Download Condensed Matter Physics

Condensed Matter Physics
763628S
Jani Tuorila
Department of Physics
University of Oulu
April 17, 2012
General
– Experimental determination of crystal structure
– Boundaries and interfaces
The website of the course can be found at:
https://wiki.oulu.fi/display/763628S/Etusivu
It includes the links to this material, exercises, and to their
solutions. Also, the possible changes on the schedule below
can be found on the web page.
– Complicated structures
• Electronic structure
– Single-electron model
Schedule and Practicalities
– Schrödinger equation and symmetry
– Nearly free and tightly bound electrons
All lectures and exercise sessions will be held at room
TE320.
• Lectures:
– Electron-electron interactions
– Band structure
Monday 12-14
Wednesday 12-14
• Mechanical properties
• Exercises:
Thursday
12-14
– Cohesion
– Phonons
By solving the exercises you can earn extra points for the
final exam. The more problems you solve, the higher is the
raise. Maximum raise is one grade point. It is worthwhile
to remember that biggest gain in trying to solve the exercise is, however, in the deeper and more efficient learning
of the course!
• Electronic transport phenomena
– Bloch electrons
– Transport phenomena and Fermi liquid theory
• (Graphene)
Literature
This course is based mainly on (selected parts)
• M. Marder, Condensed Matter Physics (MM).
In addition, you can read this course material.
Other literature:
• E. Thuneberg, Solid state physics, lecture notes (2012,
only in Finnish) (ET).
• C. Kittel, Introduction to Solid State Physics, old but
still usable.
• F. Duan and J. Guojum, Introduction to Condensed
Matter Physics
• X. G. Wen, Quantum Field Theory of Many-body Systems, way too complicated but has an intriguing introduction!
• N. W. Ashcroft and N. D. Mermin, Solid State
Physics, a classic! Previously used in this course.
• P. Pietiläinen, previously used lecture notes based on
the book above, only in Finnish.
Contents
The course will cover the following:
• Atomic structure
– 2-and 3-dimensional crystals
1
1. Introduction
• Mesoscopic physics and the realisation of quantum
computation (research done in the University of Oulu)
The purpose of this course is to give the basic introduction on condensed matter physics. The subject has a wide
range and many interesting phenomena are left for other
courses, or to be learned from the literature.
• Fermi liquid theory (Oulu)
• Graphene
• Topological insulators
History
The term Condensed matter includes all such large
Many-Body Problem
groups of particles that condense into one phase. Especially, the distances between particles have to be small
All known matter is formed by atoms. The explaining
compared with the interactions between them. Examples: of the properties of single atoms can be done with quantum mechanics, with an astounding accuracy. When one
• solids
adds more atoms to the system, the number of the relevant
degrees of freedom in Schrödinger equation grows exponen• amorphous materials
tially. In principle, all properties of the many atom system
• liquids
can still be found by solving the Schrödinger equation, but
in practise the required computational power grows very
• soft materials (foams, gels, biological systems)
rapidly out of reach. As an example: the computers in
• white dwarfs and neutron stars (astrophysics)
1980s could solve for the eigenvalues of the system consisting of 11 interacting electrons. Two decades later the
• nuclear matter (nuclear physics)
computational power had increased hundred-fold but it allowed to include only two extra electrons! A typical manyCondensed matter physics originates from the study of
body problem in condensed matter physics includes 1023 ,
solids. Previously, the field was called solid state physics,
or so, electrons. Therefore, it is clear that it is extremely
until it was noticed that one can describe the behaviour of
impractical to study such physics starting from the basic
liquid metals, Helium and liquid crystals with same conprinciples.
cepts and models. This course concentrates mainly on perBased on the above, the condensed matter theories are
fect crystals, mainly due to the historical development of
the field and the simplicity of the related theoretical mod- so-called effective theories. In principle, they have to be
results of averaging the Schrödinger equation properly, but
els.
in practise their form has been guessed based on symmeApproximately, one third of the physicists in the US
tries and experimental results. In this way, the theories of
considers themselves as researchers of condensed matter
the condensed matter have become simple, beautiful, and
physics. In the last 50 years (in 2011), there has been 22
one can use them to obtain results that are precise (an
condensed-matter related Nobel prizes in physics, and also
effective theory does not have to be imprecise) and give
5 in chemistry:
detailed explanations.
• Bardeen, Cooper and Schrieffer, Theory of low temperature superconductivity (1972)
The modest goal of condensed matter physics is to explain the whole material world. It overlaps with statistical
physics, material physics and liquid and solid mechanics.
Due to its diversity, the coherent treatment of the subject
is blurred.
• Josephson, Josephson effect (1973)
• Cornell, Ketterle and Wieman, Bose-Einstein condensation in dilute gases of alkali atoms (2001)
“The ability to reduce everything to simple fundamental
laws
does not imply the ability to start from those laws
• Geim and Novoselov, Graphene (2010)
and reconstruct the universe. ... The constructionist
The field has been the source of many practical applica- hypothesis breaks down when confronted by the twin difficulties of scale and complexity. The behaviour of large and
tions:
complex aggregates of elementary particles, it turns out,
is not to be understood in terms of a simple extrapolation
• Transistor (1948)
of the properties of a few particles. Instead, at each
• Magnetic recording
level of complexity entirely new properties appear, and
the understanding of the new behaviours require research
• Liquid crystal displays (LCDs) . . .
which I think is as fundamental in its nature as any other.”
Therefore, condensed matter physics is both interesting
and beneficial for basic physics research and also for ap- - P. W. Anderson, 1972
plications!
Finally, a short listing on ”hot topics” in the current
condensed matter research:
2
2. Atomic Structure
crystal. We refer to this as basis.
MM, Chapters 1 and 2, excluding 2.3.3-2.3.6 and
2.6.2.
2) Group of points in space where one has to place the
basis in order to form the crystal. Such a group is
presented in form
r = n1 a1 + n2 a2 + n3 a3 .
(1)
Here n1 , n2 and n3 are integers. Vectors a1 , a2 and
a3 called primitive vectors (they have to be linearly
independent). The group of points (1) is referred to
as Bravais lattice and its members as lattice points.
a3
Scanning-tunnelling-microscopic image (page 17)
on NbSe2 surface in atomic resolution.
The distance between nearest neighbours is 0.35 nm.
(http://www.pma.caltech.edu/GSR/condmat.html)
a1
a2
The volume defined by the primitive vectors is called
primitive cell. The primitive cells contain the complete
information on the whole crystal. Primitive cells are not
unique, but they all have to have the same volume. A
primitive cell of a Bravais lattice contains exactly one lattice point and, thus, the volume of the cell is inverse of the
density of the crystal.
+
=
An example in two dimensions.
Fluoride chrystal on top of quartz chrystal (image by
Chip Clark).
a'2
The simplest way to form a macroscopic solid is to organize atoms into small basic units that repeat periodically.
This is called crystal structure. Let us recall the definitions
in the course of Solid State Physics (763333A).
a'1
a2
a1
2.1 Crystal Structure (ET)
The choice of the primitive vectors is not unique. In the
figure, one can use also the vectors a01 and a02 as primitive
Many crystals appear in forms where flat surfaces meet vectors. They reproduce also all lattice points.
each other with constant angles. Such shapes can be understood based on the organization of atoms. Solids are
often formed of many crystals. This means that they are
yksikkökoppi
composed of many joined crystals, oriented in different di18
rections. A single crystal can contain e.g. ∼ 10 atoms,
a'2
whereas the whole macroscopic body has ∼ 1023 atoms.
Generally speaking, the structure of a solid can be exa2
a'1
tremely complicated. Even if it was formed by basic units
a1
comprising of same atoms, it does not necessarily have a
alkeiskoppi
repeating structure. Glass is an example of such a mateIn certain symmetric lattices, it is more practical to use
rial, formed by SiO2 -units. This kind of material is called
orthogonal vectors instead of primitive ones (even if they
amorphous.
do not span the whole lattice). The volume defined by such
Let us consider an ideal case, where we ignore all possible
vectors is called unit cell. The lengths of the sides of a unit
imperfections of lattice. The description of such crystal can
cell are called lattice constants.
be divided in two parts.
1) Group of atoms that form a repeating object in the
3
2.2 Two-Dimensional Lattice
Let us first restrict our considerations into two dimensional lattices, because they are easier to understand and
visualize than their three dimensional counter-parts. However, it is worthwhile to notice that there exists genuinely
two-dimensional lattices, such as graphene that is presented later in the course. Also, the surfaces of crystals
and interfaces between two crystals are naturally two dimensional.
• oblique, arbitrary choice of primitive vectors a1 and
a2 without any specific symmetries, only inversion
symmetry r → −r.
Bravais Lattice
In order to achieve two-dimensional Bravais lattice, we
set a3 = 0 in Definition (1). One can show using group
theory that there exist five essentially different Bravais lattices.
• square, symmetric in reflections with respect to both
x and y -axes, and with respect to 90◦ -rotations.
The gray areas in the above denote the so called WignerSeitz cells. Wigner-Seitz cells are primitive cells that are
conserved in any symmetry operation that leaves the whole
lattice invariant. The Wigner-Seitz cell of a lattice point
is the volume that is closer to that point than any other
lattice point (cf. Figures).
• rectangular, when square lattice is squeezed it loses
its rotational symmetry and becomes rectangular.
Example: Hexagonal lattice A choice for the primitive vectors of the hexagonal lattice are, for example,
a1 = a 1 0
√ a2 = a 12 − 23 ,
where a is the lattice constant. Another option is
√
a01 = a 23 12
a02 = a
• hexagonal (trigonal), symmetric with respect two x
and y -reflections, and 60◦ -rotations.
√
3
2
− 21 .
Lattice and a Basis
A structure is a Bravais lattice only if it is symmetric
with respect to translations with a lattice vector (cf. the
definition in a later section). In nature, the lattices are
seldom Bravais lattices, but lattices with a basis. As an
example, let us consider the honeycomb lattice which is
the ordering for the carbon atoms in graphene.
Example: Graphene
Geim and Kim, Carbon Wonderland, Scientific
American 90-97, April 2008
• centered rectangular, squeezed hexagonal, no rotational symmetry. By repeating the boxed structure
one obtains the lattice, hence the name.
Graphene is one atom thick layer of graphite, in which
the carbon atoms are ordered in the honeycomb structure
resembling a chicken wire
4
(Wikipedia).
Graphene has been used as a theoretical tool since the
1950’s, but experimentally it was ”found” only in 2004.
A. Geim and K. Novoselov were able to separate thin layers of graphite (the material of pencils, consists of stacked
layers of graphene), some of which were only one atom
thick. Therefore, graphene is the thinnest known material
in the Universe. It is also the strongest ever measured material (200 times stronger than steel). It is flexible, so it
is easy to mold. Graphene supports current densities that
are six times of those in copper. Its charge carriers behave
like massless fermions, that are described with the Dirac
equation. This allows the study of relativistic quantum
mechanics in graphene. These and many other interesting properties are the reason why graphene is used as an
illustrative tool in this course.
In each cell, the neighbours of the left- and right-hand
atoms are found in different directions. Anyhow, if one
allows π/3-rotations, the surroundings of any atom are
identical to any other atom in the system (exercise). The
dashed vertical line in the figure right is the so-called glide
line. The lattice remains invariant when it is translated
vertically by a/2 and then reflected with respect to this
line. Neither operation alone is enough to keep the lattice
invariant. Let us return to to the properties of graphene
later.
2.3 Symmetries
Let us then define the concept of symmetry in a more
consistent manner. Some of the properties of the crysAs was mentioned in the above, graphene takes the form tals observed in scattering experiments (cf. the next chapof the honeycomb lattice, which is a Bravais lattice with a ter) are a straight consequence of the symmetries of the
basis. The starting point is the hexagonal lattice whose crystals. In order to understand these experiments, it is
important to know which symmetries are possible. Also,
primitive vectors are
the behaviour of electrons in periodic crystals can only be
explained by using simplifications in the Schrödinger equa
√
a01 = a 23 12
tion, permitted by the symmetries.
Space Group
a02 = a
√
3
2
We are interested in such rigid operations of the crystal
that leave the lattice points invariant. Examples of those
include translations, rotations and reflections. In Bravais
lattices such operations are:
− 12 .
• Operations defined by a (Bravais) lattice vector (translations)
• Operations that map at least one lattice point onto
itself (point operations)
• Operations that are obtained as a sequence of translations and point operations.
These can be described as a mapping
Each lattice point is then replaced with the basis, defined
by
v1 = a
1
√
2 3
1
v2 = a − 2√
3
0
y = a + Rx,
(2)
which first rotates (or reflects or inverses) an arbitrary vector x with a matrix R, and then adds a vector a to the
result. In order to fulfil the definition of the symmetry operation, this should map the whole Bravais lattice (1) onto
itself. The general lattice (with a basis) has symmetry operations that are not of the above mentioned form. They
0 .
5
are known as the glide line and screw axis. We will return
to them later.
The goal is to find a complete set of ways to transform
the lattice in such way, that the transformed lattice points
are on top of the original ones. Many of such transformations can be constructed from a minimal set of simpler
When one transforms the rectangular into the centered
transformations. One can use these symmetry operations
rectangular,
the reflection symmetry with respect to the
in the classification of different lattice structures and, for
y-axis
is
destroyed.
example, to show, using group theoretical arguments, that
there are only five essentially different Bravais-lattices in
two dimensions, a result stated earlier.
2.4 3-Dimensional Lattice
Space group (or symmetry group) G is the set of operations that leave the crystal invariant (why such a set is a
group?).
One has to study 3-dimensional lattices in order to describe the crystals found in Nature. Based on symmetries,
one can show that there exists 230 different lattices with a
basis, and those have 32 different point groups. The complete listing of all of them is, of course, impossible to do
here. Therefore, we will restrict ourselves in the classification of the 3-dimensional Bravais lattices. Let us first
introduce some of the structures found in Nature.
Translations and Point Groups
Let us consider two sub-groups of the space group. The
elements in the translation group move all the points in the
lattice by a vector
Simple cubic lattice (sc) is the simplest 3-dimensional
lattice. The only element that has taken this form as its
m1 a1 + m2 a2 + m3 a3 ,
ground state is polonium. This is partly due to the large
”empty” space between the atoms: the most of the eleand thus leave the lattice invariant, according to the defi- ments favour more efficient ways of packing.
nition of the lattice.
Point group includes rotation-like operations (rotations,
reflections, inversions), that leave the structure invariant
and, in addition, map one point onto itself. The space
group is not just a product of the point and translation
groups. For example, the glide line (see the definition before) and screw axis (translation and rotation), are both
combinations whose parts are not elements of the space
group.
a
a
a
Face-centered cubic lattice (fcc) is formed by a simple cubic lattice, with an additional lattice points on each
face of the cube.
Does the point group define the lattice? No: the lattices with the same point group belong to the same crystal
system, but they do not necessarily have the same lattice
structure nor the space group. The essential question is
whether the lattices can be transformed continuously to
one another without breaking the symmetries along the
process. Formally, this means that one must be able to
make a linear transformation S between the space groups
G and G0 of the crystals, i.e.
a2
a1
a3
a
Primitive vectors defining the lattice are, for example,
SGS
−1
0
=G.
a1 =
Then there exists a set of continuous mappings from the
unit matrix to matrix S
a
1
2
1
0
a
1 0 1
2
a
0 1 1 ,
a3 =
St = (1 − t)I + St,
2
where a is the lattice constant giving the distance bewhere t is between [0, 1]. Using this, one obtains such a con- tween the corners of the cube (not with the nearest neightinuous mapping from one lattice to the other that leaves bours!). Face-centered cubic lattice is often called cubic
the symmetries invariant.
closed-packed structure.√ If one takes the lattice points as
For example, there does not exist such a set of transfor- spheres with radius a/2 2, one obtains the maximal packmations between the rectangular and centered rectangular ing density. Fcc-lattice can be visualized as layered trigonal
lattices.
lattices, even though they share the same point group.
a2 =
6
Both of the closed-packing structures are common especially among the metal elements. If the atoms behaved like
hard spheres it would be indifferent whether the ordering
was fcc or hcp. Nevertheless, the elements choose always
either of them, e.g. Al, Ni and Cu are fcc, and Mg, Zn and
Co are hcp. In the hcp, the ratio c/a deviates slightly from
the value obtained with the hard sphere approximation.
a
Body-centered cubic lattice (bcc) is formed by inIn addition to the closed-packed structures, the bodyserting an additional lattice point into the center of prim- centered cubic lattice is common among elements, e.g. K,
itive cell of a simple cubic lattice.
Cr, Mn, Fe. The simple cubic structure is rare with crystals. (ET)
a
3
a1
a2
Diamond lattice is obtained by taking a copy of an fcclattice and by translating it with a vector (1/4 1/4 1/4).
a
An example of a choice for primitive vectors is
a
1 1 −1
a1 =
2
a
−1 1 1
a2 =
2
a
1 −1 1 ,
a3 =
2
The most important property of the diamond structure
is that every lattice point has exactly four neighbours (compare with the honeycomb structure in 2-D, that had three
neighbours). Therefore, diamond lattice is quite sparsely
packed. It is common with elements that have the tendency of bonding with four nearest neighbours in such a
way that all neighbours are at same angles with respect to
one another (109.5◦ ). In addition to carbon, also silicon
(Si) takes this form.
Hexagonal lattice is cannot be found amongst the elements. Its primitive vectors are
a1 = a 0 0
√
a2 = a2 a 2 3 0
a
0 0 c .
a3 =
2
Compounds
The lattice structure of compounds has to be described
with a lattice with a basis. This is because, as the name
says, they are composed of at least two different elements.
Let us consider as an example two most common structures
for compounds.
Hexagonal closed-packed lattice (hcp) is more interesting than the hexagonal lattice, because it is the ground
state of many elements. It is a lattice with a basis, formed
by stacking 2-dimensional trigonal lattices, like in the case
of fcc-closed packing. The difference to fcc is that in hcp
the lattice points of layer are placed on top of the centres
of the triangles in the previous layer, at a distance c/2. So,
the structure is repeated in every other layer in hcp. This
should be contrasted with the fcc-closed packing, where the
repetition occurs in every third layer.
Salt - Sodium Chloride
The ordinary table salt, i.e. sodium chloride (NaCl),
consists of sodium and chlorine atoms, ordered in an alternating simple cubic lattice. This can be seen also as an
fcc-structure (lattice constant a), that has a basis at points
(0 0 0) (Na) and a/2(1 0 0).
c
a
Hcp-lattice is formed by a hexagonal lattice with a basis
v1 = 0 0 0
a
c
.
v2 = a2 2√
2
3
The p
lattice constants c and a are arbitrary, but by choosing
Many compounds share the same lattice structure
c = 8/3a one obtains the closed-packing structure.
(MM):
7
Crystal
AgBr
AgCl
AgF
BaO
BaS
BaSe
BaTe
CaS
CaSe
CaTe
CdO
CrN
CsF
FeO
a
5.77
5.55
4.92
5.52
6.39
6.60
6.99
5.69
5.91
6.35
4.70
4.14
6.01
4.31
Crystal
KBr
KCl
KF
KI
LiBr
LiCl
LiF
LiH
LiI
MgO
MgS
MgSe
MnO
MnS
a
6.60
6.30
5.35
7.07
5.50
5.13
4.02
4.09
6.00
4.21
5.20
5.45
4.44
5.22
Crystal
MnSe
NaBr
NaCl
NaF
NaI
NiO
PbS
PbSe
PbTe
RbBr
RbCl
RbF
RbI
SnAs
a
5.49
5.97
5.64
4.62
6.47
4.17
5.93
6.12
6.45
6.85
6.58
5.64
7.34
5.68
Crystal
SnTe
SrO
SrS
SrSe
SrTe
TiC
TiN
TiO
VC
VN
ZrC
ZrN
a
6.31
5.16
6.02
6.23
6.47
4.32
4.24
4.24
4.18
4.13
4.68
4.61
• Monoclinic The orthorhombic symmetry can be reduced by tilting the rectangles perpendicular to the
c-axis (cf. the figure below). The simple and basecentered lattices transform into the simple monoclinic
lattice. The face- and body-centered orthorhombic
lattices are transformed into body-centered monoclinic
lattice.
Lattice constants a (10−10 m). Source: Wyckoff (1963-71),
vol. 1.
• Triclinic The destruction of the symmetries of the
cube is ready when the c-axis is tilted so that it is no
longer perpendicular with the other axes. The only
remaining point symmetry is that of inversion. There
is only one such lattice, the triclinic lattice.
Cesium Chloride
In cesium chloride (CsCl), the cesium and chlorine atoms
alternate in a bcc lattice. One can see this as a simple
cubic lattice that has basis defined by vectors (0 0 0) and
a/2(1 1 1). Some other compounds share this structure
(MM):
Crystal
AgCd
AgMg
AgZn
CsBr
a
3.33
3.28
3.16
4.29
Crystal
CsCl
CuPd
CuZn
NH4 Cl
a
4.12
2.99
2.95
3.86
Crystal
NiAl
TiCl
TlI
TlSb
By torturing the cube, one has obtained five of the seven
crystal systems, and 12 of the 14 Bravais lattices. The
sixth and the 13th are obtained by distorting the cube in
a different manner:
a
2.88
3.83
4.20
3.84
• Rhobohedral or Trigonal Let us stretch the cube
along its diagonal. This results in the trigonal lattice, regardless of which of the three cubic lattices was
stretched.
Lattice constants a (10−10 m). Source: Wyckoff (196371), vol. 1.
The last crystal system and the last Bravais lattice are
not related to the cube in any way:
2.5 Classification of Lattices by Symmetry
• Hexagonal Let us place hexagons as bases and perpendicular walls in between them. This is the hexagonal point group, that has one Bravais lattice, the
hexagonal lattice.
Let us first consider the classification of Bravais lattices.
3-dimensional Bravais lattices have seven point groups that
are called crystal systems. There 14 different space groups,
meaning that on the point of view of symmetry there are
14 different Bravais lattices. In the following, the cyrstal It is not in any way trivial why we have obtained all possystems and the Bravais lattices belonging to them, are sible 3-D Bravais lattices in this way. It is not necessary,
however, to justify that here. At this stage, it is enough
listed:
to know the existence of different classes and what belongs
• Cubic The point group of the cube. Includes the sim- in them. As a conclusion, a table of all Bravais lattice
presented above:
ple, face-centered and body-centered cubic lattices.
• Tetragonal The symmetry of the cube can be reduced
by stretching two opposite sides of the cube, resulting
in a rectangular prism with a square base. This eliminates the 90◦ -rotational symmetry in two directions.
The simple cube is, thus, transformed into a simple
tetragonal lattice. By stretching the fcc and bcc lattices one obtains the body-centered tetragonal lattice
(can you figure out why!).
• Orthorombic The symmetry of the cube can be
further-on reduced by pulling the square bases of the
tetragonal lattices to rectangles. Thus, the last 90◦ rotational symmetry is eliminated. When the simple
tetragonal lattice is pulled along the side of the base
square, one obtains the simple orthorhombic lattice.
When the pull is along the diagonal of the square,
one ends up with the base-centered orthorhombic lattice. Correspondingly, by pulling the body-centered
tetragonal lattice, one obtains the body-centered and (Source:
"http://www.iue.tuwien.ac.at/phd/
face-centered orthorhombic lattices.
karlowatz/node8.html")
8
On the Symmetries of Lattices with Basis
meaning that the cell is not identical with its mirror image
The introduction of basis into the Bravais lattice com- (in translations and rotations).
plicates the classification considerably. As a consequence,
the number of different lattices grows to 230, and number
of point groups to 32. The complete classification of these 2.6 Binding Forces (ET)
is not a subject on this course, but we will only summarise
Before examining the experimental studies of the lattice
the basic principle. As in the case of the classification of structure, it is worthwhile to recall the forces the bind the
Bravais lattices, one should first find out the point groups. lattice together
It can be done by starting with the seven crystal systems
At short distances the force between two atoms is always
and by reducing the symmetries of their Bravais lattices,
repulsive.
This is mostly due to the Pauli exclusion prinin a similar manner as the symmetries of the cube were reciple
preventing
more than one electron to be in the same
duced in the search for the Bravais lattices. This is possible
quantum
state.
Also,
the Coulomb repulsion between elecdue to the basis which reduces the symmetry of the lattice.
trons
is
essential.
At
larger distances the forces are often
The new point groups found this way belong to the origiattractive.
nal crystal system, up to the point where their symmetry
is reduced so far that all of the remaining symmetry operations can be found also from a less symmetrical system.
E(r)
Then, the point group is joined into the less symmetrical
system.
The space groups are obtained in two ways. Symmorphic lattices result from placing an object corresponding
to every point group of the crystal system into every Bravais lattice of the system. When one takes into account
that the object can be placed in several ways into a given
lattice, one obtains 73 different space groups. The rest of
the space groups are nonsymmorphic. They contain operations that cannot be formed solely by translations of the
Bravais lattice and the operations of the point group. E.g.
glide line and screw axis.
0
r
0
A sketch of the potential energy between two atoms as
a function of their separation r.
Covalent bond. Attractive force results because pairs
of atoms share part of their electrons. As a consequence,
the electrons can occupy larger volume in space, and thus
their average kinetic energy is lowered. (In the ground state
and according to the uncertainty principle, the momentum
p ∼ ~/d, where d is the region where the electron can be
found, and the kinetic energy Ekin = p2 /2m. On the other
hand, the Pauli principle may prevent the lowering of the
energy.)
Macroscopic consequences of symmetries
Sometimes macroscopic phenomena reveal symmetries
that reduce the number of possible lattice structures. Let
us look more closely on two such phenomenon.
Metallic bond. A large group of atoms share part of
their electrons so that they are allowed to move throughout the crystal. The justification otherwise the same as in
covalent binding.
Pyroelectricity
Some materials (e.g. tourmaline) have the ability of producing instantaneous voltages while heated or cooled. This
is a consequence of the fact that pyroelectric materials have
non-zero dipole moments in a unit cell, leading polarization
of the whole lattice (in the absence of electric field). In a
constant temperature the electrons neutralize this polarization, but when the temperature is changing a measurable
potential difference is created on the opposite sides of the
crystal.
Ionic bond. In some compounds, e.g. NaCl, a sodium
atom donates almost entirely its out-most electron to a
chlorine atom. In such a case, the attractive force is due
to the Coulomb potential. The potential energy between
two ions (charges n1 e and n2 e) is
E12 =
1 n1 n2 e2
.
4π0 |r1 − r2 |
(3)
In the equilibrium the polarization is a constant, and
therefore the point group of the pyroelectric lattice has to
In a NaCl-crystal, one Na+ -ion has six Cl− -ions as its
leave its direction invariant. This restricts the number of nearest neighbours. The energy of one bond between a
possible point groups. The only possible rotation axis is nearest neighbour is
along the polarization, and the crystal cannot have reflection symmetry with respect to the plane perpendicular to
1 e2
Ep1 = −
,
(4)
the polarization.
4π0 R
Optical activity
where R is the distance between nearest neighbours. The
neighbours are 12 Na+ -ions, with a distance
Some crystals (like SiO2 ) can rotate the plane of polar- next-nearest
√
ized light. This is possible only if the unit cells are chiral, d = 2R.
9
6 kpl
d=R
In the table below, the melting temperatures of some
solids are listed (at normal pressure), leading to estimates
8 kpl
of the strengths between interatomic forces. The lines did= 3R
vide the materials in terms of the bond types presented
above.
12
kpl
material melting temperature (K) lattice structure
tutkitaan
d= 2R
tämän
Si
1683
diamond
naapureita
C
(4300)
diamond
GaAs
1511
zincblende
ClSiO2
1670
Na+
Al
O
2044
2
3
The interaction energy of one Na+ ion with other ions is
Hg
234.3
obtained by adding together the interaction energies with
Na
371
bcc
neighbours at all distances. As a result, one obtains
Al
933
fcc
Cu
1356
fcc
8
e2 α
e2
12
Ep = −
6 − √ + √ − ... = −
. (5)
Fe
1808
bcc
4π0 R
4π0 R
2
3
W
3683
bcc
Here α is the sum inside the brackets, which is called the
CsCl
918
Madelung constant. Its value depends on the lattice, and
NaCl
1075
in this case is α = 1.7627.
H2 O
273
He
In order to proceed, on has to come up with form for the
Ne
24.5
fcc
repulsive force. For simplicity, let us assume
Ar
83.9
fcc
βe2
H2
14
.
