine functions. The effects aretwofold:• The energy of all the states is slightly lowered.

We will neglect this small change in whatfollows.

• The energy of states close to kx = ±¹/a is af-fected in a manner similar to the way in whichstates with exactly kx = ±¹/a were affected inthe first-order calculation.

These results are graphed in Figure W1.11. Nowwe see that the potential has introduced a gap(known as a band gap or energy gap) in the spec-trum of possible electron energies. This is the keyresult of this nearly-free electron analysis and theresult is quite general. It tells us that if we considerthe amount of energy required to move an electronfrom one quantum state to another, then normallythis quantity is infinitesimally small. However ifwe consider quantum states on two sides of a Bril-lioun zone boundary then:• To cross the Brillioun zone boundary associ-

ated with a potential V x ao cos /2π( )we mustprovide an additional 2Vo of energy.

Before we can see why this is so important, weneed to take two more steps: we need to see theeffect of a more realistic potential, and then weneed to consider the effect of a periodic potentialin more than one dimension.



Figure W1.10 Illustration of the meaning of the results of the nearly-free electron calculation summarised in EquationsW1.76 and W1.77

(a)The top curve shows the cosine potential (Equation W1.38) which has the periodicity of the lattice (a). We imaginethat the minima of the potential correspond to the atomic positions. The lower curves show the wave functions andcharge densities associated with solutions A and B (Equation W1.39 and W1.40) when the wave vector is small, faraway from a Brillioun zone boundary. The example shown corresponds to k ≈ 0.17 π/a. Notice that in this case thecharge density is large in some unit cells and small in others. On average this causes the perturbation to the energy ofthe both states A and B to remain nearly unaffected by the potential.

Position within crystal (x)

V(x)

ΨA(x)

ΨB(x)

|ΨA(x)|2

|ΨB(x)|2

a

(b) This is the equivalent graph to (a) for the case where k ≈ π/a. Notice that now (in contrast with the case in (a)) thecharge density for each solution A or B is exactly the same in every unit cell of the crystal. The detailed nature of thefunctions are shown for a single unit cell in (c). In case A the charge density in every unit cell is concentrated in theregion of lowest potential, and so this solution has lower energy than case B, where the charge density in every unitcell is concentrated in the region of highest potential.

(b)


V(x)

ΨA(x)

ΨB(x)

|ΨA(x)|2

|ΨB(x)|2

a (c)


V(x)

ΨA(x)

ΨB(x)

|ΨA(x)|2

|ΨB(x)|2



Figure W1.11 Illustration of the effect of a weak periodic potential on the one-dimensional E(k).

(a) The free electron parabola (corre-sponding to the kinetic energy of anelectron) is shown along with the poten-tial energy contributions for each solu-tion A and B (Equation W1.39 andW1.40).

Ene

rgy

kx

π/a–π/a 0

+Vo

–Vo

Kinetic energy

Potential energyΨA

ΨB

( a )

(b) The result of the correct calculationof the energy is shown together with thefree electron parabola (dotted). In thiscalculation, the composition of the wavefunctions changes across the curve.

Ene

rgy

kx

π/a–π/a 0

+Vo

–Vo

+Vo

–Vo

( b )

Total energy

(c) In the parts of the curve which are‘free-electron-like’, the wave functionsare composed of equal amounts of sineand cosine functions. However, near theBrillioun zone boundary, the functionsare either like or sine or cosine func-tions.

Ene

rgy

kx

π/a–π/a 0

mainlycosine (Ψ

B)

sine (ΨA)

+cosine (Ψ

B)

( c )

sine (ΨA)

+cosine (Ψ

B)

mainlysine (Ψ

A)

mainlysine (Ψ

A)



Making the potential more realisticDoes the nearly-free electron model allow us touse a ‘model’ potential closer to the ‘actual’ po-tential? Fortunately, the answer to this is ‘Yes’. Todo this we write the potential as a Fourier sum ofcosine (and sine) terms. For example, consider theone dimensional potential:

V x Vx

aV

x

a( ) cos cos= π

+ π

o 1

2 6 (W1.78)

This potential (Figure W1.12) is more ‘pointed’than the simple sinusoidal potential and representsat least a step towards the atomic potentials shownon the bottom of Figure W1.8.

The first thing to notice about Equation W1.78 isthat the new wave vector we have added (k = 6¹/a)belongs to the reciprocal lattice. In fact, as wesynthesise the new potential, we can only includewave vectors which have the same periodicity asthe lattice i.e. those which belong to the reciprocallattice.

V x Vn x

an( ) cos

,= +[ ] π

∑

= … n2 1 2

1 2(W1.79)

Now, in the same way that the main component ofthe potential (k = 2¹/a) only affects the energy ofthe quantum states with wave vectors (k = ±¹/a),so the additional component (k = 6¹/a) only affectsthe energy of the quantum states with wave vectors(k = ±3¹/a). (These bisectors of the successive

reciprocal lattice vector are known as the second,third, fourth etc. Brillioun zone boundaries.) Thisresult dramatically simplifies our analysis.

In metals, electrons only occupy quantum stateswith wave-vectors less than (or of the order of) kF.So, if kF is less than (for example) 3¹/a, then (inthis analysis) we can simply ignore the effect ofthe higher order components of the potential. Inother words, no matter how complex the potentialenergy curve, (in the nearly-free electron approxi-mation) it will simply make no difference to theenergy of the quantum states. This remains true aslong as the components of the potential are not toolarge.

PotassiumRecall that for potassium, kF was small enough thatthere were no occupied states in the region of theBrillioun zone boundary (Equation W1.33, FigureW1.7), i.e. kF < ±¹/a. So now we can understandwhy the free-electron theory gave such anoma-lously good predictions of the basic electronicproperties of potassium. Because the occupiedquantum states have wave vectors smaller thaneven the first Brillioun Zone boundary, there is nocomponent of the electronic potential that affectstheir energy. Another way of stating this is to saythat the wavelengths of the occupied electron wavefunctions are all so long that the details of the po-tential are effectively averaged across many unitcells. This is exactly the result illustrated in FigureW1.10

Figure W1.12 Illustration of the potentialin Equation W1.78 for the case whenV1 = 0.3 Vo. The two ‘component’ poten-tials are shown in grey and their sum isshown bold. Also shown are positions ofthe ion cores.

