Solid State Physics Lecture notes

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1 Solid State Physics Semiclassical motion in a magnetic ¯eld 16 Lecture notes by Quantization of the cyclotron orbit: Landau levels 16 Michael Hilke Magneto-oscillations 17 McGill University Phonons: lattice vibrations 17 (v. 10/25/2006) Mono-atomic phonon dispersion in 1D 17 Optical branch 18 Experimental determination of the phonon Contents dispersion 18 Origin of the elastic constant 19 Quantum case 19 Introduction 2 Transport (Boltzmann theory) 21 The Theory of Everything 3 Relaxation time approximation 21 H O - An example 3 2 Case 1: F =¡eE 21 Di®usion model of transport (Drude) 22 Binding 3 Case 2: Thermal inequilibrium 22 Van der Waals attraction 3 Physical quantities 23 Derivation of Van der Waals 3 Repulsion 3 Semiconductors 24 Crystals 3 Band Structure 24 Ionic crystals 4 Electron and hole densities in intrinsic (undoped) Quantum mechanics as a bonder 4 semiconductors 25 Hydrogen-like bonding 4 Doped Semiconductors 26 Covalent bonding 5 Carrier Densities in Doped semiconductor 27 Metals 5 Metal-Insulator transition 27 Binding summary 5 In practice 28 p-n junction 28 Structure 6 Illustrations 6 One dimensional conductance 29 Summary 6 More than one channel, the quantum point contact 29 Scattering 6 Scattering theory of everything 7 Quantum Hall e®ect 30 1D scattering pattern 7 Point-like scatterers on a Bravais lattice in 3D 7 superconductivity 30 General case of a Bravais lattice with basis 8 BCS theory 31 Example: the structure factor of a BCC lattice 8 Bragg's law 9 Summary of scattering 9 Properties of Solids and liquids 10 single electron approximation 10 Properties of the free electron model 10 Periodic potentials 11 Kronig-Penney model 11 Tight binging approximation 12 Combining Bloch's theorem with the tight binding approximation 13 Weak potential approximation 14 Localization 14 Electronic properties due to periodic potential 15 Density of states 15 Average velocity 15 Response to an external ¯eld and existence of holes and electrons 15 Bloch oscillations 162 INTRODUCTION derived based on a periodic lattice. However it appeared that most of the properties are very similar independent of the presence or not of a lattice. But there are many What is Solid State Physics? exceptions, for example, localization due to disorder or Typically properties related to crystals, i.e., period- the disappearance of the periodicity. icity. What is Condensed Matter Physics? Even in crystals, liquid-like properties can arise, such Properties related to solids and liquids including crys- asaFermiliquid,whichisaninteractingelectronsystem. tals. For example: liquids, polymers, carbon nanotubes, The same is also true the other way around since liquids rubber,... can form liquid crystals. Historically, SSP was considers as he basis for the un- derstanding of solids since many of the properties were Hence,thestudyofSSPandCMParestronglyinterre- lated and can not be separated. These inter-correlations are illustrated below. The Big Bang of Condensed Matter Physics Crystal Crystal Supraconductivity Opto-effects Magnetism Metals Insulators Semiconductors properties 3 THE THEORY OF EVERYTHING See ppt notes H O - AN EXAMPLE 2 See ppt notes BINDING What holds atoms together and also keeps them from FIG. 1: Graphical representation of a H molecule 2 collapsing? We will start with the simplest molecule: H to ask 2 what holds it together. The total Hamiltonian can then be written as the sum betweenthenoninteractingHatomsandthecrossterms Van der Waals attraction due to Coulomb interactions. Hence, H =(H )+(H ) (1) TheVanderWaalsforcearisessimplyfromthechange 0 int in energy due to the cross Coulomb interactions between atomaandb,whichissimplyadipole-dipoleinteraction. with µ ¶ µ ¶ 2 2 2 ¡ ¢ e e 1 1 1 1 2 2 2 H = ¡ r +r ¡ ¡ +e + ¡ ¡ (2) 1 2 2m r r r r r r 1a 2b ab 12 1b 2a Using standard perturbation theory it is then possible which leads to a total potential of the form to evaluate the gain in energy due to H . This is left µ ¶ int 6 12 ¾ ¾ as an exercise. The result is Á(r)=¡4² ¡ ; (4) 6 12 r r ¡® ® a b ¢E' ; (3) which is usually referred to as the Lennard-Jones po- 6 r ab tential. The choice of the repulsive term is somewhat arbitrary but it re°ects the short range nature of the where ® are the atomic polarizabilities. x interaction and represents a good approximation to the full problem. The parameters ² and ¾ depend on the molecule. Derivation of Van der Waals Crystals Problem 1 Whataboutcrystals? Let's¯rstthinkaboutwhatkind ofenergyscalesareinvolvedintheproblem. Ifweassume Repulsion that the typical distance between atoms is of the order º of 1A we have We just saw that there is an attractive potential of 2 º e =1A'14:4eV for the Coulomb energy and (5) 6 the form ¡1=r . If there were only a Coulomb repul- sionofstrength1=r thiswouldleadtothecollapseofour µ ¶ 2 molecule. In fact there is a very strong repulsion, which 1 2 º '3:8eV for the potential in a 1A quantum box comes from the Pauli principle. For the general pur- º 1A 12 posethisrepulsivepotentialisoftentakentobe»1=r , (6)4 0 In comparison to room temperature (300K' 25meV) the ground state energy is then given by E =-13.6 eV. + these energies are huge. Hence ionic Crystals like NaCl Whathappensifweaddoneprotonor H tothesystem are extremely stable, with binding energies of the order which is R away. The potential energy for the electron is of 1eV. then 2 2 e e U(r)=¡ ¡ (10) r jr¡Rj Ionic crystals The lowest eigenfunction with eigenvalue -13.6eV is Somesolidsorcrystalsaremainlyheldtogetherbythe µ ¶ 3=2 1 1 ¡r=a electrostatic potential and they include the alkali-halides 0 »(r)= p e ; (11) + ¡ ¼ a like (NaCl ¡ Na Cl ). 