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Condensed Matter Physics J. Ellis (10 Lectures) Periodic Systems: Overview of crystal structures, the reciprocal lattice. Phonons: Phonons as normal modes – classical and quantum picture. 1D monatomic chain, 1D diatomic chain, examples of phonons in 3D. Debye theory of heat capacity, thermal conductivity of insulators. Electrons in solids: Free electron model: Fermi-Dirac statistics, concept of Fermi level, electronic contribution to heat capacity. Bulk modulus of a nearly free electron metal. Electrical and thermal conductivity. Wiedemann-Franz law. Hall effect. Nearly free electron model: Derivation of band structure by considering effect of periodic lattice on 1-D free electron model. Bloch’s theorem. Concept of effective mass. The difference between conductors, semiconductors and insulators explained by considering the band gap in 2D. Hole and electron conduction. Doping of semiconductors, p and n types, pn junctions – diodes, LEDs and solar cells. Books In general the course follows the treatment in Solid State Physics, J.R. Hook and H.E. Hall (2nd edition, Wiley, 1991). Introduction to Solid State Physics, Charles Kittel (8th edition, Wiley, 2005) is highly recommended. (need not be the latest edition) Another book, generally available in College libraries and may usefully be consulted is The Solid State, Rosenberg H M (3rd edn OUP 1988) Webpage http://www-sp.phy.cam.ac.uk/~je102/ 1 Condensed Matter Physics: Periodic Structures • Course deals with crystalline materials – can be extended later to amorphous materials. • Crystalline structure characterised by set of lattice points – each in equivalent environment, but not necessarily at the position of an atom. • Each lattice point will have associated with it one or more atoms - the ‘basis’. e.g. NaCl Structure Lattice = Basis * • (Mathematically, the lattice would be represented by an array of delta functions, and the crystal described by a convolution of the lattice with a function that described e.g. the electron density associated with the basis.) 2 Condensed Matter Physics: Unit Cells • Lattice described by a unit cell – which may have one lattice point per unit cell (a ‘primitive’ unit cell) or more than one (‘non-primitive). e.g. for cubic: • Primitive Cubic • Face Centred Cubic (fcc) Non primitive unit cell, 4 lattice points per unit cell Primitive unit cell, 1 lattice point per unit cell • Body Centred Cubic (bcc) Non primitive, 2 lattice points per unit cell • How many lattice points per unit cell? Either count those at corners and face centres with weight 1/8 and ½ respectively, or move whole cell so that no lattice points are on the sides/corners, and count lattice points inside the cell. 3 Bravais Lattices In 3D there are 14 different lattices – know as ‘Bravais’ lattices. P=primitive I= body centred F=Face centred on all faces A,B,C = centred on a single face Need to remember the P,I, and F forms of the cubic unit cells 4 Directions • Unit cells characterised by the 3 ‘lattice vectors’ (a,b, and c) that define their edges. e.g. for a face centred cubic (fcc) lattice Unit Cell Lattice Vectors c Non primitive b a Primitive c b a • Directions given in terms of basis vectors – a direction: r ua vb wc would be writen as [u,v,w]. _ _ _ • In a cubic lattice [100], [010], [001], [1 00], [01 0], [001] are all related by symmetry. They are together denoted by 100 . 5 Planes • The notation describing a set of uniformly spaced planes within a crystal is defined as follows: • Assume one of the planes passes through the origin • Look at where the next plane cuts the three axes that are defined by the three lattice vectors. • If the plane cuts the three axes at a/h , b/k , c/l , then the set of planes is described by the Miller indices (h,k,l), and {h,k,l} indicates all planes related to (h,k,l) by symmetry. • If h,k, or l is zero it indicates that the plane is parallel to the respective axis. e.g. (showing only the plane next to one that contains the origin) (111) z c/ (120) z l y y b/ b/ l a/ l x 2 a/ l x 6 Fourier Transforms and The Reciprocal Lattice • 1D periodic functions • A 1D periodic function, f(x)=f(a+x), can be represented as a Fourier series: ik x Ch e f ( x) h h where : k h 2h a The wave vectors used, kh are a uniformly separated set of points in 1D wave vector (k) space. • To illustrate how a 3D Fourier series is built up consider the orthorhombic case (a ≠ b ≠ c, 90° between axes) • In 2D, the coefficients Ch vary with y: Ch y eikh x f ( x, y ) • h But the function is periodic in y, so represent Ch(y) as a Fourier series: Ch ( y ) • C k hk eikk y where : k k 2k In 3D the Chk vary with z: f ( x, y, z ) b ikh x k k y C z e hk h , k • And again since the function is periodic in z, Chk (z) can be written as a Fourier series: Chk ( z ) • C l Hence the 3D series is: hkl eikl z where : kl 2l f ( x, y, z ) C h , k ,l • hkl c e i ( k h x k k y kl z ) k vectors, (kh,kk,kl), needed for the Fourier transform form a lattice in 3D reciprocal space known as the RECIPROCAL LATTICE. 7 The Reciprocal Lattice: the General Case • For a periodic function in 3D with a lattice described by lattice vectors a, b and c, all the wavevectors you need in 3D k space for a 3D Fourier transform representation are: G hkl hA kB lC (integer h, k , l ) Where : bc ca ab , B 2π , C 2π abc a bc a bc and : A a 2 , b 0, A c 0 etc : hence ' reciprocal ' A 2π (Always use primitive unit cells.) • The set of G vectors given by all possible integer values of h,k, and l is known as the reciprocal lattice. The G vectors are know as reciprocal lattice vectors. • A periodic function f(r) can then be expressed as the 3D Fourier series. iG hkl r f (r) C h , k ,l , hkl e • Since the dot product of a lattice vector (ua+vb+wc) with a reciprocal lattice vector Ghkl is 2(uh+vk+wl) – an integer multiple of 2 - if you move by a lattice vector the phase of the exponentials remains unchanged giving the same value for f(r) and the correct periodicity in real space. 8 The Reciprocal Lattice. An Orthorhombic Example • Orthorhombic: a ≠ b ≠ c, 90° between axes, a, b, c form a right handed set. • Reciprocal lattice vectors: bc bc 2 2 a a, a bc abc a 2π 2π B b, C c b c A 2π • View structure down ‘c’ axis: Reciprocal Space Lattice Real Space Lattice b a B=2/b A=2/a • If the angles between the a,b, and c axes are not 90° then a axis in real space will not necessarily be parallel to the A axis in reciprocal space. (See hexagonal example later.) 9 The Reciprocal Lattice and Miller Index Planes • The first plane (after the plane going through the origin) with a Miller index (h, k, l) goes through the points: a , b , c h k l • The normal to this plane is parallel to the cross product of two vectors in this plane, and hence to Ghkl : z c/ Ghkl l a h b k a h c l hkl1 hb c kc a la b a bc hA kB lC hkl G hkl y b/ k a/ h x • For a plane wave, wavevector Ghkl, the difference in phase between a point on the plane that goes through the origin, and a point in the plane shown in the diagram above is: G hkl a hA kB lC a 2 h h (The phase difference between two points separation r is k.r. Ghkl is perpendicular to the planes and so any vector, r joining a pair of points, one in each plane, will do.) • Thus the set of planes with Miller indices (h,k,l) are perpendicular Ghkl, and the phase of a wave, wavevector Ghkl changes by 2 between one plane and the next. The set of planes have the same spacing, therefore, as wavefronts of the wave with wavevector Ghkl 10 Reciprocal Lattice and Miller Index Planes: Orthorhombic Example • Examples of Miller indices and G vectors Real Space 3rd index undefined as we are looking in 2D Reciprocal Space (01•) planes G01• b B a A G000 Reciprocal Space Real Space (02•) planes G02• b B a A 11 Reciprocal Lattice: Orthorhombic and Hexagonal Examples Reciprocal Space Real Space (12•) planes G12• b B a Real Space A Reciprocal Space (10•) planes G000 B b A a G10• Note:[1] G vector perpendicular to planes, and of length inversely proportional to the plane spacing. [2] If the lines (=planes in 3D) drawn in the figure were ‘wave crests’ then the wavevector of that wave would be the associated G vector 12 X-ray and Neutron Diffraction • The diffraction of x-rays and neutrons from a solid is used to study structure. • The phase of a wave changes by k.r over distance r • The condition for diffraction from a crystal relates to the scattering wave vector ks, which is the difference: ks = kf - ki between the wavevectors of the outgoing (kf) and incoming (ki) beams. Scattering objects Extra phase ki.r incoming wave wavevector ki phase difference (kf –ki).r=ks.r r Extra phase kf.r • If the scattering wavevector is equal to a reciprocal lattice vector then since the product of a lattice vector with a reciprocal lattice vector is an integer multiple of 2, all equivalent points within the crystal (e.g. all identically located atoms) will scatter in phase and give a strong outgoing beam. i.e. the diffraction condition is: and: ks = Ghkl kf = ki + Ghkl 13 X-ray and Neutron Diffraction: Energy Conservation • Conservation of energy requires that the incoming and outgoing wavevectors (once the scattering particle is free of the crystal) must be of equal magnitude. • Condition for diffraction neatly represented by Ewald’s sphere construction: both ki and kf must lie on the surface of a sphere, and be separated by a G vector. Ewald’s Construction kf G ki • It is clear that it is quite possible that for a particular incident condition there is no diffraction from a crystal – the Ewald construction is quite specific on ki and kf . • Diffraction from powders overcomes this by having a large number of crystals in different orientations. 