Lecture Notes Condensed Matter Theory

lecture notes on field theory in condensed matter physics , condensed matter theory and experiment and condensed matter theory applications
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Lecture Notes: Condensed Matter Theory I (TKM1) J. Schmalian (Dated: December 1, 2011) 1I. INTRODUCTION Condensedmatterphysicsisconcernedwiththebehaviorof largeaggregatesofatomsor molecules in liquid or solid form. It is one of the largest branches of physics, with a wide variety of di¤erent systems, approaches, challenges and concepts. Often, it is subdivided in soft condensed matter physics and hard condensed matter physics. While the transi- tion between the two branches is gradual, one way to distinguish them is by the role of quantum mechanics for the elementary excitations of the systems. Soft condensed matter physics (the physics of polymers, liquid crystals, the statistical mechanics of bio-molecules etc.) is frequently termed " = 0"-physics, stressing that classical dynamics su¢ ces for an understanding of the motion and aggregation of these systems. In distinction, for hard condensed matter physics, i.e. " = 1"-physics, the motion of electrons, lattice vibrations etc. is determined by Schrödinger’s equation. Physicsisabasicscienceanditsultimatepurposeistheaccumulationofnewknowledge. In addition, condensed matter physics is closely connected to materials science as well as mechanical, chemical, and electric engineering that focus on the design of novel materials and devices, ranging from better batteries, thermoelectric devices for waste heat conversion to superconductors, magnets, all the way to better agents in drug delivery. This applied aspect of condensed matter physics is exciting and important. Still, we should not forget that the …eld also contributes to the accumulation of fundamental knowledge and to major philosophical issues of our times. The value of Planck’s quantum, , or of the electron charge, e, are de…ned via solid state e¤ects in semiconductors and metals (the quantum Hall e¤ect and the Josephson e¤ect). It is quite amazing that these fundamental constants of nature are not determined in an experiment at a particle accelerator but rather in a solid state laboratory. Other frequently cited examples about the fundamental importance of condensed matter physics are that the famous Higgs particle was …rst proposed in the context of superconducting phase transitions, that asymptotic freedom (important for our current understanding of hadronization of quarks) occurs in case of the Kondo e¤ect of a magneticimpurityinametal, thatfractional charges emergenaturallyinthecontext of the fractional Quantum Hall e¤ect etc. etc. The beauty of condensed matter physics is that it combineshands-onapplicationswiththedevelopmentoffundamentallynewconcepts,often even in the same material 2Lets discuss one epistemological issue that is heavily debated these days: Particle and string theorists search to …nd a better way to formulate the fundamental laws of physics. Thisisveryexcitingresearch. Butsupposeforamoment,weknewthefundamental"theory ofeverything"(TOE).Doesthismeanthatthephysicswouldstopexistingasabasicscience? Well,incondensedmatterphysicswehaveaTOEsince1927,yetmajordiscoveriescontinue to take place. The TOE of condensed matter physics is the many particle Schrödinger equation i =H (1) t with Hamiltonian H =T +T +V +V +V (2) e i ee ii ei with individual terms: N N e i 2 2 X X 2 2 T = r and T = r e i j l 2m 2M l j=1 l=1 N N e i 2 2 X X e e ZZ0 l l V = andV = ee ii jr r 0j jRR0j j j l l 0 0 j;j =1 l;l =1 N ;N e i 2 X e Z l V = : (3) ei jr Rj j l j;l=1 Here r (R) refers to the coordinates of the N (N) electrons (nuclei) with mass m (M). j l e i l Z is the corresponding nuclear charge. The wave function depends in …rst quantization on l all coordinates (r ; ;r ;R ; ;R ) = (frg;fRg): (4) 1 N 1 N j l e i With the exception of radiation e¤ects and spin-orbit interaction (both can easily be in- cluded into the formalism) can all phenomena of condensed matter physics be described by this Hamiltonian and the corresponding equation of motion. At least there is no experi- ment (or Gedankenexperiment) that is in con‡ict with this assertion. Thus, the "theory of