Condensed Matter Physics

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Condensed Matter Physics
763628S
Jani Tuorila
Department of Physics
University of Oulu
April 17, 2012
General
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.
Schedule and Practicalities
All lectures and exercise sessions will be held at room
TE320.
• Lectures: Monday 12-14
Wednesday 12-14
• Exercises: Thursday 12-14
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 exer-
cise is, however, in the deeper and more efficient learning
of the course!
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 Sys-
tems, way too complicated but has an intriguing in-
troduction!
• N. W. Ashcroft and N. D. Mermin, Solid State
Physics, a classic! Previously used in this course.
• P. Pietila¨inen, 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
– Experimental determination of crystal structure
– Boundaries and interfaces
– Complicated structures
• Electronic structure
– Single-electron model
– Schro¨dinger equation and symmetry
– Nearly free and tightly bound electrons
– Electron-electron interactions
– Band structure
• Mechanical properties
– Cohesion
– Phonons
• Electronic transport phenomena
– Bloch electrons
– Transport phenomena and Fermi liquid theory
• (Graphene)
1
1. Introduction
The purpose of this course is to give the basic introduc-
tion 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.
History
The term Condensed matter includes all such large
groups of particles that condense into one phase. Espe-
cially, the distances between particles have to be small
compared with the interactions between them. Examples:
• solids
• amorphous materials
• liquids
• soft materials (foams, gels, biological systems)
• white dwarfs and neutron stars (astrophysics)
• nuclear matter (nuclear physics)
Condensed matter physics originates from the study of
solids. Previously, the field was called solid state physics,
until it was noticed that one can describe the behaviour of
liquid metals, Helium and liquid crystals with same con-
cepts and models. This course concentrates mainly on per-
fect crystals, mainly due to the historical development of
the field and the simplicity of the related theoretical mod-
els.
Approximately, one third of the physicists in the US
considers themselves as researchers of condensed matter
physics. In the last 50 years (in 2011), there has been 22
condensed-matter related Nobel prizes in physics, and also
5 in chemistry:
• Bardeen, Cooper and Schrieffer, Theory of low tem-
perature superconductivity (1972)
• Josephson, Josephson effect (1973)
• Cornell, Ketterle and Wieman, Bose-Einstein conden-
sation in dilute gases of alkali atoms (2001)
• Geim and Novoselov, Graphene (2010)
The field has been the source of many practical applica-
tions:
• Transistor (1948)
• Magnetic recording
• Liquid crystal displays (LCDs) . . .
Therefore, condensed matter physics is both interesting
and beneficial for basic physics research and also for ap-
plications!
Finally, a short listing on ”hot topics” in the current
condensed matter research:
• Mesoscopic physics and the realisation of quantum
computation (research done in the University of Oulu)
• Fermi liquid theory (Oulu)
• Graphene
• Topological insulators
Many-Body Problem
All known matter is formed by atoms. The explaining
of the properties of single atoms can be done with quan-
tum mechanics, with an astounding accuracy. When one
adds more atoms to the system, the number of the relevant
degrees of freedom in Schro¨dinger equation grows exponen-
tially. In principle, all properties of the many atom system
can still be found by solving the Schro¨dinger equation, but
in practise the required computational power grows very
rapidly out of reach. As an example: the computers in
1980s could solve for the eigenvalues of the system con-
sisting of 11 interacting electrons. Two decades later the
computational power had increased hundred-fold but it al-
lowed to include only two extra electrons! A typical many-
body problem in condensed matter physics includes 1023,
or so, electrons. Therefore, it is clear that it is extremely
impractical to study such physics starting from the basic
principles.
Based on the above, the condensed matter theories are
so-called effective theories. In principle, they have to be
results of averaging the Schro¨dinger equation properly, but
in practise their form has been guessed based on symme-
tries and experimental results. In this way, the theories of
the condensed matter have become simple, beautiful, and
one can use them to obtain results that are precise (an
effective theory does not have to be imprecise) and give
detailed explanations.
The modest goal of condensed matter physics is to ex-
plain 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.
“The ability to reduce everything to simple fundamental
laws does not imply the ability to start from those laws
and reconstruct the universe. ... The constructionist
hypothesis breaks down when confronted by the twin diffi-
culties of scale and complexity. The behaviour of large and
complex aggregates of elementary particles, it turns out,
is not to be understood in terms of a simple extrapolation
of the properties of a few particles. Instead, at each
level of complexity entirely new properties appear, and
the understanding of the new behaviours require research
which I think is as fundamental in its nature as any other.”
- P. W. Anderson, 1972
2
2. Atomic Structure
MM, Chapters 1 and 2, excluding 2.3.3-2.3.6 and
2.6.2.
Scanning-tunnelling-microscopic image (page 17)
on NbSe2 surface in atomic resolution. The dis-
tance between nearest neighbours is 0.35 nm.
(http://www.pma.caltech.edu/GSR/condmat.html)
Fluoride chrystal on top of quartz chrystal (image by
Chip Clark).
The simplest way to form a macroscopic solid is to orga-
nize 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).
2.1 Crystal Structure (ET)
Many crystals appear in forms where flat surfaces meet
each other with constant angles. Such shapes can be un-
derstood based on the organization of atoms. Solids are
often formed of many crystals. This means that they are
composed of many joined crystals, oriented in different di-
rections. A single crystal can contain e.g. ∼ 1018 atoms,
whereas the whole macroscopic body has ∼ 1023 atoms.
Generally speaking, the structure of a solid can be ex-
tremely complicated. Even if it was formed by basic units
comprising of same atoms, it does not necessarily have a
repeating structure. Glass is an example of such a mate-
rial, formed by SiO2-units. This kindof material is called
amorphous.
Let us consider an ideal case, where we ignore all possible
imperfections of lattice. The description of such crystal can
be divided in two parts.
1) Group of atoms that form a repeating object in the
crystal. We refer to this as basis.
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 = n1a1 + n2a2 + n3a3. (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.
a1 a2
a3
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 lat-
tice point and, thus, the volume of the cell is inverse of the
density of the crystal.
+ =
An example in two dimensions.
a1
a2
a'1
a'2
The choice of the primitive vectors is not unique. In the
figure, one can use also the vectors a′1 and a
′
2 as primitive
vectors. They reproduce also all lattice points.
a1
a2 a'1
a'2
alkeiskoppi
yksikkökoppi
In certain symmetric lattices, it is more practical to use
orthogonal vectors instead of primitive ones (even if they
do not span the whole lattice). The volume defined by such
vectors is called unit cell. The lengths of the sides of a unit
cell are called lattice constants.