14 Strong Scattering of Waves in Crystals • Neutrons (provided the sample is typically thinner than 1cm) and x-rays are scattered at most once as they pass through a crystal. • If you try to send a beam of electrons through a crystal it is very strongly scattered – the mean free path depends on energy, but takes a minimum at 50-100eV of about 6Å in a typical metal. • If you imagine starting a beam of electrons inside a crystal with a particular wave vector k, it will quickly be scattered into a set of waves travelling with wavevectors k + Ghkl . • There will then be more scattering, but now, since diffraction simply adds a G vector to the intial wave vector, it will be from one of these new set of waves to another – indeed some may be scattered back into the original wave with wave vector k. • After a while a sort of equilibrium is reached with the rate of scattering out of a particular set of waves equalling the rate of scattering into it. Once this has happened, no further effect of the scattering can be seen, and this explains why despite the large scattering cross sections, as we shall see later, electrons can behave as if they move through a crystal unimpeded. • We shall see later that because electrons moving through a crystal with a certain wave vector (k) can in fact have some of their ‘probability amplitude’ in whole set of associated waves (wavevectors k + Ghkl ), one may have to allow for this by making a correction to the ‘effective mass’ that they seem to have. 15 Condensed Matter Physics: Phonons • Aims: • Lattice vibrations ‘normal modes’/’phonons’ • Establish concepts by considering modes of a 1dimensional, harmonic chain, both monatomic and diatomic. • Examples of phonons in a 3D lattice. • Debye theory of heat capacity • 3kB/atom at ‘high’ temperatures. a T3 at low temperatures. • What are high/low temperatures, concept of Debye temperature. • Thermal Conductivity of insulating crystals a T3 at low temperatures. a T-1 at high temperatures. • Strong effect of defects and specimen dimensions at low temperatures. Thermal Conductivity of Ge v. Temperature Thermal Conductivity/W/cmK 100 10 1 1 10 100 1000 0.1 Temperature/K 16 Atomic Motion in a Lattice • In a solid, the motion of every atom is coupled to that of its neighbours – so cannot describe motion atom by atom – use ‘normal mode’ approach instead. • The motion of a ‘harmonic’ system (objects connected by ‘Hook’s law’ springs), can be described as a sum of independent ‘normal modes’ in which the coordinates all oscillate at same frequency and maintain fixed ratios to each other. • Motion of atoms must be described quantum mechanically, but we will use the results that: • The displacement patterns of the classical normal modes are the same as the ratios of the coordinates in the quantum mechanical ones. • The energy of the quantum mechanical modes is expressed in terms of the frequency (w) of the classical mode: 1 E w n 2 • In a solid these quantised normal modes are called phonons. 17 Normal Modes - Classical View: 2 Coordinate Example k k k m m x1 x2 Spring tensions, left to right : F1 kx1 , F2 k x2 x1 , F3 kx2 Equations of motion : mx1 F2 F1 2kx1 kx2 mx2 F3 F2 kx1 2kx2 (1) (2) mx1 x2 k x1 x2 mu1 ku1 Subtacting (1) (2) : mx1 x2 3k x1 x2 mu2 3ku2 Adding (1) (2) : 1/√2 for normalisation u1 1 x1 x2 and u2 1 x1 x2 are the normal coordinate s 2 2 with independen t equations of motion, solutions : u1 A1 cosw1t 1 , u2 A2 cosw2t 2 where w1 k , w2 3k and A1, A2 are independen t amplitudes . m m In mode 1, u2 0 so x1 x2 . • Mode 1: In mode 2, u1 0 so x1 x2 . Mode 2: x1 1 1 1 1 u1 u2 x 1 2 2 1 2 1 1 1 1 cosw1t 1 A2 cosw2t 2 A1 1 2 2 1 18 • General solution. Normal Modes - Quantum View: 2 Coordinate Example Schroeding er' s equation : 2 2 x1 , x2 2 2 x1 , x2 V x1 , x2 x1 , x2 E x1 , x2 2 2 2m 2 m x1 x2 where : V x1 , x2 1 2 1 1 2 2 kx1 k x2 x1 kx2 2 2 2 however V x1 , x2 can be written as V u1 , u2 1 2 1 2 ku1 3ku2 2 2 i.e. Schroeding er' s equation separates in u1 , u2 coordinate s : 2 2 u1 , u2 1 2 2 2 u1 , u2 1 2 ku u , u 3 ku u1 , u2 1 1 2 2 2 2 2m 2 2m 2 u1 u2 and we can write u1 , u2 1 u1 2 u2 and : 1 2 2 1 u1 1 2 ku u 1 1 1 1 u1 2m u12 2 E u1 , u2 Each term has only one variable but their sum is constant, so each must be constant giving two Independent equations 1 2 2 2 u2 1 2 3 ku u 2 2 2 E1 E2 E 2 2 u2 2m u2 2 Where the standard SHO solution applies and : 1 1 E1 w1 n1 and E2 w2 n2 2 2 • The frequencies and amplitude ratios are the same as for the classical case, but the energy is quantized 19 Lattice vibrations • 1-D harmonic chain • Take identical masses, m, separation a connected by springs (spring constant, a): • This is a model limited to “nearest-neighbour” interactions. Equation of motion for the nth atom is: mun a un 1 un un un 1 mun a un 1 un 1 2un • We have N coupled equations (for N atoms). • Take cyclic boundary conditions – N+1th atom equivalent to first (will be discussed later). • All masses equivalent – so the normal mode solutions must reflect this symmetry and all have the same amplitude (u0) and phase relation to their neighbours, i.e. un1 un exp i , un1 un exp i 20 Lattice Vibrations: Frequency of Modes • Look for normal mode solutions • Each coordinate has time dependence: exp iwt • Substitute into equation of motion: mw 2un exp iwt a ei e i 2un exp iwt mw 2 a 2 2 cos 4a sin 2 2 w 4a sin m 2 • Phase, , only has unique meaning for a range of 2: makes most sense to consider w as a function of over the range – to , giving: - Phase 21 Lattice Vibrations: Nature of Modes • Can write the amplitude of the nth atom as:` un u0 exp in wt • Can write the phase difference between successive atoms in terms of a wavevector, conventionally written as q for phonons: qa • Now it is clear that the modes are waves travelling along the chain of atoms: na is the distance x along the chain un u0 exp iqna wt u0 exp iqx wt • The dispersion relation for these waves is: 4a qa sin m 2 w q • Since the phase, , only has unique meaning for the range –/2 to /2, q only has a unique value over the range: a q a 1 E w n • Energy stored in mode is 2 i.e. a ground state of energy ħw/2 plus: n phonons each of energy ħw. • Momentum of a phonon turns out to be ħq . • Velocity = w/q (If you can see the wave move you must have formed a wavepacket, so velocity is group velocity) 22 The Meaning of phonon wavevector q. • The wavevector q gives the phase shift between successive unit cells. • q is defined on the range a q a i.e. G G q 2 2 where G =2/a is the smallest reciprocal lattice vector. • q has no meaning between lattice points, so is equivalent to q+G. q+G Amplitude 0.8 q 0.3 0 1 2 3 4 5 6 7 8 -0.2 -0.7 -1.2 n Phase shift between lattice points is meaningless • Remember: a wave vector that is a reciprocal lattice vector gives a phase shift of 2n between two points separated by a lattice vector. • Free space is uniform, so a phase shift along a wave given by f=kr works for any r. In a crystal, space is not uniform - equivalent points are separated by a lattice vector, and f=kr only has meaning if r is a lattice vector. 23 Phonon dispersion • Dispersion curves w versus q gives the wave dispersion • Key points • The periodicity in q (reciprocal space) is a consequence of the periodicity of the lattice in real space. Thus the phonon at some wavevector, say, q1 is the same as that at q1+nG, for all integers n, where G=2/a (a reciprocal lattice vector). • In the long wavelength limit (q→0) we expect the “atomic character” of the chain to be unimportant. 24 Limiting behaviour • Long wavelength limit • dispersion formula 4a qa w sin m 2 q→0 sin qa 2 qa 2 • leads to the continuum result (see IB waves course) w q q 0 aa ma ; w q Continuum result Y - Young’s modulus - density Y • These are conventional sound-waves. • Short wavelength limit • “Atomic character” is evident as the wavelength approaches atomic dimensions q→/a. l=2a is the shortest, possible wavelength. • Here we have a standing wave w/q=0 wmax 4a m 25 Momentum of a Phonon: ħq • Need to extend our concept of momentum to something that works for phonons – a so called ‘crystal momentum’. • If, for example, a neutron hits a crystal and creates a phonon, we want a definition of phonon momentum such that momentum will be conserved in the scattering/phonon creation process. • For a static lattice we simply have diffraction, and to get a large scattered intensity all the scatterers have to scatter in phase, i.e. (kf - ki).r = 2n (r is a lattice vector: separation of identical atoms) and for this to be true kf - ki = G. • If the lattice is now distorted by a phonon, the way each atom scatters will be modified by an extra phase term, q.r, so, if the scattered amplitudes are all to add up, the scattering wavevector will have to give an extra phase difference between lattice positions of q.r. i.e: (kf - ki).r = 2n + q.r and kf - ki = G + q . • This means that on scattering the crystal changes momentum by ħ(G + q). ħG is the momentum transfer due to diffraction from the lattice causing the whole crystal to recoil, and so it is sensible to define the momentum of the phonon as ħq – after scattering either you have created a phonon momentum -ħq or you have annihilated one of momentum ħq. • But you say, you can’t just define momentum anyhow you like – surely it is something that exists and we have to measure it. Not at all. Momentum and energy entered physics as constructions created to make the maths of doing physics easy. Consider potential energy – to what measurable ‘real’ quantity can you add an arbitrary offset and everything is still ok? Why is energy conserved ?– because we carefully define all forms of energy so that it is, at least that is how the idea started. 26 Momentum of a Phonon: ħq,but is it reasonable? (non examinable) • The problem is that if you really do have a infinite uniform wave in an infinite lattice, there are as many atoms going forwards as backwards and it carries no momentum. • The total (classical) momentum is carried somehow by all the atoms in the crystal – what we are trying to do is divide it notionally between momentum carried by the phonon and that associated with motion of the centre of mass – so its complicated. • However, if you make a wavepacket out of a small spread of wavevectors, then if you give the wavepacket an energy ħw and add up the momentum associated with all the vibrating particles they don’t quite cancel out, but do indeed give ħq. • Effectively, after the neutron scatters the whole crystal starts to move, carrying momentum ħG, and inside the crystal is a wavepacket of vibrations travelling through the crystal that carries a net extra momentum of ħq. • If you are considering scattering of a neutron, you are not considering an infinite crystal, so one can reconcile normal momentum with crystal momentum. • To understand the infinite case (non quantum mechanically) – you have to take the limit of the wavepacket going to infinite length – which is approaching infinitiy in a different way from saying that we have uniform oscillations throughout the crystal and let the crystal size go to infinity, so you get a different result for the momentum when you go to the limit in a different way. • If an inifinte lattice has to supply ħG or ħq of momentum it does not change the state of motion of the lattice (i.e. the lattice does not start to move) because it has infinite mass. 27 1st Brillouin Zone • Periodicity: All the physically distinguishable modes lie within a single span of 2/a. • First Brillouin Zone (BZ) • we chose the range of q to lie within |q| /a. This is the 1st BZ. 1st Brillouin zone (shaded) • Number of modes must equal the number of atoms, N, in the chain and for finite N the allowed q values are discrete, separation 2/Na (see ‘waves in a box’ later). • To Summarise: Each mode (at particular q) is a quantised, simpleharmonic oscillator, E= ħw(n+1/2). Phonons have particle character – bosons: each mode can have any number of phonons in it with: Energy=ħw, Mom.