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Phys560Notes-3

Course: PHYS 560, Spring 2012
School: University of Illinois,...
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Drude Review model (free electron approximation) + (independent electron approximation) + (Maxwell Boltzmann statistics) Sommerfeld model (free electron approximation) + (independent electron approximation) + (Fermi Dirac statistics) Next: Band structure (crystal potential) + (independent electron approximation) + (Fermi Dirac statistics) Crystal potential arises from the crystal structure made of atoms --...

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Drude Review model (free electron approximation) + (independent electron approximation) + (Maxwell Boltzmann statistics) Sommerfeld model (free electron approximation) + (independent electron approximation) + (Fermi Dirac statistics) Next: Band structure (crystal potential) + (independent electron approximation) + (Fermi Dirac statistics) Crystal potential arises from the crystal structure made of atoms -- important for understanding the differences among metals, semiconductors, and insulators. Crystal Structure Crystal structure is evident from naturally formed facets of many mineral samples. Based on x-ray diffraction: Crystal structure = periodic,(b) infinite(a) repetition of structural units(c) (a) Infinite: N 1023 , bulk properties are independent of surfaces. (b) Periodic: translational symmetry Define Bravais lattice with lattice vector R n1a1 n2 a 2 n3 a 3 ; n1,2,3 = integers a1,2,3 primitive lattice vectors (noncoplanar); they generate the lattice. 2d example: a2 a1 (c) Structural unit: basis = a set of atoms associated with each lattice point Summarize : crystal structure = Bravais lattice + basis Theorem: A crystal looks the same from every Bravais lattice point. TH T 1 H (system is invariant under translation) Theorem: Primitive vectors are not unique; there are ways of choosing them. 2d example: a2 a2' a1 a2" Definition: primitive unit cell = a region in space, when translated using all lattice vectors R, fills neatly the whole space (w/o overlapping regions or voids) 2d examples: 2 possible choices out of 3d example: (obvious choice) a3 a2 a1 Theorem: volume of primitive cell c a1 a 2 a 3 lattice points One unit cell contains one lattice point (on average). Definition: Wigner-Seitz unit cell = the region of space that is closer to a selected lattice point (origin O) than to any other lattice points; it is a primitive unit cell. Method of construction: 1. Draw lines from O to all nearby lattice points. 2. Draw bisectors (planes in 3-d). 3. Find the smallest volume enclosed. V ; V = volume of crystal, N = # of N 2d example: Proof: Every point in space can be uniquely assigned to a W-S cell associated with a lattice point (except for those on the boundaries, but those have zero volume). Translational symmetry all W-S cells are the same. Since each point in space (other than those on boundaries) is accounted for once and only once, we have no overlapping regions or voids. W-S cell is a primitive cell. It does not depend on the choice of a1,2,3 . As symmetric as the original lattice (e.g., 3-fold, 4-fold, mirror planes, ...). Definition: conventional unit cells (non-primitive cells): volume n volume of primitive cell; filling the whole space when translated using a subset of R ; containing n lattice points per cell; often chosen for convenience. Examples of Bravais lattices: 1. Simple cubic (sc) a1,2,3 mutually perpendicular; a 1 a 2 a 3 a a = lattice constant or lattice parameter Primitive cell = cube = conventional cell W-S cell is also cubic. Po, the only one in the periodic table (under typical conditions) a3 a2 a1 2. Body centered cubic (bcc) Conventional cell = sc + another atom at center of cube body Primitive lattice vectors (symmetric form): z a a ^ ^ ^ a1 x y z 1,1,1 a 2 2 a a ^ ^ ^ a 2 x y z 1, 1,1 2 2 a2 a a a1 ^ ^ ^ a 3 x y z 1,1, 1 2 2 a = lattice constant a3 x Make sure all points are generated: a a R n1 a1 n2 a 2 n3 a 3 n2 n3 n1 , n3 n1 n2 , n1 n2 n3 1 , 2 , 3 2 2 1,2,3 = integers but not arbitrary; n1,2,3 arbitrary. 1 2 2 n3 even, etc. So, 1,2,3 either all even or all odd. a R 0,0,0 corner; 2,0,0 corner; 0, 2,0 corner; ... 1,1,1 body-center; ... 2 Wigner-Seitz cell: "truncated octahedron" (see textbook) 3 Conventional cell: v a a3 v = a1 a 2 a 3 Primitive cell: 8 1 1 a3 1 1 1 2 1 1 1 1 Conventional cell = 2 x primitive cell bcc = sc + a two-atom basis (useful for X-ray diffraction). a1 a 1,0,0 , a 2 a 0,1,0 , a 3 a 0,0,1 a 1,1,1 2 Basis: d0 a 0,0,0 and d1 Examples: Ba, Cr, Li, etc. 3. Face centered cubic (fcc) Conventional cell = sc + 6 atoms at faces centers Bravais lattice a a1 0,1,1 2 a a 2 1,0,1 2 a a 3 1,1,0 2 R ni a i generates the lattice (verify yourself). 