(6)
Ep,repulsive =
O2
54.7
4π0 Rn
The total potential energy is, thus,
α
β
e2
− n .
Ep (R) = −
4π0 R R
(7)
By finding its minimum, we result in β = αRn−1 /n and in
energy
1
αe2
1−
.
(8)
Ep = −
4π0 R
n
The binding energy U of the whole lattice is the negative
of this multiplied with the number N of the NaCl-pairs
N αe2
1
U=
1−
.
(9)
4π0 R
n
In addition to diamond structure, carbon as also another
form, graphite, that consists of stacked layers of graphene.
In graphene, each carbon atom forms a covalent bond between three nearest neighbours (honeycomb structure), resulting in layers. The interlayer forces are of van der Waals
-type, and therefore very weak, allowing the layers to move
easily with respect to each other. Therefore, the ”lead” in
pencils (Swedish chemist Carl Scheele showed in 1779, that
graphite is made of carbon, instead of lead) crumbles easily
when writing.
By choosing n = 9.4, this in correspondence with the measurements. Notice that the contribution of the repulsive
part to the binding energy is small, ∼ 10%.
Hydrogen bond. Hydrogen has only one electron.
When hydrogen combines with, for example, oxygen, the
main part of the wave function of the electron is centered
near the oxygen, leaving a positive charge to the hydrogen.
With this charge, it can attract some third atom. The resulting bond between molecules is called hydrogen bond.
This is an important bond for example in ice.
A model of graphite where the spheres represent the carbon atoms (Wikipedia)
Covalent bonds are formed often into a specific direction. Covalent crystals are hard but brittle, in other words
they crack when they are hit hard. Metallic bonds are
not essentially dependent on direction. Thus, the metallic
atoms can slide past one another, still retaining the bond.
Therefore, the metals can be mold by forging.
Van der Waals interaction. This gives a weak attraction also between neutral atoms. Idea: Due to the circular
motion of the electrons, the neutral atoms behave like vibrating electric dipoles. Instantaneous dipole moment in
2.7 Experimental Determination of Crysone atom creates an electric field that polarizes the other
tal
Structure
atom, resulting in an interatomic dipole-dipole force. The
interaction goes with distance as 1/r6 , and is essential beMM, Chapters 3.1-3.2, 3.3 main points, 3.4-3.4.4
except equations
tween, e.g., atoms of noble gases.
10
In 1912 German physicist Max von Laue predicted that
the then recently discovered (1895) X-rays could scatter
from crystals like ordinary light from the diffraction grating1 . At that time, it was not known that crystals have
periodic structures nor that the X-rays have a wave character! With a little bit of hindsight, the prediction was
reasonable because the average distances between atoms
in solids are of the order of an Ångström (Å=10−10 m),
and the range of wave lengths of X-rays settles in between
0.1-100 Å.
scattered wave. Note, that it depends only on the direction
of the r-vector.
The form factor gives the differential cross section of the
scattering
dσ
Iatom ≡
= |f (r̂)|2 .
(11)
dΩatom
The intensity of the scattered wave at a solid angle dΩ at
distance r from the sample is dΩ × Iatom /r2 .
The idea was objected at first. The strongest counterargument was that the inevitable wiggling of the atoms due
to heat would blur the required regularities in the lattice
and, thus, destroy the possible diffraction maxima. Nowadays, it is, of course, known that such high temperatures
would melt the crystal! The later experimental results have
also shown that the random motion of the atoms due to
heat is only much less than argued by the opponents. As
an example, let us model the bonds in then NaCl-lattice
with a spring. The measured Young’s modulus indicates
that the spring constant would have to be of the order
k = 10 N/m. According to the equipartition theorem
this would
p result in average thermal motion of the atoms
hxi = 2kB T /k ≈ 2 · 10−11 m, which is much less than
the average distance between atoms in NaCl-crystal (cf.
(Source: MM) Scattering from a square lattice (25
previous table).
atoms). When the radiation k0 comes in from the right
direction, the waves scattered from different atoms interfere constructively. By measuring the resulting wave k, one
Scattering Theory of Crystals
observes an intensity maximum.
The scattering experiment is conducted by directing a
There are, naturally, many atoms in a lattice, and it is
plane wave towards a sample of condensed matter. When
the wave reaches the sample, there is an interaction be- thus necessary to study scattering from multiple scatterers.
tween them. The outgoing (scattered) radiation is mea- Let us assume in the following that we know the form factor
sured far away from the sample. Let us consider here a f . The angular dependence of the scattering is due to two
simple scattering experiment and assume that the scatter- factors:
ing is elastic, meaning that the energy is not transferred
• Every scatterer emits radiation into different direcbetween the wave and the sample. In other words, the fretions with different intensities.
quencies of the incoming and outgoing waves are the same.
This picture is valid, whether the incoming radiation con• The waves coming from different scatterers interfere,
sists of photons or, for example, electrons or neutrons.
and thus the resultant wave contains information on
The wave ψ, scattered from an atom located at origin,
the correlations between the scatterers.
takes the form (cf. Quantum Mechanics II):
#
"
eik0 r
−iωt
ik0 ·r
ψ ≈ Ae
e
+ f (r̂)
.
(10)
r
In fact, this result holds whether the scattered waves are
described by quantum mechanics or by classical electrodynamics.
One assumes in the above that the incoming radiation is
a plane wave with a wave vector k0 , defining the direction
of propagation. The scattering is measured at distances r
much greater than the range of the interaction between the
atom and the wave, and at angle 2θ measured from the k0 Let us assume in the following, that the origin is placed
axis. The form factor f contains the detailed information at a fixed lattice point. First, we have to find out how
on the interaction between the scattering potential and the Equation (10) is changed when the scatterer is located at a
1 von Laue received the Nobel prize from the discovery of X-ray
distance R from the origin. This deviation causes a phase
diffraction in 1914, right after his prediction!
difference in the scattered wave (compared with a wave
11
scattered at origin). In addition, the distance travelled by When we divide with the intensity of the incoming ray |A|2 ,
the scattered wave is |r − R|. Thus, we obtain
we obtain
"
#
X
eik0 |r−R|
−iωt
ik0 ·r
ik0 ·R
(17)
I=
fl fl∗0 eiq·(Rl −Rl0 ) .
ψ ≈ Ae
e
+e
f (r̂)
.
(12)
0
|r − R|
l,l
We have assumed in the above, that the point of obser- We have utilized here the property of complex numbers:
vation r is so far, that the changes in the scattering angle | P C |2 = P C C ∗ .
l l
ll0 l l0
can be neglected. At such distances (r R), we can approximate (up to first order in r/R)
r
k0 |r − R| ≈ k0 r − k0 · R.
r
Lattice Sums
(13)
Let us then define
k = k0
r
r
Let us first study scattering from a Bravais lattice. We
can, thus, assume that the scatterers are identical and that
the intensity (or the scattering cross section)
2
X
eiq·Rl ,
I = Iatom q = k0 − k.
(18)
l
The wave vector k points into the direction of the measurement device r, has the magnitude of the incoming radiation
(elastic scattering). The quantity q describes the difference
between the momenta of the incoming and outgoing rays.
Thus, we obtain
"
#
eik0 r+iq·R
−iωt
ik0 ·r
ψ ≈ Ae
e
+ f (r̂)
.
(14)
r
where Iatom is the scattering cross section of one atom,
defined in Equation (11).
In the following, we try to find those values q of the
change in momentum, that lead to intensity maxima. This
is clearly occurs, if we can choose q so that exp(iq·Rl ) = 1
at all lattice points. In all other cases, the phases of the
complex numbers exp(iq · Rl ) reduce the absolute value
The second term in the denominator can be neglected in of the sum (destructive interference). Generally speaking,
Equation (13). However, one has to include into the ex- one ends up with similar summations whenever studying
ponential function all terms that are large compared with the interaction of waves with a periodic structure (like conduction electrons in a lattice discussed later).
2π.
The magnitude of the change in momentum is
q = 2k0 sin θ,
We first restrict the summation in the intensity (18) in
one dimension, and afterwards generalize the procedure
(15) into three dimensions.
where 2θ is the angle between the incoming and outgoing
One-Dimensional Sum
waves. The angle θ is called the Bragg’s angle. Assuming
specular reflection (Huygens’ principle, valid for X-rays),
The lattice points are located at la, where l is an integer
the Bragg angle is the same as the angle between the in- and a is the distance between the points. We obtain
coming ray and the lattice planes.
Finally, let us consider the whole lattice and assume
N
−1
X
again that the origin is located somewhere in the middle of
Σq =
eilaq ,
the lattice, and that we measure far away from the sample.
l=0
By considering that the scatterers are sparse, the observed
radiation can be taken to be sum of the waves produced
by individual scatterers. Furthermore, let us assume that where N is the number of lattice points. By employing the
the effects due to multiple scattering events and inelastic properties of the geometric series, we get
processes can be ignored. We obtain
"
#
sin2 N aq/2
X
|Σq |2 =
.
(19)
eik0 r+iq·Rl
−iωt
ik0 ·r
sin2 aq/2
ψ ≈ Ae
e
+
fl (r̂)
,
(16)
r
l
where the summation runs through the whole lattice.
When the number of the lattice points is large (as is
When we study the situation outside the incoming ray the case in crystals generally), the plot of Equation (19)
(in the region θ 6= 0), we can neglect the first term. The consists of sharp and identical peaks, with a (nearly) zero
intensity is proportional to |ψ|2 , like in the one atom case. value of the scattering intensity in between.
12
where V = L3 is the volume of the lattice.
Why do the wave vectors obeying (22) form a lattice?
Here, we present a direct proof that also gives an algorithm
for the construction of the reciprocal lattice. Let us show
that the vectors
b1
b2
b3
Plot of Equation (19). Normalization with N is introduced
so that the effect of the number of the lattice points on
the sharpness of the peaks is more clearly seen.
2π
(24)
are the primitive vectors of the reciprocal lattice. Because
the vectors of the Bravais lattice are of form
R = n1 a1 + n2 a2 + n3 a3 ,
we obtain
The peaks can be found at those values of the momentum
q that give the zeroes of the denominator
q = 2πl/a.
a2 × a3
a1 · a2 × a3
a3 × a1
= 2π
a2 · a3 × a1
a1 × a2
= 2π
a3 · a1 × a2
=
bi · R = 2πni
(20)
where i = 1, 2, 3.
Thus, the reciprocal lattice contains, at least, the vectors
These are exactly those values that give real values for the bi . In addition, the vectors bi are linearly independent. If
exponential function (= 1). As the number of lattice points
grows, it is natural to think Σq as a sum of delta functions
eiK1 ·R = 1 = eiK2 ·R ,
∞
X
Σq =
l0 =−∞
2πl0 ,
cδ q −
a
then
ei(K1 +K2 )·R = 1.
Therefore, the vectors K can be written as
where
Z
π
a
c=
dqΣq =
−π
a
2πN
.
L
K = l1 b1 + l2 b2 + l3 b3
(25)
In the above, L is length of the lattice in one dimension. where l1 , l2 and l3 are integers. This proves, in fact, that
the wave vectors K form a Bravais lattice.
Thus,
∞
0
X
2πN
2πl
The requirement
Σq =
δ q−
.
(21)
L 0
a
K·R=0
l =−∞
It is worthwhile to notice that essentially this Fourier trans- defines a Bragg’s plane in the position space. Now, the
forms the sum in the intensity from the position space into planes K · R = 2πn are parallel, where n gives the distance
the momentum space.
from the origin and K is the normal of the plane. This
kind of sets of parallel planes are referred to as families
Reciprocal Lattice
of lattice planes. The lattice can be divided into Bragg’s
Then we make a generalization into three dimensions. In planes in infinitely many ways.
Equation (18), we observe sharp peaks whenever we choose
Example By applying definition (24), one can show that
q for every Bravais vectors R as
reciprocal lattice of a simple cubic lattice (lattice constant
(22) a) is also a simple cubic lattice with a lattice constant 2π/a.
Correspondingly, the reciprocal lattice of an fcc lattice is a
where l is an integer whose value depends on vector R. bcc lattice (lattice constant 4π/a), and that of a bcc lattice
Thus, the sum in Equation (18) is coherent, and produces is an fcc lattice (4π/a).
a Bragg’s peak. The set of all wave vectors K satisfying
Miller’s Indices
Equation (22) is called the reciprocal lattice.
q · R = 2πl,
The reciprocal lattice allows the classification of all posReciprocal lattice gives those wave vectors that result in
coherent scattering from the Bravais lattice. The magni- sible families of lattice planes. For each family, there exist
tude of the scattering is determined analogously with the perpendicular vectors in the reciprocal lattice, shortest of
which has the length 2π/d (d is separation between the
one dimensional case from the equation
planes). Inversely: For each reciprocal lattice vector, there
X
(2π)3 X
iR·q
e
=N
δ(q − K),
(23) exists a family of perpendicular lattice planes separated
V
with a distance d from one another (2π/d is the length of
R
K
13
2a3
the shortest reciprocal vector parallel to K). (proof: exercise)
Based on the above, it is natural to describe a given latyksikkötice plane with the corresponding reciprocal lattice vector.
koppi
a3
The conventional notation employs Miller’s indices in the
(263)
description of reciprocal lattice vectors, lattice planes and
a2
a1
lattice points. Usually, Miller’s indices are used in lattices
with a cubic or hexagonal symmetry. As an example, let
3a1
us study a cubic crystal with perpendicular coordinate vectors x̂, ŷ and ẑ, pointing along the sides of a typical unit
If the lattice plane does not intersect with an axis, it corcell (=cube). Miller’s indices are defined in the following
responds
to a situation where the intersection is at infinity.
way:
1
= 0.
The inverse of that is interpreted as ∞
One can show that the distance d between adjacent parallel lattice planes is obtained from Miller’s indices (hkl)
by
a
,
(26)
d= √
2
h + k 2 + l2
where a is the lattice constant of the cubic lattice.
• [ijk] is the direction of the lattice
ix̂ + j ŷ + kẑ
where i, j and k are integers.
• (ijk) is the lattice plane perpendicular to vector [ijk].
It can be also interpreted as the reciprocal lattice vector perpendicular to plane (ijk).
• {ijk} is the set of planes perpendicular to vector [ijk],
and equivalent in terms of the lattice symmetries.
(100)
(110)
(111)
In the figure, there are three common lattice planes.
Note, that due to the cubic symmetry, the planes (010)
and (001) are identical with the plane (100). Also, the
planes (011), (101) and (110) are identical.
• hijki is the set of directions [ijk] that are equivalent
in terms of the lattice symmetries.
In this representation the negative numbers are denoted
with a bar −i → ī. The original cubic lattice is called
direct lattice.
Scattering from a Lattice with Basis
The condition of strong scattering from a Bravais lattice was q = K. This is changed slightly when a basis is
The Miller indices of a plane have a geometrical property introduced to the lattice. Every lattice vector is then of
that is often given as an alternative definition. Let us con- form
sider lattice plane (ijk). It is perpendicular to reciprocal
R = ul + vl0 ,
lattice vector
where ul is a Bravais lattice vector and vl0 is a basis vector.
K = ib1 + jb2 + kb3 .
Again, we are interested in the sum
!
!
X
X
X
Thus, the lattice plane lies in a continuous plane K · r = A,
eiq·R =
eiq·ul
eiq·vl0 ,
where A is a constant. This intersects the coordinate axes
0
R
l
l
of the direct lattice at points x1 a1 , x2 a2 and x3 a3 . The
coordinates xi are determined by the condition K·(xi ai ) = determining the intensity
!
!
A. We obtain
X
X
X
I∝|
eiq·R |2 =
eiq·(uj −uj0 )
eiq·(vl −vl0 ) .
A
A
A
0
0
jj
R
ll
, x2 =
, x3 =
.
x1 =
2πi
2πj
2πk
(27)
The intensity appears symmetric with respect to the lattice
We see that the points of intersection of the lattice plane and basis vectors. The difference arises in the summations
and the coordinate axes are inversely proportional to the which for the lattice have a lot of terms (of the order 1023 ),
values of Miller’s indices.
whereas for the basis only few. Previously, it was shown
Example Let us study a lattice plane that goes through that the first term in the intensity is non-zero only if q
points 3a1 , 1a2 and 2a3 . Then, we take the inverses of the belongs to the reciprocal lattice. The amplitude of the
coefficients: 13 , 1 and 12 . These have to be integers, so scattering is now modulated by the function
2
we multiply by 6. So, Miller’s indices of the plane are
X
(263). The normal to this plane is denoted with the same
Fq = eiq·vl ,
(28)
numbers, but in square brackets, [263].
l
14
caused by the introduction of the basis. This modulation
can even cause an extinction of a scattering peak.
Example: Diamond Lattice The diamond lattice is
formed by an fcc lattice with a basis
v1 = (0 0 0), v2 =
a
(1 1 1).
4
It was mentioned previously that the reciprocal lattice of
an fcc lattice is a bcc lattice with a lattice constant 4π/a.
Thus, the reciprocal lattice vectors are of form
K = l1
4π
4π
4π
(1 1 − 1) + l2 (−1 1 1) + l3 (1 − 1 1).
2a
2a
2a
Two-dimensional cross-cut of Ewald sphere. Especially,
we see that in order to fulfil the scattering condition (22),
the surface of the sphere has to go through at least two
reciprocal lattice points. In the case above, strong scattering is not observed. The problem is, thus, that the points
in the reciprocal lattice form a discrete set in the three dimensional k-space. Therefore, it is extremely unlikely that
any two dimensional surface (e.g. Ewald sphere) would go
through them.
Therefore,
v1 · K = 0
and
v2 · K =
π
(l1 + l2 + l3 ).
2
Ewald sphere gives also an estimate for the necessary
wave length of radiation. In order to resolve the atomic
structure, the wave vector k has to be larger than the lattice constant of the reciprocal lattice. Again, we obtain the
estimate that the wave length has to be of the order of an
Ångström, i.e. it has to be composed of X-rays.
The modulation factor is
FK
=
=
2
1 + eiπ(l1 +l2 +l3 )/2 
 4 l1 + l2 + l3 = 4, 8, 12, . . .
2 l1 + l2 + l3 is odd

0 l1 + l2 + l3 = 2, 6, 10, . . .
(29)
One can also use Ewald sphere to develop solutions for
the problem with monochromatic radiation. The most conventional ones of them are Laue method, rotating crystal
method and powder method.
Laue Method
Let
us
use
continuous
spectrum.
2.8 Experimental Methods
Next, we will present the experimental methods to test
the theory of the crystal structure presented above. Let
us first recall the obtained results. We assumed that we
are studying a sample with radiation whose wave vector is
k0 . The wave is scattered from the sample into the direction k, where the magnitudes of the vectors are the same
(elastic scattering) and the vector q = k0 − k has to be in
the reciprocal lattice. The reciprocal lattice is completely
determined by the lattice structure, and the possible basis
only modulates the magnitudes of the observed scattering
peaks (not their positions).
Problem: It turns out that monochromatic radiation
does not produce scattering peaks! The measurement device collects data only from one direction k. This forces
us to restrict ourselves into a two-dimensional subspace of
Rotating Crystal Method
the scattering vectors. This can be visualized with Ewald
Let
us
use
monochromatic
sphere of the reciprocal lattice, that reveals all possible
but
also
rotate
the
values of q for the give values of the incoming wave vector. ation
15
radicyrstal.
The neutrons interact only with nuclei. The interaction depends on the spin of the nucleus, allowing the study
of magnetic materials. The lattice structure of matter is
studied with so called thermal neutrons (thermal energy
ET ≈ 32 kB T , with T = 293 K), whose de Broglie wave
length λ = h/p ≈ 1 Å. It is expensive to produce neutron
showers with large enough density.
Electrons
The electrons interact with matter stronger than photons
and neutrons. Thus, the energies of the electrons have to
be high (100 keV) and the sample has to be thin (100 Å),
in order to avoid multiple scattering events.
Charge
Mass
Energy
Wave length
Attenuation length
Form factor, f
Powder Method (Debye-Scherrer Method)
X-rays
0
0
12 keV
1 Å
100 µm
10−3 Å
Neutrons
0
1.67 · 10−27 kg
0.02 eV
2 Å
5 cm
10−4 Å
Electrons
-e
9.11 · 10−31 kg
60 keV
0.05 Å
1 µm
10 Å
Similar to the rotating crystal method.
We use
monochromatic radiation but, instead of rotating, a sample consisting of many crystals (powder). The powder is Typical properties of different sources of radiation in scatfine-grained but, nevertheless, macroscopic, so that they tering experiments. (Source: MM; Eberhart, Structural
are able to scatter radiation. Due to the random orienta- and Chemical Analysis of Materials (1991).)
tion of the grains, we observe the same net effect as when
rotating a single crystal.
2.10 Surfaces and Interfaces
MM, Chapter 4, not 4.2.2-4.2.3
2.9 Radiation Sources of a Scattering Experiment
In addition to the X-ray photons we have studied so far,
the microscopic structure of matter is usually studied with
electrons and neutrons.
X-Rays
The interactions between X-rays and condensed matter
are complex. The charged particles vibrate with the frequency of the radiation and emit spherical waves. Because
the nuclei of the atoms are much heavier, only their electrons participate to the X-ray scattering. The intensity of
the scattering depends on the number density of the electrons, that has its maximum in the vicinity of the nucleus.
Production of X-rays
The traditional way to produce X-rays is by colliding
electrons with a metal (e.g. copper in the study of structure
of matter, wolfram in medical science). Monochromatic
photons are obtained when the energy of an electron is
large enough to remove an electron from the inner shells
of an atom. The continuous spectrum is produced when
an electron is decelerated by the strong electric field of the
nucleus (braking radiation, bremsstrahlung). This way of
producing X-ray photons is very inefficient: 99 % of the
energy of the electron is turned into heat when it hits the
metal.
In a synchrotron, X-rays are produced by forcing electrons accelerate continuously in large rings with electromagnetic field.
Only a small part of the atoms of macroscopic bodies
are lying on the surface. Nevertheless, the study of surfaces is important since they are mostly responsible for
the strength of the material and the resistance to chemical attacks. For example, the fabrication of circuit boards
requires good control on the surfaces so that the conduction of electrons on the board can be steered in the desired
manner.
The simplest deviations from the crystal structure occur when the lattice ends, either to another crystal or to
vacuum. This instances are called the grain boundary and
the surface. In order to describe the grain boundary, one
needs ten variables: three for the relative location between
the crystals, six for their interfaces and one for the angle in
between them. The description of a crystal terminating to
vacuum needs only two variables, that determine the plane
along which the crystal ends.
In the case of a grain boundary, it is interesting to know
how well the two surfaces adhere, especially if one is forming a structure with alternating crystal lattices. Coherent
interface has all its atoms perfectly aligned. The growing of such structure is called epitaxial. In a more general
case, the atoms in the interfaces are aligned in a larger
scale. This kind of interface is commensurate.
Experimental Determination and Creation of
Surfaces
Low-Energy Electron Diffraction
Low-energy electron diffraction (LEED) was used in the
demonstration of the wave nature of electrons (Davisson
and Germer, 1927).
Neutrons
16
Reflection
RHEED
High-Energy
Electron
Diffraction,
Electrons (with energy 100 keV) are reflected from the
surface and they are studied at a small angle. The lengths
of the wave vectors (∼ 200 Å−1 ) are large compared with
the lattice constant of the reciprocal lattice and, thus, the
scattering patterns are streaky. The sample has to be rotated in order to observe strong signals in desired directions.
In the experiment, the electrons are shot with a gun towards the sample. The energy of the electrons is small (less
than 1 keV) and, thus, they penetrate only into at most few
atomic planes deep. Part of the electrons is scattered back,
which are then filtered except those whose energies have
changed only little in the scattering. These electrons have
been scattered either from the first or the second plane.
Molecular Beam Epitaxy, MBE
Molecular Beam Epitaxy enables the formation of a solid
material by one atomic layer at a time (the word epitaxy
has its origin in Greek: epi = above, taxis=orderly). The
Technique allows the selection and change of each layer
based on the needs.
The scattering is, thus, from a two-dimensional lattice.
The condition of strong scattering is the familiar
eiq·R = 1
where R is now a lattice vector of the surface and l is an
integer that depends on the choice of the vector R. Even
though the scattering surface is two dimensional, the scattered wave can propagate into any direction in the three
dimensional space. Thus, the strong scattering condition
is fulfilled with wave vectors q of form
q = (Kx , Ky , qz ),
(30)
where Kx and Ky are the components of a reciprocal lattice vector K. On the other hand, the component qz is
A sample, that is flat on the atomic level, is placed in
a continuous variable because the z-component of the two a vacuum chamber. The sample is layered with elements
dimensional vector R is zero.
that are vaporized in Knudsen cells (three in this example).
When the shutter is open the vapour is allowed to leave
the cell. The formation of the structure is continuously
monitored using the RHEED technique.
Scanning Tunnelling Microscope
The (Scanning Tunnelling Microscope) is a thin metallic needle that can be moved in the vicinity of the studied
surface. In the best case the tip of the needle is formed
by only one atom. The needle is brought at a distance less
than a nanometer from the conducting surface under study.
In order to observe strong scattering, Ewald sphere has By changing the electric potential difference between the
to go through some of the rods defined by condition (30). needle and the surface, one can change the tunnelling probThis occurs always, independent on the choice of the in- ability of electrons from the needle to the sample. Then
coming wave vector or the orientation of the sample (com- the created current can be measured and used to map the
pare with Laue, rotating crystal and powder methods).
surface.
17
Atomic force microscope is a close relative to the scanning tunnelling microscope. A thin tip is pressed slightly
on the studied surface. The bend in the lever is recorded
as the tip is moved along the surface. Atomic force microscope can be used also in the study of insulators.
Example: Graphene
The tunnelling of an electron between the needle and the
sample can be modelled with a potential wall, whose height
U (x) depends on the work required to free the electron from
the needle. In the above figure, s denotes the distance
between the tip of the needle and studied surface. The
potential wall problem has been solved in the course of
quantum mechanics (QM II), and the solution gave the
wave function outside the wall to be
i
hi Z x
p
dx0 2m(E − U (x0 )) .
ψ(x) ∝ exp
~
(Source: Novoselov et al. PNAS 102, 10451 (2005))
Atomic force microscopic image of graphite. Note the color
scale, partly the graphite is only one atomic layer thick, i.e.
graphene (in graphite the distance between graphene layers
is 3.35 Å).
Inside the wall, the amplitude of the wave function drops
with a factor
h
i
p
exp − s 2mφ/~2 .
The current is dependent on the square of the wave function, and thus depend exponentially on the distance between the needle and the surface. By recording the current
and simultaneously moving the needle along the surface,
one obtains a current mapping of the surface. One can
reach atomic resolution with this method (figure on page
3).
neulan
kärki
(Source: Li et al. PRL 102, 176804 (2009)) STM image
of graphene.
Graphene can be made, in addition to previously mentioned tape-method, by growing epitaxially (e.g. on a
metal). A promising choice for a substrate is SiC, which
couples weakly with graphene. For many years (after its
tutkittava
discovery in 2004), graphene has been among the most expinta
pensive materials in the world. The production methods
of large sheets of graphene are still (in 2012) under develThe essential part of the functioning of the device is how opment in many research groups around the world.
to move the needle without vibrations in atomic scale.
2.11 Complex Structures
MM, Chapters 5.1-5.3, 5.4.1 (not Correlation
functions for liquid), 5.5 partly, 5.6 names, 5.8
main idea
The crystal model presented above is an idealization and
rarely met in Nature as such. The solids are seldom in a
thermal equilibrium, and equilibrium structures are not
always periodic. In the following, we will study shortly
other forms of condensed matter, such as alloys, liquids,
Pietzoelectric crystal can change its shape when placed glasses, liquid crystals and quasi crystals.
in an electric field. With three pietzoelectric crystals one
can steer the needle in all directions.
Alloys
The development of metallic alloys has gone hand in
Atomic Force Microscope
18
hand with that of the society. For example, in the Bronze
Age (in Europe 3200-600 BC) it was learned that by mixing
tin and copper with an approximate ratio 1:4 one obtains
an alloy (bronze), that is stronger and has a lower melting
point than either of its constituents. The industrial revolution of the last centuries has been closely related with the
development of steel, i.e. the adding of carbon into iron in
a controlled manner.
