Lecture notes for Quantum Physics pdf

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Quantum Condensed Matter Physics - Lecture Notes Chetan Nayak November 5, 2004CHAPTER1 Conventions, Notation, Reminders 1.1 Mathematical Conventions Vectors will be denoted in boldface, x,E, or with a Latin subscript x , E , i i i=1,2,...,d. Unless otherwise specified,we will work ind=3 dimensions. Occasionally, we will useGreek subscripts, e.g. j ,μ=0,1,...,d wherethe μ 0-component is the time-component as in x =(t,x,y,z). Unless otherwise μ noted,repeatedindicesaresummedover,e.g. ab =a b +a b +a b =a·b i i 1 1 2 2 3 3 We will use the following Fourier transform convention: Z ∞ dω −iωt ˜ f(t)= f(ω)e 1/2 (2π) −∞ Z ∞ dt iωt ˜ f(ω)= f(t)e (1.1) 1/2 (2π) −∞ 1.2 Plane Wave Expansion A standard set of notations for Fourier transforms does not seem to exist. The diversity of notations appear confusing. The problem is that the nor- malizations are often chosen differently for transforms defined on the real space continuum and transforms defined on a real space lattice. We shall dothe same, so that thereader is not confused whenconfronted with varied choices of normalizations. 34 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS 1.2.1 Transforms defined on the continuum in the interval −L/2,L/2 Consider a function f(x) defined in the interval −L/2,L/2 which we wish to expand in a Fourier series. We shall restrict ourselves to the commonly used periodic boundary condition, i. e., f(x)=f(x+L). We can write, X 1 iqx √ f(x)= f e , (1.2) q L q BecausethefunctionhastheperiodL,q mustbegivenby2πn/L,wherethe integer n=0,±1,±2,.... Note that n takes all integer values between−∞ and +∞. The plane waves form a complete orthogonal set. So the inverse is Z L/2 1 iqx f = √ dxe f(x). (1.3) q L −L/2 Let us now take the limit L → ∞, so that the interval between the successive values ofq, Δq =2π/L then tend to zero, and wecan convert the q-sum to an integral. For the first choice of the normalization we get Z ∞ √ dq iqx f(x)= lim L f e , (1.4) q L→∞ 2π −∞ and Z ∞ √ iqx lim Lf = dxe f(x). (1.5) q L→∞ −∞ √ ˜ If we definef(q)=lim Lf , everything is fine, but note the asymme- L→∞ q try: the factor (1/2π) appears in one of the integrals but not in the other, althoughwecouldhavearranged,withasuitablechoiceofthenormalization at the very beginning, so that both integrals would symmetrically involve a √ factor of (1/ 2π). Note also that ′ lim Lδ ′ →2πδ(q−q ). (1.6) q,q L→∞ These results are simple to generalize to the multivariable case. 1.2.2 Transforms defined on a real-space lattice Consider now the case in which the function f is specified on a periodic lattice in the real space, of spacing a, i. e., x =na; x =L/2, x = n N/2 −N/21.2. PLANE WAVE EXPANSION 5 −L/2, andNa =L. Theperiodicboundarycondition implies thatf(x )= n f(x +L). Thus, the Fourier series now reads n X 1 iqx n f(x )= f e . (1.7) q n L q Note that the choice of the normalization in Eq. (1.7) and Eq. (1.2) are different. Because of the periodic boundary condition, q is restricted to 2πm q = , (1.8) Na buttheintegersm constitute afiniteset. Toseethis notethatourcomplete set of functions are invariant with respect to the shiftq→q+G, where the smallest such reciprocal vectors, G, are ±(2π/a). Thus the distinct set of q’s can be chosen to be within the 1st Brillouin zone −(π/a) q ≤ (π/a); accordingly, the distinct set of integers m can be restricted in the interval −N/2 m ≤ N/2. Therefore the number of distinct q’s is equal to N, exactly the same as the number of the lattice sites in the real space. What about the orthogonality and the completeness of these set of plane waves? It is easy to see that N X ′ 2π N→∞ i(q−q )x ′ n e =Nδ ′ −→ δ(q−q ). (1.9) q,q a n=0 Note the consistency of Eq. (1.6) and Eq. (1.9). The completeness can be written as X iqx n e =Nδ . (1.10) n,0 q∈1stBZ In the limit that N →∞, this equation becomes Z π a dq 1 iqx n e = δ . (1.11) n,0 π 2π a − a Theintegration runsover a finiterangeofq, despitethefact thatthelattice is infinitely