The sum potential has a little more of thecharacter of the sharp potential seen inFigure W1.8.

-1.5

-1

-0.5

0

0.5

1

1.5

Pot

entia

l Ene

rgy

V(r

)



Moving to more than one dimensionTwo-dimensionsIt is only when we consider a hypothetical two-dimensional crystal that we can appreciate fullyhow the nearly-free electron model allows us todescribe a much richer variety of physics than doesthe free-electron model. Remember that the free-electron model describes a substance which is ametal. The nearly-free electron model allows us todescribe a material which can be either a metal oran insulator depending on the strength of the inter-action Vo.Consider a two-dimensional square lattice with an‘egg box’ potential:

V x y V x a y a( , ) cos / cos /= π[ ] π[ ]o 2 2 (W1.80)

This is illustrated in Figure W1.13. The variationof energy across the first Brillioun zone of the re-ciprocal lattice for this two-dimensional lattice isshown in Figure W1.14. Notice that the distancefrom the origin to the boundary of the first Bril-lioun zone does not have a unique value.

• If one considers the quantum states of elec-trons travelling in the x- or y-directions i.e.k = (k, 0) or k = (0, k) then one reaches theBrillioun zone boundary at the point wherek = ¹/a. At this point the energy of the quan-tum state is approximately

h h2 2 2 2

2 2

k

m ma

e e= ( )π

• If one considers quantum states of electronstravelling in between the x- and y-directionsi.e. k = (k, k) then one reaches the Brillioun

zone boundary at the point where k = 2π / a .At this point the energy of the quantum stateis approximately.

22 2

2 2 2 22

× = ( )πh hk

m m

a

e e

Figure W1.15 shows how the energy varies as oneproceeds from the origin along lines joining theorigin to the point (0, ¹/a) and to the point ( ¹/a,¹/a). Notice that the magnitude of the parameter Vo

in Equation 1.80 now becomes very important.• If Vo is large compared with the maximum

energy of a quantum state in the first Brilliounzone, then all the quantum states in the secondBrillioun Zone will have energies higher thanany quantum state in the first Brillioun Zone.In this situation there is what is known as aband gap between states in the two zones.

• However, if Vo is small compared with themaximum energy of a quantum state in thefirst Brillioun zone, then some of the quantumstates in the second Brillioun Zone will haveenergies lower than some quantum states inthe first Brillioun Zone. In this situation thereis said to be band overlap between states inthe two zones.

The significance of this comes when we imaginepopulating the quantum states.

Populating the quantum statesOne electron per lattice siteIn Section W1.1.4 we saw that the Brillioun zonecontains enough quantum states for exactly twoelectrons per lattice site in the crystal. So if weconsider the case of a monovalent substance whichdonates one electron per atom to the electron gasthen the first zone will always be large enough toaccommodate all the electrons. There will thenalways be regions where the highest-energy occu-pied quantum states have empty quantum statesimmediately above them.

This is illustrated for the two dimensional case inFigure W1.14. In two dimensions, donating oneelectron per lattice site corresponds to filling halfthe area of the zone. It is clear from the figure thateven if E (k) was such that the occupied statesreached the zone boundary, there would always be

Figure W1.13 The two dimensional ‘egg-box’ potential

V(x, y)

x

y



occupied states with adjacent empty states. Thusan excitation of the electron system (thermal orelectrical) would allow electrons to move into un-occupied states with infinitesimal expenditure ofenergy. This describes a substance that we wouldcall a metal.

Two electrons per lattice siteIf a material donates exactly two electrons per lat-tice site in the crystal, then the situation is morecomplex. The properties of the resulting substancethen depend on the magnitude of the perturbationparameter Vo.

If Vo is large, then the lowest energy states in thesecond Brillioun zone have energies which aregreater than the highest energy state in the firstBrillioun zone. In this case, all the states in thefirst Brillioun zone will be occupied and all thestates in the second Brillioun zone will be empty.We say that such a substance has a band gap. Thiscase is illustrated in Figure W1.15(a), (b) and (c).

We cannot tell much about this material from itsE(k) relationship until we draw in the Fermi En-ergy. At T = 0, this has an energy half way be-tween the last occupied level and the next unoccu-pied level. Drawing on the Fermi energy for thecase of two electrons per quantum state, we cansee that a material such as this would not be ametal, but an insulator.

To appreciate this, consider how an electron in astate in the first Brillioun zone could respond to anapplied electric field. In order to respond, it wouldneed to move into an empty quantum state, and theonly empty states are in the second Brillioun zone.In order to move into this empty state it wouldtherefore need ≈ 2Vo of energy. If we consider thecase where Vo is of the order a few electron volts,then such large energies are extremely unlikely tobe available from thermal sources. Thus, such asubstance would be an electrical insulator. (§7.7.4to §7.7.9)

If Vo is small, then the lowest energy states in thesecond Brillioun zone have energies which are lessthan the highest energy state in the first Brilliounzone. In this case, most, but not all the states in the

first Brillioun zone will be occupied and most, butnot all the states in the second Brillioun zone willbe empty. We say that such a substance has a bandover lap . This case is illustrated in FiguresW1.15(d) (e) and (f). A material such as this wouldstill be a metal, but probably with a lower conduc-tivity than a monovalent metal.

To understand this, notice that if the bands onlyjust overlap, although there are a large number ofoccupied states in the first zone, few of these oc-cupied states have adjacent empty states. Simi-larly, there are very few occupied states in the sec-ond zone. Thus if an electric field were applied tothis material, a relatively small number of elec-trons would be able to change quantum state inresponse to the field. The precise number woulddepended on precisely the extent of the overlapbetween the bands. Such a material might be calleda metal if the overlap were quite large, or a semi-metal if the overlap were small.