0 where a is the Bohr radius. But now we have two 0 protons. If the protons were in¯nitely apart then the d general solution to potential 10 is a linear superposi- tion ,i.e., Ã(r) = ®»(r) + ¯»(r ¡ R) with a degener- 0 0 ate lowest eigenvalue of E = E =-13.6 eV. When R is not in¯nite, the two eigenfunctions corresponding to the lowest eigenvalues E and E can be approximated b a by à = »(r)+»(r¡R) and à = »(r)¡»(r¡R). See b a ¯gure 3. - + Ψ (r) b FIG. 2: A simple ionic crystal such as NaCl Ψ (r) V(r) a 0 The energy per ion pair is 2 Energy e C 14:4eV C =¡® + =¡® + with 6n·12; R n n º N d d d d=A ionpair + + (7) where ® is the Madelung constant and can be calcu- r 0 lated from the crystal structure. C can be extracted ex- perimentally from the minimum in the potential energy and typically n = 12 is often used to model the e®ect of the Pauli principle. Hence, from the derivative of the FIG. 3: The potential for two protons with the bonding and anti-binding wave function of the electron potential we obtain: µ ¶ 1=11 12C d = (8) 0 The average energy (or expectation value of E ) is 2 b e ® ¤ ¤ E = hà Hà i=hà à i (12) b b b b b so that A+B 0 2 = E ¡ with (13) Energy 11®e 1+¢ = (9) Z N 12d ionpair 0 2 2 A = e dr» (r)=jr¡Rj (14) How good is this model? See table below: Z 2 B = e dr»(r)»(jr¡Rj)=r (15) Z Quantum mechanics as a bonder ¢ = dr»(r)»(jr¡Rj); (16) Hydrogen-like bonding R whereh¢i´ ¢dr and similarly, Let's start with one H atom. We ¯x the proton at A¡B 0 E =E ¡ (17) a r =o then we know form basic quantum mechanics that 1¡¢5 Hence, the total energy for state à is now b This¯gureillustrateswhytheyarecalledbondingand total 2 E =E +e =R (18) b b anti-bonding,sinceinthebondingcasetheenergyislow- ered when the distance between protons is reduced as total When plugging in the numbers E has a minimum at b long as RR . 0 º 1.5A. See ¯gure 4 -8 Covalent bonding Covalent bonds are very similar to Hydrogen bonds, tot E onlythatwehavetoextendtheproblemtoalinearcom- a Antibonding bination of atomic orbitals for every atom. N atoms -12 would lead to N levels, in which the ground state has a E a bonding wave-function. tot -13.6 eV E b -16 E Bonding b R 0 Metals R distance between protons In metals bonding is a combination of the e®ects dis- cussedabove. Theideaistoconsideracloudofelectrons FIG. 4: The energies as a function of the distance between only weakly bound to the atomic lattice. The total elec- them for the bonding and anti-binding wave functions of the electron trostatic energy can then be written as Z Z 2 2 2 X X e e e n(r )n(r ) 1 2 E =¡ drn(r) + +1=2¢ dr dr (19) el 1 2 0 r¡R R¡R jr ¡r j 1 2 0 R RR which corresponds to an ionic contribution of the form This last expressions leads to a minimum at r =a =1:6. s 0 We can now compare this with experimental values and µ ¶ 1=3 2 ®e 3 the result is o® by a factor between 2 and 6. What went E =¡ where r = (20) el s 2r 4¼n s wrong. Well, we treated the problem on a semiclassi- cal level, without incorporating all the electron-electron and ® is the Madelung constant. Deriving this requires interactions in a quantum theory. This is very very di±- quite a bit of e®ort. On top of this one has to add the cult, but in the large density case this can be estimated kinetic energy of the electrons, which is of the form: and a better agreement with experiments is obtained. µ ¶ 2=3 2 9¼ 3 E = (21) kin 2 4 10mr s Binding summary And ¯nally we have to add the exchange energy, with is a consequence of the Pauli principle. The expression for There are essentially three e®ects which contribute to this term is given by the binding of solids: µ ¶ 1=3 9¼ 3 ² Van der Waals (a dipole-dipole like interaction) E =¡ (22) ex 4 4¼r s ² ionic (Coulomb attraction between ions) Putting all this together we obtain in units of the Bohr ² Quantummechanics(overlapofthewave-function) radius: µ ¶ 2 24:35a 30:1a 12:5a Inadditionwehavetwoe®ectswhichpreventthecollapse 0 0 0 E = ¡ + ¡ eV/atom (23) 2 of the solids: r r r S s S E eV6 ² Coulomb Miller indices can be obtained through the construc- tion illustrated in the ¯gure below: ² Quantum mechanics (Pauli) STRUCTURE FIG. 6: Plane intercepts the axes at (3a;2a;2a). The in- 1 2 3 verse of these numbers are (1=3;1=2;1=2), hence the smallest integers having the same ratio are 2,3,3, i.e., the Miller in- ¹ dices are (233). For a negative intercept the convention is 1. FIG. 5: Pyrite, FeS crystal with cubic symmetry. 2 (Picture from Ashcroft and Mermin) Illustrations SCATTERING See ppt notes. See supplement on di®raction. In order to determine the structure of a crystal it Summary is possible to observe the interference pattern produced by scattering particles with wave-length comparable to Periodic solids can be classi¯ed into two main classes: º the lattice spacing, i.e., of the order if 1A. There are three main classes of particles, which can be used: ² Bravaislattices: EverypointofthelatticeRcanbe reachedfromaalinearcombinationoftheprimitive ² Photons, in particular X-rays, whose wave lengths vectors: R = n a + n a + n a, where n are º 1 1 2 2 3 3 i are around 1A. The probability to scatter o® the integers. crystal is not that large, hence they can penetrate quite deeply into the crystal. The main scattering ² Latticeswithbasis: Hereeverypointinaprimitive occurs with the electrons. Hence, what is really cell is described by a basis vector B so that any i observed with X-raysis the periodic distribution of point of the lattice can be reached through: R = electrons. n a +n a +n a+B . 