everything"ofcondensedmatterphysicsiswellknownandestablished. Itwouldhoweverbe foolishtobelievethatphenomenalikesuperconductivity,thefractionalquantumHalle¤ect, electron localization etc. could be derived from Eqs.1,2. Instead a combination of experi- mental ingenuity, symmetry based reasoning, and a clever analysis of the relevant time and length scales of a given problem (formalized in terms of the renormalization group theory) ultimately allows for such conclusions and lead to conceptually new insights. The message 3isthattheknowledgeofafundamental"theoryofeverything"hasverylittlebearingonthe fascinatingpossibilitiesthatemergewhenmanyparticlesinteractwitheachotherandorga- nizeintonewstatesofmatter. Thereisnoreason,otherthanhabitandacceptedcustom,to believethatthesituationismuchdi¤erentinotherareasofphysics, suchasparticlephysics or quantum gravity. These issues have been lucidly discussed by by Phillip W. Anderson, Science 177, 393–396 (1972) and, more recently, by Robert B. Laughlin and David Pines (Proceedings of the National Academy of Sciences 97, 28-31 (2000). II. QUANTUM THEORY OF SOLIDS A. The Born-Oppenheimer approximation We start our analysis of Eqs.1,2. Great progress in our understanding can be made if one recognizes that the two kinetic energy parts of the Hamiltonian are very di¤erent. The ratio of the masses of the electrons and nuclei is small. 3 m=M ' 10 =Z  1: (5) l l Thus, the motion of the nuclei is much slower compared to that of the electron and we can decoupletheirdynamics. ThisdecouplinggoesbacktoMaxBornandRobertOppenheimner (1927). One assumes that on the time scale of the electrons the nuclei are frozen. To this end we assume that the positions of the nuclei are …xed and play for the electronic wave function, (r ; ;r ;R ; ;R ) = (frg ;fRg) (6) 1 N 1 N j l e i solely the role of given parameters. This leads to the simpli…ed Schrödinger equation i =H (7) el t with electronic Hamiltonian H =T +V +V : (8) el e ee ei From the perspective of the electrons the electron-ion Coulomb interaction plays a role of an "external" potential: N e X V = U (r ) (9) ei j j=1 4where U (r ) is the single particle potential that originates from the nuclei: j N i 2 X e Z l U (r ) = : (10) j jr Rj j l l=1 The time independent Schrödinger equation H =E (11) el el;n n n determines the eigenvalues of the electronic system that depend parametrically on the ion positions E =E (R ; ;R ): (12) el;n el;n 1 N i Next we check whether this approximate treatment is indeed appropriate for m=M  1. l We use the fact that, no matter whether our decoupling is correct or not, the form n a complete set of states for the electronic variables. This allows to expand the full wave function X (frg;fRg) = (frg ;fRg)  (fRg) (13) j l j l n l n n Formally, the  (fRg) are the coe¢ cients of this expansion into a complete set of states. n l Physically, theycorrespondtotheamplitudeoftheionstobefoundatpositionsfRgifthe l electrons are in the state . We insert this ansatz into the full Schrödinger equation n X H = (T +T +V +V +V )  e i ee ii ei n n n X = (T +T +V +V +V )  e i ee ii ei n n n X = (E +T +V )  (14) el;n i ii n n n It holds immediately: V  = V  : (15) ii n ii n n n In addition we have N i 2 X 2 T  = r  i n n n l n 2M l l=1 N i 2 X = r ((r )  + r  ) l l n l n n n 2M l l=1 N i 2 X   2 2 = r  + 2 (r )r  + r  (16) n l l n n l n n n l 2M l l=1 5Suppose we are allowed to ignore the …rst two terms. Then follows: (E +T +V )  ' (E +T +V )  : (17) el;n i ii n el;n i ii n n n Thus, after the solution of the electronic problem is accomplished, we have to solve the purely ionic Schrödinger equation   eff T +V  =E (18) i n n ii eff withe¤ectiveion-ioninteractionV =V +E . Thisdeterminestheionicwavefunction ii el;n ii and…nallythetotalenergyeigenvalueE. Acoupledproblemofionsandelectronsistherefore simpli…ed into two separate problems of the two subsystems. To justify this approach it must obviously hold that the two terms 2   2 r  + 2 (r )r  (19) n l l n l n n 2M l 2 2 are negligible compared to r  . This must be checked a posteriori, after the wave n n l 2M l functions and  aredetermined. Atthispointitisalittlebittoearlytodothisanalysis. n n As the course proceeds, we will be able to perform this calculation and we will …nd:   2 1=2 m 2 r  ' "  ; n F n n