Equilibrium Structures
One can always mix another element into a pure crystal.
This is a consequence of the fact that the thermodynamical
free energy of a mixture has its minimum with a finite
impurity concentration. This can be seen by studying the
entropy related in adding of impurity atoms. Let us assume
that there are N points in the lattice, and that we add
M N impurity atoms. The adding can be done in
NM
N
N!
≈
=
M !(N − M )!
M!
M
different ways. The macroscopic state of the mixture has
the entropy
S = kB ln(N M /M !) ≈ −kB N (c ln c − c),
Generally, the mixing can occur in all ratios. Intermetallic compound is formed when two metals form a crystal
structure at some given concentration. In a superlattice
atoms of two different elements find the equilibrium in the
vicinity of one another. This results in alternating layers
of atoms, especially near to some specific concentrations.
On the other hand, it is possible that equilibrium requires
the separation of the components into separate crystals,
instead of a homogeneous mixture. This is called phase
separation.
Superlattices
The alloys can be formed by melting two (or more) elements, mixing them and finally cooling the mixture. When
the cooling process is fast, one often results in a similar random structure as in high temperatures. Thus, one does not
observe any changes in the number of resonances in, e.g.
X-ray spectroscopy. This kind of cooling process is called
quenching. It is used for example in the hardening of steel.
At high temperatures the lattice structure of iron is fcc
(at low temperatures it is bcc). When the iron is heated
and mixed with carbon, the carbon atoms fill the centers
of the fcc lattice. If the cooling is fast, the iron atoms do
not have the time to replace the carbon atoms, resulting
in hard steel.
Slow cooling, i.e. annealing, results in new spectral
where c = M/N is the impurity concentration. Each impupeaks.
The atoms form alternating crystal structures, surity atom contributes an additional energy to the crystal,
perlattices.
With many combinations of metals one obtains
leading to the free energy of the mixture
superlattices, mostly with mixing ratios 1:1 and 3:1.
F = E − T S = N [c + kB T (c ln c − c)].
(31)
Phase Separation
In equilibrium, the free energy is in its minimum. This
occurs at concentration
c ∼ e−/kB T .
Let us assume that we are using two substances whose
free energy is of the below form when they are mixed ho(32) mogeneously.
So, we see that at finite temperatures the solubility is nonzero, and that it decreases exponentially when T → 0. In
most materials there are ∼ 1% of impurities.
The finite solubility produces problems in semiconductors, since in circuit boards the electrically active impurities disturb the operation already at concentrations 10−12 .
Zone refining can be used to reduce the impurity concentration. One end of the impure crystal is heated, simultaneously moving towards the colder end. After the process
(Source: MM) The free energy F(c) of a homogeneous
the impurity concentration is larger the other end of the mixture of two materials as a function of the relative concrystal, which is removed and the process is repeated.
centration c. One can deduce from the figure that when
the concentration is between ca and cb , the mixture tends
Phase Diagrams
to phase separate in order to minimize the free energy This
Phase diagram describes the equilibrium at a given con- can be seen in the following way: If the atoms divide becentration and temperature. Let us consider here espe- tween two concentrations ca < c and cb > c (not necessarily
cially system consisting of two components. Mixtures with the same as in the figure), the free energy of the mixture
two substances can be divided roughly into two groups. is
The first group contains the mixtures where only a small
Fps = f F(ca ) + (1 − f )F(cb ),
amount of one substance is mixed to the other (small concentration c). In these mixtures the impurities can either where f is the fraction of the mixture that has the conreplace a lattice atom or fill the empty space in between centration ca . Correspondingly, the fraction 1 − f has the
the lattice points.
concentration cb . The fraction f is not arbitrary because
19
the total concentration has to be c. Thus,
c = f ca + (1 − f )cb ⇒ f =
c − cb
.
ca − cb
Therefore, the free energy of the phase separated mixture
is
Fps =
c − cb
ca − c
F(ca ) +
F(cb ).
ca − cb
ca − cb
(33)
Phase separation can be visualized geometrically in the
following way. First, choose two points from the curve F(c)
and connect the with a straight line. This line describes
the phase separation of the chosen concentrations. In the
figure, the concentrations ca and cb have been chosen so
that the phase separation obtains the minimum value for
the free energy. We see that F(c) has to be convex in order
to observe a phase separation.
A typical phase diagram consists mostly on regions with
a phase separation.
(Source: MM) The formation of a phase diagram.
Dynamics of Phase Separation
The heating of a solid mixture always results in a homogeneous liquid. When the liquid is cooled, the mixture
remains homogeneous for a while even though the phase
separated state had a lower value of the free energy. Let
us then consider how the phase separation comes about in
such circumstances as the time passes.
The dynamics of the phase separation can be solved from
the diffusion equation. It can be derived by first considering the concentration current
j = −D∇c.
(Source: MM) The phase separation of the mixture of
copper and silver. In the region Ag, silver forms an fcc lattice and the copper atoms replace silver atoms at random
lattice points. Correspondingly in the region Cu, silver replaces copper atoms in an fcc lattice. Both of them are
homogeneous solid alloys. Also, the region denoted with
”Liquid” is homogeneous. Everywhere else a phase separation between the metals occurs. The solid lines denote the
concentrations ca and cb (cf. the previous figure) as a function of temperature. For example in the region ”Ag+L”,
a solid mixture with a high silver concentration co-exists
with a liquid with a higher copper concentration than the
solid mixture. The eutectic point denotes the lowest temperature where the mixture can be found as a homogeneous
liquid.
20
(34)
The solution of this equation gives the atom current j,
whose direction is determined by the gradient of the concentration. Due to the negative sign, the current goes from
large to small concentration. The current is created by the
random thermal motion of the atoms. Thus, the diffusion
constant D changes rapidly as a function of temperature.
Similarly as in the case of mass and charge currents, we can
define a continuity equation for the flow of concentration.
Based on the analogy,
∂c
= D∇2 c.
∂t
(35)
This is the diffusion equation of the concentration. The
equation looks innocent, but with a proper set of boundary
conditions it can create complexity, like in the figure below
The gradient of the concentration gives the current density
j = −D∇c = DR(c∞ − ca )∇
r̂
1
= −DR(c∞ − ca ) 2 .
r
r
For the total current into the droplet, we have to multiply the current density with the surface area 4πR2 of the
droplet. This gives the rate of change of the concentration
inside the droplet
∂c
= 4πR2 (−jr ) = 4πDR(c∞ − ca ).
∂t
On the other hand, the chain rule of derivation gives the
rate of change of the volume V = 4πR3 /3 of the droplet
dV
∂V ∂c
=
.
dt
∂c ∂t
By denoting v ≡ ∂V /∂c, we obtain
Ṙ =
p
vD
(c∞ − ca ) ⇒ R ∝ 2vD(c∞ − ca )t
R
We see that small nodules grow the fastest when measured
in R.
Simulations
(Source: MM) Dendrite formed in the solidification process of stainless steel.
Let us look as an example a spherical drop of iron carbide
with an iron concentration ca . We assume that it grows in
a mixture of iron and carbon, whose iron concentration
c∞ > ca . The carbon atoms of the mixture flow towards
the droplet because it minimizes the free energy. In the
simplest solution, one uses the quasi-static approximation
When the boundary conditions of the diffusion equation
are allowed to change, the non-linear nature of the equation often leads to non-analytic solutions. Sometimes, the
calculation of the phase separations turns out to be difficult
even with deterministic numerical methods. Then instead
of differential equations, one has to rely on descriptions on
atomic level. Here, we introduce two methods that are in
common use: Monte Carlo and molecular dynamics.
Monte Carlo
∂c
≈ 0.
∂t
Monte Carlo method was created by von Neumann,
Ulam and Metropolis in 1940s when they were working
Thus, the concentration can be solved from the Laplace on the Manhattan Project. The name originates from the
equation
casinos in Monte Carlo, where Ulam’s uncle often went to
∇2 c = 0.
gamble his money. The basic principle of Monte Carlo reWe are searching for a spherically symmetric solution and, lies on the randomness, characteristic to gambling. The
assumption in the background of the method is that the
thus, we can write the Laplace equation as
atoms in a solid obey in equilibrium at temperature T the
1 ∂ 2 ∂c
Boltzmann distribution exp(−βE), where E is position der
=
0.
r2 ∂r
∂r
pendent energy of an atom and β = 1/kB T . If the energy
This has a solution
difference between two states is δE, then the relative occupation probability is exp(−βδE).
B
c(r) = A + .
r
The Monte Carlo method presented shortly:
At the boundary of the droplet (r = R), the concentration
is
c(R) = ca
and far away from the droplet (r → ∞)
1) Assume that we have N atoms. Let us choose their
positions R1 . . . RN randomly and calculate their energy E(R1 . . . RN ) = E1 .
lim c(r) = c∞ .
2) Choose one atom randomly and denote it with index
l.
These boundary conditions determine the coefficients A
and B leading to the unambiguous solution of the Laplace
equation
R
c(r) = c∞ + (ca − c∞ ).
r
3) Create a random displacement vector, e.g. by creating
three random numbers pi ∈ [0, 1] and by forming a
vector
1
1
1
Θ = 2a(p1 − , p2 − , p3 − ).
2
2
2
r→∞
21
In the above, a sets the length scale. Often, one uses As was mentioned in the beginning, the initial state dethe typical interatomic distance, but its value cannot termines the energy E, that is conserved in the process.
affect the result.
The temperature can be deduced only at the end of the
calculation, e.g. from the root mean square value of the
4) Calculate the energy difference
velocity.
The effect of the temperature can be included by adding
terms
Fl
The calculation of the difference is much simpler than
− bṘl + ξ(t),
R̈l =
ml
the calculation of the energy in the position configuration alone.
into the equation of motion. The first term describes the
δE = E(R1 . . . Rl + Θ . . . RN ) − E1 .
5) If δE < 0, replace Rl → Rl + Θ. Go to 2).
6) If δE > 0, accept the displacement with a probability
exp(−βδE). Pick a random number p ∈ [0, 1]. If p
is smaller than the Boltzmann factor, accept the displacement (Rl → Rl + Θ and go to 2). If p is greater,
reject the displacement and go to 2).
dissipation by the damping constant b that depends on the
microscopic properties of the system. The second term illustrates the random fluctuations due to thermal motion.
These additional terms cause the particles to approach the
thermal equilibrium at the given temperature T . The thermal fluctuations and the dissipation are closely connected,
and this relationship is described with the fluctuationdissipation theorem
In low temperatures almost every displacement that are
2bkB T δαβ δ(t)
accepted lower the systems energy. In very high temper.
hξα (0)ξβ (t)i =
ml
atures almost every displacement is accepted. After sufficient repetition, the procedure should generate the equilibThe angle brackets denote the averaging over time, or alrium energy and the particle positions that are compatible
ternatively, over different statistical realizations (ergodic
with the Boltzmann distribution exp(−βE).
hypothesis). Without going into any deeper details, we
obtain new equations motion by replacing in Equation (36)
Molecular Dynamics
p
Rn − Rn−1
l
Molecular dynamics studies the motion of the single
Fnl → Fnl − bml l
+ Θ 6bml kB T /dt,
dt
atoms and molecules forming the solid. In the most general case, the trajectories are solved numerically from the where Θ is a vector whose components are determined by
Newton equations of motion. This results in solution of random numbers p picked from the interval [0, 1]:
i
the thermal equilibrium in terms of random forces instead
of random jumps (cf. Monte Carlo). The treatment gives
1
1
1
,
p
−
,
p
−
.
Θ
=
2
p
−
2
3
1
the positions and momenta of the particles and, thus, pro2
2
2
duces more realistic picture of the dynamics of the system
approaching thermal equilibrium.
Liquids
Let us assume that at a given time we know the positions
of the particles and that we calculate the total energy of
Every element can be found in the liquid phase. The
the system E. The force Fl exerted on particle l is obtained passing from, e.g., the solid into liquid phase is called the
as the gradient of the energy
phase transformation. Generally, the phase transformations are described with the order parameter. It is defined
∂E
Fl = −
.
in
such way that it non-zero in one phase and zero in all
∂Rl
the other phases. For example, the appearance of Bragg’s
According to Newton’s second law, this force moves the peaks in the scattering experiments of solids can be thought
as the order parameter of the solid phase.
particle
d 2 Rl
Let us define the order parameter of the solid phase more
ml 2 = Fl .
dt
rigorously. Consider a crystal consisting of one element.
In order to solve these equations numerically (l goes Generally, it can be described with a two particle (or van
through values 1, . . . , N , where N is the number of parti- Hove) correlation function
cles), one has to discretize them. It is worthwhile to choose
*
+
X
the length of the time step dt to be shorter than any of the
n2 (r1 , r2 ; t) =
δ(r1 − Rl (0))δ(r2 − Rl0 (t)) ,
scales of which the forces Fl move the particles consider0
n
l6
=
l
ably. When one knows the position Rl of the particle l
after n steps, the position after n + 1 steps can be obtained
where the angle brackets mean averaging over temperature
by calculating
and vectors Rl denote the positions of the atoms. If an
n
atom is at r1 at time t1 , the correlation function gives the
F
Rn+1
= 2Rnl − Rn−1
+ l dt2 .
(36) probability of finding another particle at r2 at time t1 + t.
l
l
ml
22
window glass (SiO2 ) is 10 K/s, whereas for nickel it is 107
K/s.
Then, we define the static structure factor
S(q) ≡
I
1 X D iq·(Rl −Rl0 ) E
,
e
=
N Iatom
N 0
(37)
Liquid Crystals
ll
Liquid crystal is a phase of matter that is found in cerwhere the latter equality results from Equation (18). The
tain
rod-like molecules. The mechanical properties of liqexperiments take usually much longer than the time scales
uid
crystals
are similar to liquids and the locations of the
describing the movements of the atoms, so they measure
rods
are
random,
but especially the orientation of the rods
automatically thermal averages. We obtain
displays
long-range
order.
Z
D
E
1 X
iq·(r1 −r2 )
S(q) =
dr1 dr2 e
δ(r1 − Rl )δ(r2 − Rl0 )
Nematics
N 0
ll
Z
Nematic liquid crystal has a random distribution for the
1
= 1+
dr1 dr2 n(r1 , r2 ; 0)eiq·(r1 −r2 )
centers
of its molecular rods. The orientation has, neverN
theless,
long-range order. The order parameter is usually
V
= 1 + n2 (q),
(38)
defined
in
terms of quadrupole moment
N
D
E
where
O = 3 cos2 θ − 1 ,
Z
0
1
n2 (q) =
drdr0 n2 (r + r0 , r; 0)eiq·r
(39) where θ is the deviation from the optic axis pointing in the
V
direction of the average direction of the molecular axes.
The average is calculated over space and time. One cannot
and V is the volume of the system.
use the dipole moment because its average vanishes, when
We see that the sharp peaks in the scattering experiment one assumes that the molecules point ’up’ and ’down’ rancorrespond to strong peaks in the Fourier spectrum of the domly. The defined order parameter is practical because it
correlation function n2 (r1 , r2 ; 0). When one wants to define obtains the value O = 1 when the molecules point exactly
the order parameter OK that separates the solid and liquid to the same direction (solid). Typical liquid crystal has
phases, it suffices to choose any reciprocal lattice vector O ∼ 0.3 . . . 0.8 and liquid O = 0.
K 6= 0 and set
OK
V
= lim
n2 (K).
N →∞ N 2
Cholesterics
(40)
Cholesteric liquid crystal is made of molecules that are
chiral, causing a slow rotation in the direction n̂ of the
molecules in the liquid crystal
In the solid phase, the positions Rl of the atoms are located at the lattice points, and because K belongs to the
nx = 0
reciprocal lattice, the Fourier transformation (39) of the
correlation function is of the order N (N − 1)/V. Thus,
ny = cos(q0 x)
the order parameter OK ≈ 1. The thermal fluctuations of
nz = sin(q0 x).
the atoms can be large, as long as the correlation function
preserves the lattice symmetry.
Here the wave length of the rotation λ = 2π/q0 is much
The particle locations Rl are random in liquids and the larger than the size of the molecules. In addition, it can
integral vanishes. This is in fact the definition of a solid! vary rapidly as a function of the temperature.
I.e., there is sharp transition in the long-range order. It is
Smectics
worthwhile to notice that locally the surroundings of the
atoms change perpetually due to thermal fluctuations.
Smectic liquid crystals form the largest class of liquid
crystals. In addition to the orientation, they show longrange order also in one direction. They form layers that can
Glasses
be further on divided into three classes (A,B,C) in terms
Glasses typically lack the long range order, which sepa- of the direction between their mutual orientation and the
rates them from solids. On the other hand, they show a vector n̂.
short-range order similar to liquids. The glasses are different to liquids in that the atoms locked in their positions,
Quasicrystals
like someone had taken a photograph of the liquid. The
glassy phase is obtained by a fast cooling of a liquid. In
Crystals can have only 2-, 3-, 4- and 6-fold rotational
this way, the atoms in the liquid do not have the time to symmetries (cf. Exercise 1). Nevertheless, some materials
organize into a lattice, but remain disordered. This results have scattering peaks due to other, like 5-fold, rotational
in, e.g., rapid raise in viscosity. Not a single known mate- symmetries. These peaks cannot be due to periodic lattice.
rial has glass as its ground state. On the other hand, it is The explanation lies in the quasiperiodic organization of
believed that all materials can form glass if they are cooled such materials. For example, the two-dimensional plane
fast enough. For example, the cooling rate for a typical can be completely covered with two tiles (Penrose tiles),
23
resulting in aperiodic lattice, but whose every finite area
repeats infinitely many times.
3. Electronic Structure
MM, Chapter 6
A great part of condensed matter physics can be encapsulated into a Hamiltonian operator that can be written in
one line (notice that we use the cgs-units!)
Ĥ =
X P̂ 2
1 X ql ql0
+
.
2Ml
2 0 |R̂l − R̂l0 |
(41)
l6=l
l
(Source: MM) Penrose tiles. The plane can be filled by
Here, the summation runs through all electrons and nuclei
matching arrow heads with same color.
in the matter, Ml is the mass and ql the charge of lth
particle. The first term describes the kinetic energy of
the particles and the second one the Coulomb interactions
between them. It is important to notice that R̂ and P̂
are quantum mechanical operators, not classical variables.
Even though the Hamiltonian operator looks simple, its
Schrödinger equation can be solved, even numerically with
modern computers, for only 10 to 20 particles. Normally,
the macroscopic matter has on the order of 1023 particles
and, thus, the problem has to be simplified in order to be
solved in finite time.
Let us start with a study of electron states of the matter.
Consider first the states in a single atom. The positively
charge nucleus of the atom forms a Coulomb potential for
the electrons, described in the figure below. Electrons can
occupy only discrete set of energies, denoted with a, b and
c.
c
b
energia
a
(Source: ET)
When two atoms are brought together, the potential barrier between them is lowered. Even though the tunnelling
of an electron to the adjacent atom is possible, it does not
happen in state a, because the barrier is too high. This is a
significant result: the lower energy states of atoms remain
intact when they form bonds.
In state b the tunnelling of the electrons is noticeable.
These states participate to bond formation. The electrons
in states c can move freely in the molecule.
In two atom compound, every atomic energy state splits
24
in two. The energy difference between the split states grows
as the function of the strength of the tunnelling between
the atomic states. In four atom chain, every atomic state is
split in four. In a solid, many atoms are brought together,
N ∼ 1023 . Instead of discrete energies one obtains energy
bands, formed by the allowed values of energy. In between
them, there exist forbidden zones which are called the energy gaps. In some cases, the bands overlap and the gap
vanishes.
of single electron wave functions. Correspondingly, the
eigenenergy of the many electron system is a sum of single
electron energies.
In order to solve for the differential equation (43), one
must set the boundary conditions. Natural choice would
be to require that the wave function vanishes at the boundaries of the object. This is not, however, practical for the
calculations. It turns out that if the object is large enough
its bulk properties do not depend on what is happening
The electrons obey the Pauli exclusion principle: there at the boundaries (this is not a property just for a free
can be only one electron in a quantum state. Another Fermi gas). Thus, we can choose the boundary in a way
way to formulate this is to say that the wave function of that is the most convenient for analytic calculations. Typthe many fermion system has to be antisymmetric with ically, one assumes that the electron is restricted to move
respect exchange of any two fermions (for bosons it has in a cube whose volume V = L3 . The insignificance of the
to be symmetric). In the ground state of an atom, the boundary can be emphasized by choosing periodic boundelectrons fill the energy states starting from the one with ary conditions
lowest energy. This property can be used to, e.g., explain
Ψ(x1 + L, y1 , z1 , . . . zN ) = Ψ(x1 , y1 , z1 , . . . , zN )
the periodic table of elements.
Ψ(x1 , y1 + L, z1 , . . . zN ) = Ψ(x1 , y1 , z1 , . . . , zN )(44)
Similarly in solids, the electrons fill the energy bands
..
starting from the one with lowest energy. The band that is
.
only partially filled is called conduction band. The electric
conductivity of metals is based on the existence of such which assume that an electron leaving the cube on one side
a band, because filled bands do not conduct as we will returns simultaneously at the opposite side.
see later. If all bands are completely filled or empty, the
With these assumptions, the eigenfunctions of a free elecmaterial is an insulator.
tron (43) are plane waves
The electrons on the conduction band are called conduction electrons. These electrons can move quite freely
1
ψk = √ eik·r ,
through the metal. In the following, we try to study their
V
properties more closely.
where the prefactor normalizes the function. The periodic
boundaries make a restriction on the possible values of the
3.1 Free Fermi Gas
wave vector k
2π
Let us first consider the simplest model, the free Fermi
k=
(lx , ly , lz ),
(45)
gas. The Pauli exclusion principle is the only restriction
L
laid upon electrons. Despite the very raw assumptions, the where l are integers. By inserting to Schrödinger equation,
i
model works in the description of the conduction electron one obtains the eigenenergies
of some metals (simple ones, like alkali metals).
~2 k 2
The nuclei are large compared with the electrons and,
Ek0 =
.
2m
therefore, we assume that they can be taken as static particles in the time scales set by the electronic motion. Static
The vectors k form a cubic lattice (reciprocal lattice),
nuclei form the potential in which the electrons move. In
the free Fermi gas model, this potential is assumed to with a lattice constant 2π/L. Thus, the volume of the
3
be constant independent on the positions of the electrons Wigner-Seitz cell is (2π/L) . This result holds also for
(U (rl ) ≡ U0 , vector rl denote the position of the electron electrons in a periodic lattice, and also for lattice vibrations
l) and thus sets the zero of the energy. In addition, we (phonons). Therefore, the density of states, presented in
assume that the electrons do not interact with each other. the following, has also broader physical significance.
Thus, the Schrödinger equation of the Hamiltonian (41) is
Ground State of Non-Interacting Electrons
reduced in
N
As was mentioned, the wave functions of many non~2 X 2
∇l Ψ(r1 . . . rN ) = EΨ(r1 . . . rN ).
(42) interacting electrons are products of single electron wave
−
2m
l=1
functions. The Pauli principle prevents any two electrons
Because the electrons do not interact, it suffices to solve to occupy same quantum state. Due to spin, the state ψk
can have two electrons. This way, one can form the ground
the single electron Schrödinger equation
state for N electron system. First, we set two electrons into
~2 ∇2
−
ψl (r) = El ψl (r).
(43) the lowest energy state k = 0. Then, we place two elec2m
trons to each of the states having k = 2π/L. Because the
The number of conduction electrons is N , and the total energies Ek0 grow with k, the adding of electrons is done by
wave function of this many electron system is a product filling the empty states with the lowest energy.
25
Let us then define the occupation number fk of the state
k. It is 1 when the corresponding single electron state
belongs to the ground state. Otherwise it is 0. When there
are a lot of electrons, fk = 1 in the ground state for all k <
kF and fk = 0 otherwise. Fermi wave vector kF defines
a sphere into the k-space with a radius kF . The ground
state of the free Fermi gas is obtained by occupying all
states inside the Fermi sphere. The surface of the sphere,
the Fermi surface, turns out to be a cornerstone of the
modern theory of metals, as we will later see.
where the density of states
Dk = 2
1
(2π)3
and the prefactor is a consequence of including the spin
(σ).
The most important one is the energy density of states
D(E), which is practical when one deals with sums that
depend on the wave vector k only via the energy Ek
Z
X
F (Ek ) = V
dED(E)F (E).
(48)
Density of States
kσ
The calculation of thermodynamic quantities requires
The energy density of states is obtained by using the
sums of the form
properties of the delta function
X
Fk ,
Z
X
k
F (Ek ) = V
dkDk F (Ek )
kσ
where F is a function defined by the wave vectors k (45). If
Z
Z
the number of electrons N is large, then kF 2π/L, and
= V
dE dkDk δ(E − Ek )F (E)
the sums can be transformed into integrals of a continuous
Z
function Fk .
2
dkδ(E − Ek ). (49)
⇒ D(E) =
(2π)3
Results of Free Electrons
The dispersion relation for single electrons in the free
Fermi gas was
~2 k 2
Ek0 =
.
2m
Because the energy does not depend on the direction of
the wave vector, by making a transformation into spherical
coordinates in Equation (49) results in
Z
2
D(E) =
dkδ(E − Ek0 )
3
(2π)
The integrals are defined by dividing the space into
Z ∞
2
Wigner-Seitz cells, by summing the values of the function
= 4π
dkk 2 δ(E − Ek0 ).
(50)
(2π)3 0
inside the cell, and by multiplying with the volume of the
cell
Z
The change of variables k → E gives for the free Fermi gas
X 2π 3
dkFk =
Fk .
(46)
m √
L
(51)
D(E) = 3 2 2mE.
k
~ π
We obtain
Z
X
V
The number of electrons inside the Fermi surface can be
Fk =
dkFk ,
(47)
(2π)3
calculated
by using the occupation number
k
X
N =
fk
where V = L3 . Despite this result is derived for a cubic
kσ
object, one can show that it holds for arbitrary (large)
Z
2V
volumes.
=
dkfk .
(52)
(2π)3
Especially, the delta function δ should be interpreted
kq
as
Because fk = 0 when k > kF , we can use the Heaviside
step function in its place
1 k≥0
θ(k) =
0 k<0
(2π)3
δ(k − q),
V
in order that the both sides of Equation (47) result in 1.
Density of states can be defined in many different ways.
For example, in the wave vector space it is defined according to Equation (47)
Z
X
Fk = V
dkDk Fk ,
We obtain
N
=
=
kσ
26
2V
(2π)3
V kF3
.
3π 2
Z
dkθ(kF − k)
(53)
(54)
So, the Fermi wave number depends on the electron density When the electrons do not interact with each other, it
n = N/V as
is again sufficient to solve the single electron Schrödinger
kF = (3π 2 n)1/3 .
(55) equation
!
One often defines the free electron sphere
~2 ∇2
−
+ U (r) ψl (r) = El ψl (r).
(58)
2m
4π 3
V
rs = ,
3
N
By ordering the single electron energies as
where V /N is the volume per an electron.
E0 ≤ E1 ≤ E2 . . .
The energy of the highest occupied state is called the
Fermi energy
the ground state of N electron system can still be formed by
~2 kF2
.
(56) filling the lowest energies E0 . . . EN . The largest occupied
2m
energy is still called the Fermi energy.
The Fermi surface is, thus, formed by the wave vectors k
with k = kF , and whose energy is EF .
Temperature Dependence of Equilibrium
Fermi velocity is defined as
The ground state of the free Fermi gas presented above
~kF
is the equilibrium state only when the temperature T = 0.
.
vF =
m
Due to thermal motion, the electrons move faster which
leads to the occupation of the states outside the Fermi
Density of States at Fermi Surface
surface. One of the basic results of the statistic physics is
Almost all electronic transport phenomena, like heat the Fermi-Dirac distribution
conduction and response to electric field, are dependent
1
.
(59)
f (E) = β(E−µ)
on the density of states at the Fermi surface D(EF ). The
e
+1
states deep inside the surface are all occupied and, therefore, cannot react on disturbances by changing their states. It gives the occupation probability of the state with energy
The states above the surface are unoccupied at low tem- E at temperature T . In the above, β = 1/kB T and µ is the
peratures and cannot, thus, explain the phenomena due to chemical potential, which describes the change in energy in
external fields. This means that the density of states at the the system required when one adds one particle and keeps
Fermi surface is the relevant quantity, and for free Fermi the volume and entropy unchanged.
gas it is
3n
D(EF ) =
2EF
EF =
1- and 2- Dimensional Formulae
When the number of dimensions is d, one can define the
density of states as
1 d
Dk = 2
.