3
2.2 Two-Dimensional Lattice
Let us first restrict our considerations into two dimen-
sional lattices, because they are easier to understand and
visualize than their three dimensional counter-parts. How-
ever, it is worthwhile to notice that there exists genuinely
two-dimensional lattices, such as graphene that is pre-
sented later in the course. Also, the surfaces of crystals
and interfaces between two crystals are naturally two di-
mensional.
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 lat-
tices.
• square, symmetric in reflections with respect to both
x and y -axes, and with respect to 90◦ -rotations.
• rectangular, when square lattice is squeezed it loses
its rotational symmetry and becomes rectangular.
• hexagonal (trigonal), symmetric with respect two x
and y -reflections, and 60◦ -rotations.
• centered rectangular, squeezed hexagonal, no ro-
tational symmetry. By repeating the boxed structure
one obtains the lattice, hence the name.
• oblique, arbitrary choice of primitive vectors a1 and
a2 without any specific symmetries, only inversion
symmetry r→ −r.
The gray areas in the above denote the so called Wigner-
Seitz 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).
Example: Hexagonal lattice A choice for the primi-
tive vectors of the hexagonal lattice are, for example,
a1 = a
(
1 0
)
a2 = a
(
1
2 −
√
3
2
)
,
where a is the lattice constant. Another option is
a′1 = a
(√
3
2
1
2
)
a′2 = a
(√
3
2 − 12
)
.
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
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 lay-
ers 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 ma-
terial (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 interest-
ing properties are the reason why graphene is used as an
illustrative tool in this course.
As was mentioned in the above, graphene takes the form
of the honeycomb lattice, which is a Bravais lattice with a
basis. The starting point is the hexagonal lattice whose
primitive vectors are
a′1 = a
(√
3
2
1
2
)
a′2 = a
(√
3
2 − 12
)
.
Each lattice point is then replaced with the basis, defined
by
v1 = a
(
1
2
√
3
0
)
v2 = a
(
− 1
2
√
3
0
)
.
In each cell, the neighbours of the left- and right-hand
atoms are found in different directions. Anyhow, if one
allows pi/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 crys-
tals observed in scattering experiments (cf. the next chap-
ter) are a straight consequence of the symmetries of the
crystals. In order to understand these experiments, it is
important to know which symmetries are possible. Also,
the behaviour of electrons in periodic crystals can only be
explained by using simplifications in the Schro¨dinger equa-
tion, permitted by the symmetries.
Space Group
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:
• Operations defined by a (Bravais) lattice vector (trans-
lations)
• Operations that map at least one lattice point onto
itself (point operations)
• Operations that are obtained as a sequence of trans-
lations and point operations.
These can be described as a mapping
y = a +Rx, (2)
which first rotates (or reflects or inverses) an arbitrary vec-
tor x with a matrix R, and then adds a vector a to the
result. In order to fulfil the definition of the symmetry op-
eration, this should map the whole Bravais lattice (1) onto
itself. The general lattice (with a basis) has symmetry op-
erations that are not of the above mentioned form. They
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 transfor-
mations can be constructed from a minimal set of simpler
transformations. One can use these symmetry operations
in the classification of different lattice structures and, for
example, to show, using group theoretical arguments, that
there are only five essentially different Bravais-lattices in
two dimensions, a result stated earlier.
Space group (or symmetrygroup) G is the set of opera-
tions that leave the crystal invariant (why such a set is a
group?).
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
m1a1 +m2a2 +m3a3,
and thus leave the lattice invariant, according to the defi-
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 be-
fore) and screw axis (translation and rotation), are both
combinations whose parts are not elements of the space
group.
Does the point group define the lattice? No: the lat-
tices 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 G′ of the crystals, i.e.
SGS−1 = G′.
Then there exists a set of continuous mappings from the
unit matrix to matrix S
St = (1− t)I + St,
where t is between [0, 1]. Using this, one obtains such a con-
tinuous mapping from one lattice to the other that leaves
the symmetries invariant.
For example, there does not exist such a set of transfor-
mations between the rectangular and centered rectangular
lattices, even though they share the same point group.
When one transforms the rectangular into the centered
rectangular, the reflection symmetry with respect to the
y-axis is destroyed.
2.4 3-Dimensional Lattice
One has to study 3-dimensional lattices in order to de-
scribe 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 com-
plete listing of all of them is, of course, impossible to do
here. Therefore, we will restrict ourselves in the classifi-
cation of the 3-dimensional Bravais lattices. Let us first
introduce some of the structures found in Nature.
Simple cubic lattice (sc) is the simplest 3-dimensional
lattice. The only element that has taken this form as its
ground state is polonium. This is partly due to the large
”empty” space between the atoms: the most of the ele-
ments favour more efficient ways of packing.
a a
a
Face-centered cubic lattice (fcc) is formed by a sim-
ple cubic lattice, with an additional lattice points on each
face of the cube.
a1
a2
a3
a
Primitive vectors defining the lattice are, for example,
a1 =
a
2
(
1 1 0
)
a2 =
a
2
(
1 0 1
)
a3 =
a
2
(
0 1 1
)
,
where a is the lattice constant giving the distance be-
tween the corners of the cube (not with the nearest neigh-
bours!). Face-centered cubic lattice is often called cubic
closed-packed structure. If one takes the lattice points as
spheres with radius a/2
√
2, one obtains the maximal pack-
ing density. Fcc-lattice can be visualized as layered trigonal
lattices.
6
a
Body-centered cubic lattice (bcc) is formed by in-
serting an additional lattice point into the center of prim-
itive cell of a simple cubic lattice.
a1
a2
a3
a
An example of a choice for primitive vectors is
a1 =
a
2
(
1 1 −1)
a2 =
a
2
(−1 1 1)
a3 =
a
2
(
1 −1 1) ,
Hexagonal lattice is cannot be found amongst the el-
ements. Its primitive vectors are
a1 =
(
a 0 0
)
a2 =
(
a
2
a
√
3
2 0
)
a3 =
a
2
(
0 0 c
)
.
Hexagonal closed-packed lattice (hcp) is more inter-
esting 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.
a
c
Hcp-lattice is formed by a hexagonal lattice with a basis
v1 =
(
0 0 0
)
v2 =
(
a
2
a
2
√
3
c
2 .