n l 2M M   2 3=4 m (r )r  ' "  ; l l n F n n n M M 2  m 2 r  ' "  ; (20) n F n l n n 2M M l For m=M 1 follows that terms that contain a derivativer can be neglected relative l n 2 2 to the leading term r  . n n l 2M We conclude, that the Born-Oppenheimer approximation, valid in the limit m=M 1, justi…es the investigation of a purely electronic Hamiltonian N N N e e e 2 2 X X X e 2 H = r + U (r ) + : (21) el j j 2m jr r 0j j j 0 j=1 j=1 j;j =1 Still,theproblemisonlyde…nedifweknowthepositionsofthenuclei. Incaseofcrystalline solids, where the nuclei are arranged on a periodic lattice, the potential U (r) is periodic with respect to the discrete translations of the crystal, which simpli…es this aspect of the problem signi…cantly. 6After we established H as e¤ective theory of the electronic system only, we study the el impact of the potential U (r) in case of periodic lattices. To this end we …rst ignore the 1 2 electron-electron interaction e jr r 0j . This seems at …rst glance a foolish thing to do, j j as the Coulomb interaction between electrons is neither small nor irrelevant. At this point wemakethisassumptionwithoutfurtherjusti…cationandcomebacktoitlater. Wewillsee thattherearenumeroussituationwherethebehavioroftheelectronsise¤ectivelydescribed by non-interacting fermions, while in other cases (quasi-one dimensional systems, systems where plasma excitations of the electrons matter, systems that order magnetically etc.) the neglect of the Coulomb interaction cannot be justi…ed. Once electrons are noninteracting the Hamiltonian is a sumover single particle Hamilto- nians N e X H = H (p ;r ) el j j j=1 with 2 2 H (p;r) = r +U (r): 2m Since we assumed that the electrons are non-interacting, their thermodynamic behavior is that of an ideal Fermi gas. The remaining task, needed to analyze this Fermi gas, is to determine the single particle eigenfunctions ' (r) and the corresponding eigenvalues " of n n H. Before we discuss these issues we repeat the statistical mechanics of quantum gases and the formalism of second quantization. III. QUANTUM STATISTIC OF IDEA GASES We consider consequences of indistinguishability in quantum statistics. The quantity of interest in statistical mechanics is the partition function P H Z = tre = exp ( E ) (22) 1 where the sum is over all many body statesj i with energy E and = . Once we k T B knowZ we can determine all thermodynamic quantities of a given equilibrium system. For example, the Boltzmann probability 1 p = exp ( E ) Z 7P is normalized ( p = 1) because of the denominator Z. The mean energy is P P 1 E U = hHi = E p = E e Z = logZ (23) For the entropy holds P P k B E S = k p logp = e ( E logZ) B Z 1 = U +k logZ (24) B T From thermodynamics we know that F =UTS is the free energy. Thus, it follows: F (T;V;N) =k T logZ: (25) B In some situations one can also include variable particle numbers and one has to obtain the grand canonical partition function P H Z = tre = exp ( (E N )) (26) g whereN isthenumberofparticlesinthestatej i. Inthisgasholdsforthegrandpotential (or Gibbs free energy) (T;V;) =k T logZ : (27) B g and the mean particle number is 1 P hNi = N exp ( (E N )) = : (28) Z  g For indistinguishable particles it would be necessary to introduce the symmetrized or antisymmetrizedstatesinordertoperformtheabovesum,whichistechnicallyverycompli- cated. Awayoutofthissituationisthesocalledsecondquantization,whichsimplyrespects thefactthatlabelingparticleswasastupidthingtobeginwithandthatoneshouldcharac- terizeaquantummanyparticlesystemdi¤erently. If thelabel of aparticlehasnomeaning, a quantum state is completely determined if one knows which states of the system are oc- cupied by particles and which not. The states of an ideal quantum gas are obviously the momenta since the momentum operator pb = r (29) l l i 8commutes with the Hamiltonian of an ideal quantum system N 2 X pb l H = : (30) 2m l=1 In case of interacting systems the set of allowed momenta do not form the eigenstates of the system, but at least a complete basis the eigenstates can be expressed in. Thus, we characterize a quantum state by the set of numbers n ;n ;:::n (31) M 1 2 which determine how many particles occupy a given quantum state with momentum p ,p , 1 2 ...p . In a one dimensional system of size L those