= ħq, Velocity = w/q. The unique modes lie within the first B.Z.. 28 Measurement of Phonons • Basic principle: • Need a probe with a momentum and energy comparable to that of the phonons e.g. thermal energy neutrons for bulk, and He atoms at surfaces. X-rays can have correct wavelength, but the energy is so high it is hard to resolve the small changes induced by phonon interactions. (At l=1Å, energy is 12.4keV – typical phonon energies are up to 40meV) • Particle hits the lattice and creates/annihilates phonons. • Illuminate sample with a monochromatic beam – incident wavevector ki • Energy analyse scattered signal – peaks in signal correspond to single phonon creation/annihilation occurring at a particular kf. • Use of conservation laws • Energy of phonon (+ = creation, - = annihilation): 2 2 w ki k 2f 2m • Crystal momentum conservation for phonon creation: ki k f q G • Crystal momentum conservation for phonon annihilation: ki q G k f 29 Measurement of Phonons II • To measure energy of probe can use time of flight techniques – e.g. helium atom scattering (HAS): Rotating Disk Chopper • Time flight of individual atoms through apparatus – to determine energy transfer on scattering. HAS ToF data for Cu(100) surface Relative Intensity Single phonon creation peaks Elastic peak Phys. Rev. B 48, 4917, (1993) Energy Transfer/meV 30 Diatomic lattice • Technically a lattice with a basis mA mB • proceeding as before. Equations of motion are: m Au2 n a u2 n 1 u2 n 1 2u2 n mB u2 n 1 a u2 n 2 u2 n 2u2 n 1 Trial solutions: u2 n U1 exp i2nqa w t u2 n 1 U 2 exp i 2n 1 qa w t substituting gives m 2 w 2a U1 2a cos qa U 2 0 A 2a cos qa U1 mB w 2 2a U 2 0 homogeneous equations require determinant to be zero giving a quadratic equation for w2. w a 2 m A mB m mA mB mB 4m A mB sin qa 2 A Two solutions for each q 2 12 31 Acoustic and Optic modes • Solutions • q→0: • Optic mode (higher frequency) w a 2m A mB m A mB 2a Effective mass • Acoustic mode (lower frequency) w2 a m A mB m A mB 12 m A mB 2 m A mB 1 4 qa 2 m A m B 2aa 2 w q m A mB w2a/mB w2a/mA Periodic: all distinguishable modes lie in |q|</2a 32 Displacement patterns • Displacements shown as transverse to ease visualisation. • Acoustic modes: Neighbouring atoms in phase • Optical modes: Neighbouring atoms out of phase • Zone-boundary modes • q=/2a; l=2/q=4a (standing waves) • Higher energy mode – only light atoms move • Lower energy mode – only heavier atoms move 33 Origin of optic and acoustic branches • Effect of periodicity • The modes of the diatomic chain can be seen to arise from those of a monatomic chain. Diagrammatically: Monatomic chain, period a period in q is /a for diatomic chain Modes with q outside new BZ period ‘backfolded’ into new BZ by adding ± G=/a Energy of optical Acoustic and and acoustic modes optical modes split if alternating masses different 34 Diatomic chain: summary • Acoustic modes: • correspond to sound-waves in the longwavelength limit. Hence the name. w→0 as q→0 • Optical modes: • In the long-wavelength limit, optical modes interact strongly with electromagnetic radiation in polar crystals. Hence the name. • Strong optical absorption is observed. (Photons annihilated, phonons created.) w→finite value as q→0 • Optical modes arise from folding back the dispersion curve as the lattice periodicity is doubled (halved in q-space). • Zone boundary: • All modes are standing waves at the zone boundary, w/q = 0: a necessary consequence of the lattice periodicity. • In a diatomic chain, the frequency-gap between the acoustic and optical branches depends on the mass difference. In the limit of identical masses the gap tends to zero. 35 Phonons in 3-D crystals: Monatomic lattice • Example: Neon, an f.c.c. solid: • Inelastic neutron scattering results in different crystallographic directions (00) () ( 0) Phys. Rev. B 11, 1681, (1975) • Many features are explained by our 1-D model: • Dispersion is sinusoidal (nearest neighbour. interactions). • All modes are acoustic (monatomic system) 36 Neon: a monatomic, f.c.c. solid • Notes: (continued) • There are two distinct types of mode: • Longitudinal (L), with displacements parallel to the propagation direction, • These generally have higher energy • Transverse (T), with displacements perpendicular to the propagation direction • These generally have lower energy • They are often degenerate in high symmetry directions (not along (0)) • Minor point (demonstrating that real systems are subtle and interesting, but also complicated): • L mode along (0) has 2 Fourier components, suggesting next-n.n. interactions (see Q 3, sheet 1). In fact there are only n.n. interactions • The effect is due to the (110) fcc structure. Nearestneighbour interactions from atom, A (in plane I) join to atom C (in plane II) and to B C atom B (in plane III) thus linking nearest- and next-nearest-planes. A C I II III 37 Phonons in 3-D crystals: Diatomic lattice • Example: NaCl, has sodium chloride structure! • Two interpenetrating f.c.c. lattices • Main points: Phys. Rev. 178 1496, (1969) • The 1-D model gives several insights, as before. There are: • Optical and acoustic modes (labels O and A); • Longitudinal and transverse modes (L and T). • Dispersion along () is simplest and most like our 1-D model • () planes contain, alternately, Na atoms and Cl atoms (other directions have Na and Cl mixed) 38 NaCl phonons • Notes, continued… • Note the energy scale. The highest energy optical modes are ~8 THz (i.e. approximately 30 meV). Higher phonon energies than in Neon. The strong, polar bonds in the alkali halides are stronger and stiffer than the weak, van-der-Waals bonding in Neon. • Minor point: • Modes with same symmetry cannot cross, hence the avoided crossing between acoustic and optical modes in (00) and (0) directions. • Ignore the detail for present purposes 39 Conservation Laws and Symmetry • • Lagrangian Mechanics • Newton 2 normally considered in Cartesian coordinates: ‘F=ma’ • Can generalise to non-Cartesian coordinates, but now write equations of motion in terms of derivatives of the Lagrangian: L=K.E. – P.E. . Field of ‘analytical dynamics’ based on this idea. Conservation Laws • A key result of analytical dynamics is Noether’s theorem – for every symmetry in the Lagrangian (i.e. in the system), there is an associated conservation law. e.g. it turns out that: • If the system’s behaviour is independent of the time you set it going, energy is conserved. • If the system’s behaviour is independent of where it is in space , momentum is conserved. • If the system’s behaviour is independent of the its angular orientation, angular momentum is conserved. • In a crystal – space is no longer uniform but has a new symmetry – its periodic, so the law of conservation of momentum is replaced by a new law – the conservation of ‘crystal momentum’ in which momentum is conserved to within a factor of ħG. E.g. • Diffraction: wavevector allowed to change by factors of G • Phonon creation: k f ki G ki k f q G • Adding a G vector to a phonon’s wavevector does not change its properties, but its crystal momentum changes by ħG 40 A mathematical aside (for interest - non examinable). • L=K.E.-P.E (e.g. S.H.O. L 1 2 mx kx2 ) 2 • Equations of motion: Euler-Lagrange equations: SHO: d mx kx 0 dt mx kx d L L 0 dt qi qi • Conjugate momentum: pi L qi (SHO: d pi L 0 dt qi L mx qi ) Noether’s theorem • If L is independent of qi: dpi L 0 and hence 0 and pi does not vary w ith time. qi dt • Energy conservation? Under many circumstances the Hamiltonian H (defined as Nf H qi L L qi ) is the energy, and since dH L , if L does not have explicit dt t time dependence , H and hence energy is conserved. i 1 41 The Use of Conservation Laws • What do conservation laws tell you? • Conservation laws tell you what is allowed to happen – it is not possible to have an outcome of an event that violates a valid conservation law. • Unless conservation laws permit only one outcome, they do not tell you what will actually happen, nor how fast it will happen. e.g. conservation of crystal momentum inside a periodic solid tells you what possible outgoing momenta a diffracted particle may have, (k f k i G ) but they do not tell you how intense the outgoing beams will be. 42 Thermal Properties of Insulating Crystals: Heat Capacity • Thermal energy is stored in the phonons • Need to know how much energy is stored in each mode. • Need to know how many phonon modes there are. • Need to sum the thermal energy over all modes • Heat capacity is then the derivative of the thermal energy. • Energy stored in a phonon ‘normal mode’ • Each mode has an energy E=ħw(n+ ½) where n is the number of phonons in the mode. • The factor of ½ is the ‘zero point’ energy – it cannot be removed. Since thermal energies are taken to be zero in the ground state, it will be ignored in this treatment. • It the solid is in thermal contact with some fixed temperature ‘reservoir’ then the probability of the mode having n phonons relative to the chance of it having none is given by a Boltzmann factor: Pn= exp(-n ħw/kBT) 43 Energy/normal mode, continued • Calculate average energy stored in a particular normal mode (ith ) by averaging over all possible values of n (0 to ∞). Ei nw n 0 i exp( exp( nw i nw i n 0 k BT k BT ) nw n 0 ) i exp( nw i ) exp( nw ) n 0 where β 1 k BT i Denominato r is a geometric series ratio exp( w i ) : exp( nwi ) n 0 Numerator is 1 1 exp( w i ) of denominato r : w exp( w i ) n w exp( n w ) exp( n w ) i i i 2 n 0 n 0 1 exp( w i ) Hence average energy stored in a particular mode is : Ei w i exp( w i ) w i 1 exp( w i ) exp( w i ) 1 k BT Planck’s formula for a single oscillator 44 Heat Capacity at High Temperatures • Low temperature (kBT<<ħw) limit of energy: w i Ei exp( w i k BT ) 1 w i exp( w i k BT ) for small T • High temperature limit of energy (1>>ħw/ kBT) w i Ei exp( w i k BT ) 1 w i 1 w i k BT 1 k BT • How many phonon modes? • If a crystal contains N atoms, you need 3N coordinates to describe position of all N atoms and so there will be 3N normal modes. • Thermal behaviour of whole crystal at high temperatures: • Since each mode stores kBT of energy at ‘high’ temperatures, and there are 3N modes, then the total energy stored at high temperatures is 3NkBT and the heat capacity for kBT >>ħwi is 3NkB. (Basis of Dulong and Petit’s law, 1819 – heat capactiy/atomic weight constant.) 45 Debye Theory: The Aim • Thermal behaviour of whole crystal at intermediate and low temperatures. • At ‘non-high’ temperatures, Ei depends on ħwi so we need a way of summing the contribution all the modes: N w i Etotal i 1 exp( w i ) 1 k BT • The first step is to convert the sum to an integral: Etotal w w 0 exp( k BT ) 1 g w dw where g(w)=dN/dw is the ‘density of states and g(w)w gives the number of phonon states N with energies lying between w and w+w. • The actual g(w) is complicated – the Debye theory of heat capacity works by producing a simplified model for g(w) so that the integral for Etotal can be performed. 