2π
(Source: MM) Fermi-Dirac probabilities at different
temperatures. The gas of non-interacting electrons is at
classical limit when the occupation probability obeys the
Boltzmann distribution
In two dimensions, the energy density of states is
D(E) =
m
π~2
and in one dimension
r
D(E) =
f (E) = Ce−βE .
2m
.
π 2 ~2 E
This occurs when
f (E) 1 ⇒ eβ(E−µ) 1.
General Ground State
Let us assume for a moment that the potential due to
nuclei is not a constant, but some position dependent (e.g.
periodic) function U (r). Then, the Schrödinger equation
of the system is
!
N
X
~2 2
∇ + U (rl ) Ψ(r1 . . . rN ) = EΨ(r1 . . . rN ).
−
2m l
l=1
(57)
27
Due to spin, every energy state can be occupied by two
electrons at maximum. When this occurs the state is degenerate. At the classical limit, the occupation of every
energy state is far from double-fold and, thus, the electrons are referred to as non-degenerate. According to the
above condition, this occurs when kB T µ. When the
temperature drops (or the chemical potential µ, i.e. the
density grows) the energy states start degenerate starting
where f (E) is the Fermi-Dirac distribution. When one assumes that the integrand vanishes at the infinity, we obtain
by partial integration
"Z
#
Z ∞
E
h ∂f i
0
0
dE H(E )
hHi =
dE
,
∂µ
−∞
−∞
from the lowest. Also, the quantum mechanical phenomena begin to reveal themselves.
When the temperature T → 0,
f (E) → θ(µ − E).
In other words, the energy states smaller than the chemical
potential are degenerate, and the states with larger enerwhere −∂f /∂E = ∂f /∂µ. Having this function in the ingies are completely unoccupied. We see that at the zero
tegral is an essential part of the expansion. It contains the
temperature the equilibrium state is the ground state of
idea that only the electrons inside the distance kB T from
the free Fermi gas, and that µ = EF .
the Fermi surface are active.
It is impossible to define generally what do the ”high”
and ”low” temperatures mean. Instead, they have to be
determined separately in the system at hand. For example,
for aluminium one can assume that the three electrons of its
outmost electron shell (conduction electrons) form a Fermi
gas in the solid phase. This gives the electron density n,
together with the density of aluminium ρ = 2.7·103 kg/m3 .
Thus, we obtain an estimate for the Fermi temperature of
aluminium
EF
≈ 135000 K,
TF =
kB
which is much larger than its melting temperature. Fermi
temperature gives a ball park estimate for the energy
needed to excite the Fermi gas from ground state. In metals, the Fermi temperatures are around 10000 K or larger.
(Source: MM) We see that the function ∂f /∂µ is nonThis means that at room temperature the conduction elec- zero only in the range kB T . Thus, it is sufficient to consider
trons in metals are at very low temperature and, thus, form the integral in square brackets only in the vicinity of the
a highly degenerate Fermi gas. Therefore, only a small point E = µ. Expansion into Taylor series gives
fraction of the electrons, located near the Fermi surface, is Z
Z µ
E
active. This is the most important single fact of metals,
dE 0 H(E 0 ) ≈
dE 0 H(E 0 )
(60)
and it is not changed when the theory is expanded.
−∞
−∞
1
+ H(µ)(E − µ) + H 0 (µ)(E − µ)2 + · · · .
Sommerfeld expansion
2
Before the quantum age, in the beginning of 20th century,
there was a problem in the theory of electrons in metal.
Thomson estimated (1907) that every electron, proton and
neutron increases the specific heat of the metal with the
factor 3kB T according to the equipartition theorem. The
measured values for specific heats were only half of this
value. The correction due to quantum mechanics, and especially the Pauli principle, is presented qualitatively in
the above. Only the electrons near the Fermi surface contribute to the specific heat.
The specific heat is a thermodynamic property of matter.
In metals the melting temperatures are much smaller than
the Fermi temperature, which suggests a low temperature
expansion for thermodynamic quantities. This expansion
was derived first by Sommerfeld (1928), and it is based on
the idea that at low temperatures the energies of the thermodynamically active electrons deviate at most by kB T
from the Fermi energy.
By inserting this into the expectation value hHi, we see
that the terms with odd powers of E − µ vanish due to
symmetry, since ∂f /∂µ an even power of the same function.
We obtain
Z µ
hHi =
dEH(E)
(61)
−∞
+
π2
7π 4
[kB T ]2 H 0 (µ) +
[kB T ]4 H 000 (µ) + · · · .
6
360
The expectation value can be written also in an algebraic
form, but in practice one seldom needs terms that are beyond T 2 . The above relation is called the Sommerfeld expansion.
Specific Heat at Low Temperatures
Let us apply the Sommerfeld expansion in the determination of the specific heat due to electrons at low temperatures. The specific heat cV describes the change in the
average electron energy density E/V as a function of temFormal Derivation
perature, assuming that the number of electrons N and the
Assume that H(E) is an arbitrary (thermodynamic) volume V are kept fixed
function. In the Fermi gas, its expectation value is
Z ∞
1 ∂E cV =
.
hHi =
dEH(E)f (E),
V ∂T −∞
NV
28
According to the previous definition (48) of the energy density of states, the average energy density is
Z
E
=
dE 0 f (E 0 )E 0 D(E 0 )
V
Z µ
d µD(µ)
2
π
dE 0 E 0 D(E 0 ) +
=
(kB T )2
. (62)
6
dµ
−∞
If there are deviations in the measured specific heats, one
interpret them by saying that the matter is formed by effective particles, whose masses differ from those of electrons.
This can be done by replacing the electron mass with the
effective mass m∗ in the specific heat formula. When the
behaviour as a function of temperature is otherwise similar, this kind of change of parameters allows the use of the
simple theory. What is left to explain is the change in the
∗
The latter equality is obtained from the two first terms of mass. For example, for iron m /mel ∼ 10, ∗and for some
the Sommerfeld expansion for the function H(E) = ED(E). heavy fermion compounds (UBe13 , UPt3 ) m /mel ∼ 1000.
By making a Taylor expansion at µ = EF , we obtain
3.2 Schrödinger Equation and Symmetry
Z EF
E
MM, Chapters 7-7.2.2
dE 0 E 0 D(E 0 ) + EF D(EF ) µ − EF
=
V
−∞
The Sommerfeld theory for metals includes a fundamen
π2
(63) tal problem: How can the electrons travel through the lat(kB T )2 D(EF ) + EF D0 (EF ) .
+
6
tice without interacting with the nuclei? On the other
An estimate for the temperature dependence of µ − EF is hand, the measured values for resistances in different maobtained by calculating the number density of electrons terials show that the mean free paths of electrons are larger
than the interatomic distances in lattices. F. Bloch solved
from the Sommerfeld expansion
these problems in his Ph.D. thesis in 1928, where he showed
Z
N
=
dEf (E)D(E)
(64)that in a periodic potential the wave functions of electrons
V
deviate from the free electron plane waves only by periZ EF
π2
odic modulation factor. In addition, the electrons do not
=
(kB T )2 D0 (EF )scatter from the lattice itself, but from its impurities and
dED(E) + D(EF ) µ − EF +
6
−∞
thermal vibrations.
where the last equality is again due to Taylor expanding.
We assumed that the number of electrons and the volume
Bloch Theorem
are independent on temperature and, thus,
We will still assume that the electrons in a solid do not
0
interact
with each other, but that they experience the peπ2
D
(E
)
F
µ − EF = − (kB T )2
.
riodic potential due to nuclei
6
D(EF )
By inserting this into the energy density formula, and then
derivating, we obtain the specific heat
π2
2
cV =
D(EF )kB
T.
3
U (r + R) = U (r).
(67)
The above relation holds for all Bravais lattice vectorsR.
As before, it is sufficient to consider a single electron whose
(65)
Hamiltonian operator is
The specific heat of metals has two major components.
In the room temperature, the largest contribution comes
from the vibrations of the nuclei around some equilibrium
(we will return to these later). At low temperatures, these
lattice vibrations behave as T 3 . Because the conduction
electrons of metals form a free Fermi gas, we obtain for the
specific heat
!
π 2 kB
π2
T
cV =
nkB T =
nkB
,
(66)
2 EF
2
TF
Ĥ =
P̂ 2
+ Û (R̂)
2m
and the corresponding Schrödinger equation is
!
~2 ∇2
−
+ Û (r) ψ(r) = Eψ(r).
2m
(68)
(69)
Again, the wave function of the many electron system is
a product of single electron wave functions. One should
notice, that even though the Hamiltonian operator Ĥ is
periodic, it does not necessarily result in periodic eigenwhere n is the density of conduction electrons in metal. functions ψ(r).
At the temperatures around 1 K, one observes a linear
In order to describe a quantum mechanical system comdependence on temperature in the measured specific heats
of metals. This can be interpreted to be caused by the pletely, one has to find all independent operators that
commute with the Hamiltonian operator. One can aselectrons near the Fermi surface.
sign a quantum number for each such an operator, and
Because the Fermi energy is inversely proportional to together these number define the quantum state of the syselectron mass, the linear specific heat can be written as
tem. Thus, we will consider the translation operator
mkF 2
k T.
cV =
T̂R = e−iP̂ ·R/~ ,
3~2 B
29
where P̂ is the momentum operator and R is the Bravais
lattice vector (cf. Advanced Course in Quantum Mechanics). The translation operator shifts the argument of an
arbitrary function f (r) with a Bravais lattice vector R
Thus, we see that the periodic lattice causes modulation
in the amplitude of the plane wave solutions of the free
electrons. The period of this modulation is that of the
lattice.
T̂R f (r) = f (r + R).
Allowed Bloch Wave Vectors
Due to the periodicity of the potential, all operators T̂R
commute with the Hamiltonian operator Ĥ. Thus, they
have common eigenstates |ψi. One can show, that there
are no other essentially different operators that commute
with the Hamiltonian. Therefore, we need two quantum
numbers: n for the energy and k for the translations.
Consider the translations. We obtain
†
T̂R
|ψi = CR |ψi.
The finite size of the crystal restricts the possible values
of the Bloch wave vector k. In the case of a cubic crystal
(volume V = L3 ), we can use the previous values for the
wave vectors of the free electrons (45). The crystals are
seldom cubes, and it is more convenient to look at things
in a primitive cell of the Bravais lattice.
Let us first find the Bloch wave vectors with the help of
the reciprocal lattice vectors bi
k = x1 b1 + x2 b2 + x3 b3 .
When this is projected into position space (inner product
with the bra-vector hr|), we obtain
The coefficients xi are at this point arbitrary. Let us then
assume that the lattice is formed by M = M1 M2 M3 primitive cells. Again, if we assume that the properties of the
ψ(r + R) = CR ψ(r).
(70) bulk do not depend on the boundary conditions, we can
generalize the periodic boundary conditions leading to
On the other hand, if we calculate the projection into
ψ(r + Mi ai ) = ψ(r),
(76)
the momentum space, we get
eik·R hk|ψi = CR hk|ψi
where i = 1, 2, 3. According to the Bloch theorem, we have
for all i = 1, 2, 3
⇒ joko CR = eik·R tai hk|ψi = 0.
ψnk (r + Mi ai ) = eiMi k·ai ψnk (r).
So, we see that the eigenstate |ψi overlaps only with one
eigenstate of the momentum (|ki). The vector k is called
the Bloch wave vector, and the corresponding momentum
~k the crystal momentum. The Bloch wave vector is used
to index the eigenstates ψk .
Because bl · al0 = 2πδll0 , then
e2πiMi xi = 1
and
mi
For a fixed value of a Bloch wave vector k, one has many
,
xi =
Mi
possible energy eigenvalues. Those are denoted in the following with the band index n. Thus, the periodicity has where mi are integers. Thus, the general form of the alallowed the classification of the eigenstates
lowed Bloch wave vectors is
Ĥ|ψnk i = Enk |ψnk i
†
T̂R
|ψnk i = eik·R |ψnk i.
(71)
(72)
k=
3
X
ml
bl .
Ml
(77)
l=1
The latter equation is called the Bloch theorem. Let us We can make the restriction 0 ≤ ml < Ml , because we
study that and return later to the determining of the energy are denoting the eigenvalues (72) of the operator T̂R . By
quantum number.
doing this, the wave vectors differing by a reciprocal lattice
The Bloch theorem is commonly represented in two al- vector have the same eigenvalue (exp(iK·R) = 1), and they
ternative forms. According to Equation (70), one obtains can be thought to be physically identical. In addition, we
obtain that the eigenstates and eigenvalues of the periodic
ψnk (r + R) = eik·R ψnk (r).
(73) Hamiltonian operator (68) are periodic in the reciprocal
lattice
On the other hand, if we define
ψnk+K (r) = ψnk (r)
(78)
unk (r) = e−ik·r ψnk (r),
(74)
E
= E .
(79)
nk+K
we see that u is periodic
u(r + R) = u(r)
and
ψnk (r) = eik·r unk (r).
nk
We have, thus, obtained that in a primitive cell in the
reciprocal space we have M1 M2 M3 different states, which
is also the number of the lattice points in the whole crystal.
In other words, we have obtained a very practical and gen(75) eral result: The number of the physically different Bloch
30
wave vectors is the same as the number of the lattice points
in the crystal.
Calculation of Eigenstates
It turns out that due to the periodicity, it is enough to
solve the Schrödinger equation in only one primitive cell
with boundary conditions
Brillouin Zone
An arbitrary primitive cell in the reciprocal space is not
unique, and does not necessarily reflect the symmetry of
the whole crystal. Both of these properties are obtained by
choosing the Wigner-Seitz cell of the origin as the primitive
cell. It is called the (first) Brillouin zone.
ψnk (r + R)
=
eik·R ψnk (r)
n̂ · ∇ψnk (r + R)
=
eik·R n̂ · ∇ψnk (r).
(82)
This saves the computational resources by a factor 1023 .
Uniqueness of Translation States
Crystal Momentum
Because ψnk+K = ψnk , the wave functions can be inThe crystal momentum ~k is not the momentum of an
dexed in several ways:
electron moving in the periodic lattice. This is a consequence of the fact that the eigenstates ψnk are not eigen1) Reduced zone scheme Restrict to the first Brillouin
states of the momentum operator p̂ = −i~∇
zone. Then, for each Bloch wave vector k exists an
infinite number of energies, denoted with the index n.
− i~∇ψnk = −i~∇ eik·r unk (r)
2) Extended zone scheme The wave vector k has values
= ~k − i~eik·r ∇unk (r).
(80)
from the entire reciprocal space. We abandon the index n and denote ψnk → ψk+Kn .
In the following, we will see some similarities with the
momentum p, especially when we study the response of
3) Repeated zone scheme Allow the whole reciprocal
the Bloch electrons to external electric field. For now, the
space and keep the index n.
Bloch wave vector k has to be thought merely as the quantum number describing the translational symmetry of the In the first two cases, we obtain a complete and linearly
periodic potential.
independent set of wave functions. In the third option,
each eigenstate is repeated infinitely many times.
Eigenvalues of Energy
We showed in the above that the eigenvalues of the translation operator T̂R are exp(ik·R). For each quantum number k of the translation operator, there are several eigenvalues of energy. Insert the Bloch wave function (75) into the
Schrödinger equation, which leads to an eigenvalue equation for the periodic function unk (r + R) = unk (r)
Ĥk unk =
2
~2 − i∇ + k unk + Û unk = Enk unk . (81)
2m
Because u is periodic, we can restrict the eigenvalue equation into a primitive cell of the crystal. In a finite volume,
we obtain an infinite and discrete set of eigenvalues, that
we have already denoted with the index n.
It is worthwhile to notice, that although the Bloch wave
vectors k obtain only discrete values (77), the eigenenergies
(Source: MM) The classification of the free electron kEnk = En (k) are continuous functions of the variable k.
This is because k appears in the Schrödinger equation (81) states in one dimensional space. The energies are of the
form
only as a parameter.
~2 (k + nK)2
0
E
=
,
nk
Consider a fixed energy quantum number n. Because
2m
the energies are continuous and periodic in the reciprocal where K is a primitive vector of the reciprocal lattice.
lattice (En (k + K) = En (k)), they form a bound set which
is called the energy band. The energy bands Enk determine
Density of States
whether the material is a metal, semiconductor or an insuAs in the case of free electrons, one sometimes need to
lator. Their slopes give the velocities of the electrons, that
calculate sums over wave vectors k. We will need the volcan be used in the explaining of the transport phenomena.
ume per a wave vector in a Brillouin zone, so that we can
In addition, one can calculate the minimum energies of the
transform the summations into integrals. We obtain
crystals, and even the magnetic properties, from the shapes
of the energy bands. We will return to some of these later
b1 · (b2 × b3 )
(2π)3
=
.
.
.
=
.
in the course.
M1 M2 M3
V
31
Thus, the sums over the Brillouin zone can be transformed
Z
X
Fk = V
dkDk Fk ,
By generalizing the previous results for the one dimensional
sumΣq , we obtain the relation
X
X
e−iq·R = N
δqK .
kσ
R
K
where the density of states Dk = 2/(2π)3 , similar to free Thus, we can write
electrons. One should notice, that even though the denZ
X
sities of states are the same, one calculates the sums over
dre−iq·r U (r) = V
δqK UK ,
(86)
the vectors (77) in the Brillouin zone of the periodic latK
tice but for free electrons they are counted over the whole
where V = N Ω. The inverse Fourier transform gives
reciprocal space.
X
Correspondingly, one obtains the energy density of states
U (r) =
eiK·r UK ,
(87)
Z
K
2
dkδ(E − Enk ).
Dn (E) =
(2π)3
where the sum is calculated over the whole reciprocal lattice, not just over the first Brillouin zone.
Earlier, we stated that the eigenstate of the Hamiltonian
operator (68) is not necessarily periodic. Nevertheless, we
can write, by employing the periodicity of the function u,
Energy Bands and Electron Velocity
Later in the course, we will show that the electron in the
energy band Enk has a non-zero average velocity
vnk
1
= ∇k Enk .
~
ψnk (r)
(83)
eik·r unk (r)
X
=
(unk )K ei(k+K)·r .
=
(88)
K
This is a very interesting result. It shows that an electron
in a periodic potential has time-independent energy states,
in which the electron moves forever with the same average
velocity. This occurs regardless of, but rather due to, the
interactions between the electron and the lattice. This is
in correspondence with the definition of group velocity v =
∂ω/∂k that is generally defined for the solutions of the
wave equation.
Bloch Theorem in Fourier Space
Thus, we see that the Bloch state is a linear superposition of the eigenstates of the momentum with eigenvalues
~(k + K). In other words, the periodic potential mixes the
momentum states that differ by a reciprocal lattice vector ~K. This was seen already before, when we studied
scattering from a periodic crystal.
Let us study this more formally. The wave function can
be presented in the V as linear combination of plane waves
satisfying the periodic boundary condition (76)
Let us assume that the function U (r) is periodic in the
Bravais lattice, i.e.
U (r + R) = U (r)
for all vectors R in the lattice. A periodic function can
always be represented as a Fourier series. Due to the periodicity, each Fourier component has to fulfil
eik·(r+R) = eik·r .
Thus, the only non-zero components are obtained when
k = K is taken from the reciprocal lattice.
ψ(r) =
1 X
ψ(q)eiq·r .
V q
Then, the Schrödinger equation for the Hamiltonian operator (68) can be written in the form
1 X
0
0 =
Ĥ − E
ψ(q0 )eiq ·r
V 0
q
h
i
0
1 X 0
=
Eq0 − E + U (r) ψ(q0 )eiq ·r
V 0
q
i
X
0
0
1 Xh 0
0
=
(Eq0 − E)eiq ·r +
ei(K+q )·r UK ψ(q(89)
).
V 0
q
K
Consider the same property in a little more formal manner. We count the Fourier transform of the function U (r)
Next, we will multiply the both sides of the equation
with the factor exp(−iq · r) and integrate over the position
Z
Z
X
dre−iq·r U (r) =
dre−iq·R U (r + R)e−iq·r space
unitcell
i
R
X
X Z dr h
0
0
X
0 =
(Eq00 − E)ei(q −q)·r +
ei(K+q −q)·r UK ψ(q0 )
−iq·R
= Ω
e
Uq ,
(84)
V
q0
K
R
i
Xh
X
0
=
(Eq0 − E)δq0 q +
δK+q0 −q,0 UK ψ(q0 )
where Ω is the volume of the unit cell and
q0
K
Z
X
0
1
UK ψ(q − K).
(90)
Uq ≡
dre−iq·r U (r).
(85) ⇒ 0 = (Eq − E)ψ(q) +
Ω unitcell
K
32
We have used the property
Z
dreiq·r = V δq0 .
Such points are called the degeneracy points.
When we consider that k belongs to the first Brillouin zone, we have obtained an equation that couples the
Fourier components of the wave function that differ by a
reciprocal lattice vector. Because the Brillouin zone contains N vectors k, the original Schrödinger equation has
been divided in N independent problems. The solution of
each is formed by a superposition of such plane waves that
contain the wave vector k, and wave vectors that differ
from k by a reciprocal lattice vector.
We can write the Hamiltonian using the Dirac notation
A degeneracy point of a one-dimensional electron in the
X
X
Eq00 |q0 ihq0 | +
Ĥ =
UK |q0 ihq0 − K|.
(91) repeated zone scheme.
q0
q0 K
Above discussion gives us the reason to expect that
the interaction between the electron and the lattice is the
strongest at the degeneracy points. Let us consider this
Representing the Schrödinger equation in the Fourier formally by assuming, that the potential experienced by
space turns out to be one of the most beneficial tools of the electron can be taken as a weak perturbation
condensed matter physics. An equivalent point of view
with the Fourier space is to think the wave functions of
UK = ∆wK ,
(93)
electrons in a periodic lattice as a superposition of plane
waves.
where ∆ is a small dimensionless parameter. Then, we
solve the Schrödinger (90)
When we calculate hq|Ĥ − E|ψi, we obtain Equation (90).
3.3 Nearly Free and Tightly Bound Electrons
(Eq0 − E)ψ(q) +
X
UK ψ(q − K) = 0
K
MM, Chapter 8
In order to understand the behaviour of electrons in a
periodic potential, we will have to solve the Schrödinger
equation (90). In the most general case, the problem is numerical, but we can separate two limiting cases that can be
solved analytically. Later, we will study the tight binding
approximation, in which the electrons stay tightly in the
vicinity of one nucleus. However, let us first consider the
other limit, where the potential experienced by the electrons can be taken as a small perturbation.
by assuming that the solutions can be presented as polynomials of the parameter ∆
ψ(q)
=
ψ (0) (q) + ψ (1) (q)∆ + . . .
E
=
E (0) + E (1) ∆ + . . .
(94)
The Schrödinger equation should be, then, fulfilled for each
power of the parameter ∆.
Zeroth Order
We insert Equations (94) into the Schrödinger equation
(90), and study the terms that are independent on ∆
When we study the metals in the groups I, II, III and
h
i
IV of the periodic table, we observe that their conduction
(0)
0
(0)
⇒
ψ
(q)
E
−
E
= 0.
q
electrons behave as they were moving in a nearly constant
potential. These elements are often called the ”nearly free
electron” metals, because they can be described as a free In the extended zone scheme, the wave vector can have
values from the entire reciprocal lattice. In addition, for
electron gas perturbed with a weak periodic potential.
each value of the wave vector k, we have one eigenvalue of
Let us start by studying an electron moving in a weak pe- energy. Remembering that
riodic potential. When we studied scattering from a crystal
(0)
(0)
lattice, we noticed that strong scattering occurs when
T̂ † ψ (q) = eik·R ψ (q)
Nearly Free Electrons
R
k
k
k0 − k = K.
and
1 X (0)
(0)
In the above, k0 is the wave vector of the incoming ray and
ψk (r) =
ψ (q)eiq·r ,
V q k
k that of the scattered one. The vector K belongs to the
reciprocal lattice. If we consider elastic scattering, we have
we must have
that k0 = k and thus
0
Ek0 = Ek+K
.
(92)
33
(0)
(0)
(0)
ψk (q) = δkq ⇒ ψk (r) = eik·r ⇒ Ek = Ek0 .
Correspondingly in the reduced zone scheme, the wave By calculating the projections to the states |ψ1 i and |ψ2 i,
vector k is restricted into the first Brillouin zone, and thus we obtain
X
the wave function ψ needs the index n for the energy. We
hψi |(Ĥ − E)|ψj icj = 0.
obtain
j
(0)
(0)
(0)
0
This
is a linear group of equations that has a solution when
ψnk (q) = δk+Kn ,q ⇒ ψnk (r) = ei(k+Kn )·r ⇒ Enk = Ek+K
.
n
the determinant of the coefficient matrix
In the above, the reciprocal lattice vector Kn is chose so
ef f
Ĥij
= hψj |(Ĥ − E)|ψj i
(96)
that q − Kn is in the first Brillouin zone.
is zero. A text-book example from this kind of a situation
is the splitting of the (degenerate) spectral line of an atom
Consider the terms linear in ∆ in the perturbation theory in external electric or magnetic field (Stark/Zeeman effect).
h
i
X
We will now study the disappearance of the degeneracy of
(0)
(1)
(0)
(1) (0)
Eq0 − Ek ψk (q) +
wK ψk (q − K) − Ek ψk (q) = 0. an electron due to a weak periodic potential.
First Order
K
In this case, the degenerate states are |ψ1 i = |ki and
|ψ2 i = |k + Ki. According to Equation (91), we obtain the
matrix representation
(1)
Ek = w0 .
hk|Ĥ − E|ki
hk|Ĥ − E|k + Ki
By using this, we obtain the first order correction for the
hk + K|Ĥ − E|ki hk + K|Ĥ − E|k + Ki
wave function
0
U−K
Ek + U0 − E
X
δk,q−K
=
.
(97)
(1)
0
UK
Ek+K
+ U0 − E
ψk (q) =
wK 0
Ek − E0k+K
When q = k, according to the zeroth order calculation
K6=0
∗
We have used the relation U−K = UK
, that follows from
Equation
(85),
and
the
fact
that
the
potential
U (r) is real.
(95)
⇒ ψk (q) ≈ δqk +
When
we
set
the
determinant
zero,
we
obtain
as a result
K6=0
of straightforward algebra
Equation (95) is extremely useful in many calculations,
s
0
0
in which the potential can be considered as a weak pertur(Ek0 − Ek+K
)2
Ek0 + Ek+K
±
+ |UK |2 . (98)
E = U0 +
bation. The region where the perturbation theory does not
2
4
work is, also, very interesting. As one can see from Equation (95), the perturbative expansion does not converge
0
= Ek0 ), we obtain
At the degeneracy point (Ek+K
when
0
Ek0 = Ek+K
.
E = Ek0 + U0 ± |UK |.
The perturbation theory, thus, shows that the interaction between an electron and the periodic potential is the Thus, we see that the degeneracy is lifted due to the weak
strongest at the degeneracy points, as was noted already periodic potential and there exists an energy gap between
when we studied scattering. Generally speaking, the states the energy bands
with the same energy mix strongly when they are perEg = 2|UK |.
(99)
turbed and, thus, one has to choose their superposition as
the initial state and use the degenerate perturbation theory.
X
δk,q−K
.
UK 0
0
Ek − Ek+K
Degenerate Perturbation Theory
Study two eigenstates ψ1 and ψ2 of the unperturbed
Hamiltonian Ĥ0 (= P̂ 2 /2m in our case), that have (nearly)
same energies E1 ∼ E2 (following considerations can be generalized also for larger number of degenerate states). These
states span a subspace
ψ = c1 ψ1 + c2 ψ2 ,
whose vectors are degenerate eigenstates of the Hamiltonian Ĥ0 , when E1 = E2 . Degenerate perturbation theory
diagonalizes the Hamiltonian operator Ĥ = Ĥ0 + Û (R̂) in
this subspace.
Write the Schrödinger equation into the form
Ĥ|ψi = E|ψi ⇒ (Ĥ − E)|ψi.