)
The lattice constants c and a are arbitrary, but by choosing
c =
√
8/3a one obtains the closed-packing structure.
Both of the closed-packing structures are common espe-
cially 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.
In addition to the closed-packed structures, the body-
centered cubic lattice is common among elements, e.g. K,
Cr, Mn, Fe. The simple cubic structure is rare with crys-
tals. (ET)
Diamond lattice is obtained by taking a copy of an fcc-
lattice and by translating it with a vector (1/4 1/4 1/4).
The most important property of the diamond structure
is that every lattice point has exactly four neighbours (com-
pare 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 ten-
dency 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.
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.
Salt - Sodium Chloride
The ordinary table salt, i.e. sodium chloride (NaCl),
consists of sodium and chlorine atoms, ordered in an al-
ternating 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).
Many compounds share the same lattice structure
(MM):
7
Crystal a Crystal a Crystal a Crystal a
AgBr 5.77 KBr 6.60 MnSe 5.49 SnTe 6.31
AgCl 5.55 KCl 6.30 NaBr 5.97 SrO 5.16
AgF 4.92 KF 5.35 NaCl 5.64 SrS 6.02
BaO 5.52 KI 7.07 NaF 4.62 SrSe 6.23
BaS 6.39 LiBr 5.50 NaI 6.47 SrTe 6.47
BaSe 6.60 LiCl 5.13 NiO 4.17 TiC 4.32
BaTe 6.99 LiF 4.02 PbS 5.93 TiN 4.24
CaS 5.69 LiH 4.09 PbSe 6.12 TiO 4.24
CaSe 5.91 LiI 6.00 PbTe 6.45 VC 4.18
CaTe 6.35 MgO 4.21 RbBr 6.85 VN 4.13
CdO 4.70 MgS 5.20 RbCl 6.58 ZrC 4.68
CrN 4.14 MgSe 5.45 RbF 5.64 ZrN 4.61
CsF 6.01 MnO 4.44 RbI 7.34
FeO 4.31 MnS 5.22 SnAs 5.68
Lattice constants a (10−10m). Source: Wyckoff (1963-71),
vol. 1.
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 a Crystal a Crystal a
AgCd 3.33 CsCl 4.12 NiAl 2.88
AgMg 3.28 CuPd 2.99 TiCl 3.83
AgZn 3.16 CuZn 2.95 TlI 4.20
CsBr 4.29 NH4Cl 3.86 TlSb 3.84
Lattice constants a (10−10m). Source: Wyckoff (1963-
71), vol. 1.
2.5 Classification of Lattices by Symme-
try
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, thecyrstal
systems and the Bravais lattices belonging to them, are
listed:
• Cubic The point group of the cube. Includes the sim-
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 elim-
inates the 90◦-rotational symmetry in two directions.
The simple cube is, thus, transformed into a simple
tetragonal lattice. By stretching the fcc and bcc lat-
tices 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 lat-
tice. Correspondingly, by pulling the body-centered
tetragonal lattice, one obtains the body-centered and
face-centered orthorhombic lattices.
• Monoclinic The orthorhombic symmetry can be re-
duced by tilting the rectangles perpendicular to the
c-axis (cf. the figure below). The simple and base-
centered lattices transform into the simple monoclinic
lattice. The face- and body-centered orthorhombic
lattices are transformed into body-centered monoclinic
lattice.
• 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.
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:
• Rhobohedral or Trigonal Let us stretch the cube
along its diagonal. This results in the trigonal lat-
tice, regardless of which of the three cubic lattices was
stretched.
The last crystal system and the last Bravais lattice are
not related to the cube in any way:
• Hexagonal Let us place hexagons as bases and per-
pendicular walls in between them. This is the hexag-
onal point group, that has one Bravais lattice, the
hexagonal lattice.
It is not in any way trivial why we have obtained all pos-
sible 3-D Bravais lattices in this way. It is not necessary,
however, to justify that here. At this stage, it is enough
to know the existence of different classes and what belongs
in them. As a conclusion, a table of all Bravais lattice
presented above:
(Source: "http://www.iue.tuwien.ac.at/phd/
karlowatz/node8.html")
8
On the Symmetries of Lattices with Basis
The introduction of basis into the Bravais lattice com-
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
is not a subject on this course, but we will only summarise
the basic principle. As in the case of the classification of
Bravais lattices, one should first find out the point groups.
It can be done by starting with the seven crystal systems
and by reducing the symmetries of their Bravais lattices,
in a similar manner as the symmetries of the cube were re-
duced in the search for the Bravais lattices. This is possible
due to the basis which reduces the symmetry of the lattice.
The new point groups found this way belong to the origi-
nal crystal system, up to the point where their symmetry
is reduced so far that all of the remaining symmetry oper-
ations can be found also from a less symmetrical system.
Then, the point group is joined into the less symmetrical
system.
The space groups are obtained in two ways. Symmor-
phic lattices result from placing an object corresponding
to every point group of the crystal system into every Bra-
vais 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 oper-
ations 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.
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.
Pyroelectricity
Some materials (e.g. tourmaline) have the ability of pro-
ducing 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 polariza-
tion, but when the temperature is changing a measurable
potential difference is created on the opposite sides of the
crystal.
In the equilibrium the polarization is a constant, and
therefore the point group of the pyroelectric lattice has to
leave its direction invariant. This restricts the number of
possible point groups. The only possible rotation axis is
along the polarization, and the crystal cannot have reflec-
tion symmetry with respect to the plane perpendicular to
the polarization.
Optical activity
Some crystals (like SiO2) can rotate the plane of polar-
ized light. This is possible only if the unit cells are chiral,
meaning that the cell is not identical with its mirror image
(in translations and rotations).
2.6 Binding Forces (ET)
Before examining the experimental studies of the lattice
structure, it is worthwhile to recall the forces the bind the
lattice together
At short distances the force between two atoms is always
repulsive. This is mostly due to the Pauli exclusion prin-
ciple preventing more than one electron to be in the same
quantum state. Also, the Coulomb repulsion between elec-
trons is essential. At larger distances the forces are often
attractive.
r00
E(r)
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 = p
2/2m. On the other
hand, the Pauli principle may prevent the lowering of the
energy.)
Metallic bond. A large group of atoms share part of
their electrons so that they are allowed to move through-
out the crystal. The justification otherwise the same as in
covalent binding.