46 Boundary Conditions and Models for g(w): Permitted k values. • Reflecting B.C. • • Reflecting boundary conditions give standing wave states. At boundary may have node (photons, electrons) or antinode (phonons). Cyclic B.C. Reflecting B.C. n=3 n=3 2A n n kn A A n 2n kn A ln ln • n=1 n=2 n=1 A Cyclic boundary conditions • • • • • n=2 N+1th atom equivalent to 1st. Travelling wave solutions. 1D you can wrap into a circle, but cyclic b.c.harder to justify in 2 and 3D. Can consider repeating your block on N atoms with identical units to fill infinite space and requiring all blocks to have identical atomic displacement patterns. Ok classically, but hard to get the quantum mechanics correct. ‘Infinite’ extent: • As soon as you use the modes derived you make a wavepacket of some sort with zero amplitude at infinity, which fits any b.c. at infinity, but this method does not give density of states. 47 Debye Model: g(w) for ‘Waves in a Box’ • For small values of k (long wavelengths) phonons look like sound waves – with a linear dispersion relation w=vsk where vs is a mean speed of sound (see discussion later). • The Debye model assumes this is true for all wavelengths – not just long ones – i.e. it ignores the structure in g(w) due to the atomic nature of the material. • g(w) is calculated by assuming that the crystal is a rectangular box of side lengths A,B,C. We use reflecting b.c., though cyclic b.c. give same results • In each dimension the there must be a whole number of half wavelengths across the box so as to fit the boundary conditions, i.e. in each direction A=nl/2 and k=n/A. The total wavevector of the phonon must be: n k x A n y B nz C • Volume per state in k space is 3/(ABC) i.e. 3/V where V is the volume of the box. • (k not q is used in this derivation because the idea of waves in a box applies to many problems in physics – including black body radiation and the free electron model of a solid, and by convention k is used in these derivation.) 48 g(w) for ‘Waves in a Box’ II All states in the shell have same |k| • First, work out g(k) from no. of states, N, that have a wavevector of magnitude between k and k +k.These states lie within the positive octant of a spherical shell of radius k and thickness k. (k +ve for standing waves.) States uniformly distributed in k-space • For each phonon mode there are two transverse and one longitudinal polarisation, i.e. 3 modes per point in k space. 3 Polarisations/k state Volume of shell 1 state “occupies a volume” (3/ABC) Vol. of one state 4k 2 N g k k 3 k 8 3 . g k 3 ABCk 2 2 2 ABC Since N g k k g w w, g w g k dk dw 2 3 V w w dk For a sound wave k so 1 and g w 3 vs dw vs 2 2 vs (V ABC is the volume of the box) • For cyclic BC, states 2x as far apart –vol/state = 83/V but require full shell, not just +ve octant, – net result same g(k). • Can show (Wigner) results independent of shape of box. 49 Internal Energy in Debye Model • Heat capacity follows from differentiating the internal energy (as usual). • For the present we will ignore the zero point motion. • Need to make sure you integrate over the correct number of modes – use the fact that if there are N atoms in the crystal (volume V) then there are 3N modes. Debye suggested simply stopping the integral (at the ‘Debye frequency’, wD) once 3N modes have been covered, i.e. wD wD Vw D3 3Vw 2 3N g w dw 2 3 dw 2 2 v s3 0 0 2 v s • Internal energy U wD 0 w exp w kT 1 w D3 6 2 v s3 N / V g w dw No. of phonons in dw at w Energy per phonon (Planck formula) • Hence: U wD 0 3Vw 2 1 w dw 2 3 2 vs exp w kT 1 3V w D 3 1 2 3 w dw 0 2 vs exp w kT 1 (We can now see that the appropriate mean velocity is 1 1 1 2 where vL and vT are the longitudinal and v s3 3 v L3 vT3 transverse sound wave velocities) 50 Heat Capacity Within the Debye Model • Differential U to get heat capacity C: U C 3V 2 2 v s3 wD 0 1 dw exp w k B T 1 w3 U 3V T 2 2 v s3 wD 0 w exp w k B T 2 3 kT w dw 2 exp w k BT 1 1 9 Nk B w D3 w D 3 k T B 0 1 V k BT 3 2 3 6 Nvs w w exp w k B T d 2 k B T exp w k B T 1 k B T 4 3 T D T 4 e x 9 Nk B x dx 2 0 x e 1 D where the Debye temperatu re D is given by w D k B D C/3NkB T/D 51 Debye Temperatures Element Li Na D/K 344 158 91 Element C Si D/K 2230 645 374 200 105 V K Ge Cr Rb Cs 56 38 Sn Mn Pb Element Sc Ti Fe Co Ni Cu Zn D/K 360 420 380 630 410 470 445 450 343 437 • Debye frequency controlled by mass of atoms and stiffness of lattice. • As you go down a group in the periodic table (e.g. Li to Cs or C to Pb) the mass increases and the atoms become bigger and more deformable –so the rigidity goes down. • Transition metals tend to lie between 200-600K – so Dulong and Petit’s law works roughly at room temperature. • Note very high value for carbon – a light atom and a very rigid lattice. 52 Heat Capacity at Low Temperature • Check high temperature behaviour: T C 9 Nk B D 3 D T x 0 for T D e T C 9 Nk B D 3 ex 1 x D 0 T 2 4 e ex x 1 2 dx 1 1 x 1 2 1 x2 and : x 2 dx 3 Nk B • Limiting behaviour as T → 0. • At low temperature the higher frequency modes are not excited. Thus contributions to the integral for large w (>wD) can be ignored and wD replaced by . 3 T 4 ex C 9 Nk B x dx 2 0 x e 1 D 4/15 3 Integral = 4 4 T 12 C Nk B 5 D C T3 Debye, T3 Law • Note similarity to heat capacity of a vacuum – photons in black body radiation. 53 Measured Density of States • Example: Aluminium (shows common features) • Measured density of states compared with Debye approximation. • Both measured and Debye density of states are similar at low w, as expected (w q). • Debye frequency chosen to give same total number of modes (i.e. equal area under both curves) • Largest deviations where phonon modes approach zone boundary. • Measured curve is complex because the 3-D zone has a relatively complicated shape, and the transverse and longitudinal modes have different dispersions (as we have seen earlier). 54 Thermal Conductivity • Phonons and thermal conductivity • Phonons are travelling waves that carry energy and can therefore conduct heat. • Kinetic theory gives the thermal conductivity l z z = -l cos • Excess temperature of phonons crossing plane T dT dT z l cos dz dz • Excess energy in each phonon mode heat capacity of a phonon mode c ph T c ph dT l cos dz • l = mean free path – phonons assumed to thermalise at each collision. 55 Thermal Conductivity II • Number density of phonon modes, n number with speed c to c+dc fraction with angles to +d n f c dc 2 sin d d 2 sin d 4 sin d 2 speed normal to plane net heat per mode • Heat flux across plane H nf c dc sin d 2c cos c phl cos dT dz 0 0 1 dT 2 H c ph nl sin cos d c f c dc 0 0 2 dz 1 1 dT H c ph nl c cos 2 d cos c 1 2 dz 1 dT dT H c ph nl c 3 dz dz c ph nl c 3 Definition of thermal conductivity • Thermal conductivity 1 C cl 3 Mean free path Average speed Heat capacity per unit vol 56 Mean Free Path for Phonons. • Mean free path – limited by scattering processes • With many scattering processes add scattering rates: c l c l1 c l2 1 l 1 l1 1 l2 Thus, the shortest mean free path dominates. • “Geometric” scattering: • Sample boundaries (only significant for purest samples at low temperatures). • Impurities/grain boundaries: l independent of T. • Phonon-phonon scattering: • True normal modes do not interact with each other. • However, in an anharmonic lattice, phonons can scatter. As a phonon extends/compresses the bonds it changes the spring constants and one phonon can diffract from the grating of variable elastic properties produced by another phonon. LennardJones 6-12 potential V(r) Spring Const. r/r0 Mean phonon amplitude at 20K in Ne is 1% of mean nearest neighbour distance and gives significant changes in the spring constant. 57 Temperature Dependence of Thermal Conductivity of Insulators • Low temperatures: - few phonons, geometric scattering dominates, l constant. -C and hence a T3. • High temperatures: -C constant (3NkB) - no of phonons a T, so l a 1/T and a 1/T Thermal Conductivity/W/cmK • In insulators there are no contributions from free electrons. • In pure crystalline form the conductivity can be very high (larger than metals e.g. at 70K diamond has =12000W/m/K, at 300K Copper has =380W/m/K) • Non-crystalline systems have much lower conductivity l ~ local order e.g. at 300K for glass has l ~ 3Å, and rubber has l ~ 10-20Å. • Thermal conductivity shows strong temperature dependence. 1 Thermal Conductivity of Ge v. C c l 3 Temperature 100 aT3 10 a1/T 1 1 10 100 1000 0.1 Temperature/K • Intermediate temperatures – expect conductivity to be below 1/T and T3 asymptotes, (heat capacity is dropping and there is both phonon and geometric scattering) but actually get a large rise for pure, crystalline samples. 58 Thermal Conductivity at Intermediate Temperatures:Phonon-Phonon Scattering and Umklapp Processes. • A typical phonon-phonon collision process is coalescence: q3 q2 q1 • However, if the resulting phonon has a momentum that is simply the addition of the two colliding phonons – the collision is very ineffective as far as thermal conductivity goes – you still have the same energy going in roughly the same direction. • To really change the direction energy is flowing – the sum of the momenta of the incoming phonons must be outside the first Brillouin zone – so that you get an ‘umklapp’ (german for ‘fold around’) i.e. the new phonon momentum has a G vector subtracted from the sum of the old ones, which changes significantly the direction of travel. • As the temperature drops fewer phonons have enough energy to have enough momentum to give a resulting phonon that is produced by an umklapp process and so the mean free path and hence rise dramatically G q3 Umkapp q2 q1 Process 59 Umklapp Processes – a 1D example 1st Brillouin Zone q3= q1+ q2 q3-G L T q2 q1 -G/2 Direction of travel (group velocity) G/2 Phonons with enough q for umklapp poorly excited at intermediate temperatures • Direction of travel determined by group velocity dw/dq • To change direction you need to have enough momentum in the incoming phonons, q1+q2 so that q3 is outside the first Brillouin zone, so that its group velocity has a direction very different from q1 and q2. • By convention if q3 is outside the first B.Z. a G vector is subtracted to show it in the first B.Z.- hence the ‘umklapp’. 60 Condensed Matter Physics: Free Electron Model • Aims: • Assumptions of the free electron model • What states do the electrons occupy – the controlling fact is that electrons are fermions. • Success of free electron model: • Heat capacity of a free electron gas • Bulk modulus • Electrical and thermal conductivity ‘In a theory which has given results like these, there must certainly be a great deal of truth’ H.A. Lorentz • The limits of the free electron model. • Hall Effect - evidence for conduction by positively charged ‘holes’. • Does not explain differences between conductors/semiconductors/insulators, let alone properties of semiconductors. 