34
One-Dimensional Case
Assume that the electrons are in one-dimensional lattice
whose lattice constant is a. Thus, the reciprocal lattice
vectors are of form K = n2π/a. Equation (92) is fulfilled
at points k = nπ/a.
sity of states of d-dimensional system is
Z
2
dkδ(E − Ek )
D(E) =
(2π)d
Z
2
dΣ
=
,
(2π)d
|∇k Ek |
(101)
integrated over the energy surface E = Ek .
We will return to the van Hove singularities when we
study the thermal vibrations of the lattice.
In the extended zone scheme, we see that the periodic potential makes the energy levels discontinuous, and that the
magnitude of those can be calculated in the weak potential
case from Equation (99). When the electron is accelerated,
e.g. with a (weak) electric field, it moves along the continuous parts of these energy bands. The energy gaps Eg
prevent its access to other bands. We will return to this
later.
Brillouin Zones
When the potential is extremely weak, the energies of the
electron are almost everywhere as there were no potential
at all. Therefore it is natural to use the extended zone,
where the states are classified only with the wave vector
k. Anyhow, the energy bands are discontinuous at the
degeneracy points of the free electron. Let us study in the
following the consequences of this result.
At the degeneracy point,
0
Ek0 = Ek+K
⇒ k 2 = k 2 + 2k · K + K 2
K
.
2
We see that the points fulfilling the above equation form
a plane that is exactly at the midpoint of the origin and
the reciprocal lattice point K, perpendicular to the vector
K. Such planes are called the Bragg planes. In the scattering experiment, the strong peaks are formed when the
incoming wave vector lies in a Bragg plane.
⇒ k · K̂ = −
The reduced zone scheme is obtained by moving the energy levels in the extended zone scheme with reciprocal
lattice vectors into the first Brillouin zone.
van Hove Singularities
From the previous discussion, we see that the energy
bands Enk are continuous in the k-space and, thus, they
have extrema at the Brillouin zone boundaries (minima
and maxima) and inside the zone (saddle points). Accordingly, one can see divergences in the energy density
of states, called the van Hove singularities. For example,
in one dimension the density of states can be written as
Z
2
D(E) =
dk δ(E − Ek )
2π
Z
1 ∞ 2dEk
=
δ(E − Ek )
π 0 |dEk /dk|
2
1
=
,
(100)
π |dE /dk|
By crossing a Bragg plane we are closer to the point K
than the origin. When we go through all the reciprocal
lattice points, and draw them the corresponding planes,
we restrict a volume around the origin whose points are
closer to the origin than to any other point in the reciprocal
lattice. The Bragg planes enclose the Wigner-Seitz cell of
the origin, i.e. the first Brillouin zone.
The second Brillouin zone is obtained similarly as the
set of points (volume), to whom the origin is the second
closest reciprocal lattice point. Generally, the nth Brillouin
k
Ek =E
zone is the set of points, to whom the origin is the nth
closest reciprocal lattice point. Another way to say the
where we have used the relation Ek = E−k . According to
same, is that the nth Brillouin zone consists of the points
the previous section, the density of states diverges at the
that can be reached from the origin by crossing exactly
Brillouin zone boundary.
n − 1 Bragg planes. One can show that each Brillouin zone
Correspondingly, one can show (cf. MM), that the den- is a primitive cell, and that they all have the same volume.
35
The Fermi surface of a square lattice with weak periodic
potential in the reduced zone scheme. Generally, the Fermi
surfaces of the electrons reaching multiple Brillouin zones,
are very complicated in three dimensions in the reduced
zone scheme, even in the case of free electrons (see examples
on three-dimensional Fermi surfaces in MM!).
Six first Brillouin zones of the two-dimensional square
lattice. The Bragg planes are lines, that divide the plane
into areas, the Brillouin zones.
Effect of the Periodic Potential on Fermi Surface
As has been already mentioned, only those electrons that
are near the Fermi surface contribute to the transport phenomena. Therefore, it is important to understand how the
Fermi surface is altered by the periodic potential. In the
extended zone scheme, the Fermi surface stays almost the
same, but in the reduced zone scheme it changes drastically.
Tightly Bound Electrons
So far we have assumed that the electrons experience
the nuclei as a small perturbation, causing just slight deviations into the states of the free electrons. It is, thus, natural to study the ”opposite” picture, where the electrons
are localized in the vicinity of an atom. In this approximation, the atoms are nearly isolated from each other, and
Let us consider as an example the two-dimensional their mutual interaction is taken to be small perturbation.
square lattice (lattice constant a), and assume that each We will show in the following, that the two above menlattice point has two conduction electrons. We showed tioned limits complete each other, even though they are in
previously, that the first Brillouin zone contains the same an apparent conflict.
number of wave vectors k as there are points in the lattice.
Let us first define the Wannier functions as a superpoBecause each k- state can hold two electrons (Pauli princi- sition of the Bloch functions
ple + spin), the volume filled by electrons in the reciprocal
1 X −ik·R
lattice has to be equal to that of the Brillouin zone. The
e
ψnk (r),
(102)
wn (R, r) = √
2
2
volume of a Brillouin zone in a square lattice is 4π /a . If
N k
the electrons are assumed to be free, their Fermi surface is
a sphere with the volume
where N is the number of lattice points, and thus the number of points in the first Brillouin zone. The summation
runs through the first Brillouin zone. These functions can
4π 2
2 π
π
2
be defined for all solids, but especially for insulators they
πkF = 2 ⇒ kF = √
≈ 1.128 .
a
a
πa
are localized in the vicinity of the lattice points R.
The Wannier functions form an orthonormal set
Z
∗
drwn (R, r)wm
(R0 , r)
Z
XX 1
0
0
∗
=
dr
e−ik·R+ik ·R ψnk (r)ψmk
0 (r)
N
0
k k
1 X −ik·R+ik0 ·R0
=
e
δmn δkk0
N 0
Thus, we see that the Fermi surface is slightly outside the
Brillouin zone.
A weak periodic potential alters the Fermi surface
slightly. The energy bands are continuous in the reduced
zone scheme, and one can show that also the Fermi surface should be continuous and differentiable in the reduced
zone. Thus, we have alter the Fermi surface in the vicinity
kk
of Bragg planes. In the case of weak potential, we can show
= δRR0 δnm ,
(103)
that the constant energy surfaces (and thus the Fermi surface) are perpendicular to a Bragg plane. The basic idea is
where the Bloch wave functions ψnk are assumed to be
then to modify the Fermi surface in the vicinity of Bragg
planes. Then, the obtained surface is translated with re- orthonormal.
ciprocal lattice vectors into the first Brillouin zone.
With a straightforward calculation one can show that the
36
Bloch states can be obtained from the Wannier functions
1 X ik·R
ψnk (r) = √
e
wn (R, r).
(104)
N R
of the electrons between adjacent lattice points. The term
U describes the energy that is required to bring an electron
into a lattice point.
The eigenproblem of the tight binding Hamiltonian (108)
can
be solved analytically. We define the vector
In quantum mechanics, the global phase has no physical
significance. Thus, we can add an arbitrary phase into the
1 X ik·R
|ki = √
e
|Ri,
(109)
Bloch functions, leading to Wannier functions of form
N R
1 X −ik·R+iφ(k)
wn (R, r) = √
e
ψnk (r),
where k belongs to the first Brillouin zone. Because this is
N k
a discrete Fourier transform, we obtain the Wannier functions with the inverse transform
where φ(k) is an arbitrary real phase factor. Even though
1 X −ik·R
the phase factors do not influence on the Bloch states, they
|Ri = √
e
|ki.
(110)
alter significantly the Wannier functions. This is because
N k
the phase differences can interfere constructively and destructively, which can be seen especially in the interference
By inserting this into the tight binding Hamiltonian, we
experiments. Thus, the meaning is to optimize the choices obtain
X
of phases so, that the Wannier functions are as centred as
ĤT B =
Ek |kihk|,
(111)
possible around R.
k
where
Tight Binding Model
Ek = U + t
Let us assume that we have Wannier functions the decrease exponentially when we leave the lattice point R.
Then, it is practical to write the Hamiltonian operator in
the basis defined by the Wannier functions. Consider the
band n, and denote the Wannier function wn (R, r) with
the Hilbert space ket vector |Ri. We obtain the representation for the Hamiltonian
X
Ĥ =
|RihR|Ĥ|R0 ihR0 |.
(105)
RR0
The matrix elements
HRR0
= hR|Ĥ|R0 i
(106)
#
"
Z
2 2
~ ∇
+ U (r) wn (R0 , r)
=
drwn∗ (R, r) −
2m
X
eik·δ .
(112)
δ
If there are w nearest neighbours, the maximum value of
the energy Ek is U + |t|w and the minimum U − |t|w. Thus,
we obtain the width of the energy band
W = 2|t|w.
(113)
One can show that the localization of the Wannier function is inversely proportional to the size of the energy gap.
Therefore, this kind of treatment suits especially for insulators.
Example: 1-D lattice
In one dimension δ = ±a, and thus
Ek = U + t(eika + e−ika ) = U + 2t cos(ka).
(114)
tell how strongly the electrons at different lattice points This way we obtain the energy levels in the repeated zone
interact. We assume that the Wannier functions have very scheme.
small values when move over the nearest lattice points. In
E
other words, we consider only interactions between nearest
neighbours. Then, the matrix elements can be written as
HRR0 =
0
1 X
Enk eik·(R−R ) .
N
(107)
k
Chemists call these matrix elements the binding energies.
We see that they depend only on the differences between
the lattice vectors.
k
−2π/a −π/a
0
π/a
2π/a
Often symmetry implies, that for nearest neighbours
0
HRR0 = t = a constant. In addition, when R = R we
Example: Graphene
can denote HRR0 = U = another constant. Thus we obtain the tight binding Hamiltonian
The tight binding model presented above works for BraX
X
vais lattices. Let us then consider how the introduction of
|RithR + δ| +
|RiU hR|.
(108)
ĤT B =
basis changes things. We noted before, that graphene is
R
Rδ
ordered in hexagonal lattice form
√
In the above, δ are vectors that point from R towards its
3
1
a
=
a
nearest neighbours. The first term describes the hopping
1
2
2
37
a2 = a
√
− 12 ,
0
with a basis
vA = a
3
2
1
√
2 3
1
vB = a − 2√
3
The couplings vanish for all other vectors R. We see that
in the tight binding approximation, the electrons can jump
one nucleus to another only by changing the trigonal lattice.
0 .
In addition to the honeycomb structure, graphene can
thought to be formed of two trigonal Bravais lattices, located at points R + vA and R + vB .
In graphene, the carbon atoms form double bonds between each three nearest neighbours. Because carbon has
four valence electrons, one electron per a carbon atom is
left outside the double bonds. For each point in the Bravais lattice R, there exists two atoms at points R + vA
and R + vB . Assume also in this case, that the electrons
are localized at the lattice points R + vi , where they can
be denoted with a state vector |R + vi i. The Bloch wave
Based on the above, we can write the Schrödinger equation into the matrix form
functions can be written in form


"
#
U
t 1 + eik·a1 + eik·a2
X

 A
B
t 1 + e−ik·a1 + e−ik·a2
U
|ψk i =
A|R + vA i + B|R + vB i eik·R .
(115)
R

U
h
√
i
h
i
√
aky
t 1 + 2ei 3akx /2 cos
2
 A
B
U
aky
The meaning is to find such values for coefficients A and
t 1 + 2e−i 3akx /2 cos
2
B, that ψk fulfils the Schrödinger equation. Therefore, the
A
=
E
(124)
Hamiltonian operator is written in the basis of localized
B
states as
The solutions of the eigenvalue equation give the energies
X
Ĥ =
|R0 + vi ihR0 + vi |Ĥ|R + vj ihR + vj |, (116) in the tight binding model of graphene
ijRR0
s
√3ak x
where i, j = A, B. Here we assume that the localized states E = U ± 1 + 4 cos2 aky + 4 cos aky cos
.
2
2
2
form a complete and orthonormal basis in the Hilbert
(125)
space, similar to the case of Wannier functions.
Operate with the Hamiltonian operator to the wave function (115)
"
#
X
0
0
Ĥ|ψk i =
|R + vi ihR + vi |Ĥ A|R + vA i+B|R + vB i eik·R .
RR0 i
Next, we count the projections to the vectors |vA i and
|vB i. We obtain
i
Xh
hvA |Ĥ|ψk i =
AγAA (R) + BγAB (R) eik·R(117)
R
hvB |Ĥ|ψk i =
i
Xh
AγBA (R) + BγBB (R) eik·R(118)
,
R
where
γii (R)
=
hvi |Ĥ|R + vi i i = A, B
γij (R)
=
hvi |Ĥ|R + vj i i 6= j.
We see that in the (kx , ky )-plane, there are points where
the energy gap is zero. It turns out that the Fermi level of
graphene sets exactly to these Dirac points. In the vicinity
of the Dirac points, the dispersion relation of the energy is
(119) linear and, thus, the charge carriers in graphene are massless Dirac fermions.
(120)
3.4 Interactions of Electrons
In the tight binding approximation, we can set
MM, Chapter 9 (not 9.4)
γii (0)
= U
γAB (0) = γAB (−a1 ) = γAB (−a2 )
= t
γBA (0) = γBA (a1 ) = γBA (a2 )
= t.
(121)
Thus far, we have considered the challenging problem of
(122) condensed matter physics (Hamiltonian (41)) in the sin(123) gle electron approximation. Because the nuclei are heavy
38
compared with the electrons, we have ignored their motion
By plugging this in the Schrödinger equation of the conand considered them as static classical potentials (Born- densed matter, we obtain the Hartree equation
Oppenheimer approximation). In addition the interactions
~2 2
between electrons have been neglected, or, at most included
−
∇ ψl + [Uion (r) + UC (r)]ψl = Eψl ,
(128)
2m
into the periodic potential experienced by the electrons as
an averaged quantity.
that has to be solved by iteration. In practise, one guesses
Next, we will relax the single electron approximation and the form of the potential UC and solves the Hartree equaconsider the condensed matter Hamiltonian (41) in a more tion. Then, one recalculates the UC and solves, again, the
detailed manner. In the Born-Oppenheimer approximation Hartree equation Hartree. In an ideal case, the method
is continued until the following rounds of iteration do not
the Schrödinger equation can be written as
significantly alter the potential UC .
#
"
N
X
X
1
~2 2
Later, we will see that the averaging of the surrounding
∇ Ψ + Ze2
Ψ
ĤΨ = −
2m l
|rl − R|
electrons
is too harsh method and, as a consequence, the
R
l=1
wave
function
does not obey the Pauli principle.
2
e
1X
Ψ
+
2 i<j |ri − rj |
Variational Principle
=
EΨ.
(126)
In order to improve the Hartree equation, one should
find a formal way to derive it systematically. It can be
done with the variational principle. First, we show that
the solutions Ψ of the Schrödinger equation are extrema of
the functional
F(Ψ) = hΨ|Ĥ|Ψi
(129)
The nuclei are static at the points R of the Bravais lattice.
The number of the interacting electrons is N (in condensed
matter N ∼ 1023 ) and they are described with the wave
function
Ψ = Ψ(r1 σ1 , . . . , rN σN ),
We use the method of Lagrange’s multipliers, with a conwhere ri and σi are the place and the spin of the electron straint that the eigenstates of the Hamiltonian operator
i, respectively.
are normalized
hΨ|Ψi = 1.
The potential is formed by two terms, first of which describes the electrostatic attraction of the electron l
Uion (r) = −Ze2
X
R
The value λ of Lagrange’s multiplier is determined from
1
,
|r − R|
δ(F − λhΨ|Ψi)
=
hδΨ|(Ĥ − λ)|Ψi + hΨ|(Ĥ − λ)|δΨi
=
hδΨ|(E − λ)|Ψi + hΨ|(E − λ)|δΨi
= 0,
(130)
that is due to the nuclei. The other part of potential gives
the interaction between the electrons. It is the reason why
the Schrödinger equation of a condensed matter system is which occurs when λ = E. We† have used the hermiticity
generally difficult to solve, and can be done only when N . of the Hamiltonian operator Ĥ = Ĥ, and the Schrödinger
20. In the following, we form an approximative theory that equation Ĥ|Ψi = E|Ψi.
includes the electron-electron interactions, and yet gives
As an exercise, one can show that by choosing
(at least numerical) results in a reasonable time.
N
Y
Ψ=
ψl (rl )
(131)
Hartree Equations
l=1
The meaning is to approximate the electric field of the
surrounding electrons experienced by a single electron. The
simplest (non-trivial) approach is to assume that the other
electrons form a uniform, negative distribution of charge
with the charge density ρ. In such field, the potential energy of an electron is
Z
UC (r) =
dr0
e2 n(r)
,
|r − r0 |
(127)
where the number density of electrons is
n(r) =
ρ(r) X
|ψl (r)|2 .
=
e
l
The above summation runs through the occupied single
electron states.
39
the variational principle results in the Hartree equation.
Here, the wave functions ψi are orthonormal single electron eigenstates. The wave function above reveals why the
Hartree equation does not work. The true many electron
wave function must vanish, when two electrons occupy the
same place. In other words, the electrons have to obey the
Pauli principle. Clearly, the Hartree wave function (131)
does not have this property. Previously, we have noted
that in metals the electron occupy states whose energies
are of the order 10000 K, even in ground state! Thus, we
have to use the Fermi-Dirac distribution (not the classical Boltzmann distribution), and the quantum mechanical
properties of the electrons become relevant. Thus, we have
to modify the Hartree wave function.
Hartree-Fock Equations
Fock and Slater showed in 1930, that the Pauli principle
Expectation Value of Kinetic Energy
holds when we work in the space spanned by the antisymLet us first calculate the expectation value of the kinetic
metric wave functions. By antisymmetry, we mean that the energy
many-electron wave function has to change its sign when
two of its arguments are interchanged
hT̂ i = hΨ|T̂ |Ψi
"
#
X Z
X
Y
0
1
Ψ(r1 σ1 , . . . , ri σi , . . . , rj σj , . . . , rN σN )
dN r
=
(−1)s+s
ψs∗j (rj σj )
N! 0
= −Ψ(r1 σ1 , . . . , rj σj , . . . , ri σi , . . . , rN σN ). (132)
σ1 ...σN
j
ss
"
#
2
2
X
Y
The antisymmetricity of the wave function if the funda−~ ∇l
×
ψs0i (ri σi ) .
(135)
mental statement of the Pauli exclusion principle. The pre2m
i
l
vious notion, that the single electron cannot be degenerate,
follows instantly when one assigns ψi = ψj in the function We see that because ∇l operates on electron i = l, then
(132). This statement can be used only in the indepen- the integration over the remaining terms (i 6= l) gives, due
dent electron approximation. For example, we see that the to the orthogonality of the wave functions ψ, that s = s0
Hartree wave function (131) obeys the non-degeneracy con- and, thus,
dition, but is not antisymmetric. Thus, it cannot be the
XXZ
1 X ∗
−~2 ∇2l
true many electron wave function.
ψsl (rl σl )
ψsl (rl σl ).
hT̂ i =
drl
N! s
2m
The simplest choice for the antisymmetric wave function
σl
l
(136)
is
Ψ(r1 σ1 . . . rN σN )
1 X
(−1)s ψs1 (r1 σ1 ) · · · ψsN (rN σN )
= √
N! s
ψ1 (r1 σ1 ) ψ1 (r2 σ2 ) . . . ψ1 (rN σN )
..
..
..
= .
.
.
ψN (r1 σ1 ) ψN (r2 σ2 ) . . . ψN (rN σN )
The summation over the index s goes over all permutations of the numbers 1, . . . , N . In our case, the summation
depends only on the value of the index sl . Thus, it is practical to divide the sum in two
X
X X
→
.
(133)
,
s
j
s,sl =j
The latter sum produces only the factor (N − 1)!. Because
we integrate over the variables rl and σl , we can make the
changes of variables
where the summation runs through all permutations of the
numbers 1 . . . N . The latter form of the equation is called
rl → r , σl → σ.
the Slater determinant. We see that the wave function
(133) is not, anymore, a simple product of single electron We obtain
wave functions, but a sum of such products. Thus, the
XXZ
−~2 ∇2
1 X ∗
electrons can no longer be dealt with independently. For
ψl0 (rσ)
ψl0 (rσ). (137)
dr
hT̂ i =
N 0
2m
σ
example, by changing the index r1 , the positions of all
l
l
electrons change. In other words, the Pauli principle causes
The summation over the index l produces a factor N . By
correlations between electrons. Therefore, the spin has to
making the replacement l0 → l, we end up with
be taken into account in every single electron wave function
with a new index σ, that can obtain values ±1. If the
XZ
X
−~2 ∇2
hT̂ i = =
dr
ψl∗ (rσ)
ψl (rσ)
Hamiltonian operator does not depend explicitly on spin,
2m
σ
l
the variables can be separated (as was done in the single
"
#
N Z
2 2
electron case, when the Hamiltonian did not couple the
X
−~
∇
φl (r),
(138)
=
drφ∗l (r)
electrons)
2m
l=1
ψl (ri σi ) = φl (ri )χl (σi ),
where the spin function
δ1σi
χl (σi ) =
δ−1σi
(134) where the latter equality is obtained by using Equation
(134), and by summing over the spin.
Expectation Value of Potential Energy
spin ylös
spin alas
The expectation value of the (periodic) potential Uion
due to nuclei is obtained similarly
Hartree-Fock equations result from the expectation value
hΨ|Ĥ|Ψi
hÛion i =
N Z
X
drφ∗l (r)Uion (r)φl (r).
(139)
l=1
in the state (133), and by the application of the variational principle. Next, we will go through the laborious
but straightforward derivation.
What is left is the calculation of the expectation value of
the Coulomb interactions between the electrons Ûee . For
40
that, we will use a short-hand notation rl σl → l. We result
in
=
hÛee i
X Z
dN r
σ1 ...σN
×
Y
j
Y
hÛee i
Z
1 X
e2
=
dr1 dr2
(144)
2σ σ
|r1 − r2 |
1 2
i
Xh
×
|ψi (1)|2 |ψj (2)|2 − ψi∗ (1)ψj∗ (2)ψj (1)ψi (2) .
X 1 X e2 (−1)s+s0
N ! 0 |rl − rl0 |
0
ss
ψs∗j (j)
The terms with i = j result in zero, so we obtain
l<l
ψs0i (i).
i<j
(140)
Z
e2
1
dr1 dr2
2
|r1 − r2 |
Xh
×
|φi (r1 )|2 |φj (r2 )|2
i
=
The summations over indices l and l0 can be transferred
outside the integration. In addition by separating the electrons l and l0 from the product
Y
ψs∗j (j)ψs0i (i) = ψs∗l (l)ψs∗l0 (l0 )ψs0l (l)ψs0l0 (l0 )
(145)
i<j
i
−φ∗i (r1 )φ∗j (r2 )φj (r1 )φi (r2 )δσi σj .
Y
ψs∗j (j)ψs0i (i),
We see that the interactions between electrons produce two
j,i
terms into the expectation value of the Hamiltonian operator. The first is called the Coulomb integral, and it is
we can integrate over other electrons, leaving only two per- exactly the same as the averaged potential (127) used in
mutations s0 for each permutation s. We obtain
the Hartree equations. The other term is the so-called exchange integral, and can be interpreted as causing the inhÛee i
terchange between the positions of electrons 1 and 2. The
2
XXZ
X 1
negative sign is a consequence of the antisymmetric wave
e
(141) function.
=
drl drl0
N ! |rl − rl0 |
s
l<l0 σl σl0
h
i
Altogether, the expectation value of the Hamiltonian op× ψs∗l (l)ψs∗l0 (l0 ) ψsl (l)ψsl0 (l0 ) − ψsl0 (l)ψsl (l0 ) .
erator in state Ψ is
j,i6=l,l0
Again, the summation s goes through all permutations of
the numbers 1, . . . , N . The sums depend only on the values
of the indices sl and sl0 , so it is practical to write
X
→
s
XX
i
j
X
=
hΨ|Ĥ|Ψi
XXZ
i
Z
+
,
×
hÛee i
XXZ
i<j,σ1 σ2
According to the variational principle, we variate this expectation value with respect to all orthonormal single electron wave functions ψi∗ , keeping ψi fixed at the same time.
The constriction is now
XZ
Ei
dr1 ψi∗ (1)ψi (1).
e2
1
(142)
N (N − 1) |rl − rl0 |
ij
l<l0 σl σl0
h
i
× ψi∗ (l)ψj∗ (l0 ) ψi (l)ψj (l0 ) − ψj (l)ψi (l0 ) .
=
drl drl0
e2
(146)
|r1 − r2 |
i
X h
|ψi (1)|2 |ψj (2)|2 − ψi∗ (1)ψj∗ (2)ψj (1)ψi (2) .
dr1 dr2
s,sl =i,sl0 =j
where latter sum produces the factor (N − 2)!. Thus,
σ1
i
h
−~2 ∇2
ψi (1) + Uion (r1 )|ψi (1)|2
dr1 ψi∗ (1)
2m
X
σ1
As a result of the variation, we obtain the Hartree-Fock
equations
We integrate over the positions and spins and, thus, we
can replace
rl → r1 , σl → σ1
i
~2 ∇2
+ Uion (r) + UC (r) φi (r)
2m
Z
N
X
e2 φ∗j (r0 )φi (r0 )
−
δσi σj φj (r) dr0
|r − r0 |
j=1
h
rl0 → r2 , σl0 → σ2 .
In addition, the summation over indices l and l0 gives the
factor N (N − 1)/2. We end up with the result
=
×
−
= Ei φi (r).
hÛee i
Z
X
1 X
e2
dr1 dr2
(143)
2σ σ
|r1 − r2 |
ij
1 2
h
i
ψi∗ (1)ψj∗ (2) ψi (1)ψj (2) − ψj (1)ψi (2) .
(147)
Numerical Solution
Hartree-Fock equations are a set of complicated, coupled
and non-linear differential equations. Thus, their solution
is inevitably numerical. When there is an even number of
41
electrons, the wave functions can sometimes be divided in
two groups: spin-up and spin-down functions. The spatial
parts are the same for the spin-up and the corresponding
spin-down electrons. In these cases, it is enough to calculate the wave functions of one spin, which halves the computational time. This is called the restricted Hartree-Fock.
In the unrestricted Hartree-Fock, this assumption cannot
be made.
The basic idea is to represent the wave functions φi in
the basis of such functions that are easy to integrate. Such
functions are found usually by ”guessing”. Based on experience, the wave functions in the vicinity of the nucleus l
are of the form e−λi |r−Rl | . The integrals in the HartreeFock- equations are hard to solve. In order to reduce the
computational time, one often tries to fit the wave function
into a Gaussian
φi =
K
X
Blk γk ,
k=1
where
γl =
X
(Source: MM) Comparison between the experiments
and the Hartree-Fock calculation. The studied molecules
contain 10 electrons, and can be interpreted as simple tests
of the theory. The effects that cannot be calculated in the
Hartree-Fock approximation are called the correlation. Especially, we see that in the determining the ionization potentials and dipole moments, the correlation is more than
10 percent. Hartree-Fock does not seem to be accurate
enough for precise molecular calculations, and must be considered only directional. Chemists have developed more
accurate approximations to explain the correlation. Those
are not dealt in this course.
2
All0 e−al0 (r−Rl0 ) ,
Hartree-Fock for Jellium
l0
Jellium is a quantum mechanical model for the interacting
electrons in a solid, where the positively charged nuclei
and the coefficients All0 and al0 are chosen so, that the
are
consider as a uniformly distributed background (or as
shape of the wave function φi is as correct as possible.
a rigid jelly, with uniform charge density)
The functions γ1 , . . . , γK are not orthonormal. Because
Z
e2
N
one has to be able to generate arbitrary functions, the numdr0
= constant,
(148)
Uion (r) = −
V
|r − r0 |
ber of these functions has to be K N . In addition to
the wave functions, also the functions 1/|r1 − r2 | must be where N is the number of electrons and V the volume of the
represented in this basis. When such representations are system. This enables the focus on the phenomena due to
inserted into the Hartree-Fock equations, we obtain a non- the quantum nature and the interactions of the electrons
linear matrix equation for the coefficients Blk , whose size in a solid, without the explicit inclusion of the periodic
is K ×K. We see that the exchange and Coulomb integrals lattice.
produce the hardest work, because their size is K 4 . The
It turns out that the solutions of the Hartree-Fock equagoal is to choose the smallest possible set of functions γk .
tions
(147) for jellium are plane waves
Nevertheless, it is easy to see why the quantum chemists
use the largest part of the computational time of the worlds
eik·r
φi (r) = √ .