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 n1e and n2e) is
E12 =
1
4pi�0
n1n2e
2
|r1 − r2| . (3)
In a NaCl-crystal, one Na+-ion has six Cl−-ions as its
nearest neighbours. The energy of one bond between a
nearest neighbour is
Ep1 = − 1
4pi�0
e2
R
, (4)
where R is the distance between nearest neighbours. The
next-nearest neighbours are 12 Na+-ions, with a distance
d =
√
2R.
9
6 kpl
d = R
12 kpl
d = 2 R
Na+
Cl-
tutkitaan
tämän
naapureita
8 kpl
d = 3 R
The interaction energy of one Na+ ion with other ions is
obtained by adding together the interaction energies with
neighbours at all distances. As a result, one obtains
Ep = − e
2
4pi�0R
(
6− 12√
2
+
8√
3
− . . .
)= − e
2α
4pi�0R
. (5)
Here α is the sum inside the brackets, which is called the
Madelung constant. Its value depends on the lattice, and
in this case is α = 1.7627.
In order to proceed, on has to come up with form for the
repulsive force. For simplicity, let us assume
Ep,repulsive =
βe2
4pi�0Rn
. (6)
The total potential energy is, thus,
Ep(R) = − e
2
4pi�0
(
α
R
− β
Rn
)
. (7)
By finding its minimum, we result in β = αRn−1/n and in
energy
Ep = − αe
2
4pi�0R
(
1− 1
n
)
. (8)
The binding energy U of the whole lattice is the negative
of this multiplied with the number N of the NaCl-pairs
U =
Nαe2
4pi�0R
(
1− 1
n
)
. (9)
By choosing n = 9.4, this in correspondence with the mea-
surements. 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 re-
sulting bond between molecules is called hydrogen bond.
This is an important bond for example in ice.
Van der Waals interaction. This gives a weak attrac-
tion also between neutral atoms. Idea: Due to the circular
motion of the electrons, the neutral atoms behave like vi-
brating electric dipoles. Instantaneous dipole moment in
one atom creates an electric field that polarizes the other
atom, resulting in an interatomic dipole-dipole force. The
interaction goes with distance as 1/r6, and is essential be-
tween, e.g., atoms of noble gases.
In the table below, the melting temperatures of some
solids are listed (at normal pressure), leading to estimates
of the strengths between interatomic forces. The lines di-
vide the materials in terms of the bond types presented
above.
material melting temperature (K) lattice structure
Si 1683 diamond
C (4300) diamond
GaAs 1511 zincblende
SiO2 1670
Al2O3 2044
Hg 234.3
Na 371 bcc
Al 933 fcc
Cu 1356 fcc
Fe 1808 bcc
W 3683 bcc
CsCl 918
NaCl 1075
H2O 273
He -
Ne 24.5 fcc
Ar 83.9 fcc
H2 14
O2 54.7
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 be-
tween three nearest neighbours (honeycomb structure), re-
sulting 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.
A model of graphite where the spheres represent the car-
bon atoms (Wikipedia)
Covalent bonds are formed often into a specific direc-
tion. 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.
2.7 Experimental Determination of Crys-
tal Structure
MM, Chapters 3.1-3.2, 3.3 main points, 3.4-3.4.4
except equations
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 grat-
ing1. At that time, it was not known that crystals have
periodic structures nor that the X-rays have a wave char-
acter! With a little bit of hindsight, the prediction was
reasonable because the average distances between atoms
in solids are of the order of an A˚ngstro¨m (A˚=10−10 m),
and the range of wave lengths of X-rays settles in between
0.1-100 A˚.
The idea was objected at first. The strongest counter-
argument 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. Nowa-
days, 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 result in average thermal motion of the atoms
〈x〉 = √2kBT/k ≈ 2 · 10−11 m, which is much less than
the average distance between atoms in NaCl-crystal (cf.
previous table).
Scattering Theory of Crystals
The scattering experiment is conducted by directing a
plane wave towards a sample of condensed matter. When
the wave reaches the sample, there is an interaction be-
tween them. The outgoing (scattered) radiation is mea-
sured far away from the sample. Let us consider here a
simple scattering experiment and assume that the scatter-
ing is elastic, meaning that the energy is not transferred
between the wave and the sample. In other words, the fre-
quencies of the incoming and outgoing waves are the same.
This picture is valid, whether the incoming radiation con-
sists of photons or, for example, electrons or neutrons.
The wave ψ, scattered from an atom located at origin,
takes the form (cf. Quantum Mechanics II):
ψ ≈ Ae−iωt
[
eik0·r + f(rˆ)
eik0r
r
]
. (10)
In fact, this result holds whether the scattered waves are
described by quantum mechanics or by classical electrody-
namics.
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-
axis. The form factor f contains the detailed information
on the interaction between the scattering potential and the
1von Laue received the Nobel prize from the discovery of X-ray
diffraction in 1914, right after his prediction!
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
Iatom ≡ dσ
dΩatom
= |f(rˆ)|2. (11)
The intensity of the scattered wave at a solid angle dΩ at
distance r from the sample is dΩ× Iatom/r2.
(Source: MM) Scattering from a square lattice (25
atoms). When the radiation k0 comes in from the right
direction, the waves scattered from different atoms inter-
fere constructively. By measuring the resulting wave k, one
observes an intensity maximum.
There are, naturally, many atoms in a lattice, and it is
thus necessary to study scattering from multiple scatterers.
Let us assume in the following that we know the form factor
f . The angular dependence of the scattering is due to two
factors:
• Every scatterer emits radiation into different direc-
tions with different intensities.
• The waves coming from different scatterers interfere,
and thus the resultant wave contains information on
the correlations between the scatterers.
Let us assume in the following, that the origin is placed
at a fixed lattice point. First, we have to find out how
Equation (10) is changed when the scatterer is located at a
distance R from the origin. This deviation causes a phase
difference in the scattered wave (compared with a wave
11
scattered at origin). In addition, the distance travelled by
the scattered wave is |r−R|. Thus, we obtain
ψ ≈ Ae−iωt
[
eik0·r + eik0·Rf(rˆ)
eik0|r−R|
|r−R|
]
. (12)
We have assumed in the above, that the point of obser-
vation r is so far, that the changes in the scattering angle
can be neglected. At such distances (r � R), we can ap-
proximate (up to first orderin r/R)
k0|r−R| ≈ k0r − k0 r
r
·R. (13)
Let us then define
k = k0
r
r
q = k0 − k.