3 Conventional cell: v a a a1 a2 a3 3 Primitive cell: v a1 a 2 a 3 a 4 There are 4 lattice points per conventional cell Counting Lattice points 8 corner atoms 6 face atoms Weight 1/8 1/2 Total 1 3 4 Or, could one offset the cube slightly to avoid shared points. Examples: Ni, Cu, Ag, Au, Al, Pt, ... 4. Simple hexagonal (sh) 60 triangular net in 2d, stacked vertically directly atop a1 a2 a a3 c a 3 ^ ^ ^ ^ a1 a x , a 2 x a y , a3 c z a, c: lattice constants; z-axis = c-axis 2 2 c a a Examples of lattice + nontrivial basis, one kind of atom: 1. Diamond structure a 1,1,1 4 fcc lattice (lattice constant a) + 2-atom basis at 0,0,0 and Coordination number = 4; tetrahedral bonding Examples: diamond, Si, Ge, grey-Sn 2. Hexagonal close packed (hcp) hcp = (simple hexagonal lattice) 1 1 1 + 2-atom basis at 0,0,0 and a1 a 2 a3 3 3 2 For close packed hard spheres, c / a 8 (ideal structure); 3 but most are not ideal. c a a A B Green: first layer. Gold: second layer. The third layer repeats the first layer positions. Stacking sequence: ABAB Examples: Ru, Cd, Re, Y, Hf , Gd, etc. Changing the stacking sequence to ABCABC yields the fcc structure. A B C First layer at A sites; second layer at B sites, third layer at C sites; fourth layer at A sites; etc. Looking along the body diagonal, the fcc structure can be viewed as a close stacking of atomic layers. fcc and hcp have the same packing density (close packed). Examples of lattice + nontrivial basis, different kinds of atoms 1. Zinc blende (ZnS) a 1,1,1 4 Replacing Zn and S by C diamond structure fcc lattice + 2-atom basis, Zn at 0,0,0 , S at Coordination number = 4 2. NaCl structure fcc lattice + 2-atom basis, Na at 0,0,0 , Cl at Two interpenetrating fcc lattices. a 1,1,1 2 3. CsCl structure sc lattice + 2-atom basis, Cs at 0,0,0 , Cl at Replacing Cl by Cs bcc Cs Coordination # = 8 4. Perovskite structure (CaTiO3, or ABO3 generally) sc lattice + 5-atom basis Ca2+ at A site, Ti4+ at B site, and 3 O atoms Key feature: oxygen cage. Many closely related variations. Examples: complex oxides, high-temp superconductors, colossal magnetoresistive materials, ferroelectrics , .... a 1,1,1 2 Symmetry (Brief review in terms of group theory) Space group: all symmetry operations {PT} that leave the crystal invariant, where P is a point operation (leaving one point fixed) including rotation, inversion, reflection, and combos, and T is a translation. T might be a translation by R n1a1 n2 a 2 n3a 3 , but it could also involve a nontrivial fraction of R in more complicated cases: Glide planes mirror reflection + translation by, for example, of a primitive lattice vector Screw axes rotation by + translation by, for example, of a primitive lattice vector Translation group: all translations by R n1a1 n2 a 2 n3a 3 ; an invariant subgroup of the space group (invariant under all symmetry transformations). Note {IT} (I being the identity operator) is not necessary the translation group, as there can be additional translations involving a nontrivial fraction of R associated with glide planes and screw axes. Point group: all operations {PI}. Not necessarily a subgroup of the space group because of glide planes and screw axes. It leaves the point lattice & Wigner Seitz cell invariant. Factor group: (Space group) (Translation group); isomorphic to the point group. Bravais lattice 7 (crystal systems) 14 (lattice structures) Crystal structure 32 230 Point group Space group Listed in M.J. Berger, Elementary Crystallography, Wiley, NY, 1963. See textbook or http://en.wikipedia.org/wiki/Crystal_structure 7 crystal systems Triclinic Monoclinic Simple Orthorhombic Simple Hexagonal Rhombohedral Tetragonal (trigonal) Simple Cubic Simple Base-centered Base-centered Body-centered Body-centered 14 Bravais lattices Body-centered Face-centered Face-centered Bravais lattices: Consider the highest symmetry, the cubic point group Oh (full symmetry group of a cube); there are 3 Bravais lattices having this point group symmetry: sc, fcc, and bcc. Distorting the system in various ways results in additional crystal systems and lattice structures that have lower symmetries (fewer symmetry operations). The triclinic structure has the lowest symmetry. For systems with one atom per unit cell, there are only 14 possible structures (Bravais lattices). Crystal structures: The number of possible structures explodes to 230 as you add a basis (which tends to reduce the number of symmetry operations). For enumerating all possible structures, the exercise is to reduce the symmetry in a systematic manner. Make sure that all structures are distinct. It is an interesting but timeconsuming exercise.