(149)
supercomputers!
V
The equation is solved by iteration. The main principle
This is seen by direct substitution. We assume the periodic
presented shortly:
boundaries, leading to the kinetic energy
~2 ki2
φi .
2m
1. Guess N wave functions.
The Coulomb integral cancels the ionic potential. We
2. Present the HF-equations in the basis of the functions are left with the calculation of the exchange integral
γi .
Z
N
X
e2 φ∗j (r2 )φi (r2 )
Iexc = −
δσi σj φj (r) dr2
|r − r2 |
j=1
3. Solve the so formed K × K-matrix equation.
Z
N
X
eikj ·r
dr2 ei(ki −kj )·r2
= −e2
δ σi σj √
V |r − r2 |
V
j=1
4. We obtain K new wave functions. Choose N with
Z
0
N
X
dr0 ei(ki −kj )·r
the lowest energy and return to 2. Repeat until the
, (150)
= −e2 φi (r)
δσi σj
V
r0
solution converges.
j=1
42
where in the last stage we make a change of variables r0 =
r2 − r. Because the Fourier transform of the function 1/r
is 4π/k 2 , we obtain
Iexc = −e2 φi (r)
N
1 X
4π
.
δσi σj
V j=1
|ki − kj |2
Density Functional Theory
Instead of solving the many particle wave function from
the Schrödinger equation (e.g. by using Hartree-Fock), we
can describe the system of interacting electrons exactly
with the electron density
(151)
We assume then that the states are filled up to the Fermi
wave number kF , and we change the sum into an integral.
The density of states in the case of the periodic boundary
condition is the familiar 2/(2π)3 , but the delta function
halves this value. Thus, the exchange integral gives
Z
4π
dk
2
Iexc = −e φi (r)
3 k 2 + k 2 − 2k · k
(2π)
i
|k|<kF
i
!
2
2e kF
k
= −
φi (r),
(152)
F
π
kF
where the latter equality is left as an Exercise and
#
"
1 + x
1
2
F (x) =
(1 − x ) ln + 2x
1 − x
4x
(153)
n(r)
=
hΨ|
N
X
δ(r − Rl )|Ψi
(157)
l=1
Z
=
N
dr2 . . . drN |Ψ(r, r2 , . . . , rN )|2 . (158)
This results in the density functional theory (DFT), which
is the most efficient and developable approach in the description of the electronic structure of matter. The roots
of the density functional theory lie in the Thomas-Fermi
model (cf. next section), but its theoretical foundation was
laid by P. Hohenberg and W. Kohn2 in the article published
in 1964 (Phys. Rev., 136, B864). In the article, Hohenberg
and Kohn proved two basic results of the density functional
theory.
1st H-K Theorem
is the Lindhard dielectric function.
The electron density of the ground state determines the
external
potential (up to a constant).
We see that in the case of jellium, the plane waves are
the solutions of the Hartree-Fock equations, and that the
This is a remarkable result if one recalls that two differenergy of the state i is
ent many-body problems can differ by potential and num!
ber of particles. According to the first H-K theorem, both
2e2 kF
k
~2 ki2
Ei =
−
F
.
(154) of them can be deduced from the electron density which,
2m
π
kF
therefore, solves the entire many-body problem. We prove
the claim by assuming that it is not true and showing that
this results into a contradiction. Thus, we assume that
there exists two external potentials U1 and U2 , which result
in the same density n. Let then Ĥ1 and Ĥ2 be the Hamiltonian operators corresponding to the potentials, and Ψ1
and Ψ2 their respective ground states. For simplicity, we
assume that the ground states are non-degenerate (the result can be generalized for degenerate ground states, but
that is not done here). Therefore, we can write that
E1 = hΨ1 |Ĥ1 |Ψ1 i
Apparently, the slope of the energy diverges at the Fermi
< hΨ2 |Ĥ1 |Ψ2 i
surface. This is a consequence of the long range of the
= hΨ2 |Ĥ2 |Ψ2 i + hΨ2 |Ĥ1 − Ĥ2 |Ψ2 i
Coulomb interaction between two electrons, which is inZ
h
i
versely proportional to their distance. Hartree-Fock ap=
E
+
drn(r)
U
(r)
−
U
(r)
.
(159)
2
1
2
proximation does not take into account the screening due
to other electrons, which created when the electrons beCorrespondingly, we can show starting from the ground
tween the two change their positions hiding the distant
state
of the Hamiltonian Ĥ2 , that
pair from each other.
Z
h
i
The total energy of jellium is
E2 < E1 + drn(r) U2 (r) − U1 (r) .
(160)
!#
"
2
X ~2 k 2
e kF
kl
l
E =
−
F
(155) By adding the obtained inequalities together, we obtain
2m
π
kF
l
E1 + E2 < E1 + E2 ,
(161)
"
#
3
3 e2 kF
= N EF −
,
(156) which is clearly not true. Thus, the assumption we made
5
4 π
in the beginning is false, and we have shown that the elecwhere the correction to the free electron energy is only half tron density determines the external potential (up to a conof the value give by the Hartree-Fock calculation (justifi- stant). The electrons density contains in principle the same
2 Kohn received the Nobel in chemistry in 1998 for the development
cation in the Exercises). The latter equality follows by
changing the sum into an integral.
of density functional theory.
43
information as the wave function of entire electron system. Especially, the density function describing the group
of electrons depends only on three variables (x, y, z), instead of the previous 3N .
In addition, the exchange integral caused an extra term
into the energy of jellium
3 e2 kF
3
Uexc [n] = −N
= −V
4 π
4
Based on the above result, it is easy to understand that
all energies describing the many electron system can be
represented as functionals of the electron density
3
π
!1/3
e2 n4/3 .
(167)
In the Thomas-Fermi theory, we assume that in a system
whose charge density changes slowly, the above terms can
E[n] = T [n] + Uion [n] + Uee [n].
(162) be evaluated locally and integrated over the volume. The
kinetic energy is then
Z
2nd H-K Theorem
~2 3
T
[n]
=
dr
(3π 2 )2/3 n5/3 (r).
(168)
The density n of the ground state minimizes the func2m 5
tional E[n] with a constraint
Correspondingly, the energy due to the exchange integral
Z
is
!1/3
drn(r) = N.
Z
3 3
Uexc [n] = − dr
e2 n4/3 (r).
(169)
4 π
If we can assume that for each density n we can assign a
wave function Ψ, the result is apparently true.
The energy functional can then be written as
The energy functional can be written in the form
Z
Z
Z
~2 3
(3π 2 )2/3 drn5/3 (r) + drn(r)Uion (r)
E[n] =
E[n] = drn(r)Uion (r) + FHK [n],
(163)
2m 5
Z
n(r1 )n(r2 )
e2
dr1 dr2
+
where
2
|r1 − r2 |
FHK [n] = T [n] + Uee [n].
(164)
!1/3 Z
3 3
We see that FHK does not depend on the ionic potential
−
e2 drn4/3 (r).
(170)
4
π
Uion and, thus, defines a universal functional for all systems with N particles. By finding the shape of this funcWhen the last term due to the exchange integral is ignored,
tional, one finds a solution to all many particle problems
we obtain the Thomas-Fermi energy functional. The funcwith all potentials Uion . So far this has not been done,
tional including the exchange term takes into account also
and it is likely that we never will. Instead along the years,
the Pauli principle, which results in the Thomas-Fermimany approximations have been developed, of which we
Dirac theory.
will present two.
Conventional Thomas-Fermi-Dirac equation is obtained,
Thomas-Fermi Theory
when we variate the functional (170) with respect to the
The simplest approximation resulting in an explicit form electron density n. The constriction is
Z
of the functionals F [n] and E[n] is called the Thomas-Fermi
drn(r) = N,
theory. The idea is to find the energy of the electrons as
a function of the density in a potential that is uniformly
distributed (jellium). Then, we use this functional of the and Lagrange’s multiplier for the density is the chemical
potential
density in the presence of the external potential.
δE
µ=
.
(171)
We will use the results obtained from the Hartree-Fock
δn(r)
equations for jellium, and write them as functions of the
density. By using Equation (55), we obtain the result for
The variational principle results in the Thomas-Fermithe kinetic energy functional
Dirac equation
Z
X ~2 k 2
3
~2 3
~2
e2 n(r2 )
2 2/3 2/3
= N EF = V
(3π 2 )2/3 n5/3 . (165)
T [n] =
(3π
)
n
(r)
+
U
(r)
+
dr
ion
2
2m
5
2m 5
2m
|r − r2 |
kσ
!1/3
3
The Coulomb integral due to the electron-electron interac−
e2 n1/3 (r) = µ.
(172)
tions is already in the functional form
π
Z
1
e2 n(r1 )n(r2 )
UC [n] =
dr1 dr2
.
(166) When the last term on the left-hand side is neglected, we
2
|r1 − r2 |
obtain the Thomas-Fermi equation.
In the case of jellium, this cancelled the ionic potential,
The Thomas-Fermi equation is simple to solve, but not
but now it has to be taken into account, because the ionic very precise. For example for an atom, whose atomic numpotential is not assumed as constant in the last stage.
ber is Z, it gives the energy −1.5375Z 7/3 Ry. For small
44
atoms, this is two times too large, but the error decreases as
the atomic number increases. The results of the ThomasFermi-Dirac equation are even worse. The Thomas-Fermi
theory assumes that the charge distribution is uniform locally, so it cannot predict the shell structure of the atoms.
For molecules, both theories give the opposite result to the
reality: by moving the nuclei of the molecule further apart,
one reduces the total energy.
where we have defined the exchange-correlation potential
Uexc [n] = (T [n] − TN I [n]) + (Uee [n] − UC [n]).
(176)
The potential Uexc contains the mistakes originating in the
approximation, where we use the kinetic energy of the noninteracting electrons, and deal with the interactions between the electrons classically.
The energy functional has to be variated in terms of the
Despite of its inaccuracy, the Thomas-Fermi theory is
single electron wave function ψl∗ , because we do not know
often used as a starting point of a many particle problem,
how to express the kinetic energy in terms of the density
because it is easy to solve and, then, one can deduce quickly
n. The constriction is again hψl |ψl i = 1, and we obtain
qualitative characteristics of matter.
"
~2 ∇2
Kohn-Sham Equations
−
ψl (r) +
Uion (r)
2m
The Thomas-Fermi theory has two major flaws. It ne#
Z
2
0
∂Uexc (n)
glects the fact, apparent in the Schrödinger equation, that
0 e n(r )
+
+ dr
ψl (r)
the kinetic energy of an electron is large in the regions
|r − r0 |
∂n
where its gradient is large. In addition, the correlation is
= El ψl (r).
(177)
not included. Due to this shortcomings, the results of the
Thomas-Fermi equations are too inaccurate. On the other We have obtained the Kohn-Sham equations, which are
hand, the Hartree-Fock equations are more accurate, but the same as the previous Hartree-Fock equations. The
the finding of their solution is too slow. W. Kohn and L. complicated, non-local exchange-potential in the HartreeJ. Sham presented in 1965 a crossing of the two, which is Fock has been replaced with a local and effective exchangenowadays used regularly in large numerical many particle correlation potential, which simplifies the calculations. In
calculations.
addition, we include effects, that are left out of the range
As we have seen previously, the energy functional has of the Hartree-Fock calculations (e.g. screening). One
should note, that the correspondence between the many
three terms
particle problem and the non-interacting system is exact
E[n] = T [n] + Uion [n] + Uee [n].
only when the functional Uexc is known. Unfortunately,
the
functional remains a mystery and in practise one has
The potential due to the ionic lattice is trivial
to
approximate
it somehow. Such calculations are called
Z
the
ab
initio
methods.
Uion [n] = drn(r)Uion (r).
When the functional Uexc is replaced with the exchangecorrelation
potential of the uniformly distributed electron
Kohn and Sham suggested, that the interacting system
gas,
one
obtains
the local density approximation.
of many electrons in a real potential is replace by noninteracting electrons moving in an effective Kohn-Sham
single electron potential. The idea is that the potential 3.5 Band Calculations
of the non-interacting system is chosen so, that the elecMM, Chapters 10.1 and 10.3
tron density is the same as that of the interacting system
We end this Chapter by returning to the study of the
under study. With these assumptions, the wave function
ΨN I of the non-interacting system can be written as the band structure of electrons. The band structure calculaSlater determinant (133) of the single electron wave func- tions are done always in the single electron picture, where
the effects of the other electrons are taken into account with
tions, resulting in the electron density
effective single particle potentials, i.e. pseudopotentials (cf.
N
X
last section). In this chapter, we studied simple methods to
2
n(r) =
|ψl (r)|
(173)
obtain the band structure. These can be used as a starting
l=1
point for the more precise numerical calculations. For exThe functional of the kinetic energy for the non-interacting ample, the nearly free electron model works for some metals
electrons is of the form
that have small interatomic distances with respect to the
Z
range of the electron wave functions. Correspondingly, the
~2 X
drψl∗ (r)∇2 ψl (r).
(174) tight binding model describes well the matter, whose atoms
TN I [n] = −
2m
l
are far apart from each other. Generally, the calculations
are numerical, and one has to use sophisticated approximaAdditionally, we can assume that the classical Coulomb
tions, whose development has taken decades and continues
integral produces the largest component UC [n] of the
still. In this course we do not study in detail the methods
electron-electron interactions. Thus, the energy functional
used in the band structure calculations. Nevertheless, the
can be reformed as
understanding of the band structure is important, because
E[n] = TN I [n] + Uion [n] + UC [n] + Uexc [n],
(175) it explains several properties of solids. One example is the
45
division to metals, insulators and semiconductors, in terms
of electric conductivity.
(Source: MM) In insulators the lowest energy bands are
completely occupied. When an electric field is turned on,
nothing occurs because the Pauli principle prevents the degeneracy of the single electron states. The only possibility
to obtain a net current is that the electrons tunnel from the
completely filled valence band into the empty conduction
band, over the energy gap Eg . An estimate for the probability P of the tunnelling is obtained from the Landau-Zener
formula
P ∝ e−kF Eg /eE ,
(178)
where E is the strength of the electric field. This formula
is justified in the chapter dealing with the electronic transport phenomena.
Correspondingly, the metals have one band that is only
partially filled, causing their good electric conductivity.
Namely, the electrons near the Fermi surface can make a
transition into nearby k-states, causing a shift of the electron distribution into the direction of the external field,
producing a finite net momentum and, thus, a net current.
Because the number of electron states in a Brillouin zone
is twice the number of primitive cells in a Bravais lattice
(the factor two comes from spin), the zone can be completely occupied only if there are an odd number electrons
per primitive cell in the lattice. Thus, all materials, with
an odd number of electrons in a primitive cell, are metals. However, it is worthwhile to recall that the atoms in
the columns 7A and 5A of the periodic table have an odd
number of valence electrons, but they are, nevertheless, insulators. This due to the fact that they do not form a
Bravais lattice and, therefore, they can have an even number of atoms, and thus electrons, in their primitive cells.
In addition to metals and insulators, one can form two
more classes of solids based on the filling of the energy
bands, and the resulting conduction properties. Semiconductors are insulators, whose energy gaps Eg are small
compared with the room temperature kB T . This leads
to a considerable occupation of the conduction band. Naturally, the conductivity of the semiconductors decreases
with temperature.
Semimetals are metals, with very few conduction electrons, because their Fermi surface just barely crosses the
46
Brillouin zone boundary.
4. Mechanical Properties
MM, Chapters 11.1 Introduction, 13-13.3.2,
13.4-13.4.1, 13.5 main idea
We have studied the electron structure of solids by assuming that the locations of the nuclei are known and static
(Born-Oppenheimer approximation). The energy needed
to break this structure into a gas of widely separated atoms,
is called the cohesive energy. The cohesive energy itself is
not a significant quantity, because it is not easy to determine in experiments, and it bears no relation to the
practical strength of matter (e.g. to resistance of flow and
fracture).
is a symmetric 3 × 3-matrix. Because the lowest energy
state has its minimum as a function of the variables ul , the
linear term has to vanish. The first non-zero correction is,
thus, quadratic. If we neglect the higher order corrections,
we obtain the harmonic approximation.
Periodic Boundary Conditions
Similarly as in the case of moving electrons, we require
that the nuclei obey the periodic boundary conditions. In
other words, the assumption is that every physical quantity
(including the nuclear deviations u) obtains the same value
in every nth primitive cell. With this assumption, every
point in the crystal is equal at the equilibrium, and the
The cohesive energy tells, however, the lowest energy crystal has no boundaries or surfaces. Thus, the matrix
0
l
l0
state of the matter, in other words its lattice structure. Φll
αβ can depend only on the distance R − R .
Based on that, the crystals can be divided into five classes
Thus, the classical equations of motion of the nucleus l
according to the interatomic bonds. These are the molecbecomes
X 0 0
ular, ionic, covalent, metallic and hydrogen bonds. We
Φll ul ,
(183)
M ül = −∇l E = −
had discussion about those already in the chapter on the
0
l
atomic structure.
where M is the mass of the nucleus. One should note that
0
the matrix Φll has to be symmetric in such translations
4.1 Lattice Vibrations
of the crystal, that move each nucleus with a same vector.
Born-Oppenheimer approximation is not able to explain Therefore, the energy of the crystal cannot change and
X 0
all equilibrium properties of the matter (such as the speΦll = 0
(184)
cific heat, thermal expansion and melting), the transport
0
l
phenomena (sound propagation, superconductivity), nor
its interaction with radiation (inelastic scattering, the am- Clearly, Equation (183) describes the coupled vibrations of
plitudes of the X-ray scattering and the background radi- the nuclei around their equilibria.
ation). In order to explain these, one has to relax on the
assumption that the nuclei sit still at the points R of the
Normal Modes
Bravais lattice.
The classical equations of motion are solved with the
We will derive the theory of the lattice vibrations by
trial function
l
using two weaker assumptions:
ul = eik·R −iωt ,
(185)
• The nuclei are located on average at the Bravais lattice
points R. For each nucleus, one can assign a lattice
point around which the nucleus vibrates.
• The deviations from the equilibrium R are small compared with the internuclear distances.
where is the unit vector that determines the polarisation
of the vibrations. This can be seen by inserting the trial
into Equation (183)
X 0
l0
l
M ω2 =
Φll eik·(R −R ) l0
Let us then denote the location of the nucleus l with the
vector
where
rl = Rl + ul .
(179)
=
X
(186)
l
l0
0
eik·(R −R ) Φll .
(187)
l0
The purpose is to determine the minimum of the energy of
a solid
The Fourier series Φ(k) does not depend on index l,
E = E(u1 , . . . , uN )
(180) because all physical quantities depend only on the differ0
as a function of arbitrary deviations ul of the nuclei. Ac- ence Rl − Rl . In addition, Φ(k) is symmetric and real
cording to our second assumption, we can express the cor- 3 × 3-matrix, that has three orthogonal eigenvectors for
rections to the cohesive energy Ec , due to the motions of each wave vector k. Here, those are denoted with
the nuclei, by using the Taylor expansion
kν ν = 1, 2, 3,
X
X ∂E
l
l ll0 l0
u
+
u
Φ
u
+
·
·
·
,
(181)
E = Ec +
α αβ β
and the corresponding eigenvalues with
∂ul α
α,l
α
Φ(k) =
Φ(k),
αβ,ll0
2
Φν (k) = M ωkν
.
where α, β = 1, 2, 3 and
0
Φll
αβ =
∂2E
0
∂ulα ∂ulβ
(188)
As usual for plane waves, the wave vector k determines the direction of propagation of the vibrations. In
(182)
47
an isotropic crystals, one can choose one polarization vector to point into the direction of the given wave vector k
(i.e. 1 k k), when the other two are perpendicular to the
direction of propagation (2 , 3 ⊥ k). The first vector is
called the longitudinal vibrational mode and the latter two
the transverse modes.
In anisotropic matter, this kind of choice cannot always
be made, but especially in symmetric crystals one polarization vector points almost into the direction of the wave
vector. This works especially when the vibration propagates along the symmetry axis. Thus, one can still use the
terms longitudinal and transverse, even though they work
exactly with specific directions of the wave vector k.
This simple example contains general characteristics,
that
can be seen also in more complicated cases. When
Because we used the periodic boundary condition, the
the
wave
vector k is small (k π/a), the dispersion relamatrix Φ(k) has to be conserved in the reciprocal lattice
tion
is
linear
translations
ω = c|k|
Φ(k + K) = Φ(k),
(189)
where the speed of sound
r
where K is an arbitrary vector in the reciprocal lattice. So,
K
∂ω
we can consider only such wave vectors k, that lie in the
≈a
, k → 0.
(192)
c=
∂k
M
first Brillouin zone. The lattice vibrations are waves like
the electrons, moving in the periodic potential caused by We discussed earlier that the energy of lattice has to be conthe nuclei located at the lattice points Rl . Particularly, served when all nuclei are translated with the same vector.
we can use the same wave vectors k in the classification of This can be seen as the disappearance of the frequency at
the wave functions of both the lattice vibrations and the the limit k = 0, because the large wave length vibrations
electrons.
look like uniform translations in short length scales.
The difference with electrons is created in that the first
As in discrete media in general, we start to see deviations
Brillouin zone contains all possible vibrational modes of from the linear dispersion when the wave length is of the
the Bravais lattice, when the number of the energy bands order of the interatomic distance. In addition, the (group)
of electrons has no upper limit.
velocity has to vanish at the Brillouin zone boundary, due
to condition (189).
Example: One-Dimensional Lattice
Consider a one-dimensional lattice as an example and
Vibrations of Lattice with Basis
assume, that only the adjacent atoms interact with each
The treatment above holds only for Bravais lattices.
other. This can be produced formally by defining the
When the number of atoms in a primitive cell is increased,
spring constant
the degrees of freedom in the lattice increases also. The
K = −Φl,l+1 .
same happens to the number of the normal modes. If one
assumes that there are M atoms in a primitive cell, then
In this approximation, Φll is determined by the transla- there are 3M vibrational modes. The low energy modes
tional symmetry (184) of the whole crystal. The other are linear with small wave vectors k, and they are called
0
terms in the matrix Φll are zero. Thus, the classical equa- the acoustic modes, because they can be driven with sound
waves. In acoustic vibrations, the atoms in a primitive cell
tion of motion becomes
move in same directions, i.e. they are in the same phase of
l
l+1
l
l−1
M ü = K(u
− 2u + u ).
(190) vibration.
In addition to acoustic modes, there exists modes in
which
the atoms in a primitive cell move into opposite diBy inserting the plane wave solution
rections. These are referred to as the optical modes, because they couple strongly with electromagnetic (infrared)
ul = eikla−iωt
radiation.
The changes due to the basis can be formally included
into the theory by adding the index n into the equation of
motion, in order to classify the nuclei at points determined
by the basis
X
0 0
0 0
Mn üln = −
Φlnl n ul n .
(193)
one obtains the dispersion relation for the angular frequency, i.e. its dependence on the wave vector
M ω 2 = 2K(1 − cos ka) = 4K sin2 (ka/2)
r K
(191)
⇒ ω=2
sin(ka/2).
M
l 0 n0
48
As in the case of the Bravais lattice, the solution of the
classical equation of motion can be written in the form
ln
uln = n eik·R
−iωt
.
(194)
We obtain
⇒ M n ω 2 n =
X
0
0
Φnn (k)n .
(195)
n0
Example: One-Dimensional Lattice with Basis
With large wave lengths (k π/a), we obtain
Consider as an example the familiar one-dimensional lattice with a lattice constant a, and with two alternating
atoms with masses M1 and M2 .
s
ω(k)
ω(k)
=
K
ka,
2(M1 + M2 )
1 = 1; 2 = 1 + ika/2
s
2K(M1 + M2 )
=
,
M1 M2
(200)
1 = M2 ; 2 = −M1 (1 + ika/2)
When we assume that each atom interacts only with Clearly at the limit k → 0, in the acoustic mode the atoms
its nearest neighbours we obtain, similar to the one- vibrate in phase (1 /2 = 1), and the optical modes in the
opposite phase (1 /2 = −M2 /M1 ).
dimensional Bravais lattice, the equations of motion
Example: Graphene
M1 ül1
M2 ül2
= K ul2 − 2ul1 + ul−1
2
= K ul+1
− 2ul2 + ul1 .
1
Graphene has two carbon atoms in each primitive cell.
Thus, one can expect that it has six vibrational modes.
(196)
By inserting the trial function (194) into the equations of
motion, we obtain
⇒ −ω 2 M1 1 eikla
−ω 2 M2 2 eikla
= K 2 − 21 + 2 e−ika eikla
= K 1 eika − 22 + 1 eikla(197)
This can be written in matrix form as
⇒
2
ω M1 − 2K
K(1 + eika )
K(1 + e−ika )
1
=0
2
ω 2 M2 − 2K
(198)
(Source: L. A. Falkovsky, Journal of Experimental
and Theoretical Physics 105(2), 397 (2007)) The acousThe solutions of the matrix equation can be found, again, tic branch of graphene consists of a longitudinal (LA) and
from the zeroes of the determinant of the coefficient matrix, transverse (TA) modes, and of a mode that is perpendicular to the graphene plane (ZA). Correspondingly, the opwhich are
tical mode contains the longitudinal (LO) and transverse
s
(TA) modes, and the mode perpendicular to the lattice
p
√
M1 + M2 ± M12 + M22 + 2M1 M2 cos(ka)
plane (ZO). The extrema of the energy dispersion relation
.
⇒ω= K
have been denoted in the figure with symbols Γ (k = 0),
M1 M 2
(199) M (saddle point) and K (Dirac point).
49
4.2 Quantum Mechanical Treatment of
Lattice Vibrations
Especially, the Hamiltonian operator of the harmonic oscillator simplifies to
So far, we dealt with the lattice vibrations classically.
1
1
Ĥ = ~ω ↠â +
= ~ω n̂ +
.
(206)
In the harmonic approximation, the normal modes can be
2
2
seen as a group of independent harmonic oscillators. The
†
vibrational frequencies ωkν are the same in the classical The number operator n̂ = â â describes the number of
and the quantum mechanical treatments. The difference quanta in the oscillator.
between the two is created in the vibrational amplitude,
which can have arbitrary values in according to the clasPhonons
sical mechanics. Quantum mechanics allows only discrete
We return to the lattice vibrations in a solid. As previvalues of the amplitude, which is seen as the quantization
ously, we assume that the crystal is formed by N atoms.
of the energy (the energy of the oscillator depends on the
In the second order, the Hamiltonian operator describing
amplitude as E ∝ |A|2 )
the motion of the atoms can be written as
1
.
(201)
~ωkν n +
N
X
1 X l ll0 l0
P̂l2
2
û Φ û + . . . .
(207)
+
Ĥ =
2M
2 0
The excited states of a vibrational mode are called
ll
l=1
phonons 3 , analogous to the excited states (photons) of the
electromagnetic field.
The operator ûl describes the deviation of the nucleus l
l
Similar to photons, an arbitrary number of phonons can from the equilibrium R . One can think that each atom in
occupy a given k-state. Thus, the excitations of the lattice the lattice behaves like a simple harmonic oscillator and,
vibrations can described as particles that obey the Bose- thus, the vibration of the whole lattice appears as a colEinstein statistics. It turns out that some of the properties lective excitation of these oscillators. This can be made
of matter, such as specific heat at low temperatures, de- explicit by defining the annihilation and creation operator
viate from the classically predicted values. Therefore, one of the lattice vibrations, analogous to Equations (203),
has to develop formally the theory of the lattice vibrations.
âkν
=
Simple harmonic oscillator is described by the Hamiltonian operator
â†kν
=
Harmonic Oscillator
2
Ĥ =
P̂
1
+ M ω 2 R̂2 ,
2M
2
r
r
N
i
1 X −ik·Rl ∗ h M ωkν l
1
√
e
kν
û + i
P̂ l
2~
2~M ωkν
N l=1
r
r
N
h Mω
i
1
1 X ik·Rl
kν l
√
e
kν
û − i
P̂ l .
2~
2~M ωkν
N l=1
(202)
We see that the operator â†kν is a sum of terms that excite
where R̂ is the operator (hermitian, R̂ = R̂) that describes locally the atom l. It creates a collective excited state that
the deviation from the equilibrium, and P̂ the (hermitian) is called the phonon.
operator of the corresponding canonical momentum. They
Commutation relation [ûl , P̂ l ] = i~ leads in
obey the commutation relations
[âkν , â†kν ] = 1.