Summary of nearly-free electron modelBefore moving on to ever more complex models ofmaterials, it is worth pausing to summarise whatwe are the achievements and limitations of thenearly-free electron model.

Figure W1.14 Illustration of the occupancy of quantumstates. The figure shows a two dimensional E(k) relation(grey) with occupied states (dark) corresponding to ap-proximately half the area of the Brillioun zone. Note thatthe highest energy occupied states have adjacent emptystates into which they can be easily excited.

ky

kx

E(kx,ky)

Brilliounzone



Figure W1.15 Illustration of the results of two-dimensional, nearly-free electron calculation for the case of two electronsper lattice site. (a) (b) and (c) correspond to the case when V0 is large and (d) (e) and (f) to the case when V0 is small.(a) and (d) show the structure of E(k) along k,0,0 and k,k,0 directions (b) and (e) show the resulting density of states.Notice that in (b) the density of states at the Fermi energy is zero which is indicative of insulator. (c) and (f) show threedimensional views of the band structure in the two bands

(a)

Ene

rgy

kπ/a–π/a 0

Fermienergy

Bandgap

E(k) along (1,1,0)

directionE(k) along

(1,0,0) direction

E(k) along (1,0,0)

direction

1.414π/a

(b)

Ene

rgy

Density of states

Free electron density of states

Band gap

(c)

kx

E(kx, ky)

ky

upperband

lowerband

(d)

Ene

rgy

kπ/a–π/a 0

1.414π/a

Bandoverlap

Fermienergy

E(k) along (1,1,0)

directionE(k) along

(1,0,0) direction

E(k) along (1,0,0)

direction

E(k) along (1,1,0)

direction

(e)

Ene

rgy

Density of states

Free electron density of states

Band overlap

(f)

kx

E(kx, ky)

ky

upperband

lowerband

AchievementsThe most impressive achievement of this model isthat, in a relatively simple way, it allows us to de-scribe an electronic structure which can exhibiteither metallic or insulating properties. Just byaltering the strength of the parameter Vo and thefilling of the quantum states, we can describe ma-terials which are either metals, semi-metals or in-sulators.

LimitationsThe main limitation of the nearly-free electronmodel is that it is (fairly obviously) not very real-istic! It is useful pedagogically, and it can also be

used to describe some simple states in some met-als. However for the most part, the quantum statesthat solve the Schrödinger equation in real metalsare simply not at all like plane waves (or simplesums of plane waves). The nearly-free electronmodel can be useful when the kinetic energy de-rived from the regions in between the ion coresdominates the kinetic and potential energy derivedfrom the regions close to the ion cores. This ismost commonly true for states derived from va-lence states in atoms, and s-states in particular.The bands produced in this approach are all broad,and perturbed from their parabolic form only at theBrillouin zone boundaries.



The nearly-free electron approximation is not ap-plicable to electronic states derived from p-, d- andf- orbitals because for these states the bulk of thecontribution to the energy comes from the regionclose to the ion cores. To describe such states weneed to use the so-called tight-binding model.

Change in the meaning of kBefore moving onto the tight-binding model, weshould note that there has been one development inthe nearly-free electron model to which I have notspecifically drawn your attention. This is that themeaning of the term of term k has changed.

In the free electron discussion, the wave functionswere plane waves (cosine and sine), and k wastheir wave vector. In the nearly-free electron dis-cussion we have continued to use k to label thequantum states but now the wave functions are nolonger pure sine and cosine waves. Recall thatwhen we described the effect of a weak periodic

potential, we noted that wave functions whichsolved the Schrödinger equation were no longersimple sine and cosine waves. However, we keptthe label k because as long as the potential was nottoo large, the wave functions could be fairly de-scribed as being a superposition of:• mainly sine or cosine waves with wave vector

k, plus• some other parts which we did not describe in

detail.It is not difficult to see that if the perturbing po-tential became too large then the solutions to theSchrödinger equation would become quite dis-torted and not at all well-described by a wavevector k. You might think then that the meaning ofk would have been lost, but surprisingly this is notso. To see why this is so requires you to read eventhe following section on the tight-binding theory ofsolids, but I am sure you were going to do thatanyway.


W1.5 Tight-binding theory

IntroductionThe tight-binding theory of electronic states insolids starts from the premise that the quantumstates within a solid are similar to those stateswhich exist on isolated atoms.

• For so-called core states deep within an atomthis is certainly true: for these states the in-tense electric field of the ‘parent’ nucleus ismuch greater than the electric field due toother atoms in a solid.

• In contrast, for the electrons in states derivingfrom valence s-states this is probably not avery good description.

The tight-binding theory is relevant to states thatfall in between these two extremes. These aretypically states such as p- and d- valence states.

Such states are quite complicated even when theyare based on a single isolated atom. In the tight-binding theory, however we wish to examine theeffects that occur when electrons occupying suchcomplicated states are brought close to each other,as occurs with neighbouring atoms in a solid. Thismay sound tremendously difficult (and its noteasy!), but our task is made much easier by appre-ciating the results of Bloch’s theorem.

Bloch’s theoremBloch’s theorem tells us that any solution to theSchrödinger equation in a periodic potential has tohave a rather simple form. More precisely Bloch’stheorem states that:

• in a Bravais lattice: i.e. in a lattice where anylattice site can be reached at from any otherlattice site by integer multiple steps of threebasic lattice vectors:

R a a a= + +n n n1 1 2 2 3 3 (W1.81)

• in the independent electron approximation i.e.in the approximation that electrons do not in-teract strongly with each other. This is notreally a good approximation, but for a numberof complex reasons, it often gives acceptableresults

then the electron wave functions may always bewritten as a product of two parts:• A plane wave;

exp( )ik r⋅ (W1.82)

• A function un ( , )k r with the periodicity of thelattice i.e.

u un n( , ) ( , )k r k r R= + (W1.83)

where R is a Bravais lattice vector and n is a so-called band index. So, subject to the approxima-tions above, Bloch’s theorem states that electronwave functions in a lattice may always be writtenas the product of Equations W1.82 and W1.83 i.e.:

Ψ( ) exp( ) ( , )r k r k r= ⋅ ×i un (W1.84)

Functions of this type are called Bloch functions.We should note that Bloch functions solutions areexplicitly not appropriate in quite a few commonsituations. For example:

• near impurities in a lattice,• in random alloys,• in amorphous materials,• near surfaces of otherwise periodic crystals.