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University of Illinois, Urbana Champaign - PHYS - 560
Reciprocal Lattice Direct lattice given by R ni aiiDefinition: reciprocal lattice is a Bravais lattice given byK = reciprocal lattice vectors i bi , where i = integers, anda2 a3 a 3 a1 a1 a 2 b1 2 ; b2 2 ; b3 2 a1 a 2 a 3 a1 a 2 a 3 a1 a 2 a 3Useful
University of Illinois, Urbana Champaign - PHYS - 560
X-Ray Diffraction A method to determine crystal structure. Assumption: ignoring thermal vibrations of lattice (including zero-point vibrations) for now. A charged particle of charge q in an EM field:1 q H p A V r 2m c 2 1 2 q q q2 p A p p A 2 A2 V r 2m
University of Illinois, Urbana Champaign - PHYS - 560
Electronic States in a Crystal ~1023 interacting electrons and ion cores a big computational problem. p2 1 p2 Z Z e2 1 e2 n n n HT i i 2m 2 i , j ri r j n 2Mn 2 n,n ' R n R n Hspin-orbit-interaction . Hel H core H el core Hsoi . Z e2 n n,i ri R n Frozen
University of Illinois, Urbana Champaign - PHYS - 560
Band Structure Calculations using a plane wave basis set 2 2 k r U r k r k k r 2mFourier transform: U r UK e iK rK k r e ik r uk r e ik r ck-K e iK r e ik r ck+K e iK rK K2 2m c k K k-K K2ei k K r UK e iK r ck-K e K Ki k K r kcKk-Kei k
University of Illinois, Urbana Champaign - PHYS - 560
Density of States: g Definition: g d = number of states between and d (per unit volume) For a single band k , g d 1 V1 8 d3 k ,s dd23k1 4 3 dS dkddk dS = area element on constant energy surface k dk = change in k, along k k (perpendicular to d
University of Illinois, Urbana Champaign - PHYS - 560
Screening (Electron-Electron Interaction) Classical pictureEextClassical metal Ein 0The internal field is zero. Electrons are free to move; a layer of charge (screening charge) accumulates at the surface. The external field is screened. Quantum picture
University of Illinois, Urbana Champaign - PHYS - 560
Electron Dynamics (of Semiconductors) Basic assumption: independent electron approximation (single-particle picture) Question: what happens to an electron under an external field (transport properties)? Bloch state nk e ik r uk r p nk i nk e ik r p k uk
University of Illinois, Urbana Champaign - PHYS - 560
Lattice Waves Thus far, static lattice model. In reality, atoms vibrate even at T 0 because of zero-point vibration. Monatomic Crystals Basis = 1 atom.r R R u R;t Lattice: R ni aiii 1,2,3Actual atomic position = lattice position + vibrationu 0rave
University of Illinois, Urbana Champaign - PHYS - 560
Lattice Specific Heat Classical: Equi-partition theorem:1 1 1 kx x2 mvx 2 kT 2 2 26 degrees of freedom each atom 3 each from kinetic energy and potential energy. Internal energy/volume u Volume specific heat CV 33NkT ; N = total number of atoms (includ
University of Illinois, Urbana Champaign - PHYS - 560
Anharmonic Effects Assume one atom per unit cell. Potential energy U r R UR u R U U 0 harmonic potential u 2 O u 3 O u 4 Terms O u 3 , O u 4 anharmonic effects or phonon-phonon interactions uR 1 Nks2 M s (k )aks + a-ks e s (k )eik RO u 3 . a a a
University of Illinois, Urbana Champaign - PHYS - 560
Neutron Scattering Counts E E' E Spectrum at a fixed scattering geometry: elastic peak caused by defect scattering (or Bragg diffraction) one-phonon emission peaks at lower energies (Stokes peaks) one-phonon absorption peaks at higher energies (anti-Stoke
University of Illinois, Urbana Champaign - PHYS - 560
Infrared Properties (of Ionic Crystals) Assume the system is nonmagnetic, 1 . Optical properties determined by the dielectric function, D E. Assume cubic symmetry. D E Assume no free charges. D 0 DEP ik E 0 E ik E 4 ( = bound charge density) Longitudinal
University of Illinois, Urbana Champaign - PHYS - 560
Magnetism B H 4 M H M = magnetizationBi Bo Bournday conditions: B 0 4 1 D H j 0 , assuming steady state and no free current c c tH i H oFerromagnetic Paramagnetic DiamagneticM 0 for B a 0 (zero applied field); generally nonlinear BH ~ constant ~ 1 + O
University of Illinois, Urbana Champaign - PHYS - 560
Ferromagnetism A simple picture: ferromagnetism arises from "atomic magnets" aligned in the same direction. What is the interaction to align them? Each atomic magnetic has a magnetic moment ~ g0 B . Dipolar interaction energy between neighboring atomic ma