1 1 2 2 3 3 i ² Neutrons are also extensively used since they In 3D there are 14 Bravais lattices and 230 symmetry mainly interact with the nuclei and the magnetic groups for lattices with basis. In 2D there are 5 Bravais moments. lattices and in 1D only 1. Thisisthezoologyofcrystalsandtheyallhavenames. ² Electrons, have a very scattering probability with It is important to remember that many of the physical anythinginthecrystal,hencetheydonotpenetrate properties cannot be deduced from the crystal structure verydeep,buttheyarethereforeaninterestingtool directly. The same crystal structure could be a metal to probe the surface structure. (Cu) or an insulator (Ca). All forms of scattering share very similar basic prin- ciples and can be applied to the scattering's theory of everything described in the next section.7 Scattering theory of everything Physics is hidden in n(r), which contains the informa- tion on the position of scatterers and their individual scattering distribution and probability. In the following we discuss the most important implications. dV ← crystal r k 1D scattering pattern Incident Outgoing k’ φ φ φ φ beam beam O Let's suppose we have a 1D crystal along direction y ik ¢r in ik'⋅r with lattice spacing a and that the incoming wave e ik⋅r e e is perpendicular to the crystal along x. We want to cal- culate the scattered amplitude along direction k . By out The total scattered wave off V is de¯ning q = k ¡k , we can write the scattering am- out in 3 iq⋅r plitude as A(q) ∝ dr n(r) ⋅e ∫∫∫ Z V iq¢r 3 A(q)= n(r)e dr (24) q = k - k' V where n( r) is the distribution of scatterers If we assume that the crystal is composed of point-like scatterers we can write: N¡1 X C FIG. 7: Di®raction set-up (picture from G. Frossati) n(x;y;z)= ±(y¡an)±(x)±(z); (25) N n=0 This formulation can describe any scattering process where C is simply a constant. Hence, by inserting 25 in terms of the scattering amplitude A(q). The entire into 24, and de¯ning q =(0;q;0) we have ½ N¡1 iaqN X C C 1¡e C when q =2¼m=a iqan A(q)= e = = =C± (26) N1 q;G iaq 0 when q6=2¼m=a N N 1¡e n=0 G = y2¼m=a is the reciprocal lattice and m an integer. and If we had a screen along y we would see a di®raction 2 ½ N¡1 pattern along I(q) = jA(q)j , since what is measured X C C when q2G iq¢R n A(q)= e = =C± N1 experimentally is the intensity. In this simple case the q;G N 0 when q2=G n=0 reciprocal lattice is the same as the real lattice, but with (28) lattice spacing 2¼=a instead. In this case the Bravais lattice (or Real-Space) is R =n a +n a +n a (29) n 1 1 2 2 3 3 andthereciprocalspaceG (ork-space)canbededuced m Point-like scatterers on a Bravais lattice in 3D iR ¢G n m from e =1, hence We start again with the general form of A(q) from 24 G =m b +m b +m b ; (30) m 1 1 2 2 3 3 and assume that our 3D crystal is formed by point-like scatterers on a Bravais lattice R. Hence, where 2¼a £a 2¼a £a 2¼a £a 2 3 3 1 1 2 N¡1 X b = , b = and b = 1 2 3 C 3 ja ¢a £aj ja ¢a £aj ja ¢a £aj 1 2 3 1 2 3 1 2 3 n(r)= ± (r¡R ); (27) n N (31) n=08 As a simpli¯cation we suppose that f =f =f, then 1 2 (ia=2)q¢(x+y+z) e S(q)=(1+e )f(q); (37) e where f(q) is the Fourier transform of f. Hence, ½ 2± if q +q +q is even 1 2 3 q;G e A(q)=S(q)¢± =f(q) q;G 0 if q +q +q is odd 1 2 3 (38) General case of a Bravais lattice with basis The most general form for n(r), when the atoms sits on the basisfug along the Bravais lattice R is j n M;N¡1 X C n(r)= f (r¡R ¡u ); (32) j n j N j=1;n=0 where f is the scattering amplitude for the atoms on j site u and is typically proportional to the number of j electrons centered on u . Now j M;N¡1Z X C iq¢r 3 A(q) = f (r¡R ¡u )e d r j n j N j=1;n=0 M;N¡1Z X C iq¢r 3 iq¢u iq¢R j n = f (r)e d r¢e ¢e j N j=1;n=0 with r r+R +u n j = CS(q)¢± ; (33) q;G where we have de¯ned the structure factor S(q) as Z M X iq¢r 3 iq¢u j S(q)= f (r)e d r¢e (34) j j=1 This is the most general form. It is interesting to re- mark that in most case, experiments measure the inten- 2 sity I(q)=jA(q)j , rather than the amplitude. Example: the structure factor of a BCC lattice TheBCCcrystalcanbeviewedasacubiccrystalwith lattice a and a basis. Therefore, all lattice sites are de- scribed by R =n ax+n ay+n az+u ; (35) n 1 2 3 j where u =0(x+y+z) and u =(a=2)(x+y+z). 1 2 The structure factor then becomes: Z X iq¢r iq¢u 3 j S(q)= f (r)e e d r: (36) j j=1;29 Bragg's law nuclear sites. The scattered wave amplitude with wave number k is then Z Equivalence between Bragg's law for Miller planes and 3 i(k¡k )¢r in A(k¡k )» d rn(r)e (43) the reciprocal lattice. in V This leads to three typical cases: ² Mono crystal di®raction: Point-like Bragg peaks (1,0,0) (0,1,0) Miller index in out q =k -k 2 2 b out k 1 in out θ q =k -k 1 1 a in k FIG. 8: Point-like Bragg peaks from a single crystal. (Ref: lassp.cornell.edu/lifshitz) z y x ² Powder di®raction: From Bragg's law we know that for the planes perpen- dicular to x or Miller index (1,0,0) the following di®rac- tion condition applies: n¸=2asin(µ): (39) hence, 2¼ in n =2sin(µ)¢jk j=jqj; (40) 1 a in FIG. 9: Circle-like Bragg peaks from a powder with di®erent sincek =2¼=¸andwesupposedthatk isperpendicular grain sizes to z. This condition is equivalent to q = x2¼=a 2 G. 1 The same relation applies for the scattering of the plane perpendicular to y or Miller index (0,1,0) the following di®raction condition applies: ² Liquid di®raction: n¸=2bcos(µ): (41) hence, 2¼ in n =2cos(µ)¢jk j=jqj; (42) 2 b or q =y2¼=b2G. 2 FIG. 10: Pattern evolution for a complex molecule evolving Summary of scattering from a crystal-like structure to an isotropic liquid ik ¢r in We have an incoming wave e di®racting on some Theliquiddi®ractionisessentiallythelimitingcase sample with volume V and with a scattering probability of the powder di®raction when the grain size be- n(r) inside V. For X-ray n is typically given by the elec- comes comparable to the size of an atom. tron