In these situations Bloch’s theorem does not ap-ply, and the nature of the electronic states be-comes more difficult to calculate. So on the whole

Figure W1.16 The range of energies in which tight-binding solutions to the Schrödinger equation are rea-sonable. (See also Figure W1.8)

The ‘free electron’ or ‘particle in a box’ approximation to the real potential

The expected potential energy for an electron in a crystal

The potential energy in the

nearly-free electron model

States describedby the 'tight-

binding' model have energies in

this range



Bloch’s theorem should be considered as a ‘goodthing’. Figure W1.17 illustrates graphically thesimple truths of Bloch’s theorem.

Tight bindingThe wave functions of electrons in isolated atoms,χ( )r , are calculated by solving the Schrödingerequation for electrons in the potential of :

• all the other electrons on an atom, and• the nuclear charge.

The single electron (i.e. the hydrogenic) solutionsignore other electrons except in the crudest sense.These solutions are the familiar set of 1s, 2s, 2petc. functions. In complicated many-electron at-oms, the wave functions are more complicated,but are often qualitatively similar to the hydro-genic wave functions. Pictorial representations ofthese functions can be found at the web site forthis book (www.physicsofmatter.com) and, nodoubt, in several other places.

In a solid, the potential the electrons experiencecan be considered as being equal to the sum of• the original atomic potential plus• an extra contribution from the surrounding

atoms.In the tight-binding approximation, we assumethat this extra potential is relatively small com-

pared to the atomic potential. We then assume thatthe wave functions can be described by Blochfunctions that are simply linear combinations ofatomic orbitals from different atomic sites. Usingthis approach we can approximate the effect of thecrystalline environment on electronic states thatare too deep in the atomic potential for the nearly-free electron approximation to be valid. Perhapssurprisingly, such states again give rise to bands.To see how, we will proceed as follows:• First, we will make an educated guess for a

wave function which should (at least ap-proximately) solve the Schrödinger equation.We know that this function must have thegeneral form of a Bloch function.

• Second we will work out how the energy ofthe wave function depends on k.

Constructing the wave functionThe exact Schrödinger equation for the atomicorbital χ( )r in an isolated atom is:

− ( ) + ( ) = ( )h2

2mV E

eatomic atomicχ χ χr r r

(W1.85)

where Eatomic is the energy of the state χ( )r . Wetake the function un ( , )k r to be the same as the

Figure W1.17 Illustration of the nature of states in a periodic lattice. The figure shows a function u(r) which has thesame periodicity as the lattice and a modulation factor with wave vector k ≈ 0.125 π/a. Below these is plotted the Blochfunction which is the result of multiplying the two together. At the bottom of the figure is shown the charge density whichwould be associated with such a wave function. Note: I have assumed that the atomic wave functions are entirely realand shown only the real component of the modulation factor and the resulting real components of the wave functions.

Position in crystal

cos( )k r⋅

Ψ( , ) ( ) cos( )k r r k r= × ⋅u

Ψ( , ) ( ) cos( )k r r k r2 2= × ⋅u

u( )r

a

TIGHT BINDING THEORY


isolated atom wave function χ( )r . This is a simplechoice for un(k,r) in that it has no dependence onk. We then construct our approximate solution tothe Schrödinger equation for the crystal by addingtogether the atomic wave functions on neighbour-ing sites as follows:

Ψ k r k R r RR

, exp( ) = ⋅( )∑ −( )1

Ni χ (W1.86)

This function is illustrated in Figure W1.18.

Calculating E(k): Where do the ‘bands’ comefrom?Now we need to solve the Schrödinger equationusing the wave function in Equation W1.86. Whenwe do this we will see explicitly how its energydepends on the k We proceed as we did in thenearly-free electron case i.e. we evaluate the ex-pectation value for the total energy using the wavefunction we have just defined.

• In the NFE case our wave functions wereplane waves and their unperturbed energy de-pended strongly on k. The perturbation to thefree electron case was a weak periodic poten-tial. The effect of the perturbation in first or-der was to add a potential energy to electronsoccupying quantum states with wave vectorsk close to a Brillouin zone boundary.

• In the tight binding case the wave functionsare linear superpositions of atomic wavefunctions added together with a phase shiftexp(ik.r) between the functions on neigh-bouring sites. If these Bloch functions areconstructed for atoms which are far apart,then their energy does not depend upon thephase shift (i.e. upon k) and their energy isjust given by Eatomic, the energy of the atomicstate upon which the Bloch function is based.The perturbation to the atomic case is thatwhen the atoms are close together (as in acrystal), the potential is no longer Vatomic, be-cause the atoms at neighbouring sites have af-fected the potential energy. The effect of thisperturbation (in first order) is to introduce adependence of the energy on the phase shiftbetween neighbouring sites. The magnitude ofthis dependence is proportional to:• the extent of the overlap of wave func-

tions on adjacent sites, and• the extent to which (in the region of wave

functions overlap), the potential differsfrom the free atom potential energy.

This dependence on the phase shift is essen-tially a kinetic energy term associated, notwith motion around an atom, but with motionbetween atoms i.e. with motion through thelattice.