The wave vector k points into the direction of the measure-
ment 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
ψ ≈ Ae−iωt
[
eik0·r + f(rˆ)
eik0r+iq·R
r
]
. (14)
The second term in the denominator can be neglected in
Equation (13). However, one has to include into the ex-
ponential function all terms that are large compared with
2pi.
The magnitude of the change in momentum is
q = 2k0 sin θ, (15)
where 2θ is the angle between the incoming and outgoing
waves. The angle θ is called the Bragg’s angle. Assuming
specular reflection (Huygens’ principle, valid for X-rays),
the Bragg angle is the same as the angle between the in-
coming ray and the lattice planes.
Finally, let us consider the whole lattice and assume
again that the origin is located somewhere in the middle of
the lattice, and that we measure far away from the sample.
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
the effects due to multiple scattering events and inelastic
processes can be ignored. We obtain
ψ ≈ Ae−iωt
[
eik0·r +
∑
l
fl(rˆ)
eik0r+iq·Rl
r
]
, (16)
where the summation runs through the whole lattice.
When we study the situation outside the incoming ray
(in the region θ 6= 0), we can neglect the first term. The
intensity is proportional to |ψ|2, like in the one atom case.
When we divide with the intensity of the incoming ray |A|2,
we obtain
I =
∑
l,l′
flf
∗
l′e
iq·(Rl−Rl′ ). (17)
We have utilized here the property of complex numbers:
|∑l Cl|2 = ∑ll′ ClC∗l′ .
Lattice Sums
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)
I = Iatom
∣∣∣∣∣∑
l
eiq·Rl
∣∣∣∣∣
2
, (18)
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
of the sum (destructive interference). Generally speaking,
one ends up with similar summations whenever studying
the interaction of waves with a periodic structure (like con-
duction electrons in a lattice discussed later).
We first restrict the summation in the intensity (18) in
one dimension, and afterwards generalize the procedure
into three dimensions.
One-Dimensional Sum
The lattice points are located at la, where l is an integer
and a is the distance between the points. We obtain
Σq =
N−1∑
l=0
eilaq,
where N is the number of lattice points. By employing the
properties of the geometric series, we get
|Σq|2 = sin
2Naq/2
sin2 aq/2
. (19)
When the number of the lattice points is large (as is
the case in crystals generally), the plot of Equation (19)
consists of sharp and identical peaks, with a (nearly) zero
value of the scattering intensity in between.
12
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.
The peaks can be found at those values of the momentum
q that give the zeroes of the denominator
q = 2pil/a. (20)
These are exactly those values that give real values for the
exponential function (= 1). As the number of lattice points
grows, it is natural to think Σq as a sum of delta functions
Σq =
∞∑
l′=−∞
cδ
(
q − 2pil
′
a
)
,
where
c =
∫ pi
a
−pia
dqΣq =
2piN
L
.
In the above, L is length of the lattice in one dimension.
Thus,
Σq =
2piN
L
∞∑
l′=−∞
δ
(
q − 2pil
′
a
)
. (21)
It is worthwhile to notice that essentially this Fourier trans-
forms the sum in the intensity from the position space into
the momentum space.
Reciprocal Lattice
Then we make a generalization into three dimensions. In
Equation (18), we observe sharp peaks whenever we choose
q for every Bravais vectors R as
q ·R = 2pil, (22)
where l is an integer whose value depends on vector R.
Thus, the sum in Equation (18) is coherent, and produces
a Bragg’s peak. The set of all wave vectors K satisfying
Equation (22) is called the reciprocal lattice.
Reciprocal lattice gives those wave vectors that result in
coherent scattering from the Bravais lattice. The magni-
tude of the scattering is determined analogously with the
one dimensional case from the equation∑
R
eiR·q = N
(2pi)3
V
∑
K
δ(q−K), (23)
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 = 2pi
a2 × a3
a1 · a2 × a3
b2 = 2pi
a3 × a1
a2 · a3 × a1 (24)
b3 = 2pi
a1 × a2
a3 · a1 × a2
are the primitive vectors of the reciprocal lattice. Because
the vectors of the Bravais lattice are of form
R = n1a1 + n2a2 + n3a3,
we obtain
bi ·R = 2pini
where i = 1, 2, 3.
Thus, the reciprocal lattice contains, at least, the vectors
bi. In addition, the vectors bi are linearly independent. If
eiK1·R = 1 = eiK2·R,
then
ei(K1+K2)·R = 1.
Therefore, the vectors K can be written as
K = l1b1 + l2b2 + l3b3 (25)
where l1, l2 and l3 are integers. This proves, in fact, that
the wave vectors K form a Bravais lattice.
The requirement
K ·R = 0
defines a Bragg’s plane in the position space. Now, the
planes K ·R = 2pin are parallel, where n gives the distance
from the origin and K is the normal of the plane. This
kind of sets of parallel planes are referred to as families
of lattice planes. The lattice can be divided into Bragg’s
planes in infinitely many ways.
Example By applying definition (24), one can show that
reciprocal lattice of a simple cubic lattice (lattice constant
a) is also a simple cubic lattice with a lattice constant 2pi/a.
Correspondingly, the reciprocal lattice of an fcc lattice is a
bcc lattice (lattice constant 4pi/a), and that of a bcc lattice
is an fcc lattice (4pi/a).
Miller’s Indices
The reciprocal lattice allows the classification of all pos-
sible families of lattice planes. For each family, there exist
perpendicular vectors in the reciprocal lattice, shortest of
which has the length 2pi/d (d is separation between the
planes). Inversely: For each reciprocal lattice vector, there
exists a family of perpendicular lattice planes separated
with a distance d from one another (2pi/d is the length of
13
the shortest reciprocal vector parallel to K). (proof: exer-
cise)
Based on the above, it is natural to describe a given lat-
tice plane with the corresponding reciprocal lattice vector.
The conventional notation employs Miller’s indices in the
description of reciprocal lattice vectors, lattice planes and
lattice points. Usually, Miller’s indices are used in lattices
with a cubic or hexagonal symmetry. As an example, let
us study a cubic crystal with perpendicular coordinate vec-
tors xˆ, yˆ and zˆ, pointing along the sides of a typical unit
cell (=cube). Miller’s indices are defined in the following
way:
• [ijk] is the direction of the lattice
ixˆ+ jyˆ + kzˆ
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 vec-
tor perpendicular to plane (ijk).