distribution whereas for neutrons it is typically the10 PROPERTIES OF SOLIDS AND LIQUIDS where the pre-factor 2 comes from the spin degeneracy. Hence, The Theory of everything discussed in the ¯rst section 3 k 4¼ 3D F D 3 can serve as a guideline to illustrate which part of the n = in D =3 since V = k (48) e k F 2 F 3¼ 3 Hamiltonian is important for a given property. For in- 2 k 2D F D 2 stance, wheninterestedinthemechanicalproperties, the n = in D =2 since V =¼k (49) e k F F 2¼ terms containing the electrons can be seen as a pertur- 2k F 1D D bation. However,whenconsideringthermalconductivity, n = in D =1 since V =2k ; (50) F e k F ¼ for example, the kinetic terms of the ions and the elec- trons are important. where the maximum energy of the electrons is Inthefollowingwewillstartbyconsideringafewcases 2 2 k and we will start with the single electron approximation. F E = the Fermi energy (51) F 2m e This de¯nes the Fermi energy: it is the highest energy single electron approximation when all possible states with energy lower than E are F occupied, which corresponds to the ground state of the The single electron approximation can be used to de- system. This is one of the most important de¯- rive the energy and density of the electrons. This sim- nitions in condensed matter physics. The electron plest model will always serve as a reference and in some density can now be rewritten as a function of the Fermi cases the result is very close to the experimental value. energy, through eqs. (51) and (48) in 3D: Alkalimetals(Li,Na,K,..) arereasonablywelldescribed 3=2 bythismodel, whenwesupposethattheoutershellelec- (2mE ) F n(E )= : (52) F tron/atom(thereis1forLi,Na,K)isfreetomoveinside 2 3 3¼ themetal. Thispictureleadstothesimpleexampleof N We now want to de¯ne the energy density of states electrons in a box. This box can be viewed as a uniform D(E) as andpositivelychargebackgroundduetotheatomicions. WesupposethatthisboxwithsizeL£L£Lhasperiodic Z E F boundary conditions, i.e., n(E )´ D(E)dE =)D(E)= n(E); (53) F E 0 N 2 2 X r which leads to n H =¡ with ª(0)=ª(L) (44) p 2m e n=1 m 2mE D(E) = in 3D (54) 3 2 ¼ For one electron the solutions can be written as m D(E) = in 2D (55) 2 2 2 ¼ jkj 2¼ ik¢r r ª(r)=e with E = and k = (n ;n ;n ); 1 2 3 2m 2m L e D(E) = in 1D (56) 2 2 (45) ¼ E where n are the quantum numbers, which are positive i and illustrated below. or negative integers. Since electrons are Fermions we 3D cannot have two electrons in the same state, except for the spin degeneracy. Hence each electron has to have di®erent quantum numbers. This implies that 2¼=L is 2D theminimumdi®erencebetweentwoelectronsink-space, 1D which means that 1 electron uses up µ ¶ D E 2¼ (46) L of volume in k-space, if D is the dimension of the space. FIG. 11: Density of states in 1D, 2D and 3D If we now want to compute the electron density (number D of electrons per unit volume n = N=L ) in the ground e state, which have k k (Fermi sphere with radius k ), F F we obtain: Properties of the free electron model µ ¶ D L D D n =2¢V ¢ =L ; (47) e k F Physical quantities at T=0: 2¼ D(E)11 hEi 3 ² Average energy per electron: = E F n 5 ¶ 2 (hEiV) 2 1 hEi ¼ ² Pressure: P = ¡ = nE , where hEiV is 2 F V 5 C = = k TD(E ) (64) V F V T 3 the total energy. ¹;V P 2 ¡1 ² Compressibility · =¡V = nE C V F V 3 Hence, =°, which is the Sommerfeld parameter. T Case 2: T= 6 0: In equilibrium Periodic potentials 1 f = ; (57) FD (E¡¹)=kT e +1 Theperiodicityoftheunderlyinglatticehasimportant consequences for many of the properties. We will walk where the chemical potential is the energy to add one through a few of them by starting with the simplest case electron ¹ = F ¡F . ¹ = E at T=0 and F is the N+1 N F in 1D. free energy. The Sommerfeld expansion is valid for kT ¹ and Z 1 Kronig-Penney model hHi = dEH(E)F (E) (58) FD 0 Z Let us ¯rst consider the following simple periodic po- ¹ 2 ¼ 2 0 tential in 1D. ' H(E)dE+ (kT) H (¹)+::: (59) 6 0 Z 2 2 1 X r H =¡ ¡V ±(x¡na) (65) hEi = dEED(E)F (E) (60) FD 2m 0 n Z E 2 F ¼ 2 ' ED(E)dE+ (kT) D(E )+::: (61) Thesolutionsfornaxna+aaresimplyplanewaves F 6 0 and can be written as hni = n(T =0) (62) ikx ¡ikx 2 0 ¼ D (E ) Ã(x)=A e +B e : (66) F n n 2 )¹(T) ' E ¡ (kT) +::: (63) F 6 D(E ) F Now the task is to use the boundary conditions in order Physical quantities at T6=0: Speci¯c heat to determine A and B . We have two conditions: n n (1) Ã(na¡²)=Ã(na+²) when ²0 and (67) R na+² 0 0 (2) (H¡E)Ã(x)dx=0=)¡ (à (na+²)¡Ã (na¡²))=VÃ(na): na¡² 2m Inserting 66 into condition (1) yields ½ ikna ¡ikna ikna ¡ikna A e +B e =A e +B e =Ã(na)=à n¡1 n¡1 n n n (68) ikna ¡ika ¡ikna ika A e e +B e e =à n¡1 n¡1 n¡1 ½ ¡ ¢ ¡ika ¡ikan ¡ika ika e à ¡Ã =B e e ¡e n n¡1 n¡1 ¡ ¢ =) (69) ika ikan ika ¡ika e à ¡Ã =A e e ¡e n n¡1 n¡1 ½ ¡ ¢ ¡ika ¡ikan ¡ika ¡ika ika e à ¡Ã =B e e e ¡e n+1 n n ¡ ¢ and (70) ika ikan ika ika ¡ika e à ¡Ã =A e e e ¡e n+1 n n When taking the derivative of 66 then condition (2) implies 0 1 B C B C ikan ¡ikan ikan ¡ikan ¡ Bik A e ¡ik B e ¡ik A e +ik B e C=Và ; (71) n n n¡1 n¡1 n z z z z 2m A ¡ika ika ika ¡ika à ¡e à à ¡e à n n e à ¡Ã e à ¡Ã n+1 n+1 n n¡1 n n¡1 2isin(ka) ¡2isin(ka) 2isin(ka) ¡2isin(ka)12 hence, 0 1 2 k B C ¡ika ika ika ¡ika ¡ 2à +2à +à (¡e ¡e ¡e ¡e ) =Và (72) A n+1 n¡1 n n 2m2sin(ka) z ¡4cos(ka) and ¯nally, This equation is often called the tight binding equation as we will see in the next section. Finding a solution for µ ¶ 2msin(ka) this equation is trivial since à +à = ¡ V +2cos(ka) à ; (73) n+1 n¡1 n 2 k z ipan W à =e (75) n Which can be rewritten as is a solution. This is easily veri¯ed by plugging (75) into à +à =Wà (74) (74), which yields n+1 n¡1 n ipan ipa ipan ¡ipa ipan ipa ¡ipa e e +e e =We =)W =e +e =2cos(pa): (76) This equation has a solution only if Tight binging approximation ¡2·W ·2 (77) Let us consider the general potential due to the ar- rangement of the atoms on a lattice: We now recall that the Eigenstate of the original Hamil- 2 2 k ikx tonian is given by E(k) = , where e is the plane 2 2 2m X r wave between two ± functions, hence the dispersion re- H =¡ + V (r¡R ) (78) 0 n 2 2 2m k n lation is equal to E(k) = as long as ¡2· W · 2. 