large. Why shouldn’t it? No matter how large the lattice is, the lattice periodicity has not disappeared. It is only in the limita→0 that we recover the results of the continuum given above. To summarize, we started with a function which was only defined on a discrete set of lattice points; in the limit N → ∞, this discreteness does not go away but the set q approaches a bounded continuum. The function f is periodic with respect q6 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS to the reciprocal lattice vectors, i.e., the entire q space can be divided up into periodic unit cells, but clearly not in an unique manner. Finally, the inverse Fourier series is given by X −iqx n f =a e f(x ). (1.12) q n n In the limit N →∞, Z π a dq iqx n f(x )= e f , (1.13) n q π 2π − a X −iqx n f =a e f(x ). (1.14) q n n The prefactor a in front of this sum is actually the volume of the unit cell in real space. You can now generalize all this to three dimensions and work out the consequences of various normalizations. 1.3 Quantum Mechanics A quantum mechanical system is defined by a Hilbert space,H, whose vec- tors are states,ψi. Thereare linear operators,O which act on this Hilbert i space. These operators correspond to physical observables. Finally, there is an inner product, which assigns a complex number,hχψi, to any pair of states,ψi,χi. A state vector,ψi gives a complete description of a system through the expectation values, hψOψi (assuming that ψi is normalized i so thathψψi=1), which would be the average values of the corresponding physical observables if we could measure them on an infinite collection of identical systems each in the stateψi. If for all vectorsψi andχi, ∗ hχLψi=hψOχi . (1.15) then the operator L is the Hermitian adjoint of O and will be denoted † ∗ by O . Here c is the complex conjugate of the complex number c. The notation follows Dirac and tacitly uses the dualvector space of brashψ corresponding to vector space of kets ψi. Although the introduction of the dual vector space could be avoided, it is a very elegant and useful concept. Just see how ugly it would be if we were to define the scalar ∗ product of two vectors as (χi,ψi)=(ψi,χi) . An Hermitian operator satisfies † O =O (1.16)1.3. QUANTUM MECHANICS 7 while a unitary operator satisfies † † OO =O O =1 (1.17) IfO is Hermitian, then iO e (1.18) is unitary. Given an Hermitian operator,O, its eigenstates are orthogonal, ′ ′ ′ ′ hλOλi=λhλλi=λ hλλi (1.19) ′ For λ= 6 λ, ′ hλλi=0 (1.20) If there are n states with the same eigenvalue, then, within the subspace spanned by these states, we can pick a set of n mutually orthogonal states. Hence, we can usethe eigenstatesλias abasis for Hilbertspace. Any state ψi can be expanded in the basis given by the eigenstates ofO: X ψi = c λi (1.21) λ λ with c =hλψi. (1.22) λ TheHamiltonian, ortotalenergy,whichwewilldenotebyH,isapartic- ularlyimportantoperator. Schr¨odinger’sequationtellsusthatH determines how a state of the system will evolve in time. ∂ i ψi =Hψi (1.23) ∂t If the Hamiltonian is independent of time, then we can define energy eigen- states, HEi=EEi (1.24) which evolve in time according to: Et −i E(t)i=e E(0)i (1.25) An arbitrary state can be expanded in the basis of energy eigenstates: X ψi= cEi. (1.26) i i i8 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS It will evolve according to: X E t j −i ψ(t)i = c e E i. (1.27) j j j The usual route for constructing the quantum mechanical description of a physical system (Hilbert space, inner product, operators corresponding to physicalobservables)leansheavily ontheclassicaldescription. Theclassical variables p,q are promoted to quantum operators and the Poisson bracket relation p,q = 1 becomes the commutator of the corresponding opera- P.B. torsp,q: p,q=−i. Hilbertspace is then constructed as the representation space for the algebra of the operators p,q. The theory is then “solved” by finding the eigenstates and eigenvalues of the Hamiltonian. With these in hand, we can determine the state of the system at