momentum states are M 2l p = (32) l L which guarantee a periodic wave function. For a three dimensional system we have 2 (l e +l e +l e ) x x y y z z p = : (33) l ;l ;l x y z L Aconvenient waytolabel theoccupationnumbers is thereforen whichdeterminedtheoc- p cupationofparticleswithmomentumeigenvaluep. Obviously,thetotalnumberofparticles is: X N = n (34) p p whereas the energy of the system is X E = n " (p) (35) p p If we now perform the summation over all states we can just write X X P Z = exp n " (p)  (36) p N; n p p p fn g p P where the Kronecker symbol  ensures that only con…gurations with correct particle N; n p p number are taken into account. Returning to our earlier problem of noninteracting quantum gases we therefore …nd X X Z = exp n (" (p)) (37) g p p fn g p 9for the grand partition function. This can be rewritten as XX Y YX n ("(p)) n ("(p)) p p Z = ::: e = e (38) g n n p p n p p p 1 2 Fermions: In case of fermions n = 0; 1 such that p Y  ("(p)) Z = 1 +e (39) gFD p which gives (FD stands for Fermi-Dirac) X  ("(p)) =k T log 1 +e (40) FD B p Bosons: In case of bosons n can take any value from zero to in…nity and we obtain p X X  n 1 p n ("(p)) ("(p)) p e = e = (41) ("(p)) 1e n n p p which gives (BE stands for Bose-Einstein) Y  1 ("(p)) Z = 1e (42) gBE p as well as X  ("(p)) =k T log 1e : (43) BE B p A. Analysis of the ideal fermi gas We start from X  ("(p)) =k T log 1 +e (44) B p which gives ("(p)) X X X e 1 hNi = = = = hni (45) p ("(p)) ("(p))  1 +e e + 1 p p p i.e. we obtain the averaged occupation number of a given quantum state 1 hni = (46) p ("(p)) e + 1 Often one uses the symbol f (" (p)) =hni. The function p 1 f () = (47) e + 1 10is called Fermi distribution function. For T = 0 this simpli…es to 8 1 " (p) hni = (48) p : 0 " (p) States below the energy  are singly occupied (due to Pauli principle) and states above  are empty.  (T = 0) =E is also called the Fermi energy. F In many cases will we have to do sums of the type Z X V 3 I = f (" (p)) = d pf (" (p)) (49) 3 h p these three dimensional integrals can be simpli…ed by introducing the density of states Z 3 d p  () =V  (" (p)) (50) 3 h such that Z I = d ()f () (51) Wecandetermine () bysimplyperformingasubstitutionofvariables =" (p) if" (p) = " (p) only depends on the magnitudejpj =p of the momentum Z Z V 4 V 4 dp 2 2 I = p dpf (" (p)) = d p ()f () (52) 3 3 h h d such that p p 4m  () =V 2m =VA (53) 0 3 h p 4 3=2 withA = 2m . Often it is more useful to work with the density of states per particle 0 3 h p  () V  () = = A . (54) 0 0 hNi hNi We use this approach to determine the chemical potential as function ofhNi forT = 0. Z Z Z E E F F 2 3=2 1=2 hNi =hNi  ()n ()d =hNi  ()d =VA d =V A E (55) 0 0 0 0 F 3 0 0 which gives   2=3 2 2 h 6 hNi E = (56) F 2m V 2 3 h 2 If V =d N it holds that E  d . Furthermore it holds that F 2m   1=3 2 V 2m 6 hNi 3 1  (E ) = = (57) F 0 2 2 hNi V 2E 4 h F 11Equally we can analyze the internal energy X  ("(p)) U = logZ = log 1 +e g p X " (p) = (58) ("(p)) e + 1 p such that Z X 3 U = " (p)hni =  ()n ()d = hNiE (59) p F 5 p At …nite temperatures, the evaluation of the integrals is a bit more subtle. The details, which are only technical, will be discussed in a separate handout. Here we will concentrate on qualitative results. At …nite but small temperatures the Fermi function only changes in a regimek T around the Fermi energy. In case of metals for example the Fermi energy B 4 5  with d' 1 10A leads to E ' 1:::10eV i.e. E =k ' 10 :::10 K which is huge compared F F B to room temperature. Thus, metals are essentially always in the quantum regime whereas low density systems like doped semiconductors behave more classically. If we want to estimate the change in internal energy at a small but …nite temperature one can argue that there will only be changes of electrons close to the Fermi level. Their excitationenergyisk T whereastherelativenumberofexcitedstatesisonly (E )k T. B F B 0 1 Due to  (E ) it follows in metals  (E )k T 1. We therefore estimate F F B 0 0 E F 3 2 U' hNiE +hNi (E ) (k T ) +::: (60) F F B 0 5 at lowest temperature. This leads then to a speci…c heat at constant volume C 1 U V 2 c = =  2k  (E )T = T (61) V F B 0 hNi hNiT whichislinear,withacoe¢ cientdeterminedbythedensityofstatesattheFermilevel. The correct result (see handout3 and homework 5) is 2  2 = k  (E ) (62) F B 0 3 whichisalmostidenticaltotheonewefoundhere. Note, thisresultdoesnotdependonthe speci…c form of the density of states and is much more general than the free electron case with a square root density of states. 