61 Assumptions of Free Electron Model • All ‘valence’ electrons are free to move. • Atomic structure consists of full ‘shells’ or atoms + more loosely bound shells partially occupied by ‘valence’ electrons – so called because they participate in chemical bonding. • The FE theory assumes these valence electrons are free to move and explain the electrical conductivity of a metal. • Sodium, magnesium and aluminium assumed to have 1,2 and 3 free electrons/atom respectively • Positively charged ion cores are left behind. • The charge from the positive ions is represented as a uniform positive background. • Electron-Electron repulsion ignored. • The electrons are treated as independent particles. • There are a range of rather subtle reasons why these assumptions work so well… • Cyclic boundary conditions. • Best examples: Alkali metals (Li, Na, K, …); also noble metals (Cu, Ag, Au) 62 What States Do the Electrons Occupy? • Cyclic boundary conditions used as we want to consider travelling electrons. • Basic density of states in k space as before, but electrons have 2 spin states giving two states per permitted point in k space. Volume of shell Degeneracy 2 Spin up or down Vol. of one state [=(2/A)3] 8 3 . g k Vk 2 2 N g k k 2 4k k V 2 • Since thermodynamics is dominated by the energy of the states – we will need g(). dN dN dk dk g g k d dk d d 2k 2 d 2 k Since , 2m dk m 3 m Vkm V 2m 2 1 2 g 2 . 2 2 2 k 2 2 2 Vk 2 g 1 2 63 Concept of Fermi Level: Electron Distribution at T=0 • Electrons are fermions and obey Pauli exclusion principal – only one electron allowed per state. (2 spin states per k state – already included in g().) • At T = 0 electrons occupy lowest possible states and states fill up to the Fermi energy, F. (Na: 3.2eV, Cu: 7eV) Fermi energy • All states within a sphere of radius kF, the Fermi wavevector, are filled. • Determine kF and F from the requirement that we have N electrons (i.e. N filled states) 2 spin states/k state 4 3 k f N 2. 3 Volume of sphere Volume/k state 8 / V 3 k 3f 3 2 N 3 2 n V 2 2 2 2 F kF 3 2 n 3 2m 2m where n N V 64 The Fermi-Dirac Distribution: a Thermodynamical Aside (non examinable, but must know result) • What is distribution of electrons for T>0? • First law of thermodynamics: dU dq dW • Use state functions and add particle exchange: dU TdS PdV dN U N S ,V chemical potential • At const. V (keeps energy levels and g() the same). dS dU T dN T • Consider a state in equilibrium with a reservoir of particles and heat. In the ground state (electron state unoccupied) entropy of the reservoir is S0=kBln(W0) where W0 is the number of different reservoir configurations. • To occupy the electron state, 1 electron and energy must be transferred from the reservoir, whose entropy now is: S 0 dS S 0 T T k B ln W ln W ln W 0 ln W W 0 k B T • Rel. chance of state being occupied: W/W0=exp[ -(-)/kBT] • Normalise to obtain probability of state being occupied: exp k BT 1 p 1 exp k BT exp k BT 1 Fermi-Dirac Distribution 65 p 1 exp k B T 1 • At T = 0. • Low temperature limit of the Fermi-Dirac distribution (also known as the Fermi function) is a step function: • States fill up to energy starting at = 0. All higher energy states are empty. • For T ≈ 0 • At low T and for < , (-)/kBT is large and negative, so exp[(-)/kBT] is v. small and FD distribution 1. • At low T for > , (-)/kBT is large and positive, so exp[(-)/kBT] is v. large and FD distribution 0. • In thermodynamics, the value of (T) at T = 0K is known as the Fermi energy, F. • In semiconductor textbooks the value of (T) at any temperature is called the Fermi energy. • For T > 0 66 Chemical Potential, (T), for T>0 • Occupied states (F.D. dist. x g() ) • Chemical potential (T) for T>0K. • Require total number of electrons to be N: g N d 0 exp k T 1 B ~ constant an implicit equation for (T) • Solve numerically: key point – at low T, g() changes very little within kBT of , so (T) ≈ (0) at low T. (True for most practical situations – see Q2 on 2nd question sheet. 67 Thermal properties of metals • Thermal capacity • Electronic contribution to the heat capacity follows from differentiating the electronic energy w.r.t. T. g U el d Result from a straight0 exp k T 1 B C el U el T Nk B T 2 TF 2 forward, if lengthy, manipulation. See Ashcroft and Mermin p42-7 (TF = F/kB = ‘Fermi temperature’) • A similar result follows from a qualitative argument – emphasising the essential physics • Compare occupied states at T=0K with those at T0K 68 Electronic Contribution to the Heat Capacity of a Metal • cont….. • only electrons within ~kBT are active thermally. • There are nex g F kT such electrons • If we treat these excited electrons like classical electrons. That is, having kinetic energy 3kBT/2 per electron. U el nex 3k BT 2 g F 3k B2 T 2 2 A Cel • Recall: U el g F 3k B2 T T g F 3 V 2m 2 1 2 F 2 2 2 at T 0K N B 1F 2 εF εF 0 0 g d K 12 K 1F 2 2 32 d K F 3 g F 3N 2 F 3N 2k BTF • Combining A and B gives C el 3N T 3k B2T 4.5 Nk B 2kTF TF Note T dependence • Even more simply: fraction of electrons excited T/TF, energy gained per electron 3/2 kT, change in U 2/3kT2/TF, so Cel 3kT/TF 69 Electronic Heat Capacity cont….. • Absolute magnitude is much less than the classical result (3NkB/2). T/TF~10-2 at room temperature for typical metals. (The original freeelectron model (Drude model) assumed the electrons were a classical perfect gas - one of its many failures was that the heat capacity was far too large.) • Observed values are in good agreement. E.g. Sodium (Na) a “classic”, free-electron metal: (Cel)meas = 1.5T mJ mol-1 K-1 (Cel)f.e. = 1.1T mJ mol-1 K-1 • T dependence arises because, as the Fermiedge broadens, more electrons get excited. • N.B. the electronic contribution to the heat capacity of a metal is masked by a much larger contribution from vibration at all but the lowest temperatures. (See Q3 on second question sheet.) 70 Deviations from Theoretical Heat Capacity: Effective Mass 2 2 T 2 2 2 3 Cel NkB NkB T 3 n NTmn 3 2mkB 2 TF 2 2 (Valid at low T, ≈ arises from assumption that isT indep.) • Seen only at low T, otherwise masked by phonons. • At low temperatures: C = gT+T3, so plot: C/T v. T2 for Potassium • For K: gTheoretical =1.67mJmol-1K-2 gexpt =2.08mJmol-1K-2. • In free electron model, C el NTmn 2 3. Discrepancy arises because electrons move with an ‘effective mass’ m*. • For ‘nearly free electron metals’ main contribution: ions around an electron pushed around by moving electron, (‘electron-phonon coupling’) and the other electrons tend to avoid the first electron, => extra contribution to the K.E. of the system included in the FE model by increasing the effective mass of the moving electron. For ‘nearly free electron metals’ m*/m somewhat greater than 1: e.g. K: m*/m=1.25 Mg: m*/m=1.3: Al m*/m=1.48. • Other types of deviation: • Strong effects of the periodic potential of the ion lattice can get larger deviations, or e.g. for Zn and Cd m*/m <1 • Strong electron-electron correlations can give huge values (many 1000’s) for m*/m. 71 Pressure Due to Electron Gas • Consider the energy levels in a box under compression. When n the box is compressed the 4 single-particle wavelengths shorten and the K.E. rises. 3 2 Hence there must be an 1 outwards pressure. n 3 2 1 a = l/2 a=l • At T = 0, compressing the gas, we have ΔU = PΔV and hence: P U so we need U. V • Calculate <U>, the mean energy of an electron F F U g d g d 0 F 32 0 d 2 52 F 5 A F 0 1 2 d 2 32 3 F F 3 5 F N V 2 3 • for N electrons: U N U 3N F 5 P 0 U F 3 2 N F F V 5 3V P 2n F 5 N V 53 72 Bulk modulus of metals • Electron pressure in metals • The pressure of the degenerate electron gas also contributes to the mechanical properties of metals. • Isothermal bulk modulus, KT is defined as p F N V 2 3 KT V V T Eq. A, previous slide 2N p 2 N F P F 5V V 3 V2 Electron K.E. contribution 2 N F KT to the bulk modulus 3 V • Calculated values are of the right magnitude. We have neglected the attractive forces, due to the ion cores. Attractive forces make the metal more compressible (Experimental bulk modulus, Kexp, is usually smaller than KT, above). n=(N/V)/m-3 EF/eV Kexp/KT Li 4.6E+28 4.7 0.63 Na 2.5E+28 3.1 0.83 K 1.3E+28 2.0 1.03 Metal 73 Transport Properties of Electrons: Representation as Wavepackets • When considering a particle you usually want to consider a object that is located to some degree in space – combine fractions of basic k states to make a wave packet. • What about Pauli exclusion principle? • Start with N k states, and just as with normal modes, you can make N new states out of them. • The quantum mechanics does not change – its just a matter of which functions you chose a basis set to visualise/analyse what is going one, and some pictures are more helpful. Separation 2/A Start with N normal 1D k states Make a new single electron state as a wavepacket of k states You can make N new states, each by Pauli E.P. taking 1 electron 74 Dynamics of Wavepackets • dw v Electron velocity given by group velocity: dk • Its just as well that we are not using the phase velocity since in Q-mech, w has an arbitrary offset making phase velocity w/k meaningless.(only changes in w have meaning). • q of a phonon has a similar arbitrary offset- q is equivalent to q+G. • Effect of a force on a wavepacket? Work done by a force f in a time t : fvt If the electrons energy increases by its momenum also changes and k can be worked out from : d dw k k vk dk dk So fvt vk dk f dt • For an electron in a particular ‘k’ state, the force causes k to increase continuously with time – not easy to visualise with discrete k states – it would have to jump every now and again from one state to another-so consider the electron as a wave packet in k space. It moves continuously across the array of available k states: can be written : k k t 0 ft 75 Effective Mass, m*, of a Wavepacket • The velocity of a wavepacket is determined by the dispersion relation w(k): v dw dk Rate of change of velocity time : dv d dw d 2w dk 1 d 2 dk 2 dt dt dk dk dt dk 2 dt A But we would like to think of this in terms of an effective mass : dv m* f dt dk Since : f dt dv dk B we have * and equating A and B gives dt m dt dk 1 d 2 dk m * dt dk 2 dt 2 m* d 2 dk 2 • The motion of the electron wavepacket obeys the classical equations of motion for a particle of mass m* makes considering the motion mathematically much easier. (The caveat is that the wavepacket must be significantly larger than the lattice spacing) 76 Equation of Motion for the Electrons • If no collisions occur then the force on the electron is the Lorentz force and: dv f m* eE ev B dt • Need to allow for collisions – use idea that if the mean time to the next collision is t, then the probability of the electron surviving a time t without a collision is exp(-t/ t). After the collision the velocity is randomised, so its average value is 0 and hence after a time t the average velocity is: <v>=v. exp(-t/t) +0.