(208)
[R̂, R̂] = 0 = [P̂ , P̂ ]
†
[R̂, P̂ ] = i~.
In the quantum mechanical treatment of the harmonic
oscillator, one often defines the creation and annihilation
operators
r
r
Mω
1
†
R̂ − i
P̂
â =
2~
2~M ω
r
r
Mω
1
â =
R̂ + i
P̂ .
(203)
2~
2~M ω
Accordingly to their name, the creation/annihilation operator adds/removes one quantum when it operates on an
energy eigenstate. The original operators for position and
momentum can be written in form
r
~
R̂ =
(↠+ â)
(204)
2M ω
r
~M ω †
P̂ = i
(â − â).
(205)
2
3 Greek:
φων ή (phonẽ).
50
In addition, the Hamiltonian operator (207) can be written
in the simple form (the intermediate steps are left for the
interested reader, cf. MM)
Ĥ =
X
kν
1
~ωkν â†kν âkν +
.
2
(209)
If the lattice has a basis, one has to add to the above sum an
index describing the branches. On often uses a shortened
notation
X
1
,
(210)
Ĥ =
~ωi n̂i +
2
i
where the summation runs through all wave vectors k, polarizations ν and branches.
Specific Heat of Phonons
We apply the quantum theory of the lattice vibrations
to the calculation of the specific heat of the solids. We calculated previously that the electronic contribution to the
specific heat (cf. Equation (65)) depends linearly on the
temperature T . In the beginning of the 20th century, the
problem was that one had only the prediction of the classical statistical physics at disposal, stating that the specific
heat should be kB /2 for each degree of freedom. Because
each atom in the crystal has three degrees of freedom for
both the kinetic energy and the potential energy, the specific heat should be CV = 3N kB , i.e. a constant independent on temperature. This Dulong-Petit law was clearly in
contradiction with the experimental results. The measured
values of the specific heats at low temperatures were dependent on temperature, and smaller than those predicted
by the classical physics.
Similarly as one could determine the thermal properties of
the electron gas from the electron density of states, one can
obtain information on the thermal properties of the lattice
from the density of states of the phonons. The density of
states is defined as
1 X
1 X
δ(ω − ωi ) =
δ(ω − ωkν )
D(ω) =
V i
V
kν
Z
dk X
=
δ(ω − ωkν ).
(216)
(2π)3 ν
For example, one can us the density of states and write the
specific heat (215) in form
Z ∞
Einstein presented the first quantum mechanical model
∂
~ω
dωD(ω)
.
(217)
CV = V
for the lattice vibrations. He assumed that the crystal has
β~ω
∂T e
−1
0
N harmonic oscillators with the same natural frequency ω0 .
In addition, the occupation probability of each oscillator It is illustrative to consider the specific heat in two limiting
obeys the distribution
cases.
n=
1
,
eβ~ω0 − 1
High Temperature Specific Heat
which was at that time an empirical result. With these assumptions, the average energy of the lattice at temperature
T is
3N ~ω0
E = β~ω0
.
(211)
e
−1
The specific heat tells how much the average energy
changes when temperature changes
∂E 3N (~ω0 )2 eβ~ω0 /(kB T 2 )
CV =
.
(212)
=
∂T (eβ~ω0 − 1)2
When the temperature is large (kB T ~ωmax ), we obtain the classical Dulong-Petit law
CV = 3N kB ,
which was a consequence of the equipartition theorem of
statistical physics. In the point of view of the classical
physics, it is impossible to understand results deviating
from this law.
Low Temperature Specific Heat
V
At low temperatures, the quantum properties of matter
can
create large deviations to the classical specific heat.
The specific heat derived by Einstein was a major imWhen
~ωi kB T , the occupation of the mode i is very
provement to the Dulong-Petit law. However, at low temsmall
and
its contribution to the specific heat is minor.
peratures it is also in contradiction with the experiments,
Thus,
it
is
enough to consider the modes with frequencies
giving too small values for the specific heat.
of the order
In reality, the lattice vibrates with different frequencies,
kB T
νi .
≈ 0.2 THz,
and we can write, in the spirit of the Hamiltonian operator
h
(210), the average energy in the form
when one assumes T = 10 K. The speed of sound in solids
is c = 1000m/s or more. Therefore, we obtain an estimate
X
1
E=
~ωi ni +
,
(213) for the lower bound of the wave length of the vibration
2
i
λmin = c/ν ≥ 50 Å.
where the occupation number of the mode i is obtained
from the Bose-Einstein distribution
ni =
1
eβ~ωi − 1
.
This is considerably larger than the interatomic distance
in a lattice (which is of the order of an Ångström). Thus,
we can use the long wave length dispersion relation
(214)
One should note, that unlike electrons, the phonons are
ωkν = cν (k̂)k,
(218)
bosons and, thus, do not obey the Pauli principle. Therefore, their average occupation number can be any non- where we assume that the speed of sound can depend on
negative integer. Accordingly, the denominator in the the direction of propagation.
Bose-Einstein distribution has a negative sign in front of
We obtain the density of states
one.
Z
dk X
The specific heat becomes
D(ω) =
δ(ω − cν (k̂)k)
(2π)3 ν
X
∂ni
3ω 2
.
(215)
CV =
~ωi
,
(219)
=
∂T
i
2π 2 c3
51
where the part that is independent on the frequency is
Debye temperatures are generally half of the melting
denoted with (integration is over the directions)
temperature, meaning that at the classical limit the harZ
monic approximation becomes inadequate.
1 X dΣ 1
1
=
.
(220)
c3
3 ν
4π cν (k̂)
Specific Heat: Lattice vs Electrons
Because the lattice part of the specific heat decreases as
T 3 , in metals it should at some temperature go below linear
component due to electrons. However, this occurs at very
low temperatures, because of the difference in the order
of magnitude between the Fermi and Debye temperatures.
where x = β~ω.
The specific heat due to electrons goes down like T /TF (cf.
There is large region between the absolute zero and the
Equation (66)), and the Fermi temperatures TF ≥ 10000
room temperature, where low and high temperature apK. The phonon part behaves as (T /ΘD )3 , where ΘD ≈
proximations do not hold. At these temperatures one has
100 . . . 500 K. Thus, the temperature where the specific
to use the general form (217) of the specific heat. Instead
heats of the electrons and phonons are roughly equal is
of a rigorous calculation, it is quite common to use interp
polation between the two limits.
T = ΘD ΘD /TF ≈ 10 K.
Einstein model was presented already in Equation (211),
and it can be reproduced by approximating the density of
It should also be remembered, that insulators do not
states with the function
have the electronic component in specific heat at low
3N
temperatures, because their valence bands are completely
δ(ω − ω0 ).
(222)
D(ω) =
filled. Thus, the electrons cannot absorb heat, and the speV
cific heat as only the cubic component CV ∝ T 3 due to the
Debye model approximates the density of states with the lattice.
function
3ω 2
D(ω) =
θ(ωD − ω),
(223)
4.3 Inelastic Neutron Scattering from
2π 2 c3
where the density of states obtains the low temperature Phonons
value, but is then cut in order to obtain a right number of
The shape of the dispersion relations ων (k) of the normodes. The cut is done at the Debye frequency ωD , which
mal modes can be resolved by studying energy exchange
is chosen so that the number of modes is the same as that
between external particles and phonons. The most inforof the vibrating degrees of freedom
mation can be acquired by using neutrons, because they
Z ∞
3
ωD
are charge neutral and, therefore, interact with the elec3N = V
,
(224)
dωD(ω) ⇒ n =
2
3
trons in the matter only via the weak coupling between
6π c
0
magnetic moments. The principles presented below hold
where n is the number density of the nuclei.
for large parts also for photon scattering.
With the help of the Debye frequency, one can define the
The idea is to study the absorbed/emitted energy of the
Debye temperature ΘD ,
neutron when it interacts with the lattice (inelastic scatkB ΘD = ~ωD ,
(225) tering). Due to the weak electron coupling, this is caused
by the emission/absorption of lattice phonons. By studywhere all phonons are thermally active. The specific heat
ing the directions and energies of the scattered neutrons,
becomes in this approximation
we obtain direct information on the phonon spectrum.
!3 Z
ΘD /T
x4 ex
T
Consider a neutron, with a momentum ~k and an energy
dx x
(226)
CV = 9N kB
ΘD
(e − 1)2
~2 k 2 /2mn . Assume that after the scattering the neutron
0
momentum is ~k0 . When it propagates through the lattice,
the neutron can emit/absorb energy only in quanta ~ωqν ,
where ωqν is one of the normal mode frequencies of the
lattice. Thus, due to the conservation of energy we obtain
Thus, the specific heat can be written in form
Z
4
x3
2π 2 kB
∂ 3(kB T )4 ∞
dx
T 3 , (221)
=
V
CV = V
∂T 2π 2 (c~)3 0
ex − 1
5~3 c3
~2 (k 0 )2
~2 k 2
=
± ~ωqν .
2mn
2mn
(227)
In other words, the neutron travelling through the lattice can create(+)/annihilate(-) a phonon, which is absorbed/emitted by the lattice.
For each symmetry in the Hamiltonian operator, there
exists a corresponding conservation law (Noether theorem).
For example, the empty space is symmetric in translations.
52
This property has an alternative manifestation in the conMössbauer observed in 1958, that the crystal lattice can
servation of momentum. The Bravais lattice is symmetric absorb the momentum of the photon, leaving (almost) all
only in translations by a lattice vector. Thus, one can ex- of the energy into the excitation of the nucleus. This a
pect that there exists a conservation law similar to that of consequence of the large mass of the crystal
momentum, but a little more constricted. Such was seen
~2 k 2
in the case of elastic scattering, where the condition for
Erecoil =
≈ 0.
2Mcrystal
strong scattering was
k − k0 = K.
With the spectroscopy based on the Mössbauer effect, one
can study the effects of the surroundings on the atomic
This is the conservation law of the crystal momentum. nuclei.
In elastic scattering, the crystal momentum is conserved,
provided that it is translated into the first Brillouin zone 4.5 Anharmonic Corrections
with the appropriate reciprocal lattice vector K. We will
In our model of lattice vibrations, we assume that the
assume that the conservation of crystal momentum holds
deviations
from the equilibrium positions of the nuclei are
also in inelastic scattering, i.e.
small, and that we can use the harmonic approximation
k = k0 ± q + K ⇒ k − k0 = ±q + K
(228) to describe the phenomena due to vibrations. It turns out
that the including of the anharmonic terms is essential in
Here ~q is the crystal momentum of a phonon, which is the explanations of several phenomena. We list couple of
not, however, related to any true momentum in the lattice. those:
It is just ~ times the wave vector of the phonon.
• Thermal expansion cannot be explained with a theory,
By combining conservations laws (227) and (228), we
that has only second order corrections to the cohesive
obtain
~2 (k 0 )2
~2 k 2
energy. This is because the size of the harmonic crystal
=
± ~ωk−k0 ,ν .
(229)
in equilibrium does not depend on temperature.
2mn
2mn
By measuring k and k0 , one can find out the dispersion
relations ~ων (q).
The previous treatment was based on the conservation
laws. If one wishes to include also the thermal and quantum fluctuations into the theory, one should study the matrix elements of the interaction potential. Using those, one
can derive the probabilities for emission and absorption by
using Fermi’s golden rule (cf. textbooks of quantum mechanics). This kind of treatment is not done in this course.
4.4 Mössbauer Effect
Due to the lattice, one can observe phenomena that are
not possible for free atomic nuclei. One example is the
Mössbauer effect. It turns out that the photons cannot
create excited states for free nuclei. The reason is that
when the nucleus excites part of the energy ck of the photon
is transformed into the recoil energy of the nucleus
~ck = ∆E +
~2 k 2
,
2m
due to the conservation of momentum. In the above, ∆E
is the energy difference of two energy levels in the nucleus.
If the life time of the excited state is τ , the uncertainty
principle of energy and time gives an estimate to the line
width W
~
W ≈ .
τ
In nuclear transitions,
~2 k 2
> W,
2m
and the free nuclei cannot absorb photons.
53
• Heat conductivity requires also the anharmonic terms,
because that of a completely harmonic (insulating)
crystal is infinite.
• The specific heat of a crystal does not stop at the
Dulong-Petit limit as the temperature grows, but exceeds it. This is an anharmonic effect.
In this course, we do not study in more detail the anharmonic lattice vibrations.
5. Electronic Transport Phenomena
MM, Chapters 16-16.2.1, 16.2.3, 16.3.1 (Rough
calculation), 16.4 (no detailed calculations), 1717.2.2, 17.4-17.4.6, 17.4.8, 17.5.1
One of the most remarkable achievements of the condensed matter theory is the explanation of the electric conductivity of matter. All dynamical phenomena in matter
can be explained with the interaction between the electrons
and the lattice potential. The central idea is that the energy bands Enk determine the response of the matter to
the external electromagnetic field. The first derivatives of
the energy bands give the velocities of electrons, and the
second derivatives describe the effective masses. A large
part of the modern electronics is laid on this foundation.
When the electric field has been present for a long time
(compared with the relaxation time), we obtain
v=−
τe
E.
m
Based on the above, we obtain the current density created by the presence of the electric field
j = −nev =
nτ e2
E = σE,
m
(232)
where n is the density of the conduction electrons. The
above defined electric conductivity σ, is the constant of
proportionality between the electric current and field.
The electric conductivity is presented of in terms of its
reciprocal, the resistivity ρ = σ −1 . By using the typiDrude Model
cal values for the resistivity and the density of conduction
In 1900, soon after the discovery of electron4 , P. Drude electrons, we obtain an estimate for the relaxation time
presented the first theory of electric and heat conductivity
m
(233)
τ = 2 ≈ 10−14 s.
of metals. It assumes that when the atoms condense into
ne ρ
solid, their valence electrons form a gas that is responsible
for the conduction of electricity and heat. In the following,
these are called the conduction electrons, or just electrons,
Heat Conductivity
when difference with the passive electrons near the nuclei
The Drude model gets more content when it is applied
does not have to be highlighted.
to another transport phenomenon, which can be used to
The Drude model has been borrowed from the kinetic determine the unknown τ . Consider the heat conductivtheory of gases, in which the electrons are dealt with as ity κ, which gives the energy current jE as a function of
identical solid spheres that move along straight trajecto- temperature gradient
ries until they hit each other, the nuclei, or the boundaries
jE = −κ∇T.
(234)
of the matter. These collisions are described with the relaxation time τ , which tells the average time between collisions. The largest factor that influences the relaxation time Let us look at the electrons coming to the point x. Eleccomes from the collisions between electrons and nuclei. In trons travelling along positive x-axis with the velocity vx
between the collisions, the electrons propagate freely ac- have travelled the average distance of vx τ after the previous
cording to Newton’s laws, disregarding the surrounding scattering event. Their energy is E(x − vx τ ), whence the
electrons and nuclei. This is called the independent and energy of the electrons travelling into the direction of the
free electron approximation. Later, we will see that that in negative x-axis is E(x + vx τ ). The density of electrons arorder to obtain the true picture of the conduction in met- riving from both directions is approximately n/2, because
als, one has to treat, especially, the periodic potential due half of the electrons are travelling along the positive and
half along the negative x-axis. We obtain an estimate for
to nuclei more precisely.
the energy flux
In the presence of external electromagnetic field (E, B),
n ∂E
the Drude electrons obey the equation of motion
jE =
vx E(x − vx τ ) − E(x + vx τ ) ≈ −nvx2 τ
2
∂x
v
v
(230)
mv̇ = −eE − e × B − m .
∂E
∂T
2n
1
∂T
c
τ
= −nvx2 τ
=−
mv 2 cV τ
∂T ∂x
m2 x
∂x
When the external field vanishes, we see that the move2
τ
n
3k
T
∂T
B
ment of the electrons ceases. This can be seen by assuming
= −
.
(235)
m 2 ∂x
an initial velocity v0 for the electrons and by solving the
equation of motion. The result
We have used in the above the classical estimates for the
v(t) = v0 e−t/τ
(231)
shows that the relaxation time describes the decay of any
fluctuation.
specific heat cV = ∂E/∂T = 3kB /2, and for the kinetic
energy mvx2 /2 = kB T /2.
We obtain
!2
Assume then that electrons are in a static electric field
κ
3 kB
J
⇒
=
= 1.24 · 10−20
.
(236)
E. Thus, an electron with initial velocity v0 propagates as
σT
2 e
cmK2
h
i
τe
τe
v(t) = − E + v0 + E e−t/τ .
m
m
Clearly, we see that the above discussion is not plausible
4 J. J. Thomson in 1897
(electrons have the same average velocity vx but different
54
amount of kinetic energy). The obtained result is, however,
more or less correct. It says that the heat conductivity divided by the electric conductivity and temperature is a
constant for metals (Wiedemann-Franz law ). Experimentally measured value is around 2.3 · 10−20 J/cm K2 .
wave vector k grows without a limit, the average location
of the electron is a constant. Clearly, this is not a common phenomenon in metals, because otherwise the electrons would oscillate and metals would be insulators. It
turns out, that the impurities destroy the Bloch oscillaThe improvement on the Drude model requires a little tions and, therefore, they are not seen except in some special materials.
bit of work, and we do it in small steps.
Justification of Semiclassical Equations
5.1 Dynamics of Bloch Electrons
The first step in improving the Drude model is to relax
the free electron assumption. Instead, we take into consideration the periodic potential due to the lattice. In many
cases, it is enough to handle the electrons as classical particles, with a slightly unfamiliar equations of motion. We
will first list these laws and then try to justify each of them.
Semiclassical Electron Dynamics
• Band index n is a constant of motion.
• Position of an electron in a crystal obeys the equation
of motion
1 ∂Ek
(237)
ṙ =
~ ∂k
• Wave vector of an electron obeys the equation of motion
e
~k̇ = −eE − ṙ × B.
(238)
c
We have used the notation ∂/∂k = ∇k =
(∂/∂kx , ∂/∂ky , ∂/∂kz ). Because the energies and wave
functions are periodic in the k-space, the wave vectors
k and k + K are physically equivalent.
The rigorous justification of the semiclassical equations
of motion is extremely hard. In the following, the goal
is to make the theory plausible and to give the guidelines
for the detailed derivation. The interested reader can refer
to the books of Marder and Ashcroft & Mermin, and the
references therein.
Let us first compare the semiclassical equations with the
Drude model. In the Drude model, the scattering of the
electrons was mainly from the nuclei. In the semiclassical
model, the nuclei have been taken into account explicitly
in the energy bands Ek . As a consequence, the electron in
the band n has the velocity given by Equation (237), and
in the absence of the electromagnetic field it maintains it
forever. This is a completely different result to the one obtained from the Drude model (231), according to which the
velocity should decay in the relaxation time τ , on average.
This can be taken as a manifestation of the wave nature of
the electrons. In the periodic lattice, the electron wave can
propagate without decay, due to the coherent constructive
interference of the scattered waves. The finite conductivity of metals can be interpreted to be caused by the lattice imperfections, such as impurities, missing nuclei, and
phonons.
Analogy with Free Electron Equations
Bloch Oscillations
In spite of the classical nature of the equations of motion
(237) and (238), they contain many quantum mechanical
effects. This is a consequence of the fact that Enk is periodic function of the wave vector k, and that the electron
states obey the Fermi-Dirac distribution, instead of the
classical one. As an example, let us study the semiclassical
dynamics of electrons in the tight binding model, which in
one dimension gave the energy bands (114)
Ek = −2t cos ak
In a constant electric field E, we obtain
=
−eE
(240)
⇒k
=
(241)
⇒ ṙ
=
⇒r
=
−eEt/~
aeEt 2ta
−
sin
~
~
aeEt 2t
cos
.
eE
~
v=
~k
1 ∂E
=
.
m
~ ∂k
(244)
We see that the semiclassical equation of motion of the
electron (237) is a straightforward generalization of that of
the free electron.
(239)
~k̇
The first justification is made via an analogy. Consider
first free electrons. They have the familiar dispersion relation
~2 k 2
.
E=
2m
The average energy of a free electron in the state defined
by the wave vector k can, thus, be written in the form
Also, the equation determining the evolution of the wave
vector has exactly the same form for both the free and
Bloch electrons
e
~k̇ = −eE − ṙ × B.
c
(242)
It is important to remember, that in the free electron case
~k is the momentum of the electron, whereas for Bloch
electrons it is just the crystal momentum (cf. Equation
(80)). Particularly, the rate of change (238) of the crystal momentum does not depend on the forces due to the
(243)
We see that the position of the electron oscillates in time.
These are called the Bloch oscillations. Even though the
55
lattice, so it cannot describe the total momentum of an
electron.
electron is obtained from the first derivative of the energy
band.
Average Velocity of Electron
Effective Mass
The average velocity (237) of an electron in the energy
band n, is obtained by a perturbative calculation. Assume
that we have solved Equation (81)
By calculating the second order correction, we arrive at
the result
2
~2 Ĥk0 unk (r) =
−i∇+k unk (r)+ Û unk (r) = Enk unk (r)
2m
1 ∂ 2 Enk
=
(252)
~2 ∂kα ∂kβ
1
1 X hψnk |P̂α |ψn0 k ihψn0 k |P̂β |ψnk i + c.c.
δαβ + 2
m
m 0
Enk − Enk0
with some value of the wave vector k. Let us then increase
n 6=n
the wave vector to the value k + δk, meaning that the
equation to solve is
where c.c. denotes the addition of the complex conjugate
of the previous term. The sum is calculated only over the
2
~2 index n0 .
− i∇ + k + δk u(r) + Û u(r)
2m "
#
We get an interpretation for the second order correction
2
~2 by studying the acceleration of the electron
− i∇ + k + δk · δk − 2i∇ + 2k u(r)
=
2m
X ∂hv̂α i ∂kβ
d
+ Û u(r)
hv̂α i =
,
(253)
dt
∂kβ ∂t
0
1
β
= (Ĥk + Ĥk )u(r) = Eu(r),
(245)
where v̂α is the α-component of the velocity operator (in
~2
the Cartesian coordinates α = x, y or z). This acceleration
=
δk · δk − 2i∇ + 2k
(246)
2m
can be caused by, e.g. the electric or magnetic field, and as
is considered as a small perturbation for the Hamiltonian a consequence the electrons move along the energy band.
In order the perturbative treatment to hold, the field have
Ĥk0 .
to be weak and, thus, k changes slowly enough.
According to the perturbation theory, the energy eigenBy combining the components of the velocity into a sinvalues change like
gle matrix equation, we have that
(1)
(2)
En,k+δk = Enk + Enk + Enk + . . .
(247)
d
hv̂i = ~M−1 k̇,
(254)
dt
The first order correction is of the form
where
Ĥk1
(1)
Enk
=
=
=
~2 unk δk · − i∇ + k unk
(248)
m
~2 ψnk eik·r δk · − i∇ + k e−ik·r ψnk
m
~
ψnk δk · P̂ ψnk ,
(249)
m
where we have defined the effective mass tensor
(M−1 )αβ =
1 ∂ 2 Enk
.
~2 ∂kα ∂kβ
(255)
Generally, the energy bands Enk are not isotropic, which
means that the acceleration is not perpendicular with the
wave vector k. An interesting result is also that the eigenvalues of the mass tensor can be negative. For example, in
the one-dimensional tight binding model
where the momentum operator P̂ = −i~∇. We have used
the definition of the Bloch wave function
unk = e−ik·r ψnk
Ek = −2t cos ak
and the result
and the second derivative at k = π/a is negative.
− i∇ + k e−ik·r ψnk = −ie−ik·r ∇ψnk .
According to Equation (238), the wave vector k grows
in
the opposite direction to the electric field. When the
By comparing the first order correction with the Taylor
effective
mass is negative, the electrons accelerate into the
expansion of the energy En,k+δk , we obtain
opposite direction to the wave vector. In other words, the
electrons accelerate along the electric field (not to the op∂Enk
1
=
ψnk P̂ ψnk
(250) posite direction as usual). Therefore, it is more practical
∂~k
m
= hv̂i = vnk ,
(251) to think that instead of negative mass, the electrons have
positive charge, and call them holes.
where the velocity operator v̂ = P̂ /m. Thus, we have
shown that in the state ψnk , the average velocity of an
Equation of Wave Vector
56
The detailed derivation of the equation of motion of the
Assume that the energy delivered into the system by the
wave vector (238) is an extremely difficult task. How- electric field in the unit time is small compared with all
ever, we justify it by considering an electron in a time- other energies describing the system. Then, the adiabatic
independent electric field E = −∇φ (φ is the scalar poten- theorem holds:
tial). Then, we can assume that the energy of the electron
is
System whose Hamiltonian operator changes slowly in
En (k(t)) − eφ(r(t)).
(256)
time stays in the instantaneous eigenstate of the time
independent Hamiltonian.
This energy changes with the rate
∂En
· k̇ − e∇φ · ṙ = vnk · (~k̇ − e∇φ).
∂k
The adiabatic theory implies immediately that the band
index is a constant of motion in small electric fields.
(257)
What is left to estimate is what we mean by small electric
Because the electric field is a constant, we can assume that
field.
When the Hamiltonian depends on time, one generthe total energy of the electron is conserved when it moves
ally
has
to solve the time-dependent Schrödinger equation
inside the field. Thus, the above rate should vanish. This
occurs, at least, when
∂
(260)
Ĥ(t)|Ψ(t)i = i~ |Ψ(t)i.
∂t
e
(258)
~k̇ = −eE − vnk × B.
c
In the case of the periodic potential, we can restrict to the
This is the equation of motion (238). The latter term can vicinity of the degeneracy point, where the energy gap Eg
always be added because it is perpendicular with the ve- has its smallest value. One can show that in this region,
locity vnk . What is left is to prove that this is in fact the the tunnelling probability from the adiabatic state Enk is
only possible extra term, and that the equation works also the largest, and that the electron makes transitions almost
in time-dependent fields. These are left for the interested entirely to the state nearest in energy.
reader.
In the nearly free electron approximation, the instance,
where we can truncate to just two free electron states, was
Adiabatic Theorem and Zener Tunnelling
described with the matrix (97)
0
So far, we have presented justifications for Equations
Ek (t)
Eg /2
Ĥ(t) =
,
(261)
(237) and (238). In addition to those, the semiclassical
0
Eg /2 Ek−K
(t)
theory of electrons assumes that the band index n is a constant of motion. This holds only approximatively, because where E 0 (t) are the energy bands of the free (uncoupled)
k
due to a strong electric field the electrons can tunnel be- electrons, with the time-dependence of the wave vector of
tween bands. This is called the Zener tunnelling.
the form
k = −eEt/~.
Let us consider this possibility by assuming that the matter is in a constant electric field E. Based on the course
The Hamiltonian operator has been presented in the free
of electromagnetism, we know that the electric field can be
electron
basis {|ki, |k − Ki}. The solution of the timewritten in terms of the scalar and vector potentials
dependent Schrödinger equation in this basis is of the form
1 ∂A
E=−
− ∇φ.
|Ψ(t)i = Ck (t)|ki + Ck−K (t)|k − Ki.
(262)
c ∂t
One can show that the potentials can be chosen so, that
the scalar potential vanishes. This is a consequence of the
gauge invariance of the theory of electromagnetism (cf. the
course of Classical Field Theory). When the electric field
is a constant, this means that the vector potential is of the
form
A = −cEt.
C. Zener showed in 19325 , that if the electron is initially
in the state |ki it has the finite asymptotic probability
PZ = |Ck (∞)|2 to stay in the same free electron state,
even though it passes through the degeneracy point.
From now on, we will consider only the one-dimensional
case for simplicity.
According to the course of Analytic Mechanics, a particle in the vector potential A can be described by the
Hamiltonian operator
2
e 2
1 1 P̂ + A + Û (R̂) =
P̂ − eEt + Û (R̂).
2m
c
2m
(259)
The present choice of the gauge leads to an explicit time
dependence of the Hamiltonian.
H=
5 In fact, essentially the same result was obtained independently
also in the same year by Landau, Majorana and Stückelberg.