2m z This condition will create gaps inside the spectrum as V(r) illustrated in the graph below: Thisisaverygeneralformforaperiodicpotentialassum- ingthatweonlyhaveonetypeofatoms. Theperiodicity is given by the lattice index R . For a general periodic n ΔE potentialBloch's theorem (see A& M for proof) tells us that a solution to this Hamiltonian can be written as ik¢r à (r)=e u (r); (79) k k 0 1 2 3 where u is a periodic function such that u (r+R ) = k k n π/a u (r). This implies that k ik¢R n à (r+R )=e à (r) (80) k n k FIG. 12: Dispersion curve for the Kronig-Penney model Let us now suppose that the solution of the Hamiltonian 2 2 r Conclusion: a periodic potential creates gaps, H =¡ +V (r¡R )withoneatomisÁ (r¡R ) 1atom 0 n l n 2m which leads to the formation of a band structure. correspondingtotheenergylevel² ,i.e.,H Á =² Á . l 1atom l l l This is a very general statement which is true for almost Now comes the important assumption, which allows us any periodic potential. to simplify the problem: We suppose that W-2 -2W2 W-2 E(k)13 0 1 X A hÁ (r¡R )jHjÁ (r¡R )i=¡t ± +± +² ± (81) l m l n l l m; n m; n+i m; n¡i i=x; y;z Hence, only the nearest neighbor in every direction is Combining Bloch's theorem with the tight binding approximation takentobenon-zero. Further,weassumethatthereisno overlap between levels of the one atom potential, which allows us to look for a general solution of the following The tight binding approximation is very general and form for each energy level ² . can be applied to almost any system, including non- l periodic ones, where the tight binding elements can be X l assembled in an in¯nite matrix. For the periodic case, à (r)= c Á (r¡R ) (82) l l n n on the other hand, it is possible to describe the system n with a ¯nite matrix in order to obtain the full disper- with eigenvalue E (k). To calculate E (k) we plug-in l l sion relation. This is obtained by combining the Bloch this Ansatz into eq. (82) and obtain an equation for theorem for periodic potentials, where the wave-function l the coe±cients c , which leads to the following equation n from (79) is again: when using the tight binding approximation given in eq. (81). ik¢r à (r)=e u (r) k k X l l l Instead of writing (82) we write the Bloch-tight- c ² ¡t c =E (k)c (83) l l l m m m +i binding solution as i=§x; y;z X The solution of this equation are plane waves, which k ik¢R l n à (r)= e c Á (r¡R ); l n l n l ik¢R m can be veri¯ed readily by taking c = e and plug- m n ging it into the equation to obtain whichnowdependsexplicitlyonthewavevectork. Using Bloch's theorem this implies that E (k)=² ¡t (2cos(k a )+2cos(k a )+2cos(k a ); l l l x x y y z z (84) l l c =c ; where a is the lattice constant in all 3 space directions: n m a = R ¡R . The energy diagram is illustrated in m +a m whenevern and m are related by a linear combination of ¯g. . The degeneracy of each original single atomic en- Bravais vectors. Moreover, the tight binding equation in ergy level ² is lifted by the coupling to the neighboring l l l ik¢R n (83) is the same but with c replaced by c e atoms and leads to a dispersion curve or electronic band n n structure. E a ε B 3 A ε 2 ε 1 -1.0 -0.5 0.0 0.5 1.0 FIG. 14: Diatomic square crystal -π /a k π/a We apply this to the simple example of a diatomic FIG. 13: Dispersion curve for the tight binding model squarelatticeoflatticeconstantawithalternatingatoms A and B shown in ¯gure . We will further assume that This tight binding approximation is very successful we have only one band l. Hence, the Bloch-tight-binding in describing the electrons which are strongly bound to solution is written as the atoms. In the opposite limit where the electrons or X k ik¢R more plane-wave like, the weak potential approximation n à (r)= e c Á(r¡R ); n n is more accurate: n14 2 2 k 0 ik¢r 0 where c takes on only to possible values due to Bloch's à (r)=e , with corresponding energies ² = and n k k 2m 0 0 0 theorem: c or c . This leads to the following simpli¯ed Hamiltonian H , i.e., H à (r) = ² à (r). Since the ori- A B 0 0 k k k tight binding equations (assuming hÁ (r¡R )jHjÁ (r¡ gin of this energy dispersion relation can be chosen from l n l 0 0 R )i = ² or ² and t = ¡hÁ (r¡R )jHjÁ (r¡R )i any site of the reciprocal lattice, we have ² = ² , n A B l n l m k+K k when n and m are nearest neighbors): hence these energies are degenerate. This implies that we have to use a degenerate perturbation theory. The P iak¢i c ² ¡t c e = E(k)c mathematicalprocedureisverysimilartothetightbind- A A B A i=§x; y P iak¢i ing approximation, but we now expand the solution Ã(r) c ² ¡t c e = E(k)c B B A B i=§x; y ofthefull Hamiltonian H =H +V(r)in termsofa sum 0 0 0 0 of plane waves à (r). Since ² = ² we will only use It is now quite straightforward to rewrite these equa- k k+K k 0 0 two plane waves in this expansion: à (r) and à (r). tions in matrix form: k k+K µ ¶µ ¶ µ ¶ Hence, ² ¡t¢g c c A A A =E(k) ; ¤ ¡t¢g ² c c B B B 0 0 Ã(r)=®Ã (r)+¯Ã (r); (85) k k+K iak¢x ¡iak¢x iak¢y ¡iak¢y with g = e +e +e +e . The disper- sion relation or band structure is then simply given by where the coe±cients ® and ¯ have to be determined in obtaining the eigenvalues of H , where order to solve SchrÄodinger's equation: BTB µ ¶ ² ¡t¢g A H = : (H¡E)Ã(r)=0: (86) BTB ¤ ¡t¢g ² B We can ¯nd the solution by ¯rst