Figure W1.18 Illustration of the way in which atomic wave functions centred on different sites are added together tomake functions which have the form of Bloch functions (Equation W1.84). Recall that it is known (from Bloch's theorem)that solutions to the Schrödinger equation in a periodic potential must have this form. (a) Two atomic wave functions χhave the same shape, but are centred on different sites within the crystal. Also shown is the summation of all suchatomic functions, and the normalised summation which accommodates just one electron. (b) This figure is similar (a) butincludes additionally the effect of the phase shift between neighbouring atomic functions. Note: To simplify the presen-tation I have assumed that the atomic wave functions are entirely real and I have shown only the real component of themodulation factor and the resulting real components of the wave functions.

Position in crystal

1

Nχ( )R

R∑

χ( )RR∑

χ( ( , , ))R = 0 0 0

χ( ( , , ))R = a 0 0

( a )

aPosition in crystal

1

Niχ( )exp( )R k r

R⋅∑

χ( )exp( )R k rR

i ⋅∑

χ( ( , , ))exp( )R k r= ⋅a i0 0

χ( ( , , ))exp( )R k r= ⋅0 0 0 i

( b )

a



Calculating E(k).So let us begin our total energy calculation. Thekinetic energy (KE) operator is:

− ∇ = − + +

h h22

2 2

2

2

2

2

22 2m m x y z

∂∂

∂∂

∂∂ (W1.87)

and the potential energy operator is:

V V Vcrystal atomicr r r( ) = ( ) + ∆ ( ) (W1.88)

Notice that we have written the potential energy asthe sum of two terms, where the term ∆ ( )V r is theadditional potential which an electron experiencesas a result of being in the crystal (Figure W1.19(a)). Numerically this is usually negative becausethe potential energy of the valence electrons isgenerally lower in a solid.

Now we use some intelligence: notice that as afirst approximation the sum of the kinetic energyand potential energy terms deriving from theatomic potential will just result in Eatomic, the en-ergy of the unperturbed atomic state. We are inter-ested in the difference between:

• the energy of the state in an isolated atom, and• the energy of the state in the crystal.

This difference depends mainly on ∆ ( )V r , thedifference between the potential energy in thesetwo situations. So we may evaluate the total en-ergy as:

E KE V

E KE V V

E E V

E E V

E

all space

= +[ ]∫

= + + ∆

∫

= + ∆[ ]∫= + ∆∫

∗

∗

∗

∗

Ψ Ψ

Ψ Ψ

Ψ Ψ

Ψ Ψ

crystalall space

atomicall space

atomicall space

atomic

d

d

d

d

atomic

( )

( ) ( )

( )

( )

r r

r r r

r r

r r

1 244 344

(W1.89)

The final term in the last of Equations W1.89makes clear the nature of the perturbation that weneed to evaluate. It involves the term ∆V(r) andour constructed wave function. Now we recall that

our wave function (and its complex conjugate) iswritten as:

Ψ

Ψ

k r k R r R

k r k R r R

R

R

, exp

, exp

( ) = ⋅( )∑ −( )

( ) = − ⋅ ′( )∑ − ′( )∗

′

∗

1

1N

i

Ni

χ

χ

(W1.90)

Substituting for this in the expression for the totalenergy yields an expression which looks verycomplicated:E E

Ni

Vall space

= +⋅ − ′∑∑ ×

− ′∫ ∆ −′

∗

atomic

exp( ( ))

( ) ( ) ( )

1k R R

r R r r R

RR

χ χ (W1.91)

However, the double summation make things lookmuch more complicated than they really are. Wenote that for each chosen value of R', the sumover R has the same value (because each latticesite R' in a crystal is equivalent to any other). Sowe can just evaluate the sum for R' = 0,.and thenmultiply the answer by N, the number of unit cellsin the crystal. We can then replace one of thesummations by a factor N which we can subse-quently cancel:

E EN

N i

Vall space

= + ⋅∑ ×− ′∫ ∆ −∗

atomic1

exp( )

( ) ( ) ( )

k R

r R r r R

R

χ χ(W1.92)

E E i

Vall space

= + ⋅∑ ×− ′∫ ∆ −∗

atomic exp( )

( ) ( ) ( )

k R

r R r r R

R

χ χ(W1.93)

We can now write this integral in a more compactform and look at the origin of each of the terms:

E E i A( ) exp( ) ( )k k R RR

= + ⋅∑atomic (W1.94)

where A(R) is given by:

A V( ) ( ) ( ) ( )R r r r R r= ∆ −∫ ∗χ χ d (W1.95)



Let us look at more closely at the integral A(R). Itinvolves the ‘overlap’ between the atomic wavefunctions centred at R = 0 and at R . Notice thatA(R) is a function only of the overlap of wavefunctions on neighbouring sites. So it is a functionof R, but not of r. So for the overlap of:

• any particular orbital, at• a particular separation between atoms, and• with a given change in potential.

A(R) simply has a value in units of energy, typi-cally a few electron volts. So our expression forthe energy becomes

E E A i( ) ( )exp( )k R k RR

= + ⋅∑atomic (W1.96)

where the sum is over all the lattice sites in thecrystal. Now this expression looks a whole lotsimpler than Equation W1.91. In fact we can clar-ify matters a little further by noticing that one termin the summation of Equation W1.96 is generallymuch larger than all the rest. The case when R = 0involves the ‘overlap’ between an orbital and it-self. This is generally much larger than the overlapbetween functions on neighbouring sites whenR 0≠ (Figure W1.19 (b)). We thus consider thesetwo cases separately:

α χ χ= = ∆∫ ∗

= =

A Vatomic

wavefunctionat site

atomicwavefunctionat site

( ) ( ) ( ) ( )

0 r r r r

R R0 0

123 { d

(W1.97)

A Vatomic

wavefunctionat site


( ) ( ) ( ) ( )

R r r r R r

R R

= ∆ −∫ ∗

= ≠

χ χ

0 0

123 1 24 34d (W1.98)

Using this new terminology our expression for theenergy becomes:

E E A i

A i

( ) ( )exp( )