University of Illinois, Urbana Champaign - PHYS - 560
Crystallographic restriction theorem See: http:/en.wikipedia.org/wiki/Crystallographic_restriction_theorem Problem: Consider rotational symmetry operations of a 2D periodic lattice with the rotation axis perpendicular to the lattice plane. Show that only
University of Illinois, Urbana Champaign - PHYS - 560
Physics 560 Homework problem set 1 1. A&M Problem 1, Chapter 1. 2. A&M Problem 2, Chapter 1. 3. A&M Problem 3, Chapter 1. 4. Recall 1 i 4 , where is the conductivity derived from the valence electrons.This expression was derived assuming that the posit
Rutgers - FINANCE - 250
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Rutgers - FINANCE - 250
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Rutgers - FINANCE - 250
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University of Texas - PHY - 102M
Question 14 out of 4 pointsRead the derivation for acceleration of the 2 objects in Atwood machine system in your manual. If we have m1=150g, m2=155g, what's the acceleration (m1 goes up and m2 goes down)? Answer Selected Answer: 0.161m/(s^2)Question
University of Texas - PHY - 102M
Question 13 out of 3 pointsLook at Fig. 4-1 in your manual, let's say PEtot is the total potential energy of masses m1 and m2 (m1<m2). After m1 goes up by a distance d, and m2 goes down the same distance. The change in PEtot is (when something goes up
University of Texas - PHY - 102M
Question 13 out of 3 pointsPick the right descriptions for inelastic collision Answer Selected Answers:Question 23 out of 3 pointsPick the right descriptions for completely elastic collision Answer Selected Answers:Question 33 out of 3 pointsIf m
University of Texas - PHY - 102M
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University of Texas - PHY - 102M
Question 13 out of 3 pointsWhat are the necessary conditions for objects in translational and rotational equilibrium? is torque on the object(s), F is the force on the object(s), v is the translational velocity of the object(s), and is the angular velo
University of Texas - PHY - 102M
Question 13 out of 3 pointsOn graph paper draw a vector from the origin of a graph 8 cm at 0 degrees from the x-axis. Now add a vector at 90 degrees and 6 cm long to the first vector. What is the length of the addition calculated? What is the direction
Columbia State Community College - FREN - 1200
MBG*2040 Foundations in Molecular Biology and GeneticsMock Midterm Exam Winter 2012 It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: Complete the midterm in an hour and 15 minutes. Work on your own. Keep your not
Columbia State Community College - FREN - 1200
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Columbia State Community College - FREN - 1200
University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
Nom d'tudiant: Nicolas Charron Numro d'tudiant: 6390078 Nom et numro d'tudiant du partenaire: Katherine Noah 6377440 Nom du dmonstrateur: Sarah Ouanounou Date : 19 janvier 2012REMARQUE:Si l'information demande ci-dessus n'est pas CLAIRE ou n'est pas DON
University of Ottawa - BIO - 3170
Cher membre du jury, enseignent, parents, lve et amis Bonjours, comment mon optimiste m'aide surmonter mes obstacles, pour moi c'tait quand j'avais perdu mon chien aux Qubec. Sa toute commencer quand ont rapportait toute nous chose d'an dehors dans le cha
University of Ottawa - BIO - 3170
Fidle ou non!Il y a plusieurs roman et livre comme Harry-Potter qui sont fidles leur film, mais il y a aussi plusieurs romans et livre comme le voleur de foudre qui ne sont pas fidle a leur film ce qui m'amne a pos la question : Est-ce que la production
University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
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University of Ottawa - BIO - 3170
tat de condensation des chromosomes lors de la la division cellulaire est maximal lors de la mitose ET MIOSE V ou F - j'avais oubli de spcifi seulement lors*Chromatine est une molcule monocatnaire d'ADN associe des protines V ou FDans le euchromatine le
University of Ottawa - BIO - 3170