• {ijk} is the set of planes perpendicular to vector [ijk],and equivalent in terms of the lattice symmetries.
• 〈ijk〉 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 → i¯. The original cubic lattice is called
direct lattice.
The Miller indices of a plane have a geometrical property
that is often given as an alternative definition. Let us con-
sider lattice plane (ijk). It is perpendicular to reciprocal
lattice vector
K = ib1 + jb2 + kb3.
Thus, the lattice plane lies in a continuous plane K ·r = A,
where A is a constant. This intersects the coordinate axes
of the direct lattice at points x1a1, x2a2 and x3a3. The
coordinates xi are determined by the condition K·(xiai) =
A. We obtain
x1 =
A
2pii
, x2 =
A
2pij
, x3 =
A
2pik
.
We see that the points of intersection of the lattice plane
and the coordinate axes are inversely proportional to the
values of Miller’s indices.
Example Let us study a lattice plane that goes through
points 3a1, 1a2 and 2a3. Then, we take the inverses of the
coefficients: 13 , 1 and
1
2 . These have to be integers, so
we multiply by 6. So, Miller’s indices of the plane are
(263). The normal to this plane is denoted with the same
numbers, but in square brackets, [263].
a1
3a1
2a3
(263)
a3
a2
yksikkö-
koppi
If the lattice plane does not intersect with an axis, it cor-
responds to a situation where the intersection is at infinity.
The inverse of that is interpreted as 1∞ = 0.
One can show that the distance d between adjacent par-
allel lattice planes is obtained from Miller’s indices (hkl)
by
d =
a√
h2 + k2 + l2
, (26)
where a is the lattice constant of the cubic lattice.
(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.
Scattering from a Lattice with Basis
The condition of strong scattering from a Bravais lat-
tice was q = K. This is changed slightly when a basis is
introduced to the lattice. Every lattice vector is then of
form
R = ul + vl′ ,
where ul is a Bravais lattice vector and vl′ is a basis vector.
Again, we are interested in the sum∑
R
eiq·R =
(∑
l
eiq·ul
)(∑
l′
eiq·vl′
)
,
determining the intensity
I ∝ |
∑
R
eiq·R|2 =
(∑
jj′
eiq·(uj−uj′ )
)(∑
ll′
eiq·(vl−vl′ )
)
.
(27)
The intensity appears symmetric with respect to the lattice
and basis vectors. The difference arises in the summations
which for the lattice have a lot of terms (of the order 1023),
whereas for the basis only few. Previously, it was shown
that the first term in the intensity is non-zero only if q
belongs to the reciprocal lattice. The amplitude of the
scattering is now modulated by the function
Fq =
∣∣∣∣∣∑
l
eiq·vl
∣∣∣∣∣
2
, (28)
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
4
(1 1 1).
It was mentioned previously that the reciprocal lattice of
an fcc lattice is a bcc lattice with a lattice constant 4pi/a.
Thus, the reciprocal lattice vectors are of form
K = l1
4pi
2a
(1 1 − 1) + l2 4pi
2a
(−1 1 1) + l3 4pi
2a
(1 − 1 1).
Therefore,
v1 ·K = 0
and
v2 ·K = pi
2
(l1 + l2 + l3).
The modulation factor is
FK =
∣∣∣1 + eipi(l1+l2+l3)/2∣∣∣2 (29)
=
 4 l1 + l2 + l3 = 4, 8, 12, . . .2 l1 + l2 + l3 is odd
0 l1 + l2 + l3 = 2, 6, 10, . . .
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 direc-
tion 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 de-
vice collects data only from one direction k. This forces
us to restrict ourselves into a two-dimensional subspace of
the scattering vectors. This can be visualized with Ewald
sphere of the reciprocal lattice, that reveals all possible
values of q for the give values of the incoming wave vector.
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 scatter-
ing is not observed. The problem is, thus, that the points
in the reciprocal lattice form a discrete set in the three di-
mensional k-space. Therefore, it is extremely unlikely that
any two dimensional surface (e.g. Ewald sphere) would go
through them.
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 lat-
tice constant of the reciprocal lattice. Again, we obtain the
estimate that the wave length has to be of the order of an
A˚ngstro¨m, i.e. it has to be composed of X-rays.
One can also use Ewald sphere to develop solutions for
the problem with monochromatic radiation. The most con-
ventional ones of them are Laue method, rotating crystal
method and powder method.
Laue Method
Let us use continuous spectrum.
Rotating Crystal Method
Let us use monochromatic radi-
ation but also rotate the cyrstal.
15
Powder Method (Debye-Scherrer Method)
Similar to the rotating crystal method. We use
monochromatic radiation but, instead of rotating, a sam-
ple consisting of many crystals (powder). The powder is
fine-grained but, nevertheless, macroscopic, so that they
are able to scatter radiation. Due to the random orienta-
tion of the grains, we observe the same net effect as when
rotating a single crystal.
2.9 Radiation Sources of a Scattering Ex-
periment
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 fre-
quency of the radiation and emit spherical waves. Because
the nuclei of the atoms are much heavier, only their elec-
trons participate to the X-ray scattering. The intensity of
the scattering depends on the number density of the elec-
trons, 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 elec-
trons accelerate continuously in large rings with electro-
magnetic field.
Neutrons
The neutrons interact only with nuclei. The interac-
tion 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 energyET ≈ 32kBT , with T = 293 K), whose de Broglie wave
length λ = h/p ≈ 1 A˚. 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 A˚),
in order to avoid multiple scattering events.
X-rays Neutrons Electrons
Charge 0 0 -e
Mass 0 1.67 · 10−27 kg 9.11 · 10−31 kg
Energy 12 keV 0.02 eV 60 keV
Wave length 1 A˚ 2 A˚ 0.05 A˚
Attenuation length 100 µm 5 cm 1 µm
Form factor, f 10−3 A˚ 10−4 A˚ 10 A˚
Typical properties of different sources of radiation in scat-
tering experiments. (Source: MM; Eberhart, Structural
and Chemical Analysis of Materials (1991).)
2.10 Surfaces and Interfaces
MM, Chapter 4, not 4.2.2-4.2.3
Only a small part of the atoms of macroscopic bodies
are lying on the surface. Nevertheless, the study of sur-
faces is important since they are mostly responsible for
the strength of the material and the resistance to chemi-
cal attacks. For example, the fabrication of circuit boards
requires good control on the surfaces so that the conduc-
tion of electrons on the board can be steered in the desired
manner.