(1- exp(-t/t) ) = v. exp(-t/t) i.e. the collisions induce an exponential decay of the mean velocity and their effect may be written: d v dt v 1 v exp t t t t • Since we are looking at the movement of wavepackets, any velocity we use is an average velocity and so the effect of collisions looks like a velocity dependent friction and may be included in the equation of motion by adding in an extra term to the wavepacket acceleration: dv 1 v * eE ev B dt m t dv v 77 m* eE ev B dt t Electrical Conductivity • All the electrons obey the same equation of motion, and so we can work out their average velocity, the drift velocity, in the presence of a constant electric field: v m* drift eE τ • Electron mobility, , is defined as: vdrift eτ E m* • Since the current density (j) is given by the product of the number of electrons per unit volume (n), their charge and drift velocity we have: j n e v drift ne 2t * E neE m • Hence Ohm’s law, with conductivity: ne 2t m * ne • As for phonons, collisions can occur either with defects in the crystal, or with phonons, and since we add collision rates: 1 1 1 t t phonon t defect • Electron-electron ‘collisions’ have to be handled by a much more sophisticated theory that deals with electron-electron correlated behaviour. 78 Electron Scattering and Resistivity • When an electric field is applied, all the electrons in the Fermi sphere start to move continuously though k space in the –E direction at a rate: dk 1 eE dt i.e. the whole Fermi sphere moves continuously together in k space. • Phonons have a momentum (wavevector) comparable to that of electrons, but their energies are typically up to 40meV, tiny compared to F, so phonon scattering can change the direction of an electrons wavevector strongly, but only change slightly the magnitude of the wavevector. • Similarly scattering from defects does not produce large changes in energy since again only phonons could pick up energy changes. • There must be an appropriate empty state available for scattering to occur. Since there are no free states within the Fermi sphere, only electrons near the Fermi surface can be scattered. 79 Electron Scattering and Drift Velocity • The majority of electrons that can be scattered lie at the ‘front’ of the moving Fermi sphere, and the scattering process results in them being transferred to the ‘back’ of the moving sphere and so equilibrium is reached with the Fermi Sphere shifted in k space – i.e. with the electrons having a net drift velocity. • The scattering rate, t, used on overheads 77 and 78 is therefore an average rate – when the electrons are in the middle of the Fermi sphere they are not scattered at all, and are then scattered strongly when the arrive a the ‘front’ of the moving sphere. Electrons ‘thermalise’ back to Fermi sphere by more phonon collisions ky phonon scattering kx Fermi sphere shifted * by k m v drift E Blurred edge of Fermi sphere 80 Temperature Dependence of Resistivity 1 t 1 t phonon 1 t defect • Scattering from defects is temperature independent • At high temperatures, all phonon states are populated and the number density of phonons is proportional to temperature. Since tphonon << tdefect, 1/t and hence the resistivity are proportional to temperature => use of resistance thermometers (usually Pt). e.g. for a ‘Pt100’ resistance thermometer: Temp/K 136.5 273 546 Res./W 54.8 100 202.4 • Different samples have different defect densities, so have resistivity v. temperature curves related to each other by a fixed offset – Matthiessen’s rule. Resistivity -> Sample A Offset temp. independent Sample B 81 Temperature -> 82 Thermal Conductivity 1 C c l 3 • Same basic formula as for phonons. Use values for heat capacity of the free electrons, and since only electrons near the Fermi level are scattered, use <c>=vF and l = vFt, 12 T 12 T nk B vF l nk B * 2 3 2 TF 3 2 m vF v t 2 F 2 nk B2Tt 3m * 2k B • Normal pure metals have a thermal conductivity about 100x that of insulators at room temperature – electron conductivity swamps phonon conductivity. • At room temperature and above (‘high’ temperatures) phonons dominate scattering, and so t a 1/T and thermal conductivity roughly constant with temperature. ky Phonon scattering to thermalise distribution kx Electrons in +ve direction hotter than in –ve direction – so Fermi sphere more blurred on right hand side 83 Wiedemann-Franz Law • Experimental Law • ‘For metals at not too low temperatures the ratio of the thermal to electrical conductivities is directly proportional to temperature with the same constant of proportionality for all metals’. • Points to the same basic mechanism for the two conductivities. • For free electron model: 2 2 k B2Tnt 3m* 2 k B T 2 * 3 e ne t m • The Lorenz number, L, is defined as L T • Theoretical value for L is 2.45x10-8 WW/K2. • Experimental values: Element Ag Au Cd Cu Ir Mo Pb Pt Sn W L/ 10-8WW/K2 2.31 2.35 2.42 2.23 2.49 2.61 2.47 2.51 2.52 3.04 • Good agreement between experimental values and theoretical values except at low temperatures where the the different phonons required for scattering in the two conductivities have different probabilities of excitation. • Good agreement gives strong support to free electron model. 84 Hall Effect: Derivation • B field applied perpendicular to a conductor – electrons pushed sideways until charges build up on opposite faces that produce an electric field that stops any further lateral Build up of positive motion. charge on top side y + EH B z + + + + + - - + + + + v - - - j x -ev x B Build up of negative charge on lower side Balanced by force from Hall voltage • Lorentz force zero at equilibrium: dv f m* drift eE H ev drift B 0 dt • Since j n ev drift then E B v drift • Hall Coefficient: RH 1 B j RH B j ne 1 ne 85 Hall Effect: Experimental Values • Hall Coefficient: R H (n = no. density for electrons) 1 ne • Look at ratio of measured Hall coefficients to value expected for 1 electron /atom, i.e. n 1 natom natom eRH where natom is the number of atoms/unit volume. • Some values of n/natom look reasonable (ish)… Na: 0.9 K: 1.1 Mg: 1.5 Al:3.5 • and some certainly do not….. Be: -0.2 Cd: -2.2 !!! THE CHARGE CARRIERS HAVE THE WRONG CHARGE SIGN 86 Nearly Free Electron Theory: Bloch’s Theorem • Now let’s consider electrons in a periodic environment. • The translational symmetry of the lattice means that the electron probability density |2| must have the same periodicity as the lattice – i.e. if you move a lattice vector it must remain unchanged. • The amplitude of the wave function can therefore only change by a phase factor . e.g. in 1D, for a unit cell length a we have. i ( x a) x e and : ( x na) x ein • As before with phonon’s we can write = ka where k only has distinct values within the first Brillouin zone, and we have: ika ( x a) x e • Given that x has this form, we can write : x uk ( x)eikx where uk x is now a function of periodicit y a. • Hence Bloch’s theorem: The solutions of the Schrödinger equation in a periodic potential must be of the form: k r uk (r)eik.r where uk(r) has the periodicity of the potential. 87 Matrix Representation (Q.mech H09, pg 345-363) • (x) can be represented as a sum of contributions from each of a set of basis functions (e.g. as a Fourier transform). The idea is similar to the ‘normal mode’ approach where instead of using the displacement of each atom, it turns out to be easier to work with normal modes. • In quantum mechanics – the eigenfunctions of any operator form a complete set of functions that can be used as a basis set to represent the wavefunction when you are considering the effect of that operator. • This approach is particularly useful when the wavefunction is dominated by only a couple of the basis functions. • With this representation of the wavefunction, matrix methods can be used to represent, for example, the Schrödinger equation. • e.g. in the present case (and now working in 1D to illustrate theory) uk(x) is periodic and can be written: uk ( x) n - 1 inG1x e A (where G1 2π ) a A = length of crystal and k(x): k x u k ( x )e Ck,n ikx Ck,n n - 1 i ( k nG1 ) x e Ck,n fk ,n A n - 88 Matrix Representation of the Schrödinger Equation • The time independent Schrödinger equation is then: Hˆ x x Hˆ Ck,m fk ,m Ck,mHˆ fk ,m m - Ck,p fk , p m - p - Ck,p fk , p p - premultipl ying by fk ,n gives : Ck,m fk ,n Hˆ fk ,m m - Inp Ck,p fk ,n fk , p p - Hnm Which can be written as an eigenvecto r equation : H nmCk,m Ck,n i.e. HC C m - H I C 0 and the eigenvecto rs of H give the coefficien ts Ck,n of the basis functions in solutions of the Schoedinge r 89 equation. Nearly Free Electon Model in 1D • Nearly free electron model: • Ions represented by a periodic potential. • If we assume that the potential is weak we will see that we only need include states of comparable energy. • Electron-electron interaction/correlations/exchange all ignored. • To illustrate the phenomenology, consider the 1D case. Take origin of potential at an ion location: a V(x) V x V p cos pG1 x where : G1 p 2 a 90 Values of Matrix Elements • Look at values of matrix elements: A fk ,n Hˆ fk ,m 0 1 i mG k x 1 i nG1 k x 2 2 1 e x V e A 2m x 2 A e 1 inG1x 2 mG1 k 2 imG1x dx e e 2me A0 A A 1 i nG1 k x V p ipG1x ipG1x i mG1 k x dx e e e e 2 A0 p Vp 2 mG1 k 2 n,m p n,m p n ,m p 2 2me • For simplicity taking only first term in V(x), the matrix eqn is: . . . E k ,- 2 0 V / 2 1 0 0 0 0 0 0 0 0 Ck , 2 Ck , 1 Ck ,0 k Ck ,1 Ck , 2 . Ck , 2 Ck ,1 Ck ,0 Ck ,1 Ck , 2 . K.E. of a free electron of wavevecto r nG1 k 91 0 V1 / 2 Ek , 1 V1 / 2 0 0 0 where : Ek ,n 0 0 0 0 0 V1 / 2 0 0 0 Ek ,0 V1 / 2 0 0 V1 / 2 Ek ,1 V1 / 2 0 0 V1 / 2 Ek , 2 . 0 0 . . 0 0 0. . . 2 nG1 k 2 2me Choose Significant States • So: • If we are looking for a solution to the S.E. that has a wavevector k, from Bloch’s theorem it must be in the form: k x u k ( x )e ikx C n - k,n 1 i ( k nG1 ) x e C k,n fk ,n A n - • If we take the free electron case, V(x)=0, then the solution is simply the n=0 term: 1 ikx A e fk , 0 • If we then add in a small V(x), then the n=0 term will remain dominant, but now there will be small contributions (coefficients Ck,n ) from the other basis states. • The larger the energy difference between the nth state and the main, n=0 state, the smaller Ck,n (See Qmech handout 9, p357-9). Consider n=1 row of the matrix eqn. on OH 91: V1 V1 Ck , 0 Ek ,1Ck ,1 Ck , 2 k Ck ,1 2 2 V1 V Ck , 0 Ek ,1 k Ck ,1 1 Ck , 2 0 2 2 Small Ignore Large V1 Ck , 0 Ck ,1 2 k Ek ,1 Small • For small potentials, we need only consider the contribution form the basis state with the energy closest to that of the main, n=0, component state. 92 Choose Significant States • K values of possible states iin 1D: Ek Free electron dispersion curve k G1 A k • For small potentials, only states |fk,n> with energy close to |fk,0 > will have significant contributions to the overall wavefunction (See Qmech. handouts p357-9). • Include state with the nearest energy (A). Ek G1 A k k • For small k this state is also very different in energy – so contributes very little – have a plane wave state k with an energy very close to free particle energy. 