tat de condensation des chromosomes lors de la la division cellulaire est maximal lors de la mitose V ou F Chromatine est une molcule monocatnaire d'ADN associe des protines V ou F Dans le euchromatine lequel de ces noncs est faux : Liaisons non-covalente
University of Ottawa - BIO - 3170
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University of Ottawa - ANP - 1105
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University of Ottawa - ANP - 1105
Rapport de laboratoire Nicolas Charron (No 6390078) 1) Mise en situation : Bob est sous la douche 5 C pendant 2 heures Les donnes dmontrent que prendre une douche 5 C pendant deux heures peut causer une diminution de la frquence cardiaque, de 72 60 battem
University of Ottawa - ANP - 1105
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University of Ottawa - ANP - 1105
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University of Ottawa - ANP - 1105
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University of Ottawa - ANP - 1105
Fonctions principales : Drainer le surplus de liquide interstitiel: le liquide interstitiel est rcupr par les vaisseaux lymphatiques et retourn au sang. Transporter les lipides alimentaires: les vaisseaux lymphatiques transportent jusque dans le sang les
University of Ottawa - ANP - 1105
1) En observant mon graphique je peux observer que la Kalmia augustifolia a t vue dans plus ou moins dans l'cotone et le plus observ dans la tourbire comparativement dans le champ, la fort et le marais qui semble avoir environ le mme nombre faible d'obser
University of Ottawa - ANP - 1105
Recombinaison:Populations sexues: miose (brassage gntique) variations propices l'adaptation. *pas source de variations, mcanisme du gnome (erreur du miose). Accouplement au hasard: panmixie (maintien H.W.); pas d'volution, rare chez les animaux, pas de ch
University of Ottawa - ANP - 1105
Cladogramme initial. Aucune rsolution La mchoire (caractre 1) et l'estomac (caractre 12) sont prsents chez toutes les espces excluant la lamproie connue sous l'acronyme GF. Les tissus squelettiques osseux (caractre 8) et les poumons/drivs (caractre 5) son
University of Ottawa - ANP - 1105
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UGA - CLAS - 1020
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UGA - CLAS - 1020
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UGA - CLAS - 1020
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UGA - CLAS - 1020
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Coast Guard Academy - CHEM - 222
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Iowa State - STAT - 101
SECTION 15.2: Conservative Fields811Exercises 15.1In Exercises 18, sketch the given plane vector field and determine its field lines.11. v(x, y, z) = yi - xj + kxi + yj (1 + z 2 )(x 2 + y 2 ) 13. v(x, y, z) = x zi + yzj + xk1. F(x, y) = xi + xj 3. F
Iowa State - STAT - 101
916CHAPTER 17 Ordinary Differential Equations(a) Show that the solution of (B) remains between the solutions of (A) and (C) on any interval [0, X] where solutions of all three problems exist. Hint: We must have u(x) 1, y(x) 1, and v(x) 1 on [0, X]. (Why
Iowa State - STAT - 101
SECTION 16.3: Green's Theorem in the Plane8653 18.Suppose div F = 0 in a domain D any point P of which can by joined to the origin by a straight line segment in D. Let r = t xi + t yj + t zk, (0 t 1), be a parametrization of the line segment from the o
Iowa State - STAT - 101
SECTION 17.5: Linear Differential Equations with Constant Coefficients919Exercises 17.41. Show that y = e x is a solution of y - 3y + 2y = 0, andfind the general solution of this DE. y (x) 1 . y(x) = . , . yn (x) a (x)112. Show that y = e-2x is a so
Iowa State - STAT - 101
820CHAPTER 15 Vector Fields19. Show that the gradient of a function expressed in terms ofpolar coordinates in the plane is 1 ^ ^ r+ . r r (This is a repeat of Exercise 16 in Section 12.7.) (r, ) = ^ ^ 21. Show that F = r sin 2 r + r cos 2 is conservati
Iowa State - STAT - 101
824CHAPTER 15 Vector FieldsExercises 15.31. Show that the curve C given byr = a cos t sin t i + a sin2 t j + a cos t k, lies on a sphere centred at the origin. FindC12. Find the centroid of the curve in Exercise 11.(0 t z ds.3 14. 2 ), 3 13.Find