The simplest deviations from the crystal structure oc-
cur 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 form-
ing a structure with alternating crystal lattices. Coherent
interface has all its atoms perfectly aligned. The grow-
ing 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).
16
In the experiment, the electrons are shot with a gun to-
wards 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.
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 scat-
tered 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 lat-
tice vector K. On the other hand, the component qz is
a continuous variable because the z-component of the two
dimensional vector R is zero.
In order to observe strong scattering, Ewald sphere has
to go through some of the rods defined by condition (30).
This occurs always, independent on the choice of the in-
coming wave vector or the orientation of the sample (com-
pare with Laue, rotating crystal and powder methods).
Reflection High-Energy Electron Diffraction,
RHEED
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 A˚−1) are large compared with
the lattice constant of the reciprocal lattice and, thus, the
scattering patterns are streaky. The sample has to be ro-
tated in order to observe strong signals in desired direc-
tions.
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.
A sample, that is flat on the atomic level, is placed in
a vacuum chamber. The sample is layered with elements
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 metal-
lic 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.
By changing the electric potential difference between the
needle and the surface, one can change the tunnelling prob-
ability of electrons from the needle to the sample. Then
the created current can be measured and used to map the
surface.
17
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
ψ(x) ∝ exp
[ i
~
∫ x
dx′
√
2m(E − U(x′))
]
.
Inside the wall, the amplitude of the wave function drops
with a factor
exp
[
− s
√
2mφ/~2
]
.
The current is dependent on the square of the wave func-
tion, and thus depend exponentially on the distance be-
tween 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
tutkittava
pinta
The essential part of the functioning of the device is how
to move the needle without vibrations in atomic scale.
Pietzoelectric crystal can change its shape when placed
in an electric field. With three pietzoelectric crystals one
can steer the needle in all directions.
Atomic Force Microscope
Atomic force microscope is a close relative to the scan-
ning 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 micro-
scope can be used also in the study of insulators.
Example: Graphene
(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 A˚).
(Source: Li et al. PRL 102, 176804 (2009)) STM image
of graphene.
Graphene can be made, in addition to previously men-
tioned 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
discovery in 2004), graphene has been among the most ex-
pensive materials in the world. The production methods
of large sheets of graphene are still (in 2012) under devel-
opment in many research groups around the world.
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
rarelymet 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,
glasses, liquid crystals and quasi crystals.
Alloys
The development of metallic alloys has gone hand in
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 revolu-
tion 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(
N
M
)
=
N !
M !(N −M)! ≈
NM
M !
different ways. The macroscopic state of the mixture has
the entropy
S = kB ln(N
M/M !) ≈ −kBN(c ln c− c),
where c = M/N is the impurity concentration. Each impu-
rity atom contributes an additional energy � to the crystal,
leading to the free energy of the mixture
F = E − TS = N [c�+ kBT (c ln c− c)]. (31)
In equilibrium, the free energy is in its minimum. This
occurs at concentration
c ∼ e−�/kBT . (32)
So, we see that at finite temperatures the solubility is non-
zero, and that it decreases exponentially when T → 0. In
most materials there are ∼ 1% of impurities.
The finite solubility produces problems in semiconduc-
tors, since in circuit boards the electrically active impuri-
ties disturb the operation already at concentrations 10−12.
Zone refining can be used to reduce the impurity concen-
tration. One end of the impure crystal is heated, simulta-
neously moving towards the colder end. After the process
the impurity concentration is larger the other end of the
crystal, which is removed and the process is repeated.
Phase Diagrams
Phase diagram describes the equilibrium at a given con-
centration and temperature. Let us consider here espe-
cially system consisting of two components. Mixtures with
two substances can be divided roughly into two groups.
The first group contains the mixtures where only a small
amount of one substance is mixed to the other (small con-
centration c). In these mixtures the impurities can either
replace a lattice atom or fill the empty space in between
the lattice points.
Generally, the mixing can occur in all ratios. Intermetal-
lic 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) ele-
ments, mixing them and finally cooling the mixture. When
the cooling process is fast, one often results in a similar ran-
dom 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
peaks. The atoms form alternating crystal structures, su-
perlattices. With many combinations of metals one obtains
superlattices, mostly with mixing ratios 1:1 and 3:1.
Phase Separation
Let us assume that we are using two substances whose
free energy is of the below form when they are mixed ho-
mogeneously.
(Source: MM) The free energy F(c) of a homogeneous
mixture of two materials as a function of the relative con-
centration c. One can deduce from the figure that when
the concentration is between ca and cb, the mixture tends
to phase separate in order to minimize the free energy This
can be seen in the following way: If the atoms divide be-
tween two concentrations ca < c and cb > c (not necessarily
the same as in the figure), the free energy of the mixture
is
Fps = fF(ca) + (1− f)F(cb),
where f is the fraction of the mixture that has the con-
centration ca. Correspondingly, the fraction 1− f has the
concentration cb. The fraction f is not arbitrary because
19
the total concentration has to be c. Thus,
c = fca + (1− f)cb ⇒ f = c− cb
ca − cb .
Therefore, the free energy of the phase separated mixture
is
Fps = c− cb
ca − cbF(ca) +
ca − c
ca − cbF(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 phase separation of the mixture of
copper and silver. In the region Ag, silver forms an fcc lat-
tice and the copper atoms replace silver atoms at random
lattice points. Correspondingly in the region Cu, silver re-
places 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 separa-
tion between the metals occurs. The solid lines denote the
concentrations ca and cb (cf. the previous figure) as a func-
tion 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 tem-
perature where the mixture can be found as a homogeneous
liquid.
(Source: MM) The formation of a phase diagram.
Dynamics of Phase Separation
The heating of a solid mixture always results in a ho-
mogeneous 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 consider-
ing the concentration current
j = −D∇c. (34)
The solution of this equation gives the atom current j,
whose direction is determined by the gradient of the con-
centration. 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
∂t
= D∇2c. (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
20
(Source: MM) Dendrite formed in the solidification pro-
cess 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
∂c
∂t
≈ 0.
Thus, the concentration can be solved from the Laplace
equation
∇2c = 0.
We are searching for a spherically symmetric solution and,
thus, we can write the Laplace equation as
1
r2
∂
∂r
(
r2
∂c
∂r
)
= 0.