an arbitrary time t, given its state at some initial time t , according to (1.26) and (1.27). This 0 procedure is known as canonical quantization. Letuscarrythisoutexplicitlyinthecaseofasimpleharmonicoscillator. ThesolutionoftheharmonicoscillatorwillbeusefulpreparationfortheFock space construction of quantum field theory. The harmonic oscillator is defined by the Hamiltonian,  1 2 2 H = ω p +q (1.28) 2 and the commutation relations, p,q=−i (1.29) We define raising and lowering operators: √ a =(q+ip)/ 2 √ † a =(q−ip)/ 2 (1.30) The Hamiltonian and commutation relations can now be written:   1 † H =ω a a+ 2 † a,a =1 (1.31) WeconstructtheHilbertspaceofthetheorybynotingthat(1.31)implies the commutation relations, † † H,a =ωa1.3. QUANTUM MECHANICS 9 H,a=−ωa (1.32) These, in turn, imply that there is a ladder of states, † † Ha Ei =(E +ω)a Ei HaEi=(E−ω)aEi (1.33) Thisladder will continue downto negative energies (which itcan’t sincethe Hamiltonian is manifestly positive definite) unless there is an E ≥ 0 such 0 that aE i=0 (1.34) 0 † To find E , we need to find the precise action of a, a on energy eigen- 0 † statesEi. From the commutation relations, we know thata Ei∝E+ωi. † To get the normalization, we writea Ei=c E +ωi. Then, E 2 † c =hEaa Ei E ω =E + (1.35) 2 Hence, r ω † E+ E +ωi a Ei= 2 r ω aEi= E− E−ωi (1.36) 2 From the second of these equations, we see that aE i=0 if E =ω/2. 0 0 Thus, we can label the states of a harmonic oscillator by their integral † a a eigenvalues,ni, with n≥0 such that   1 Hni=ω n+ ni (1.37) 2 and √ † a ni= n+1n+1i √ ani= nn−1i (1.38) These relations are sufficient to determine the probability of any physical observation at time t given the state of the system at time t . 0 Inthisbook,wewillbeconcernedwithsystemscomposedofmanyparti- cles. Atthemostgeneralandabstractlevel, theyareformulatedinprecisely the same way as any other system, i.e. in terms of a Hilbert space with an10 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS innerproductactedonbyoperatorscorrespondingtoobservables. However, thereisonefeatureofthisdescriptionwhichispeculiartomany-particlesys- temscomposedofidentical particlesandhasnorealclassicalanalog: Hilbert space must furnish an irreducible representation of the permutation group acting on identical particles. We will briefly review this aspect of quantum many-particle systems. When we have a system with many particles, we must now specify the states of all of the particles. If we have two distinguishable particles whose Hilbert spaces are spanned by the bases i,1i (1.39) where i=0,1,... are the states of particle 1 and α,2i (1.40) where α = 0,1,2,... are the states of particle 2. Then the two-particle Hilbert space is spanned by the set: i,1;α,2i≡i,1i⊗α,2i (1.41) Suppose that the two single-particle Hilbert spaces are identical, e.g. the two particles are in the same box. Then the two-particle Hilbert space is: i,ji≡i,1i⊗j,2i (1.42) Iftheparticles areidentical, however, wemustbemorecarefulbecausei,ji andj,ii must be physically the same state, i.e. iα i,ji =e j,ii. (1.43) Applying this relation twice implies that 2iα i,ji =e i,ji (1.44) iα soe =±1. Theformercorrespondstobosons,whilethelattercorresponds to fermions. The two-particle Hilbert spaces of bosons and fermions are respectively spanned by: i,ji+j,ii (1.45) and i,ji−j,ii (1.46)1.4. STATISTICAL MECHANICS 11 Then-particleHilbertspacesofbosonsandfermionsarerespectivelyspanned by: X i ,...,i i (1.47) π(1) π(n) π and X π (−1) i ,...,i i (1.48) π(1) π(n) π Hereπ denotes a permutation of theparticles. Inposition space, this means that a bosonic wavefunction must be completely symmetric: ψ(x ,...,x,...,x ,...,x )=ψ(x ,...,x ,...,x,...,x ) (1.49) 1 i j n 1 j i n while a fermionic wavefunction must be completely antisymmetric: ψ(x ,...,x,...,x ,...,x )=−ψ(x ,...,x ,...,x,...,x ) (1.50) 1 i j n 1 j i n 1.4 Statistical Mechanics The concept of partition function is central to equilibrium statistical me- chanics. For a canonical ensemble that we shall frequently use, it is given by