( )exp( )

k R 0 k

R k RR 0

= + = ⋅+ ⋅∑

≠

atomic 0

E E A A i( ) ( ) ( )exp( )k R 0 R k RR 0

= + = + ⋅∑≠atomic

E E A iThe energy

of theoriginalatomicorbital

The changein energy

of theatomicorbital

due to thecrystalline

environment

The change in energy of theatomic orbital that dependson the overlap of between

an orbital and the same typeof orbital on

( ) ( )exp( )k R k RR 0

= + + ⋅∑≠

atomic123 123α

.a neighbouring siteIt depends on the phase shift

between the two states

1 2444 3444

(W1.99)

The term α is just the change in energy of the

Figure W1.19 Illustration of the meaning of some of the functions and integral kernels from Equations W1.88 to W1.99.(a) The difference between the potential energy of an electron within an atom and within a crystal ∆V(r). In order to showthe function clearly, the function plotted is actually minus ∆V(r). (b) Illustration of the difference in magnitude betweenthe kernels of the so-called direct integral α (Equation W1.97).and the so-called transfer integral A(R≠0) (EquationW1.98). The value of the integrals is the area underneath the curves shown. The direct kernel is shown reduced in scaleby a factor 5 and the transfer kernel is shown amplified by a factor 5.

Pot

entia

l Ene

rgy

Position in crystal

Vatomic

(r)

Vcrystal

(r)

-∆V(r)

a

( a )

Position in crystal

Ψ(r= 0)

∆V(r)

( b )

Ψ(r= a)

Ψ(r= 0) × ∆V(r) × Ψ(r= 0)

Ψ(r= 0) × ∆V(r) × Ψ(r= a)

Shown 5 × smaller than actual

Shown 5 × larger than actual

a



atomic orbital χ caused by the presence of theother ions. Since ∆V(r) is always negative (FigureW1.19(a)), the term α always represents a lower-ing of the energy level in the crystal as comparedwith the isolated atom (Figure W1.20).

It is from this final term that the bands arise i.e.the dependence of E upon k. It involves the over-lap of orbitals centred on neighbouring sites. Ex-amination of the form of A( )R shows that the in-tegral is only large when the potential differs sig-nificantly from what it was in the atomic case.Notice that E( )k can take on different values as kchanges, and shifts the phases of the contributionsto the sum i.e. the phase of the wave function inadjacent unit cells is shifted (Figure W1.20).

An exampleLet us find an expression for the energies of atight-binding band in a crystal with a simple cubiclattice and a basis of one atom placed at each lat-tice site. We will assume the atomic orbital χ(r) isreal and spherically symmetric and take the inte-gral A(R) to be 0, except for the case of nearestneighbours.

Evaluating E(k)The unit cell of the simple cubic lattice is a cubeof side a. Placing one lattice site at the origin ofour co-ordinate system we find it has nearest

neighbours at:

R

R

R

= ±= ±= ±

( , , )

( , , )

( , , )

a

a

a

0 0

0 0

0 0

(W1.100)

Our final expression for the energy of a tightly-bound state is:

E E A i( ) ( )exp( )k R k RR 0

= + + ⋅∑≠

atomic α

(*W1.99)

where A V( ) ( ) ( ) ( )R r r r R r= ∆ −∫ ∗χ χ d is an inte-gral involving the overlap of orbitals at R = 0 withorbitals at R. In this case, we consider only nearestneighbours, of which there are six (Figure W1.21).Because of the spherical symmetry of the orbitals,all six integrals are equal. Let us call the value ofthis integral A.

Evaluating k R⋅ for R = ( , , )a 0 0 we find:

k R⋅ = + +

= × + × + ×

=

k x k y k z

k a k k

k a

x y z

x y z

x

0 0 (W1.101)

Similarly, we can find results for the other fiveorbitals. Evaluating exp( )ik R⋅ for each of the sixterms in the sum we find:

Figure W1.21: Each sphere represents an s orbital cen-tre on a different lattice site.

yx

zRegion ofoverlap

Figure W1.20 Energy level diagram showing meaning ofterms in Equation W1.99

Spread is proportional tothe overlap integral A(R)α

Eatomic

Ene

rgy

Position within band determined by thephase shift between neighbouring sites

which depends on k



E E

A ik a ik a

A ik a ik a

A ik a ik a

( )

exp( ) exp( )

exp( ) exp( )

exp( ) exp( )

k = + +

+ −[ ]+ −[ ]+ −[ ]

atomic

x x

y y

z z

+

+

+

α

(W1.102)

Simplifying this using the standard definition ofthe cosine function we find:

E E

A k a

A k a

A k a

( )

cos

cos

cos

k = + +[ ]

[ ][ ]

atomic

x

y

z

+

+

+

α2

2

2

(W1.103)

which simplifies to:

E E A k a k a k a( ) cos cos cosk = + + + +[ ]atomic x y zα 2

(W1.104)

Finding the bandwidthLooking at the expression for α we see:

α χ χ= = ∆∫ ∗

= =

A Vatomic

wavefunctionat site


( ) ( ) ( ) ( )

0 r r r r

R R0 0

123 { d

(W1.105)

As shown in Figures W1.19 and W1.20, the crys-talline potential is always lower than the atomicpotential. So ∆V(r) is always negative, and since

χ χ∗∫ ( ) ( )r r rd is just equal to 1, then α is alwaysnegative i.e. it corresponds to a lowering of theenergy of an atomic state by an amount |α|.

The sign of the integral A can in general be eitherpositive or negative, but in this case it is negative.This is because the orbitals χ are real so terms likeχ 2 will be positive, and because, as we saw

above, ∆V(r) is negative. Thus A will be a nega-tive number. The expression for the energy thusbecomes:

E E

A k a k a k a

( )

cos cos cos

k = −− + +[ ]

atomic

x y z

α2

(W1.106)

This has a minimum value of:

E E Amin atomic= − −α 6 (W1.107)

when all the cosine terms are equal to 1 i.e. at k =0, and a maximum value of

E E Amax atomic= − +α 6 (W1.108)

when all the cosine terms have value –1 i.e. atk = π π π( / , / , / )a a a . The bandwidth is equal tothe difference between the maximum and theminimum energies is given by 12|A | i.e. it is di-rectly proportional to integral involving:

• The overlap between neighbouring atomicorbitals,

• The difference between the crystalline and theatomic potential energy i.e.