This has a solution
c(r) = A+
B
r
.
At the boundary of the droplet (r = R), the concentration
is
c(R) = ca
and far away from the droplet (r →∞)
lim
r→∞ c(r) = c∞.
These boundary conditions determine the coefficients A
and B leading to the unambiguous solution of the Laplace
equation
c(r) = c∞ +
R
r
(ca − c∞).
The gradient of the concentration gives the current density
j = −D∇c = DR(c∞ − ca)∇1
r
= −DR(c∞ − ca) rˆ
r2
.
For the total current into the droplet, we have to multi-
ply the current density with the surface area 4piR2 of the
droplet. This gives the rate of change of the concentration
inside the droplet
∂c
∂t
= 4piR2(−jr) = 4piDR(c∞ − ca).
On the other hand, the chain rule of derivation gives the
rate of change of the volume V = 4piR3/3 of the droplet
dV
dt
=
∂V
∂c
∂c
∂t
.
By denoting v ≡ ∂V/∂c, we obtain
R˙ =
vD
R
(c∞ − ca) ⇒ R ∝
√
2vD(c∞ − ca)t
We see that small nodules grow the fastest when measured
in R.
Simulations
When the boundary conditions of the diffusion equation
are allowed to change, the non-linear nature of the equa-
tion 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
Monte Carlo method was created by von Neumann,
Ulam and Metropolis in 1940s when they were working
on the Manhattan Project. The name originates from the
casinos in Monte Carlo, where Ulam’s uncle often went to
gamble his money. The basic principle of Monte Carlo re-
lies on the randomness, characteristic to gambling. The
assumption in the background of the method is that the
atoms in a solid obey in equilibrium at temperature T the
Boltzmann distribution exp(−βE), where E is position de-
pendent energy of an atom and β = 1/kBT . If the energy
difference between two states is δE , then the relative occu-
pation probability is exp(−βδE).
The Monte Carlo method presented shortly:
1) Assume that we have N atoms. Let us choose their
positions R1 . . .RN randomly and calculate their en-
ergy E(R1 . . .RN ) = E1.
2) Choose one atom randomly and denote it with index
l.
3) Create a random displacement vector, e.g. by creating
three random numbers pi ∈ [0, 1] and by forming a
vector
Θ = 2a(p1 − 1
2
, p2 − 1
2
, p3 − 1
2
).
21
In the above, a sets the length scale. Often, one uses
the typical interatomic distance, but its value cannot
affect the result.
4) Calculate the energy difference
δE = E(R1 . . .Rl + Θ . . .RN )− E1.
The calculation of the difference is much simpler than
the calculation of the energy in the position configu-
ration alone.
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 dis-
placement (Rl → Rl + Θ and go to 2). If p is greater,
reject the displacement and go to 2).
In low temperatures almost every displacement that are
accepted lower the systems energy. In very high temper-
atures almost every displacement is accepted. After suffi-
cient repetition, the procedure should generate the equilib-
rium energy and the particle positions that are compatible
with the Boltzmann distribution exp(−βE).
Molecular Dynamics
Molecular dynamics studies the motion of the single
atoms and molecules forming the solid. In the most gen-
eral case, the trajectories are solved numerically from the
Newton equations of motion. This results in solution of
the thermal equilibrium in terms of random forces instead
of random jumps (cf. Monte Carlo). The treatment gives
the positions and momenta of the particles and, thus, pro-
duces more realistic picture of the dynamics of the system
approaching thermal equilibrium.
Let us assume that at a given time we know the positions
of the particles and that we calculate the total energy of
the system E . The force Fl exerted on particle l is obtained
as the gradient of the energy
Fl = − ∂E
∂Rl
.
According to Newton’s second law, this force moves the
particle
ml
d2Rl
dt2
= Fl.
In order to solve these equations numerically (l goes
through values 1, . . . , N , where N is the number of parti-
cles), one has to discretize them. It is worthwhile to choose
the length of the time step dt to be shorter than any of the
scales of which the forces Fl move the particles consider-
ably. When one knows the position Rnl of the particle l
after n steps, the position after n+1 steps can be obtained
by calculating
Rn+1l = 2R
n
l −Rn−1l +
Fnl
ml
dt2. (36)
As was mentioned in the beginning, the initial state de-
termines the energy E , that is conserved in the process.
The temperature can be deduced only at the end of the
calculation, e.g. from the root mean square value of the
velocity.
The effect of the temperature can be included by adding
terms
R¨l =
Fl
ml
− bR˙l + ξ(t),
into the equation of motion. The first term describes the
dissipation by the damping constant b that depends on the
microscopic properties of the system. The second term il-
lustrates the random fluctuations due to thermal motion.
These additional terms cause the particles to approach the
thermal equilibrium at the given temperature T . The ther-
mal fluctuations and the dissipation are closely connected,
and this relationship is described with the fluctuation-
dissipation theorem
〈ξα(0)ξβ(t)〉 = 2bkBTδαβδ(t)
ml
.
The angle brackets denote the averaging over time, or al-
ternatively, over different statistical realizations (ergodic
hypothesis). Without going into any deeper details, we
obtain new equations motion by replacing in Equation (36)
Fnl → Fnl − bml
Rnl −Rn−1l
dt
+ Θ
√
6bmlkBT/dt,
where Θ is a vector whose components are determined by
random numbers pi picked from the interval [0, 1]:
Θ = 2
(
p1 − 1
2
, p2 − 1
2
, p3 − 1
2
)
.
Liquids
Every element can be found in the liquid phase. The
passing from, e.g., the solid into liquid phase is called the
phase transformation. Generally, the phase transforma-
tions are described with the order parameter. It is defined
in such way that it non-zero in one phase and zero in all
the other phases. For example, the appearance of Bragg’s
peaks in the scattering experiments of solids can be thought
as the order parameter of the solid phase.
Let us define the order parameter of the solid phase more
rigorously. Consider a crystal consisting of one element.
Generally, it can be described with a two particle (or van
Hove) correlation function
n2(r1, r2; t) =
〈∑
l 6=l′
δ(r1 −Rl(0))δ(r2 −Rl′(t))
〉
,
where the angle brackets mean averaging over temperature
and vectors Rl denote the positions of the atoms. If an
atom is at r1 at time t1, the correlation function gives the
probability of finding another particle at r2 at time t1 + t.
22
Then, we define the static structure