Z, X −βE n Z = e . (1.51) n wherethetemperatureoftheensemble,T,is1/k β,andk istheBotzmann B B constant. HereE aretheenergy eigenvalues oftheHamiltonian. Given the n partition function, the macroscopic properties can be calculated from the free energy, F, 1 F =− lnZ. (1.52) β To make sure that a system is in equilibrium, we must make the scale of observation considerably greater than all the relevant time scales of the problem; however, in some cases it is not clear if we can reasonably achieve this condition. Alternatively, we may, following Boltzmann, define entropy, S, in terms of the available phase space volume, Γ(E), which is S =k lnΓ(E). (1.53) B But how do we find Γ(E)? We must solve the equations of motion, that is, we must know the dynamics of the system, and the issue of equilibration12 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS must be addressed. In contrast, in the canonical ensemble, the calculation of the partition function is a counting problem. The Boltzmann formula can be reconciled with the ensemble approach of Gibbs. We must determine Γ(E). In general, this is impossible without computing the trajectory of the system in the phase space. The recourse is to assume that Γ(E) is the entire volume of the phase space allowed by the conservation laws. No matter how complicated the motion may be, if the system, in the course of time, visits every point in the phase space, all we need to do is to calculate the measure in the phase space corresponding to the conserved quantities. It is convenient to introduce quantum mechanics atthissteptosimplifytheargument. Accordingtoquantummechanicseach pointinthephasespacecorrespondstoaquantumstate. So,wesimplyhave to count the number of states, and we write X Γ(E)= δ(E−E ). (1.54) n n Equation (1.53), combined with Eq. (1.54), defines the microcannonical en- semble of Gibbs. But in “deriving” it, we did not have to invoke the notion of an ensemble. We can go further and ask what would happen if we replaced the above formula by the following: X ′ −β(E −E) n Γ(E)= e , (1.55) n where β, for the moment, is an unknown positive number. You can show ′ that the entropy defined by Γ(E) leads to the same thermodynamics as the one defined by Γ(E), provided β = 1/k T. We have now arrived at the B cannonical ensemble. This is curious; in Eq. (1.54) we only sum over states ofenergyE,butin(1.55)weseemtosumoverallstates. Thereasonforthis 23 miracle is the extensive nature of E and S. They are of order N (∼ 10 ). Consequently, the sum is so sharply peaked that practically all the weight is concentrated atE. Now, Eq. (1.55), combined with Eq. 1.53, leads to the same thermodynamics as you would obtain from a canonical ensemble. Although theensemble approach isquite elegant andconvenient, uncrit- ical use of it can be misleading. Suppose that you are given a Hamiltonian which has two widely separated scales, a very fast one and a very slow one. If the observation scale is longer than the shorter time scale, but smaller than the longer time scale, the slow degrees of freedom can be assumed to be constant. They cannot wander very much in the phase space. Thus, in1.4. STATISTICAL MECHANICS 13 calculating the relevant volume of the phase space we must ignore the slow degrees of freedom, otherwise we would get an answer that will not agree with observations. A simple well known example of two distinct time scales is the problem ofortho-andpara-hydrogen. Thespinsofthenucleiinahydrogenmolecule canbeeither inatripletstate, orinasingletstate. Theinteraction between the nuclei is negligible and so is the interaction between the nuclei and the electronic spins that are in a singlet state. Thus, the ortho-para conver- sion takes time, on the order of days, while the momenta of the molecules equilibrate in a microscopic time scale. Therefore, the number of nuclei in the singlet state and the number of nuclei in the triplet state are separately constants of motion on the time scale of a typical experiment, and the free energies of these two