93 Behaviour Near Zone Boundary I • Two states now comparable in energy – significant contribution from both G1 Ek k • At k=G1/2, both states equivalent , and have equal contribution, giving two standing wave contributions: k 2 φG1 2 φG1 2 x 1 φG1 2 x 1 φG1 2 2 cos a x sin x a x a x cos 2 2 Low energy 2 x sin 2 a High energy x V(x) a 94 Behaviour Near Zone Boundary II • Two states of comparable energy – significant contribution from both: Ek , 1 V1 / 2 Ck ,1 C k k , 1 V /2 E C C k ,0 k ,0 1 k ,0 V1 / 2 Ck ,1 Ek , 1 k C 0 V / 2 E 1 k ,0 k k ,0 2 V Ek , 1 εk Ek ,0 εk 1 0 2 k2 V Ek , 1 Ek ,0 εk Ek , 1Ek ,0 1 2 2 Ek ,1 Ek ,0 εk Ek , 1 Ek ,0 2 Ek ,1 Ek ,0 2 V 4 1 2 k εk Ek , 1 Ek ,0 2 V 4 Ek , 1Ek ,0 1 2 V1 2 2 Stationary States at zone boundary Free electron Calculate in this region, use symmetry for rest 2 G1/2 k/G1 95 Nearly Free Electron Model: Effect of Potential • Higher Fourier components of potential: V x Vn cosnG1x n link states at higher energy and give gaps of Vn at k=nG1/2 • Potential ‘folds back’ space – k only meaningful in 1st Brillouin Zone. Reduced Zone Scheme Extended Zone Scheme V3 k 3rd Band V2 2nd Band V1 1st Band -1.5 -1 -0.5 0 0.5 1 1.5 k/G1 -0.5 0.5 1st Brillouin Zone 96 Metals, and Insulators: 1D • The material can only conduct if there are empty states accessible to the electrons – otherwise no change of state is possible • Separation of states in k space 2/A. But, A=na, so there are n k states, and 2n electron states in the first Brillouin Zone. • 1 electron per atom? First band ½ full → conductor. • 2 electrons per atom? First band full → insulator. • 3 electons per atom? 2nd band ½ full → conductor • But – get divalent metals – need full 3D picture. 1 electron 2 electrons 3 electrons /atom /atom /atom k V2 F V1 F F -0.5 0.5 -0.5 0.5 k/G1 -0.5 0.5 97 Metals, and Insulators: 2D Model of a Divalent Element Fermi Contour k kx/G1 Free electron model – parabolic dependence of k on |k| ky/G1 Empty states in 1st BZ Filled states Weak potential: energy of 2nd band overlaps with energy of 1st, Fermi contour crosses into 2nd B.Z. – empty states left in corners of 1st B.Z. → conductor All states in 1st B.Z. full Strong potential: Gap so large enough for there to be no overlap between 1st and 2nd band – so 1st band full → insulator 98 Conduction/Resistance k No Electric field – electrons in ground state k -G0/2 G0/2 In presence of electric field all states move, but are scattered back towards ground state f eE dk dt Scattering Thermalisation k 99 Bloch Oscillations If the scattering of the electrons is weak, then a constant electric field causes the occupied states to move continuously through k space – but the group velocity, and hence the position of the electrons, oscillate. dk f eE dt Group Velocity k Increasing Time k k 1st Brillouin Zone 100 Observation of Bloch Oscillations • Observation in solid state: reduce scattering rates (very cold sample, 4K), and create huge unit cell in 1 dimension (‘super lattice’ with 34 layers GaAs and 6 layers Al0.3Ga0.7As) giving a very small first Brillouin zone in k space – reducing time for the electron states to cross the zone. Submillimeter- wave radiation (THz) emitted at a frequency dependent on electric field [Phys.Rev. Lett 70,3319 (1993) • Observed for ultra cold Cs atoms held in a lattice made by optical standing wave, use lasers to push atoms with a constant force – observe following time dependence of atoms momentum [Phys. Rev. Lett 76, 4508 (1996)]. 101 Conduction of a Nearly Full Band No electric field – a few empty states at the top of the band E f eE dk dt All states move in k space vg Full band – no net motion of charge A few positively charged states moving in direction of field. HOLES 102 Hole Bands: definition Consider energy required to create a hole: The lower the starting state the more energy you put in/gets stored in the system Measure energy of empty state e w.r.t. top of band. Total energy of band= - e Total momentum of band=-ke Define: h = - e kh = - ke i.e. so ‘hole’ carries momentum and energy due to the empty state. Measure k of empty state, ke, w.r.t. top of band. Hole band: h = ħ2ke2/2mh Corresponding hole: h , k h Missing electron: e , k e 103 Hole Bands: illustration of rational I Electron Behaviour (i.e. of the electron in a full band that is not ‘cancelled out’ by empty state)_ Equivalent Hole Behaviour Momentum: Net momentum due to electron in state on opposite side to hole, and k of band (=momentum/ ) on opposite side to k of unoccupied state K of hole gives correct k and momentum due to presence of unoccupied state in band Current: Net current due to electron moving in opposite direction to group velocity at hole position Net current due to positive hole moving as expected from group velocity I I vg vg Net current carried by to electron in state on opposite side to hole 104 Hole Bands: illustration of rational II Electron Behaviour Equivalent Hole Behaviour vg Effect of electric field: Before acceleration vg E After acceleration: Missing negative charge has accelerated in direction of field – WRONG WAY – negative mass Positively charged hole has accelerated in direction of field as expected – so positive mass vg vg 105 Status of ‘Hole Model’ 2 m * • Effective mass (overhead 76: d 2 dk 2 • At top of a band – negative curvature, and negative mass, and electrons accelerate in opposite direction to force – all rather confusing. • Much better to consider motion of holes, with positive charge and positive mass, that move in the expected way: E Hole band looks just like a free particle band. vg k • As ever – all we want is a model/picture that helps us ‘understand’ what is going on and enables us with a minimum of mental gymnastics and calculation predict how a solid will behave. 106 A 2D Model of a Divalent Metal Extended zone scheme Reduced zone scheme Holes Filled states in 1st B.Z Filled states in 2nd B.Z Back to extended zone. Pockets of holes that look like 2D free particles with same m* in x and y directions. Pockets of electrons with different m* in x and y 107 directions. Observation of holes: Hall Effect • Hall effect • Holes and electrons both conduct electricity: e ne e 2t e me* , Current density : nh e 2t h h mh j e h E • If te=th and ne=nh as is the case for a divalent metal, then, since the curvature of the bottom of the 2nd band is greater than that of the top of the first, m*e>mh and hole conduction dominates, giving a Hall coefficient characteristic of positively charged carriers. • (For an evaluation of the Hall coefficient, see, for example, Hook and Hall, pg 154.) 108 Measurement of Effective Mass: Cyclotron Resonance • Cyclotron resonance • In the presence of a B field, charged particles perform circular orbits around the lines of B of frequency w=eB/m. B Holes F=ev x B=mwv Electrons kh moves around a circle in k space • To see a resonance, charged particle must at least travel 1 radian, so wt>1. To achieve this w must be high (typically ≈ a few 10’s of GHz) – so large magnetic fields are needed (i.e. a few Tesla) (use a superconducting solenoid), and very pure samples a very low tempatures (typically 4K) to increase t. • From observed resonant frequency (i.e. a strong adsorption of radio waves), m*e or mh can be determined. • dw Remember its the group velocity dk that is important -so where the origin (k=0) is does not matter. 109 Doping of Semiconductors • So far we have considered only holes and electrons intrinsic to the material. • For e.g. a divalent metal these exist at absolute zero. • For a semiconductor the lower (valence) band is full at absolute zero and there are no conductors. Thermal excitation at non zero temperatures produces pairs of holes and electrons. • Can add impurities (‘dopants’) that replace a few of the semiconductor atoms. • e.g. adding a group V atom (e.g. P or As) leaves a spare electron after all the normal bonding (valence band) states are full, and so it must go to the conduction band. However, if the electron moves away from the group V atom, it leaves the atom positively charged – and so the electron forms hydrogen like ‘donor’ states out of the available free levels at the base of the conduction band, which will have an energy below the bottom of the conduction band and mean radius given by: En me*e 4 2 n 40 2 2 2 2 rn n 2 2 me*e 2 40 H atom formulae but using and m* • E.g. for germanium me*=0.2me and the dielectric constant =15.6 giving: me* 13.6eV 0.01eV E1 2 m e me 40A 0.53A r1 m* e Radius much larger than inter110 atomic spacing – justifies ‘effective mass’ approach – see OH76 H-Atom Like Impurity States • Before looking further at dopant states in a semiconductor – look at analogue in free space: the hydrogen atom. • The electronic wavefunction in hydrogen can be written (in Fourier transform form) as a sum of plane waves. • Before a hydrogen nucleus is introduced these plane waves can be represented as a free electron band: Free electron ‘band’ 0 k H atom energy level (only one shown) • You can consider the H atom wavefunctions to be made out of a sum of these free atom states – the presence of the electrostatic potential makes a state with an binding energy – but no k dependence since it is an isolated object, not a periodic one – so the state is represented as a flat line on an v. k diagram. • For every defect state you create – you loose one state from the band of states that it is created from. 111 n type Doping: Donor States • Replace a silicon atom by e.g. a phosphorous atom that has its core shells full and the same number of valence electrons as Si. The system has the same no of electrons as pure Si, but the P atom carries a single + charge. Si Si Si Si + P Si Si Si Si Conduction band: empty Valence band: full • All states in valence band full – a log jam, no flexibility to make new states, but the energy of all the states is slightly changed. • The spare electron goes to the conduction band and is either: ‘Bound’ ‘Ionised’ Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si + -P Si Si Si + P Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si e in H atom like Si Si e- e- far away+free or: e state made of C. band states but lower in energy because e- feels potential of P+ 112 p Type Doping:Acceptor States • Now replace a silicon atom by e.g. a Al atom that has its core shells full and the same number of valence electrons as Si. They system has the same no of electrons as pure Si, but the Al atom carries a single - charge. • The electron has to come from somewhere – takes an electron from the valence band leaving a hole that either is: ‘Bound’ V. Band lacks an electron – leaves a state to be made into a localised state (no k dep.) or: Bound state occupied by e- Si Si Si Si Si Si Si Si Si Si Si h+ Si Si Si Al Si Si Si Si Si Si Si Si Si Si - The empty electron state and full valence band are upside down to the ‘full hole state and empty hole band’ picture. ‘Ionised’ h+ Si Si Si Si Si Si Si Si Si Si Si Si Al Si Si Si Si Si Si Si Si Si Si Si Si - hole far away + free 113 p/n Type Impurity States in Si K Space Picture Conduction band EDonor: 0.045eV below C. band edge for