multiplying (86) by 0 ¤ Weak potential approximation (à ) (r)andthenintegratingtheequationoverthewhole k space which will lead to one equation, and we obtain a 0 ¤ In this case we consider the e®ect of the periodic po- second equation by multiplying (86) by (à ) (r) and k+K tentialV(r)asaperturbationontheplanewavesolution then integrating of the whole space. This leads to ½ R R 3 0 ¤ 0 3 0 ¤ 0 ® d r(à ) (r)(H¡E)à (r)+¯ d r(à ) (r)(H¡E)à (r)=0 k k k k+K R R ; (87) 3 0 ¤ 0 3 0 ¤ 0 ® d r(à ) (r)(H¡E)à (r)+¯ d r(à ) (r)(H¡E)à (r)=0 k+K k k+K k+K where (assuming normalized plane waves in the integrals) 8R R 3 0 ¤ 0 3 ¡ikr ikr 0 d r(à ) (r)(H¡E)à (r)= d re (H +V(r)¡E)e =² +0¡E 0 k k k R R R 3 0 ¤ 0 3 ¡ikr i(k+K)r 3 iKr 0 d r(à ) (r)(H¡E)à (r)= d re (H +V(r)¡E)e = d re (² +V(r)¡E)=V 0 K k k+K k+K R 3 0 ¤ 0 0 d r(à ) (r)(H¡E)à (r)=² ¡E k+K k+K k+K R : 3 0 ¤ 0 d r(à ) (r)(H¡E)à (r)=V ; ¡K k+K k (88) R R R 3 iK¢r 3 iKr 3 and V = d re V(r) (the Fourier transform), d re = 0 (for K = 6 0), and d rV(r) = 0. With coe±cients K (88), equation (87) leads to the following couple of equations: ¯ ¯ ½ q 0 0 ¯ ¯ 0 0 0 0 2 ®(² ¡E)+¯V =0 ² ¡E V ² +² (² ¡² ) K K k k+K k k+K k ¯ k ¯ ) =0 )E = § +V V : (89) K ¡K 0 0 ¯ ¯ 2 4 ®V +¯(² ¡E)=0 V ² ¡E ¡K ¡K k+K k+K 0 0 0 Finally if ² = ² , we have E = ² §jV j, which the wavevector which corresponds to the two dispersion K k k+K k leads to a splitting 2jV j of the energy levels at these curves which led to the degenerate energy level. K degenerate energies. For the example in ¯gure , this weakpotentialapproximationwouldgiveusasplittingof 2jV j at k =¡¼=a and k = K¡¼=a = ¼=a. This K=2¼=a Localization impliesthatthe¯rstordercalculationoftheenergysplit- ting due to the weak periodic potential V(r) is equal to When, instead of having a purely periodic potential twice the fourier transform of this potential evaluated at disorder is included into the system, we no more have15 Bloch wave solutions but localization of the wave func- equation Hà =E(k)à it follows that u is a solution of k tionsoccur. Thisis particularlyimportantin lowdimen- µ ¶ 2 sional systems and tends to suppress transport. 2 (k¡ir) +V ¡E(k) u =0: (93) k 2m z H ¡E(k) k Electronic properties due to periodic potential In order to calculatehvi we use ¯rst a ¯rst order pertur- bation in k + q, where q is very small. Therefore, the Density of states eigenvalue corresponding to H is E(k +q), which is k+q to ¯rst order Density of states in 1D: E(k+q) = E(k)+hu jH ¡H ju i k k+q k k 2 n ±n ikr 2 ¡ikr = E(k)+hÃe j ( q +2q(k¡ir))je Ãi D(E) = ' z 2m E ±E ¯ ¯ 0 ¡1 ¯ ¯ ±n E(k) 2 ¯ ¯ = £ 2 £ 2 ikr ¡ikr ¯ ¯ z z ±k k = E(k)+hÃe j q(k¡ir))je Ãi P m spin §k ¯ ¯ ¡1 ¯ ¯ = E(k)+hÃjq (¡ir)jÃi 1 ±N E(k) ¯ ¯ m = £4 ¯ ¯ L ±k k = E(k)+qhÃjvjÃi ¯ ¯ ¡1 ¯ ¯ 2 E(k) = E(k)+qr E(k) (Taylor expansion) k ¯ ¯ D(E) = ; (90) ¯ ¯ ¼ k 1 ) hvi= r E(k): (94) k where we used that ±E =(E=k)±k and ±k =2¼=L for Hence,theexpectationvalueofthevelocityisdetermined one electron, i.e., ±N =1. by the slope of the dispersion relation. This also implies In three dimensions (D=3) we have: that the sign of the average velocity depends on the sign of E=k. n ±n D(E) = ' E ±E X Response to an external ¯eld and existence of holes and ±n ¡1 = jr E(k)j £ 2 electrons k z ±k E(k)=E spin Z The idea is to describe the average motion of an elec- 2 ¡1 2 = d kjr E(k)j ; (91) k tron in the presence of an external ¯eld (electric, E , or 3 el (2¼) E(k)=E magnetic, B) in a semiclassical way. Hence, we want P 2 where we used ±E = (r E(k))¢ ±k, (±k) k E(k)=E d R ¤ 2 3 3 m hvi=F =(qE or qhvi£B); (95) el d k, and ±n=(±k) = 1=(2¼) . This result shows dt E(k)=E that the density of state in the presence of a periodic ¤ where m is an a®ective mass and q the charge. Using potential, i.e., for a crystal depends only on the slope of (95) and (94) we have the dispersion relation or the band structure. hvi ¤ _ m k = qE el Average velocity k 2 1 E ¤ _ m k = qE el 2 The average velocity of an electron in a periodic po- k _ tential is given by the expectation value of the velocity, )k = qE =F; (96) el i.e., ¤ Where we de¯ned the e®ective mass m as hvi=hÃjvjÃi; (92) ¯ ¯ ¡1 2 ¯ ¯ E ¤ 2 ¯ ¯ m = : (97) ¯ 2¯ k where à is the wavefunction from the Hamiltonian with 2 2 r periodic potential V, i.e., H = +V. From Bloch's 2 2m E ikr If isnegativeweneedtochangethesignofq inorder 2 theorem (79)wecanwrite Ã(r)=e u (r), whereu (r) k k k 2 E has the same periodicity as V(r). From the SchrÄodinger to remain consistent. Hence, when 0, the charge 2 k16 2 E _ of an electron is q =¡e but if 0 then q is positive Using (101) and (94) and since k is perpendicular tohvi 2 k _ (+e). In this case we describe the particles as holes. and B, (k)»E=k we obtain k They represent missing electrons. With these de¯nitions Z ¤ k 2 2 of q and m , which is also called the band mass, the dk ? ±t = semiclassical equations of motion of single electrons in a qB E=k k k 1 periodicpotentialaresimplygivenbyeqs. (95)and(96). Z k 2 2 d In general, the e®ective mass is given by a tensor de- = k dk : (103) k ? qBdE k 1 ¯ned as ¯ ¯ ¡1 2 ¯ ¯ E For a complete turn this leads to ¤ 2 ¯ ¯ m = ; (98) ®¯ ¯ ¯ k k I ® ¯ 2 d T = k dk : (104) k ? where ® and ¯ are the spatial directions. An important qBdE z consequence of this semiclassical description of the mo- S tion of electrons is the dependence of the e®ective mass on the energy and the band structure. In some cases the Here S is the area enclosed by an orbit in k-space. This 2 E e®ective mass can even diverge (when = 0). Sim- orbit corresponds to an equipotential line perpendicular 2 k ilarly the sign of the carriers also depends on the band to the magnetic ¯eld. structureandtheenergyofthecarriers. Byde¯nitionwe Let'ssupposeforsimplicitythatthee®ectivemassten- ¤ call the bottom of an energy band and electronic band sor m is diagonal and given by 2 E when 0 and a hole band when at the top of the 0 1 2 k 2 m 0 0 x E energy band 0. 