subsytems must be added rather than the partition functions. Experimental observations strikingly confirm this fact. When thereare afew discrete set of widely separated scales, it is easy to apply our formulae, because it is clear what the relevant region of the phase space is. There are instances, however, where this is not the case, and there isacontinuumofoftimescales,extendingfromveryshortmicroscopicscales to very long macroscopic scales. The common amorphous material, window glass, falls into this category. If glass is to be described by a Hamiltonian, it is not sufficient to know all the states and sum over all of them; we must examinetheactual dynamics ofthesystem. Glass is knowntoexhibitmany anomalous thermal properties, including a time dependent specific heat. In this respect, the Boltzmann formula, Eq. (1.53), can still be used. In principle,wecould calculate theactual trajectories todeterminethevolume ofthephasespacesampledduringtheobservationtime. Thereisnoneedto usethehypothesisthatΓ(E)isthetotalvolumeallowedbytheconservation laws. Of course, as far as we know, this formula is a postulate as well and is not derived from any other known laws of physics. We stillhave tounderstandwhatwemean byanensembleaverage when experiments are done on a single system. The ensemble average of an ob- servableO is defined to be hOi=TrρˆO, (1.56) where the density matrix ρ is given by X ρˆ= w nihn, n n X 1 −βE n = e nihn. (1.57) Z n14 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERS Itismorelikely, however, thatanexperimentyieldsthemostprobablevalue ofO,that is, the value shared by most members of the ensemble. However, the distribution of the members in the ensemble is so strongly peaked for a macroscopic system thatroughlyonlyonemembermatters; fluctuations are insignificant in the thermodynamic limit defined by N →∞,V →∞ such N that ρ= is a given number. The relative fluctuations inO is given by V r   2 2 O − O 1 ∼O √ (1.58) 2 O N 23 which is insignificant whenN ∼10 . Thus, the most probable value is the only value, hence the mean value. Another useful partition function is the grand canonical partition func- tion Z defined by, G −β(H−μN) Z =Tre . (1.59) G In this ensemble, the number of particles is not fixed, and the system is assumed to be in contact with a particle bath as well as a heat bath. In the definition of the trace one must also include a sum over a number of particles. The average number of particles is determined by the chemical potential μ. It is convenient to think of chemical potential as a “force” and the numberof particles as a “coordinate”, similar to a mechanical system in which a force fixes the conjugate coordinate. As in mechanical equilibrium, inwhichalltheforcesmustbalance,inastatistical equilibriumthechemical potentials for all the components must balance, that is, must be equal. It is also possibletogive asimilar interpretation toourformulafor thecanonical ensemble wherewe can take the temperatureas the “force” and theentropy as the corresponding “coordinate”. For the grand canonical ensemble, we define the grand potential, Ω: 1 Ω=− lnZ G β =F −μN. (1.60) All thermodynamic quantities can be calculated from these definitions. Actually, we could go on, and define more and more ensembles. For example, we may assume that, in addition, pressure P is not constant and define a pressure ensemble, in which we add a term−PV in the exponent. For every such extension, we would add a “force” multiplied by the corre- sponding conjugate “coordinate”. We could also consider an ensemble in which the linear momentum is not fixed etc.1.4. STATISTICAL MECHANICS 15 The definition of the free energies allow us to calculate various thermo- dynamic quantities. Since F =E−TS (1.61) and dE =TdS−PdV +μdN, (1.62) we get dF =−SdT −PdV +μdN. (1.63) Then,   ∂F S =− , (1.64) ∂T V,N   ∂F P =− , (1.65) ∂V T,N   ∂F μ=− . (1.66) ∂N T,V Similarly, from the definition of the definition of the thermodynamic poten- tial Ω, we can derive the same relations as   ∂Ω S =− , (1.67) ∂T