57
In other words, the electron tunnels from the adiabatic calculation, and that they include the nuclear part of the
state to another with the probability PZ . The derivation potential. As an exercise, one can show that Equations
of the exact form of this probability requires so much work (237) and (238) follow from Hamilton’s equations of motion
that it is not done in this course. Essentially, it depends on
∂H
the shape of the energy bands of the uncoupled electrons,
ṙ =
∂p
and on the ratio between the energy gap and the electric
∂H
field strength. Marder obtains the result:
ṗ = −
.
(267)
∂r
r
3/2
4E
2m∗ g
,
(263)
PZ = exp −
3eE
~2
where m∗ is the reduced mass of the electron in the lattice
(cf. the mass tensor). For metals, the typical value for
the energy gap is 1 eV, and the maximum electric fields
are of the order E ∼ 104 V/cm. The reduced masses for
metals are of the order of that of an electron, leading to
the estimate
3
PZ ≈ e−10 ≈ 0.
5.2 Boltzmann Equation and Fermi Liquid Theory
The semiclassical equations describe the movement of a
single electron in the potential due to the periodic lattice
and the electromagnetic field. The purpose is to develop a
theory of conduction, so it is essential to form such equations that describe the movement of a large group of electrons. We will present two phenomenological models that
For semiconductors, the Zener tunnelling is a general enable the description of macroscopic experiments with a
property, because the electric fields can reach E ∼ 106 few well-chosen parameters, without the solution of the
V/cm, reduced masses can be one tenth of the mass of an Schrödinger equation.
electron and the energy gaps below 1 eV.
Boltzmann Equation
The Boltzmann equation is a semiclassical model of conductivity. It describes any group of particles, whose conWe list the assumptions that reduce the essentially quanstituents obey the semiclassical Hamilton’s equation of motum mechanical problem of the electron motion into a semition (267), and whose interaction with the electromagnetic
classical one:
field is described with Hamilton’s function (266). The
group of particles is considered as ideal non-interacting
• Electromagnetic field (E and B) changes slowly both
electron Fermi gas, taking into account the Fermi-Dirac
in time and space. Especially, if the field is periodic
statistics and the influences due to the temperature. The
with a frequency ω, and the energy splitting of two
external fields enable the flow in the gas, which is, neverbands is equal to ~ω, the field can move an electron
theless, considered to be near the thermal equilibrium.
from one band to the other by emitting a photon (cf.
the excitation of an atom).
Continuity Equation
Restrictions of Semiclassical Equations
• Magnitude E of the electric field has to be small in
order to adiabatic approximation to hold. In other
words, the probability of Zener tunnelling has to be
small, i.e.
r
eE
Eg
Eg
.
(264)
kF
EF
• Magnitude B of the magnetic field also has to be small.
Similar to the electric field, we obtain the condition
r
Eg
~ωc Eg
,
(265)
EF
We start from the continuity equation. Assume, that
g(x) is the density of the particles at point x, and that
the particles move with the velocity v(x). We study the
particle current through the surface A, perpendicular to
the direction of propagation x. The difference between the
particles flowing in and out of the volume Adx in a time
unit dt is
dN = Adxdg = Av(x)g(x)dt − Av(x + dx)g(x + dx)dt.
(268)
We obtain that the particle density at the point x changes
with the rate
∂ ∂g
=−
v(x)g(x, t) .
(269)
∂t
∂x
where ωc = eB/mc is the cyclotron frequency.
As the final statement, the semiclassical equations can be In higher dimensions, the continuity equation generalizes
derived neatly in the Hamilton formalism, if one assumes into
that the Hamilton function is of the form
X ∂ ∂g
∂
=−
vl (r)g(r, t) = −
· ṙg(r, t) , (270)
∂t
∂xl
∂r
l
H = En (p + eA/c) − eφ(r),
(266)
where r = (x1 , x2 , x3 ).
where p is the crystal momentum of the electron. This
deviates from the classical Hamiltonian by the fact that
The continuity equation holds for any group of partithe functions En are obtained from a quantum mechanical cles that can be described with the average velocity v that
58
depends on the coordinate r. We will consider, particularly, the occupation probability grk (t) of electrons, that
has values from the interval [0, 1]. It gives the probability
of occupation of the state defined by the wave vector k
at point r at time t. Generally, the rk-space is called the
phase space. In other words, the number of electrons in the
volume dr, and in the reciprocal space volume dk is
grk (t)drDk dk =
2
dkdrgrk (t).
(2π)3
(271)
the energy Ek . Thus, the relaxation time τE can be considered as a function of only energy.
Because the distribution g is a function of time t, position
r and wave vector k, its total time derivative can be written
in the form
dg
∂g
∂g
∂g
=
+ ṙ ·
+ k̇ ·
dt
∂t
∂r
∂k
g−f
= −
,
(278)
τE
By using the electron density g, one can calculate the ex- where one has used Equations (275) and (277).
pectation value of an arbitrary function Grk by calculating
The solution can be written as
Z
Z t
0
dt0
G = dkdrDk grk Grk .
(272)
grk (t) =
f (t0 )e−(t−t )/τE ,
τ
−∞ E
(279)
which can be seen by inserting it back to Equation (278).
We have used the shortened notation
The electron density can change also in the k-space, in
f (t0 ) = f r(t0 ), k(t0 ) ,
addition to the position. Thus, the continuity equation
(270) is generalized into the form
where r(t0 ) and k(t0 ) are such solutions of the semiclassical
equations of motion that at time t0 = t go through the
∂g
∂
∂
= −
· ṙg −
· k̇g
points r and k.
∂t
∂r
∂k
The above can be interpreted in the following way. Con∂
∂
= −ṙ ·
g − k̇ ·
g,
(273)
sider
an arbitrary time t0 and assume that the electrons
∂r
∂k
obey the equilibrium distribution f (t0 ). The quantity
where the latter equality is a consequence of the semiclasdt0 /τE can be interpreted as the probability of electron
sical equations of motion.
scattering in the time interval dt0 around time t0 . One
Boltzmann added to the equation a collision term
can show (cf. Ashcroft & Mermin), that in the relaxation
time approximation
dg dt0
(274)
,
f (t0 )
dt coll
τE
that describes the sudden changes in momentum, caused is the number of such electrons, that when propagated
by, e.g. the impurities or thermal fluctuations. This way without scattering in the time interval [t0 , t] reach the phase
the equation of motion of the non-equilibrium distribution space point (r, k) at t. The probability for the propagation
g of the electrons is
without scattering is exp(−(t − t0 )/τE ). Thus, the occupa
tion
number of the electrons at the phase space point (r, k)
∂g
∂
∂
dg = −ṙ ·
g(r, t) − k̇ ·
g(r, t) + ,
(275) at time t can be interpreted as a sum of all such electron
∂t
∂r
∂k
dt coll
histories that end up to the given point in the phase space
which is called the Boltzmann equation.
at the given time.
Derivation of Boltzmann Equation
By using partial integration, one obtains
Relaxation Time Approximation
The collision term (274) relaxes the electrons towards
the thermal equilibrium. Based on the previous considerations, we know that in the equilibrium, the electrons obey
(locally) the Fermi-Dirac distribution
frk =
1
eβr (Ek −µr )
+1
.
(276)
grk (t)
Z t
0
d
=
dt0 f (t0 ) 0 e−(t−t )/τE
dt
−∞
Z t
0
df (t0 )
= frk −
dt0 e−(t−t )/τE
(280)
dt0
−∞
Z t
0
∂
∂ 0
= frk −
dt0 e−(t−t )/τE ṙt0 ·
+ k̇t0 ·
f (t ),
∂r
∂k
−∞
The simplest collision term that contains the above mentioned properties is called the relaxation time approximawhere the latter term describes the deviations from the
tion
h
i
local equilibrium.
dg 1
= − grk − frk .
(277)
dt coll
τ
The partial derivatives of the equilibrium distribution
The relaxation time τ describes the decay of the electron can be written in the form
motion towards the equilibrium (cf. the Drude model).
∂f
∂f h
∇T i
=
−
∇µ
−
(E
−
µ)
(281)
Generally, τ can depend on the electron energy and the di∂r
∂E
T
rection of propagation. For simplicity, we assume in the fol∂f
∂f
=
~v.
(282)
lowing, that τ depends on the wave vector k only through
∂k
∂E
59
By using the semiclassical equations of motion, we arrive
of the mass tensor, that
Z
∂vα
2
σαβ = e τ dkDk fk
∂~kβ
Z
2
= e τ dkDk fk (M−1 )αβ .
at
Z
grk (t)
=
t
0
dt0 e−(t−t )/τE
frk −
(283)
−∞
∂f (t0 )
Ek − µ
×vk · eE + ∇µ +
∇T
.
T
dµ
When µ, T and E change slowly compared with the relaxation time τE , we obtain
grk = frk − τE vk · eE + ∇µ +
Ek − µ
∇T
T
dµ
∂f
When the lattice has a cubic symmetry, the conductivity tensor is diagonal and we obtain a result analogous
to that from the Drude model
σ=
(284)
where
When one studies the responses of solids to the electric,
magnetic and thermal fields, it is often enough to use the
Boltzmann equation in form (284).
(289)
1
1
=
∗
m
3n
Z
ne2 τ
,
m∗
dkDk fk Tr(M−1 )
(290)
(291)
2) On the other hand, we know that for metals
∂f
≈ δ(E − EF )
∂µ
Semiclassical Theory of Conduction
The Boltzmann equation (284) gives the electron distribution grk in the band n. Consider the conduction of
electricity and heat with a fixed value of the band index.
If more than one band crosses the Fermi surface, the effects
on the conductivity due to several bands have to be taken
into account by adding them together.
Electric Current
We will consider first a solid in an uniform electric field.
The electric current density j is defined as the expectation
value of the current function −evk in k-space
Z
j = −e dkDk vk grk ,
(285)
(the derivative deviates essentially from zero inside the
distance kB T from the Fermi surface, cf. Sommerfeld
expansion). Thus, by using result (101)
Z
dΣ
τE vα vβ ,
(292)
σαβ = e2
4π 3 ~v
where the integration is done over the Fermi surface
and ~v = |∂Ek /∂k|. Thus, we see that only the electrons at the Fermi surface contribute to the conductivity of metals (the above approximation for the derivative of the Fermi function cannot be made in semiconductors).
Filled Bands Do Not Conduct
where the integration is over the first Brillouin zone. In
the Drude model, the conductivity was defined as
σ=
Let us consider more closely the energy bands, whose
highest energy is lower than the Fermi energy. When the
temperature is much smaller than the Fermi temperature,
we can set
∂f
f =1 ⇒
= 0.
∂µ
Thus,
σαβ = 0.
(293)
∂j
.
∂E
Analogously, we define the general conductivity tensor
Z
∂jα
∂f
σαβ ≡
= e2 dkDk τE vα vβ
,
(286)
∂Eβ
∂µ
Therefore, filled bands do not conduct current. The reason is that in order to carry current, the velocities of the
where we have used the Boltzmann equation (284). Thus,
electrons would have to change due to the electric field. As
the current density created by the electric field is obtained
a consequence, the index k should grow, but because the
from the equation
band is completely occupied there are no empty k-states
j = σE
(287)
for the electron to move. The small probability of the Zener
In order to understand the conductivity tensor we can tunnelling, and the Pauli principle prevent the changes in
the states of electrons in filled bands.
choose from the two approaches:
Effective Mass and Holes
1) Assume that the relaxation time τE is a constant.
Assume, that the Fermi energy settles somewhere inside
Thus, one can write
the
energy band under consideration. Then, we can write
Z
Z
∂f
σαβ = −e2 τ dkDk vα
.
(288)
σαβ = e2 τ
dkDk (M−1 )αβ
(294)
∂~kβ
occupiedlevels
Z
Because v and f are periodic in k-space, we obtain
= −e2 τ
dkDk (M−1 )αβ . (295)
with partial integration, and by using definition (255)
unoccupiedlevels
60
This is a consequence of the fact that we can write f =
1 − (1 − f ), and the integral over one vanishes contributing
nothing to the conductivity.
the electric and heat current densities, j and jQ respectively, are obtained from the pair of equations
!
∇T
11
12
When the Fermi energy is near to the minimum of the
j = L G+L
−
T
band Ek , the dispersion relation of the band can be ap!
proximated quadratically
∇T
21
22
jQ = L G + L
−
.
(300)
T
~2 k 2
.
(296)
Ek = E0 +
2m∗n
By defining the electric current density as in Equation
This leads to a diagonal mass tensor, with elements
(285) and the heat current density as
Z
1
(M−1 )αβ = ∗ δαβ .
j
=
dkDk (Ek − µ)vk grk ,
(301)
mn
Because integral (294) over the occupied states gives the we obtain (by using Boltzmann equation (284)) the matrielectron density n, we obtain the conductivity
ces Lij into the form
σ=
ne2 τ
m∗n
1
1
L11 = L(0) , L21 = L12 = − L(1) , L22 = 2 L(2) . (302)
e
e
(297)
Correspondingly, if the Fermi energy is near to the maximum of the band, we can write
~2 k 2
Ek = E0 −
.
2m∗p
pe2 τ
,
m∗p
L(ν)
= e2
Z
αβ
dkDk τE
∂f
vα vβ (Ek − µ)ν .
∂µ
(303)
(298)
We define the electric conductivity tensor
Z
σαβ (E) = τE e2 dkDk vα vβ δ(E − Ek )
The conductivity is
σ=
In the above,
(299)
(304)
where p is the density of the empty states. The energy and obtain
Z
states that are unoccupied behave as particles with a pos∂f
(ν)
(E − µ)ν σαβ (E).
(305)
L
=
dE
itive charge, and are called the holes. Because the conduc∂µ
αβ
tivity depends on the square of the charge, it cannot reveal
the sign of the charge carriers. In magnetic field the elec- When the temperature is much lower than the Fermi temtrons and holes orbit in opposite directions, allowing one perature, we obtain for metals
to find out the sign of the charge carriers in the so-called
∂f
Hall measurement (we will return to this later).
≈ δ(E − EF ).
∂µ
Electric and Heat Current
Thus,
Boltzmann equation (284)
grk
∂f
Ek − µ
= frk − τE vk · eE + ∇µ +
∇T
T
dµ
αβ
can be interpreted so, that the deviations from the equilibrium distribution frk are caused by the electrochemical
”force” and that due to the thermal gradient
G=E+
L(0)
∇µ
e
L(1)
= σαβ (EF )
=
αβ
L(2)
=
αβ
π2
0
(kB T )2 σαβ
(EF )
3
π2
(kB T )2 σαβ (EF ),
3
(306)
(307)
(308)
where in the second equation we have calculated the linear
term of the Taylor expansion of the tensor σ at the Fermi
energy EF , leading to the appearance of the first derivative
and
∇T
.
T
As usual, these forces move the electrons and create a current flow. Assume, that the forces are weak which leads to
a linear response of the electrons. An example of such an
instance is seen in Equation (287) of the electric conductivity. One should note that, e.g., the thermal gradient can
also create an electric current, which means that generally
−
61
σ0 =
∂σ .
∂E E=EF
(309)
Equations (300) and (306-309) are the basic results of the
theory of the thermoelectric effects of electrons. They remain the same even though more than one energy band is
partially filled, if one assumes that the conductivity tensor
σαβ is obtained by summing over all such bands.
flow from the hot to the cold end. Because the electrons
are responsible on the heat conduction, the flow contains
The theory presented above allows the study of several
physical situations with electric and thermal forces present. also charge. Thus, a potential difference is created in beHowever, let us first study the situation where there is no tween the ends, and it should be a measurable quantity
(cf. the justification of the Wiedemann-Franz law). A dielectric current
rect measurement turns out to be difficult, because the
!
temperature gradient causes an additional electrochemical
∇T
j = L11 G + L12 −
=0
voltage in the voltmeter. It is therefore more practical to
T
use a circuit with two different metals.
−1 12 ∇T
L
⇒ G = L11
.
(310)
T
Wiedemann-Franz law
We see that one needs a weak field G in order to cancel the
electric current caused by the thermal gradient. In a finite
sample, this is created by the charge that accumulates at
its boundaries.
We obtain the heat current
−1 12
∇T
= −κ∇T,
jQ = L21 L11
L − L22
T
(311)
where the heat conductivity tensor
kB T
L22
+O
κ=
T
EF
!2
.
(312)
Because there is no current in an ideal voltmeter, we
obtain
Because L21 and L12 behave as (kB T /EF )2 they can be
neglected.
G = α∇T
We have obtained the Wiedemann-Franz law of the
Bloch electrons
2
T
π 2 kB
κ=
σ,
(313)
3e2
where the constant of proportionality
L0 =
2
π 2 kB
= 2.72 · 10−20 J/cmK2
2
3e
⇒ α = (L11 )−1
(315)
12
L
=−
T
2
T
π 2 kB
3e
σ −1 σ 0
(316)
We see that the thermal gradient causes the electrochemical field G. This is called the Seebeck effect, and the constant of proportionality (tensor) α the Seebeck coefficient.
(314)
Peltier Effect
is called the Lorenz number.
Peltier Effect means the heat current caused by the elecThe possible deviations from the Wiedemann-Franz law
tric current
that are seen in the experiments are due to the inadequacy
of the relaxation time approximation (and not, e.g., of the
jQ = Πj
(317)
quantum phenomena that the semiclassical model cannot
12
11 −1
⇒ Π = L (L ) = T α,
(318)
describe).
where Π is the Peltier coefficient.
Thermoelectric Effect
Thermoelectric effect is:
• Electric potential difference caused by a thermal gradient (Seebeck effect)
• Heat produced by a potential difference at the interface of two metals (Peltier effect)
• Dependence of heat dissipation (heating/cooling) of a
conductor with a thermal current on the direction of
the electric current (Thomson effect).
Seebeck Effect
Because different metals have different Peltier coefficients, they carry different heat currents. At the junctions
of the metals, there has to be heat transfer between the sys-
Consider a conducting metallic rod with a temperature
gradient. According to Equations (300), this results in heat
62
tem and its surroundings. The Peltier effect can be used
in heating and cooling, e.g. in travel coolers.
addition to the magnetic field. At the limit of large quantum numbers, one can use, however, the Bohr-Sommerfeld
quantization condition
Semiclassical Motion in Magnetic Field
I
dQ · P = 2π~l,
(322)
We consider then the effect of the magnetic field on the
conductivity. The goal is to find out the nature of the
charge carriers in metals. Assume first, that the electric
where l is an integer, Q the canonical coordinate of the
field E = 0 and the magnetic field B = B ẑ is a constant.
electron, and P the corresponding momentum. The same
Thus, according to the semiclassical equations of motion
condition results, e.g., to Bohr’s atomic model.
the wave vector of the electron behaves as
In a periodic lattice, the canonical (crystal) momentum
e ∂E
corresponding
to the electron position is
× B.
(319)
~k̇ = −
c ∂~k
Thus,
p = ~k −
k̇ ⊥ ẑ ⇒ kz = vakio
and
where A is the vector potential that determines the magnetic field. Bohr-Sommerfeld condition gives
∂E
k̇ ⊥
⇒ E = vakio.
∂k
I
Based on the above, the electron orbits in the k-space can
be either closed or open.
2πl
=
In the position space, the movement of the electrons is
obtained by integrating the semiclassical equation
=
e
~k̇ = − ṙ × B.
c
=
So, we have obtained the equation
e
~ k(t) − k(0) = − r⊥ (t) − r⊥ (0) × B,
c
Z
(320)
eA dr · k −
~c
τ
dtṙ ·
k−
0
=
eA ~c
Z τ
e
dtB · r × ṙ
2~c 0
~c
eB
Ar =
Ak .
~c
eB
(323)
(324)
(325)
(326)
We have used the symmetric gauge
where only the component of the position vector that is
perpendicular to the magnetic field is relevant. By taking
the cross product with the magnetic field on both sides, we
obtain the position of the electron as a function of time
c~
r(t) = r(0) + vz tẑ −
B × k(t) − k(0) .
2
eB
eA
c
(321)
We see that the electron moves with a constant speed along
the magnetic field, and in addition it has a component
perpendicular to the field, that depends whether the kspace orbit is closed or open.
A=
1
B × r.
2
In addition, τ is the period of the orbit and Ar is the area
swept by the electron during one period. One can show
that there exists a simple relation between the area Ar
and the area Ak swept in the k-space, which is
Ar =
~c 2
eB
Ak .
When the magnetic field B increases, the area Ak has to
grow
also in order to keep the quantum number l fixed. One
It turns out that if the electron orbits are closed in kcan
show
that the magnetization has its maxima at such
space (then also in the position space the projections of the
values
of
the
magnetic field 1/B, where the quantization
electron trajectories on the plane perpendicular to the field
condition
is
fulfilled
with some integer l and the area Ak
are closed), the magnetization of the matter oscillates as a
is
the
extremal
area
of
the Fermi surface. By changing the
function of the magnetic field B. This is called the de Haasdirection
of
the
magnetic
field, one can use the de Haas-van
van Alphen effect, and is an indication of the inadequacy
Alphen
effect
to
measure
the Fermi surface.
of the semiclassical equations, because not one classical
property of a system can depend on the magnetic field at
thermal equilibrium (Bohr-van Leeuwen theorem).
Hall Effect
de Haas-van Alphen Effect
The electron orbits that are perpendicular to the magIn 1879 Hall discovered the effect that can be used in the
netic field are quantized. The discovery of these energy determination of the sign of the charge carriers in metals.
levels is hard in the general case, where the electrons ex- We study a strip of metal in the electromagnetic field (E ⊥
perience also the periodic potential due to the nuclei, in B) defined by the figure below.
63
We assume that the current j = jx x̂ in the metal is
perpendicular to the magnetic field B = Bẑ. Due to the
magnetic field, the velocity of the electrons obtains a ycomponent. If there is no current through the boundaries
of the strip (as is the case in a voltage measurement), then
the charge accumulates on the boundaries and creates the
electric field Ey ŷ that cancels the current created by the
magnetic field.
According to the semiclassical equations, we obtain
e
= −eE − v × B
c
e
⇒ B × ~k̇ + eB × E = − B × v × B
c
e 2
= − B v⊥
c
c
~c
B × E,
⇒ v⊥ = − 2 B × k̇ −
eB
B2
~k̇
(327)
(328)
(329)
The solution of the Boltzmann equation in the relaxation
time approximation was of form (283)
t
g−f =−
0
dt0 e−(t−t )/τE
−∞
=
j=−
t
0
dt0 e−(t−t )/τE k̇t0 = k − hki,
(332)
j=
−∞
where
where
1
hki =
τE
Z
t
dt e
k t0 .
(338)
(339)
(340)
nec
(E × B).
B2
(341)
(333)
is the density of holes.
−∞
64
pec
(E × B),
B2
(342)
Z
p=
0 −(t−t0 )/τE
(337)
On the other hand, if the empty states are on the closed
orbits, we have 1 − f = 0 at the Brillouin zone boundary,
and obtain (by replacing f → 1 − (1 − f ))
The latter equality is obtained by partial integration
Z
(336)
Then, we assume that all occupied states are on closed
orbits. Therefore, the states at the boundary of the Brillouin zone are empty (f = 0), and the first term in the
current density vanishes. We obtain
Z
0
~c t
∂f
dt0 e−(t−t )/τE k̇t0 · E × B (330)
B 2 −∞
dµ
~c
∂f
k − hki · E × B
.
(331)
B2
∂µ
=
(334)
where ωc is the cyclotron frequency. In other words, the
electron completes several orbits before scattering, allowing
us to estimate
hki ≈ 0.
(335)
where the electron density
Z
n = dkDk f.
∂f .
vk · eE
dµ t0
By inserting the velocity obtained above, we have
g−f
ωc τ 1,
Thus, the current density is
Z
j = −e dkDk vk grk
Z
∂Ek ∂f
e~c
dk
k · (E × B)
= − 2
B
∂~k ∂µ
Z
e~c
∂f
dkDk
k · (E × B)
=
B2
∂~k
Z
ec
∂
=
dkDk
f k · (E × B)
2
B
∂k
nec
− 2 (E × B),
B
where v⊥ is component of the velocity that is perpendicular
to the magnetic field.
Z
Again, it is possible that the electron orbits in the kspace are either closed or open. In the above figure, these
have been sketched in the reduced zone scheme. In the
following, we will consider closed orbits and assume, in
addition, that the magnetic field is so large that
dkDk (1 − f )
(343)
Current density (341) gives
nec
jx = −
Ey .
B
Thus, by measuring the voltage perpendicular to the current jx we obtain the Hall coefficient of the matter
1
Ey
− nec
electrons
=
(344)
R=
1
holes
Bjx
pec
quasiparticles near to the Fermi surface approach infinity
as (E1 − EF )−2 . Similarly, one can deduce that even strong
interactions between particles are transformed into weak
interactions between quasiparticles.
The quasiparticle reasoning of Landau works only at
temperatures kB T EF . In the description of electrons
in metals, this does not present a problem, because their
Fermi temperatures are of the order of 10000 K. The most
In other words, the sign of the Hall coefficient tells whether important consequence of the Fermi liquid theory for this
course is, that we can represent the conductivity in the
the charge carriers are electrons or holes.
single electron picture, as long as we think the electrons as
In the presence of open orbits, the expectation value hki quasielectrons whose energies and masses have been altered
cannot be neglected anymore. Then, the calculation of the by the interactions between the electrons.
magnetoresistance, i.e. the influence of the magnetic field
on conductivity, is far more complicated.
Fermi Liquid Theory
The semiclassical theory of conductivity presented above
deals with the electrons in the single particle picture, neglecting altogether the strong Coulomb interactions between the electrons. The Fermi liquid theory, presented
by Landau in 1956, gives an explanation why this kind of
model works. The Fermi liquid theory was initially developed to explain the liquid 3 He, but it has been since then
applied in the description of the electron-electron interactions in metals.
The basic idea is to study the simple excited states of the
strongly interacting electron system instead of its ground
state. One can show that these excited states behave like
particles and, hence, they are called the quasiparticles.
More complex excited states can be formed by combining
several quasiparticles.
Let us consider a thought experiment to visualize how
the quasiparticles are formed. Think of the free Fermi gas
(e.g. a jar of 3 He or the electrons in metal), where the interactions between the particles are not ”turned on”. Assume, that one particle is excited slightly above the Fermi
surface, into a state defined by the wave vector k and energy E1 . All excited states of the free Fermi gas can be
created in this way by exciting single particles above the
Fermi surface. Such excited states are in principle eternal
(they have infinite lifetime), because the electrons do not
interact and cannot, thus, be relaxed to the ground state.
It follows from the the adiabatic theorem (cf. Zener
tunnelling), that if we can turn on the interactions slowly
enough, the excited state described above evolves into an
eigenstate of the Hamiltonian of the interacting system.
These excited states of the interacting gas of particles are
called the quasiparticles. In other words, there exists a
one-to-one correspondence between the free Fermi gas and
the interacting system. At low excitation energies and temperatures, the quantum numbers of the quasiparticles are
the same as those of the free Fermi gas, but due to the interactions their dynamic properties, such as the dispersion
relation of energy and effective mass, have changed.
When the interactions are on, the excited states do not
live forever. However, one can show that the lifetimes of the
65
6. Conclusion
In this course, we have studied condensed matter in
terms of structural, electronic and mechanical properties,
and transport phenomena due to the flow of electrons. The
research field of condensed matter physics is extremely
broad and, thus, the topics handled in the course cover
only its fraction. The purpose has been, especially, to acquire the knowledge needed to study the uncovered material. Such include, e.g. the magnetic, optical and superconducting properties of matter, and the description of liquid
metals and helium. These are left upon the interest of the
reader, and other courses.
The main focus has been almost entirely on the definition
of the basic concepts, and on the understanding of the following phenomena. It is remarkable how the behaviour of
the complex group of many particles can often be explained
by starting with a small set of simple observations. The
periodicity of the crystals and the Pauli exclusion principle are the basic assumptions that ultimately explain
nearly all phenomena presented in the course, together
with suitably chosen approximations (Bohr-Oppenheimer,
harmonic, semiclassical. . .). Finally, it is worthwhile to
remember that an explanation is not a scientific theory,
unless one can organize an experiment that can prove it
wrong. Thus, even though the simplicity of an explanation is often a tempting criterion to evaluate its validity,
the success of a theory is measured by comparing it with
experiments. We have had only few of such comparisons
in this course. In the end, I strongly urge the reader of
these notes read about the experimental tests presented
in the books by Marder and Ashcroft & Mermin, and the
references therein and elsewhere.
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