2 ¤ k A m = 0 m 0 (105) y 0 0 m z Bloch oscillations andthattheenergydispersionisharmonic(whichisusu- ally true at a band extremum, i.e., In the presence of an electric ¯eld and a periodic po- 2 2 2 2 2 2 tential we can use the equation of motion (96), i.e., k k k y x z E(k)= + + : (106) 2m 2m 2m eE x y z el _ k =¡eE )k =¡ t; (99) el If we assume that the magnetic ¯eld is along z and that but in a periodic potential and in the tight binding ap- the average e®ective mass perpendicular to B is given by proximation the energy is given by E(k)=¡2t cos(ka), 0 m , we can rewrite (106) as ? whereaisthelatticeconstantandt thenearestneighbor 0 ¡1 2 2 2 2 overlapintegral. Hence,sincev =r_ andhvi= E=k k k k ? E(k)= + ; (107) we have 2m 2m ? k 2t aeE t 0 el hri= cos( ): (100) 2 2 2 wherek =k +k . Using(104)and(107wethenobtain eE el ? x y This means that the average position of the electrons 2 2 2 S = ¼k =¼(2m E= )¡¼k m =m ? ? k ? k oscillates in time (Bloch oscillations). In arti¯cial struc- dS ¼2m ? turestheseBlochoscillationsaretypicallyoftheorderof ) = 2 dE 1THz. 2¼ qB ) = = ; (108) c T m ? Semiclassical motion in a magnetic ¯eld which is the cyclotron frequency. Hence, the cyclotron frequency depends on the average e®ective mass perpen- Inthepresenceofamagnetic¯eld(B),wecandescribe dicular to the magnetic ¯eld. This allows us to measure the semiclassical trajectories in k-space using (96), i.e., the e®ective mass along di®erent directions, simply by _ k =qhvi£B: (101) changing the direction of the magnetic ¯eld and by mea- suring the cyclotron frequency. Hence,onlythevaluesofkperpendiculartothemagnetic ¯eldwillchange,whichwedenotebyk . Thecomponent ? parallel to the ¯eld, k is not a®ected by B. During a k Quantization of the cyclotron orbit: Landau levels small time di®erence Z Z t k =k(t ) 2 2 2 In quantum mechanics the energies of these cyclotron _ ±t=t ¡t = dt= dk =jkj: (102) 2 1 ? orbits become quantized. To see this we can write the t k =k(t ) 1 1 117 500 300 450 Hamiltonianofanelectroninamagnetic¯eldinthehar- 200 400 monic approximation (107) as 100 350 0 300 1.1 1.3 1.5 1.8 2.0 2.2 2.4 2.6 2.8 -1 1 1 250 1/B T 2 2 H = P + (P +qA) ; (109) ? k 200 2m 2m k ? 150 100 where B =r£A)A=¡Byx in the Landau Gauge if 50 B is along z. In analogy to the harmonic oscillator, the 0 eigenvalues of (109) are then given by 0.0 0.5 1.0 B T 2 2 k qB k E = +(n+1=2) : (110) n;k k 2m m ? k FIG. 15: Magneto-resistance oscillations in a GaAs=AlGaAs z quantum well. c These eigenvalues can be found by writing ik x ik z x k the wavefunction as à = e Á (y ¡ y )e n 0 k x with y = ¡ , which leads to Hà = 0 qB µ ¶ PHONONS: LATTICE VIBRATIONS ³ ´ 2 2 2 2 k P k qB y 1 2 + + m (y¡y ) Á (y ¡ y ). ? 0 n 0 2m 2m 2 m k ? ? In general: In the y direction this is simply the harmonic oscillator with eigenvalues (n + 1=2) and in the direction c X parallel to the ¯eld we have a plane wave so that the MuÄ =¡ Á u ; (112) l lm m total energy is given by (110). The quantized levels m (n + 1=2) due to the magnetic ¯eld are called the c Landau levels. where u are the deviations from the original lattice sites l and Á are the elastic constants who have to obey this lm P sumrule Á =0(translationinvariance). Inwords, lm m Magneto-oscillations equ. 112simplymeansthattheforceproducingthedevi- ationonlatticesitelonlydependsonthedeviationsfrom The quantization of the energy levels in the presence the other lattice sites. No deviation=no force (equilib- of a magnetic ¯eld will lead to oscillations of almost any rium). Thisequationisverygeneralanddoesnotassume experimental quantity (resistance, thermal conductivity, that we have a periodic lattice, but in order to calculate magnetization, ...) as a function of B. These oscillations things we will use a periodic lattice and start with 1D. can be understood by looking at the density of states In ¯g. 16 we illustrate two typical displacement waves in due to the quantization (110). Indeed, (110) leads to a 2D. peakinthedensityofstateswhenever E =(n+1=2) . c Therefore, whenever the component of the Fermi energy u u u u u u u u u u s s 1 1 s s s s+ +1 1 s s 2 2 s s+ +2 2 perpendicular to the magnetic ¯eld is equal to one of these quantized levels there is an extremum in the quan- tity measured. The distance between two of these ex- K K trema is given by qB qB 1 2 ? (n +1=2)=E = (n +1=2) 1 2 F m m ? ? 1 1 q ? ¡1 ) ¡ = ( E ) (n ¡n ) 2 1 F u u s s+ + 4 4 B B m z u u u u u u u u u u s s+ + 1 2 ? 3 3 s s 1 1 s s s s+ +1 1 s s+ +2 2 K K 2 ¢S 2¼m ? µ ¶ 1 2¼q ¡1 ) ¢ = S ; (111) s s 1 1 s s s s + + 1 1 s s + + 2 2 s s + + 3 3 s s + + 4 4 B H F 2 where the maximum S is S = k dk = ¼(k ) = k ? x FIG.16: Twotypesoflatticedisplacementwavesin2D,trans- 2 F 2 2 (k ) S F 2 ? x verse and longitudinal modes ¼(k ) andE = = . Hence, theMagneto- y F 2m 2¼m ? ? oscillations are periodic in 1=B and the period depends on S. An example of these oscillations in resistance is shown in ¯gure 15 as a function of B and 1=B. Rxx Ω Rxx Ω 18 Mono-atomic phonon dispersion in 1D 1. Acoustic mode: s C In 1D equ. 112 for nearest neighbors and using ' ka (117) P 2(M +M ) 1 2 Á =0 simply reduces to lm m MuÄ =K(u ¡2u +u ) (113) l l+1 l l¡1 2. Optical mode: s in the simplest approximation, where only the nearest 2C(M +M ) 1 2 neighbors are important and where the elastic constants ' (118) M M are the same. Further we assume that the equilibrium 1 2 case has a lattice constant a. In this case the solution can be written as ikla¡it u =e l 2 2 =)M =2K(1¡cos(ka))=4Ksin (ka=2) (114) q K =) =2 jsin(ka=2)j M which