P in Si Gap energy: 1.08eV for Si Valence band EAcceptor: 0.057eV above V. band edge for Al in Si T=0K Real Space Picture Conduction band: empty Acceptor/donor states Valence band: full Distance in Space 114 Chemical Potential of Doped Semiconductor p • n type 1 exp k BT 1 T=0K Conduction band: empty Donor states full 1 when 2 To see where will lie, ask where half occupied states would be T>0K Donor State Ionised Valence band: full • p type T=0K Conduction band: empty Holes left by thermal excitation moves towards middle of gap as T increases. At higher T no. of thermally excited e- increases and presence of donor states becomes less important. Valence band: full Acceptor states empty (=full of holes) T>0K Acceptor State Ionised = occupied by e- Holes left by thermal excitation 115 pn Junctions • Basis of many electronic devices. • Chemical potential, , is the effective ‘filling level’ of the states: before p and n type material contact the n type is filled to a higher level. n-type T>0K p-type e- diffusion Hole diffusion • When p and n type contact, electrons diffuse from n (high concentration) to p (low) (i.e. from high to low ) and holes diffuse the other way. • This charge separation produces an electric field at the junction that stops further diffusion and produces an electrostatic potential difference that evens out the difference in . (within the p or n type material fixed by requirement of charge neutrality, ‘contact potential’ raises energy of whole of p type region w.r.t n type region) 116 pn Junctions p-type E field in junction stops carrier diffusion n-type Ej ef Contact potential Width of depletion layer such that contact potential evens out Depletion layer ndonor nacceptor Carrier Density Charge Density E field in junction end ena f Electrostatic potential f 0 dE x x dx 0 E Ej 117 Position of junction pn Junctions: Diode: Basic idea Forward Bias Junction: Apply a field (potential difference) that forces majority carriers towards junction => conducts p-type n-type ‘minority’ carrier generated by thermal excitation V ‘majority’ carriers due to doping Applied potential EBias V Forward bias acts mainly across junction and reduces contact potential Reverse Bias Junction (not shown): Apply a field that forces majority carriers away from junction and increases width of depletion layer => conducts v. poorly since there are very few minority carriers. 118 pn Junctions: Diode: Generation and Recombination Currents n-type Electrons in this region can still cross barrier e- Recombination Current e(fV p-type e- Generation Current V h+ Generation Current h+ Recombination Current • Carriers diffuse towards junction. • Majority carrier diffusion creates a ‘recombination current’ – when they diffuse across the junction they will need to recombine with holes as (for a forward bias) there will more present than the equilibrium (F.D. distribution) number. • Minority carrier diffusion creates a ‘generation current’ – when they diffuse across the junction they enter a region depleted by diffusion the other way, and so are needed to maintain the F.D. distribution of carriers. 119 pn Junctions: Diode: Current Characteristics Virtually no electrons at top of band Occupation probability Increased by expV k BT I0,e I0,e exp[V/kBT] V No bias exp k BT exp k BT 1 So if increases to μ V then p becomes : pV exp V k BT exp k BT exp V k BT Far above : p0 1 p0 exp V k BT For no bias the recombination current ( ) balances the generation current ( ). The bias does not does not change the generation current I0,e. Number density of electrons in the n type valence band and hence the recombination current is increased by exp[V/kBT], so net current due to electrons is: . I e I 0,e exp V k BT 1 An identical argument applies to the holes, so the total current through the diode is given by: I I 0 exp V k BT 1 120 (Note: we are assuming that the density of state g() is roughly constant within the bands.) Breakdown Under Reverse Bias: Zener Diodes • pn diode I/V: I I 0 exp V k BT 1 I Tiny reverse current due to thermally excited minority carriers V • A Zener Breakdown • Caused by tunnelling at large reverse bias (typically above 3V) • Used as a reference voltage in circuits Tunnelling through barrier I p-type V n-type 121 Breakdown Under Reverse Bias: Avalanche Breakdown • Avalanche Breakdown Field from reverse bias so strong that thermally excited carriers within junction gain enough energy between collisions to create electron hole pairs that then create more pairs etc p-type n-type 122 Light Emitting Diodes p-type n-type Electrons and holes recombine in depletion layer – if ‘direct band gap’ then light emission can be most favourable process– get light emitting diode with colour determined by band gap Direct Band Gap: Indirect Band Gap: Photon emission easy (e.g. GaAs) Photon emission requires momentum from phonon – slow (e.g. Si) q k Photon k =w/c so very small – almost vertical transition 123 p-n Junction Lasers Basic Laser Principle n2 Occupation probability Electrons driven up and down by light n1 Photon energy matches energy between levels If n2>n1 (‘population inversion’) then more light emitted than adsorbed – get Light Amplification by Stimulated Emission of Radiation Diode Laser Heavily doped semi-conductor chemical potential in conduction/ valence bands. For strong forward bias, get large no of carriers injected into other side of junction – hence population inversion and possibility of laser action. p-type n-type Carrier injection Below , so n2>0.5 Above , so n1<0.5 124 pn Junctions as Solar Cells Ej p-type n-type Carriers pairs created by photon adsorption are swept out of the junction by the electric field, creating a photo current IL. Equivalent Circuit IL Current source Output Impedance ID IS Shunt Resistor I Diode Characteristics- key ID to V/I of cell IL-IS-I I = IL-ID-IS Open circuit voltage drops as I increases V 125 126 Appendix:Tight Binding Model (non examinable) • Nearly Free Electron Model • Derived as a deviation from free electron model. • Basic view: electrons free to move through a metal as plane waves, but perturbed by a modest potential caused by the ion cores. • Reasonable starting point model for many metals, where outer most electrons relatively weakly bonded to the atoms. • The structure of the ion lattice is dominated by packing of the ions and a trade off between potential energy gain (the closer the ions the stronger the average potential experienced by the electrons) and kinetic energy of the electrons – some metals (potassium, iron at room temperature) are b.c.c. not the close packed f.c.c. . • Tight Binding Model • For many materials the bonding is clearly covalent not metallic – e.g. Si, Ge, organic conductors. • Bonding and structure dominated by overlap/ alignment of atomic orbitals. • An expectation that the electrons are localised in covalent bonds. 127 (This presentation of the TBM is from a previous course by Dr. W. Allison.) Spatial Extent of Orbitals • Spatial extent of the wavefunctions is determined by their l-value. • For example: hydrogen: • Spatial extent increases with n, an obvious consequence of orthogonality. • Higher l-values are more compact than lower l, for a given n. This too is a consequence of orthogonality. The angular variation in higher l-values allows orthogonality with the lower n-states without the need to expand in r. • The spatial extent of an orbital determines the overlap with neighbouring atoms 128 Tight Binding Model: the basis states. • Tight binding model • The simplest and most intuitive model describing the electronic structure of a solid comes from tight-binding theory. • Atomic orbits are used to construct Bloch states. • First write the wavefunction, for a Bloch state, labelled, q, as a sum of functions on lattice sites, j. q r exp iq.r j r r j j • The expression defines f(r), known as a Wannier function. • It is easy to show this has the correct, Bloch form q r R exp iq.r j r R r j j exp iq.R exp iq. r j R r r j R j exp iq.R k r simply another lattice vector, rm, say 129 Tight-binding model f(r) can be represented as a sum over atomic states r cnlmnlm r o r nlm • The approximation, above, uses a single atomic o state r . It is the simplest and is useful provided the state is well localised. • Then the energy of the state, k is q q H q exp iq.r j rm mo H oj j m • For simplicity take a 1-D solid with atoms at r = ma. state, f o, at site m • We can neglect all terms except the nearest neighbours j=m, m±1, since the matrix elements decrease rapidly with the separation of the wavefunction centres. Hence we get 3 terms q a exp iq.0 exp iqa exp iqa a mo H mo overlap integral mo H oj 130 Energy bands • The energy band becomes number of nearest neighbours q a 2 cosqa • Width of band directly proportional to the orbital overlap, , and the bandwidth = 4. • Note the factor 2 arises because we have two neighbours contributing. The analysis can easily be extended to 3-D. Assuming a cubic crystal we will have 6 neighbours (2 in x-direction, 2 in y and 2 in z). The band structure is q a 2 cosq x a cosq y a cosqz a • The bandwidth is now = 12. • We note that the other factor in determining the bandwidth is the number of neighbours (coordination number). 131 Shells and bands • Valence band • For s-states < 0; p-states > 0. (N.B. in H, V<0) • Spatially extended atomic states gives overlap and hence broad bands. • s-p bands typically 10-20eV • d bands usually narrower since d -states are more compact (less overlap) • Core states • Highly compact with negligible overlap • bandwidth is negligible and the states are essentially the same as atomic states. 132 Scattering and Coalescence I • A typical phonon-phonon collision process is coalescence: Real Space q3 Outgoing phonon q2 q1 Lattice distorted by a phonon Incoming phonon • Anharmonicity allows the incoming phonon q1 to diffracts off another phonon, q2 producing an outgoing phonon q3. • Usual diffraction condition obeyed: q3 = q1 + q2 133 A1 (c.f. kf = ki + G ) Scattering and Coalescence II • The diffraction condition means that the outgoing phonon picks up momentum both from the incoming phonon and the phonon from which in incoming phonon is diffracted: q3 q3 q2 q3 q1 q 2 q1 q1 q2 so one quantum has been taken from the incoming (1) and the scattering (2) phonon state and combined in the process of coalescence into the outgoing phonon (3), that will also have to take an energy that is the sum of the energies of the incoming and scattering phonon quanta: w3 w1 w2 • If a diffraction process occurs that diffracts the orignal phonon in the direction –q2, the momentum equation becomes: q3 q1 q 2 q1 q 2 q3 and the energy: w3 w1 w2 w1 w2 w3 i.e. the incoming phonon has decomposed into two outgoing phonons. q2 q3 q2 q1 q1 q3 • Remember the zero point motion – even if there are no quanta of energy in a mode – the zero point motion 134 A2 still distorts the lattice and so can induce diffraction.