A V( ) ( ) ( ) ( )R r r r R r= ∆ −∫ ∗χ χ d (*W1.95)

We can see that this makes sense.

• Firstly, notice that if the crystalline potentialis the same as the atomic potential then thequantum states will be the same in the crystalas they were in the atom. This makes∆V(r) = 0 and hence A(R) = 0 and hence thebandwidth is zero. This describes a return nar-row atomic-like states.

• Similarly, if the overlap between the wavefunctions is zero i.e. if χ∗ =( )r 0 wheneverχ( )r R− ≠ 0 and vice versa then the band-width will also be zero.

These two cases describe the situation of the deep-est ‘core’ states on atoms within a solid. However,if the crystalline potential differs from the atomicpotential, then the corresponding quantum stateswill be different in the crystal and the atom. Thusthe Bloch functions we constructed will not beaccurate. In this tight-binding approximation wehave neglected this effect and evaluated only thechange in energy of the original quantum states.



How to visualise E(k)The total energy E(k) that we have calculated(Equation W1.104) is a function that takes a sin-gle value at each point in the three-dimensional(kx, ky, kz) ‘volume’ of the Brillioun zone. Thismakes visualising how the energy varies with kquite taxing. There has evolved a standard solutionto this problem. The energy is calculated alongspecial paths through the Brillioun zone. Thesepaths are chosen to follow certain high symmetrydirections within the zone. Graphs then show howthe energy varies along the route.

Figure W1.22 Illustration of special 'representative' pathswithin the Brillioun zone as described in EquationsW1.109 to W1.114

kx

kz

π

a

, ,0 0

π π

a a

, ,0

π π π

a a a

, ,

Let us calculate how the energy varies as onemoves out from the centre of the zone k = (0,0,0)along lines joining the origin to:

• the centre of the square face of the Brilliounzone ( / , , )π a 0 0

• the midpoint of the line joining two adjacentcorners ( / , / , )π πa a 0

• a corner of the zone ( / , / , / )π π πa a a

How do we set about this problem?

E(k) along the (1,0,0) directionThe general expression for the energy anywherewithin the Brillioun zone is given by:

E E

A k a k a k a

( )

cos cos cos

k = −− + +[ ]

atomic

x y z

α2

(*W1.106)

If we now substitute ky = 0 and kz = 0 we find:

E E

A k a

( )

cos cos( ) cos( )

k = −− + +[ ]

atomic

x

α2 0 0

E E A k a( ) cosk = − − +[ ]atomic xα 2 2

(W1.109)

so the energy increases as cosk ax . The energy atthe Brillouin Zone Boundary is therefore:

E E Aa

a

E A

E A

( ) cos

k = − − π +

= − − − +[ ]= − −

atomic

atomic

atomic

ααα

2 2

2 1 2

2(W1.110)

E(k) along the (1,1,0) directionWe start again with the general expression for theenergy anywhere within the Brillioun zone:

E E

A k a k a k a

( )

cos cos cos

k = −− + +[ ]

atomic

x y z

α2

(*W1.106)

Along this special direction we now substitutekz = 0 and kx = ky= k to find:

E E A ka( ) cosk = − − +[ ]atomic α 2 2 1

(W1.111)

and the energy increases as 2coska . The energyat the zone boundary is therefore:

Ea a

E Aa

a

E A

E A

E A

π π

= − − π +

= − − − +[ ]= − − −[ ]= − +

, , cos

( )

0 2 2 1

2 2 1 1

2 1

2

atomic

atomic

atomic

atomic

αααα

(W1.112)

E(k) along the (1,1,1) directionWe start again with the general expression for theenergy anywhere within the Brillioun zone:



E E

A k a k a k a

( )

cos cos cos

k = −− + +[ ]

atomic

x y z

α2

(*W1.106)

Along this special direction we now substitute andkx = ky = ky= k to find:

E E A ka( ) cosk = − − [ ]atomic α 2 3 (W1.113)

and the energy increases as 3coska. The energy atthe zone boundary is therefore:

Ea a a

E Aa

a

E A

E A

E A

π π π

= − − π

= − − −[ ]= − − −[ ]= − +

, , cos

( )

atomic

atomic

atomic

atomic

αααα

2 3

2 3 1

2 3

6

(W1.114)

Visualisation and interpretationSo now we can calculate how the energy ofelectronic states varies with the character of thequantum state (as described by its position in theBrillioun zone). For many advanced calculations,this kind of calculation is exactly what is requiredto understand some particular property of a solid.However, for many solids the details required as

inputs to the band structure calculation (such asthe value of overlap integrals), are not wellknown. This makes the detailed results ofcalculations unreliable, but such calculations canstill reveal the general form of the band structure.In particular, the position of the Fermi energyrelative to the calculated band structure is centralto understanding a material’s properties.

Before proceeding to analyse this situation inmore depth, we will just pause to remind ourselvesof the status of concepts such as ‘band structures’and ‘Brillioun zones’. They are explanatoryconcepts which (while they exist only in ourheads) will allow us to calculate and discuss theproperties of widely different solids using aunified framework.

A band structure diagramFigure W1.23 illustrates what we mean by a ‘bandstructure’ diagram. It is a graph showing how theenergy of electron quantum states varies withposition in the Brillioun zone. It does not show theenergy of all the states in the zone, rather it showshow the energy of states varies as a standard paththrough the zone is traversed.

While plotting the band structure is in itself anachievement, is it is not until we add a line indi-cating the position of the Fermi energy that the

Figure W1.23 Illustration of the band structure described by Equation W1.106. (a) A graph of energy versus k along thepath through the Brillioun zone shown in (b). Notice that (b) is not the same as Figure W1.22.