is illustrated below In this case there is only a 1.2 ω Mono-atomic dispersion 1.0 2 (K M ) 0.8 0.6 0.4 0.2 0 k -π/a π/a 0 2π/a FIG. 17: Phonon dispersion for a mono-atomic lattice singledispersionmode,whichiscalledtheacousticmode or branch. Optical branch The situation is quite di®erent when there is an addi- FIG. 18: Phonon dispersion for a di-atomic lattice and the tion symmetry breaking in the problem. In the 1D case more general 3D case thisoccurswhenthereisasecondspecieswithadi®erent mass, forexample, orwhentherearetwodi®erentelastic constants. Let us discuss the case where there are two di®erent masses but only one elastic constant C: Experimental determination of the phonon ½ dispersion M uÄ =C(u ¡2u +u ) 1 l l+1 l l¡1 (115) M vÄ =C(v ¡2v +v ) 2 l l+1 l l¡1 In order to determine the dispersion curve of the To solve this set of equation we use a solution of the phonons there is essentially one trick. The idea is to ikla¡it ikla¡it form u = ue and v = ve , which inserted look for inelastic scattering of other particles, which are l l into eq. 115leads to typically photons or neutrons. The incoming particles scatter with phonons, thereby transferring some of their s energyandmomentumtophonons. Fromtheenergyand p 2 2 p M +M § M ¡2M M cos(ka)+M momentumoftheout-comingparticlesitisthenpossible 1 2 1 2 1 2 = C to infer the phonon absorbtion. This process can also be M M 1 2 (116) inverted in that the energy of the out-coming particle in- Thefollowingtwolimitingcasescaneasilybeobtained creasesbyabsorbtionofaphonon. Ingeneral,theenergy in the limit of small k. conservation can be expressed as19 FIG. 19: Inelastic neutron scattering experiment in Chalk River used to determine the phonon dispersion 0 0 (k)¡ (k )=§­(K) (119) in out z z in/out particle phonon and the momentum conservation as 0 k¡k =§K +G; (120) where G is a reciprocal lattice vector. A neutron scat- tering experiment is illustrated below. Origin of the elastic constant From the TOE we have the potential term due to the interactions between ions. N N ions ions 2 2 X X r q q i j i H =¡ + (121) ions 2m jr ¡r j i i j i ij This leads to a potential of the form 1 displacment 2 X X z Á 1 Á 0 0 A Á(r ;:::;r )= Á(r ;:::;r ) + u + u u +¢¢¢ (122) 1 n i i j 1 n z r 2 r r i i j 0 i i;j r 0 i r i Cohesive energy z z 0 Equilibrium pos. Á i;j The linear term has to be zero for stability reasons (no trated in ¯g. 18. From 123 we have minimum in energy otherwise). The classical equation X 2 ik¢(R ¡R ) l m of motion is then simply given by equ. 112, because m ²= Á e ¢² (124) l;m m F = ¡rÁ. To solve this equation in general, one can z write a solution of the form Á(k) , where Á(k) is a 3£3 matrix, since Á too. This leads lm 2 to an eigenvalue equation for with three eigenvalues: ik¢R ¡it l u =²¢e ; (123) 2 2 l (for k k ²) and (for k ? ²). Hence, we have L T(1;2) on longitudinal mode and two transverse modes and all phonon modes can be described by a superposition of where R are the lattice sites. One then has to solve these. l for the dispersion relation (k) in all directions, as illus- If we have two di®erent masses, we have two addi-20 tional equations of the form (124). This leads to the general case, where one has two branches the optical one Z 1 dS and the acoustic one and each is divided in longitudinal D()= (127) 3 (2¼) jr j k =const and transverse modes. The acoustic branch has always a dispersion going to zero at k = 0, whereas for the op- Let us now suppose that we have 3 low energy modes tical mode at k = 0 the energy is non-zero. The acous- with linear dispersion, one longitudinal mode (k) = L tic phonons have a gapless excitation spectrum and can c k and two transverse modes (k) = c k, hence L T T therefore be created at very low energy as opposed to R 2 r (k)=c , which implies that dS =4¼k . This L;T L;T optical phonons which require a miniumum energy to be leads to a density of state excited. 2 2 k Quantum case D ()= = (128) L;T 2 2 3 2¼ c 2¼ c L;T L;T While many of the properties, like the dispersion rela- or tion,canbeexplainedinclassicalterms,whicharesimply vibrations of the crystal ions, others, such as statistical µ ¶ properties need a quantum mechanical treatment. 2 1 2 Phonons can simply be seen as harmonic oscillators D ()= + (129) tot 3 3 2 2¼ c c L T which carry no spin (or spin 0). They can, therefore, be z 1 accurately described by bosons with energy 3 c in the isotropic case. E (k)=(k)(n+1=2): (125) n From statistics we know that the probability to ¯nd a state at E = E is given by the Boltzmann distribution n P We can now calculate the density of states of phonons, ¡E =k T n B P »e , with normalization P =1. Hence, n n or the number of states between and +d, i.e., Z ¡n=k T ¡=k T B B +d P =e (1¡e ) (130) n 1 D()d = dk (126) 3 (2¼) simply from the normalization condition. We can now but d =r dk, hence calculate the average energy at , k 1 X X ¡=k T ¡=k T n B B E()= E P = (1¡e ) (n+1=2)(e ) n n n n=0 0 1 B C 1 1 B C = + (131) B C =k T B 2 e ¡1A z hni herehniistheexpectationvalueofquantumnumbern zero energy state is totally degenerate (this does sup- at T, which is nothing else but the Bose Einstein distri- pose that there are no interactions between bosons, with bution. Asiswellknown, BosonsobeytheBose-Einstein interactions the delta function will be a little broader). statistics,whereheBose-Einsteindistributionfunctionis Thetotalaverageenergycanthenbecalculatedfromthe given by density of states as: Z 1 f = (132) BE E=k T B hEi= dD()E() (133) e ¡1 Important: at T =0, f is simply a delta function BE ±(E). This is the Bose-Einstein condensation, where the This also allows us to calculate the speci¯c heat C = V

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