V,μ   ∂Ω P =− , (1.68) ∂V T,μ   ∂Ω μ=− . (1.69) ∂N T,V16 CHAPTER 1. CONVENTIONS, NOTATION, REMINDERSPart II Basic Formalism 17CHAPTER2 Phonons and Second Quantization 2.1 Classical Lattice Dynamics Consider the lattice of ions in a solid. Suppose the equilibrium positions of the ions are the sites R . Let us describe small displacements from these i sites by a displacement fieldu(R ). We will imagine that the crystal is just i a system of masses connected by springs of equilibrium length a. At length scales much longer than its lattice spacing, a crystalline solid can be modelled as an elastic medium. We replace u(R ) by u(r) (i.e. we i replacethelattice vectors,R ,byacontinuousvariable,r). Suchanapprox- i imation is valid at length scales much larger than the lattice spacing, a, or, equivalently, at wavevectors q≪ 2π/a. r+u(r) R + u(R ) i i Figure 2.1: A crystalline solid viewed as an elastic medium. 1920 CHAPTER 2. PHONONS AND SECOND QUANTIZATION The potential energy of the elastic medium must be translationally and rotationally invariant (at shorter distances, these symmetries are broken to discrete lattice symmetries, but let’s focus on the long-wavelength physics for now). Translational invariance implies Vu+u =Vu, so V can only 0 be a function of the derivatives, ∂u . Rotational invariance implies that it i j can only be a function of the symmetric combination, 1 (∂u +∂ u ) (2.1) u ≡ ij i j j i 2 2 Thereare only two possiblesuch terms,u u andu (repeated indices are ij ij kk summed). A third term, u , is a surface term and can be ignored. Hence, kk the action of a crystalline solid to quadratic order, viewed as an elastic medium, is: Z Z h i 1 3 3 2 2 S = dtd rL= dtd r ρ(∂ u ) − 2μu u − λu (2.2) 0 t i ij ij kk 2 whereρisthemassdensityofthesolidandμandλaretheLam´ecoefficients. Underadilatation,u(r)=αr,thechangeintheenergydensityoftheelastic 2 medium isα (λ+2μ/3)/2; under a shear stress, u =αy,u =u =0, it is x y z 2 α μ/2. In a crystal – which has only a discrete rotational symmetry – there may be more parameters than just μ and λ, depending on the symmetry of the lattice. In a crystal with cubic symmetry, for instance, there are, in general, three independent parameters. We will make life simple, however, and make the approximation of full rotational invariance. 2.2 The Normal Modes of a Lattice Let us expand the displacement field in terms of its normal-modes. The equations of motion which follow from (2.2) are: 2 ρ∂ u =(μ+λ)∂∂ u +μ∂ ∂ u (2.3) i i j j j j i t The solutions, i(k·r−ωt) u (r,t)=ǫ e (2.4) i i where is a unit polarization vector, satisfy 2 2 −ρω ǫ =−(μ+λ)k (k ǫ )−μk ǫ (2.5) i i j j i For longitudinally polarized waves, k =kǫ , i i s 2μ+λ l ω =± k≡±vk (2.6) l k ρ2.3. CANONICAL FORMALISM, POISSON BRACKETS 21 while transverse waves, k ǫ =0 have j j r μ t ω =± k≡±v k (2.7) s k ρ Hence, the general solution of (2.3) is of the form:   X 1 s s † s i(k·r−ω t) −i(k·r−ω t) k k u (r,t)= p ǫ a e + a e (2.8) i i k,s s k,s 2ρω k k,s s=1,2,3 corresponds to the longitudinal and two transverse polarizations. p s The normalization factor, 1/ 2ρω , was chosen for later convenience. k The allowed k values are determined by the boundary conditions in a finite system. For periodic boundary conditions in a cubic system of size 3 2π V = L , the allowed k’s are (n ,n ,n ). Hence, the k-space volume 1 2 3 L 3 per allowed k is (2π) /V. Hence, we can take the infinite-volume limit by making the replacement: X X 1 3 f(k)(Δk) f(k)= 3 (Δk) k k Z V 3 = d kf(k) (2.9) 3 (2π) It would be natural to use this in defining the infinite-volume limit, but we will, instead, use the following, which is consistent with our Fourier transform convention: Z   3 X d k 1 s s † s i(k·r−ω t) −i(k·r−ω t) k k u (r,t)= p ǫ a e + a e i i s k,s 3/2 k,s (2π) 2ρω k s (2.10) 2.3 Canonical Formalism, Poisson Brackets The canonical conjugate to our classical field, u , is i ∂L π ≡ =ρ∂ u (2.11) i t i ∂(∂ u ) t i The Hamiltonian is given by Z 3 H = d rπ∂ u −L i t i

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