(a) (b)

0

1

2

3

4

5

6

7

En

erg

y

Position within Brillioun Zone

(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

kx

kz

π π

a a

, ,0

π π π

a a a

, ,

π

a

, ,0 0



band structure acquires any physical signifi-cance. There are three cases to consider:

• The Fermi energy is lower than the lowestenergy in this band.

• The Fermi energy is greater than the highestenergy in this band.

• The Fermi energy lies in the energy rangespanned by this band.

Let us consider each of these cases in turn. If theFermi energy is lower than the lowest energy inthis band, then no states within this band are oc-cupied (Figure W1.24(a)). The band can stillaffect the optical properties of the solid by pro-viding empty quantum states into which elec-trons (from lower energy bands) may be excitedby photons of an appropriate energy. Typicallythis energy would be of the order of a few elec-tron volts which would correspond to light in thenear infra red, optical or ultra violet regions ofthe spectrum.

Normally a material described by a band diagramsuch as that in Figure W1.24 (a) would corre-spond to that of an insulator. However, if theband gap is relatively small (≈ 1 electron volt)then it is possible for a population of electrons tobe thermally excited from quantum states in thelower occupied band to empty quantum states inthe band under consideration. These electronswould then be able to conduct electricity. Al-though the number density of this excited popu-lation is tiny in comparison with a metal, it isvery large in comparison with insulators. Suchmaterials are described as semiconductors(§7.7.8 and §7.7.9).

Alternatively, if the Fermi energy is higher thanthe highest energy in this band, then all stateswithin this band are occupied (Figure W1.24(b)).Perhaps surprisingly, this means that the band ofoccupied states contributes nothing to the conduc-tivity. This is because there are no easily accessi-ble quantum states into which electrons maymove. Thus when an electric field is applied to thesubstance, electrons in this band of states are un-able to easily change quantum state in response tothe electric field. The band can still affect the

optical properties of the solid by providing asource of electrons which may be excited by pho-tons of an appropriate energy into empty, higherenergy quantum states. Typically this energywould be of the order of a few electron voltswhich would correspond to light in the near infrared, optical or ultra violet regions of the spectrum.

As for the previous case, we should add the quali-fication that if a band of empty states is availablewith energies within ≈ 1 electron volt of the high-est energy in the band, then again the materialwould be described as a semiconductor.

Figure W1.24 Illustration of the cases described in text inwhich the Fermi energy lies (a) below the band of statesunder consideration and (b) above the band of statesunder consideration. The Figures show two bands ofquantum states based on different atomic orbitals. Theupper band is the one considered in the text and plottedin Figure W1.23.

(a)

-6

-4

-2

0

2

4

6

8

En

erg

y


(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

Fermi Energy

Optical transition

Band gap

quantum states occupied

quantum states empty

(b)

-4

0

4

8

1 2

En

erg

y


(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

Fermi Energy

Optical transition

quantum states occupied

quantum states empty



Figure W1.25 Illustration of the effect of changing the Fermi energy for the fixed hypothetical band structure describedby Equation W1.106. (a) When the Fermi energy is just above the low edge of the band, the occupied states are allnear the zone centre and the Fermi surface is only slightly distorted froma sphere. (b) When the Fermi energy is in themiddle of the band, the occupied states are all near the zone centre and the Fermi surface (i.e. a surface of constantenergy with energy EF) is complex in shape. (c) When the Fermi energy is near the top of the band, nearly all thestates are occupied and the Fermi surface is complex in shape is very much reduced in area.

(a)

0

1

2

3

4

5

6

7

En

erg

y


(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

Fermi Energy

(b)

0

1

2

3

4

5

6

7

En

erg

y


(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

Fermi Energy

(c)

0

1

2

3

4

5

6

7

En

erg

y


(0,0,0) (π/a,0,0) (0,0,0)(π/a,π/a,0) (π/a,π/a,π/a)

Fermi Energy



Finally, we consider the case when Fermi energylies in the energy range spanned by this band. Inthis case the material will be a metal. We can seethis because the Fermi energy separates occupiedfrom empty states (at T = 0 K). Since the Fermienergy lies within a band of states, electrons willbe able to find empty quantum states only infini-tesimally above the Fermi energy. Thus even aweak applied electric field will be able to cause

electrons to change states (i.e. conduct). Since thesubstance is a metal, it will possess a Fermi sur-face (which is the energy contour in k-space sepa-rating occupied from empty states (at T = 0 K).However, as Figure W1.25 makes clear, depend-ing on where the Fermi energy lies within theband, the Fermi surface can be quite different inshape from the Fermi spheres that we have dis-cussed previously.

W1.5 Summary

W1.5.1 So what?Let me explain again why I embarked on writingthese 36 theoretical pages to accompany the 400or so decidedly phenomenological pages of Un-derstanding the properties of matter. I sought toexplain how we can develop a language for de-scribing electronic quantum states in solids To thisend, we developed the nearly-free electron modelthat allowed us to see how a material might be aninsulator or a metal depending on the strength of aperturbing potential energy. We also developedthe tight binding model which was considerablymore complex, but generally more plausible. Thisagain allowed us to understand how a substancecould display insulating or metallic properties de-pending on some calculable properties of theelectronic orbitals of the atoms of the substance.

So, in terms of the problem that I stated at the startof this chapter, we can now discuss the propertiesof solid argon and solid potassium using the sametheoretical model. In solid argon, the Fermi en-ergy lies in between bands of states and so thematerial is an insulator. In solid potassium, theFermi energy lies within a band of states and sothe material is a metal. This may seem an unspec-tacular conclusion the Chapter, but that is what ithas all been for.

W1.5.2 What next?If you wish to find out more about the band theoryof solids then there is no better introductory textthan Solid State Physics by N.W. Ashcroft andN.D. Mermin. (ISBN: 0030839939).

the band theory of solids · the band theory of solids w1.1 introduction this is a special section...

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