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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. 66