12Similarly one can analyze the magnetic susceptibility of a metal. Here the energy of the up ad down spins is di¤erent once a magnetic …eld is applied, such that a magnetization M =  (hNihNi) " B   Z E F =  hNi  ( + B) ( B) d (63) B 0 B 0 B 0  () 0 For small …eld we can expand  ( + B)' () +  B which gives 0 B 0 B Z E F  () 2 0 M = 2 hNiB d B 0 2 = 2 hNiB (E ) (64) F B 0 This gives for the susceptibility M 2  = = 2 hNi (E ): (65) F B 0 B Thus, one can test the assumption to describe electrons in metals by considering the ratio of  andC which are both proportional to the density of states at the Fermi level. V B. The ideal Bose gas EvenwithoutcalculationisitobviousthatidealBosegasesbehaveverydi¤erentlyatlow temperatures. In case of Fermions, the Pauli principle enforced the occupation of all states up to the Fermi energy. Thus, even at T = 0 are states with rather high energy involved. The ground state of a Bose gas is clearly di¤erent. At T = 0 all bosons occupy the state with lowest energy, which is in our casep =0. An interesting question is then whether this macroscopic occupation of one single state remains at small but …nite temperatures. Here, a macroscopic occupation of a single state implies hni p lim 0. (66) hNi1 hNi We start from the partition function X  ("(p)) =k T log 1e (67) BE B p which gives for the particle number X 1 hNi = = : (68) ("(p))  e 1 p 13Thus, we obtain the averaged occupation of a given state 1 hni = : (69) p ("(p)) e 1 Remember that Eq.68 is an implicit equation to determine  (hNi). We rewrite this as Z 1 hNi = d () : (70) () e 1 The integral diverges if  0 since then for ' Z  () hNi d 1 (71) () if  () =6 0. Since  () = 0 if 0 it follows  0: (72) 1=2 The case  = 0 need special consideration. At least for  () , the above integral is convergent and we should not exclude  = 0. Lets proceed by using p  () =VA (73) 0 p 4 3=2 withA = 2m . Then follows 0 3 h Z p 1 hNi = A d 0 () V e 1 0 p Z 1 A d 0 e 1 0 Z 1 1=2 x 3=2 = A (k T ) dx (74) 0 B x e 1 0 It holds Z p   1 1=2 x  3 dx = & ' 2:32 (75) x e 1 2 2 0 We introduce   2=3 2 h hNi k T = a (76) B 0 0 m V with 2 a = ' 3:31: (77) 0  2=3 3 & 2 The above inequality is then simply: T T: (78) 0 14Our approach clearly is inconsistent for temperatures belowT (Note, except for prefactors, 0 k T is a similar energy scale than the Fermi energy in ideal fermi systems). Another way B 0 to write this is that   3=2 T hNihNi : (79) T 0 Note, the right hand side of this equation does not depend onhNi. It re‡ects that we could not obtain all particle states below T . 0 The origin of this failure is just the macroscopic occupation of the state with p =0. It has zero energy but has been ignored in the density of states since  ( = 0) = 0. By introducing the density of states we assumed that no single state is relevant (continuum limit). This is obviously incorrect for p =0. We can easily repair this if we take the state p =0 explicitly into account. X 1 1 hNi = + (80) ("(p))  e 1 e 1 p0 for all …nite momenta we can again introduce the density of states and it follows Z 1 1 hNi = d () + (81) ()  e 1 e 1 The contribution of the last term 1 N = (82) 0  e 1 is only relevant if N 0 lim 0: (83) hNi1hNi N 0 If  0, N is …nite and lim = 0. Thus, below the temperature T = T the 0 hNi1 0 hNi chemical potential must vanish in order to avoid the above inconsistency. For T T 0 follows therefore   3=2 T hNi =hNi +N (84) 0 T 0 which gives us the temperature dependence of the occupation of the p =0 state: IfT T 0   3=2 T N =hNi 1 : (85) 0 T 0 andN = 0 forT T . Then 0. 0 0 For the internal energy follows Z 1 U = d () (86) () e 1 15which has no contribution from the "condensate" which has = 0. The way the existence of the condensate is visible in the energy is via  (T T ) = 0 such that for T T 0 0 Z Z 3=2 1 3=2 x 5=2 U =VA d =VA (k T ) dx (87) 0 0 B x e 1 e 1 0 R p 3=2 1 x 3 It holds again dx = & (5=2)' 1:78. This gives x 0 e 1 4   3=2 T U = 0:77hNik T (88) B T 0 5=2 leading to a speci…c heat (use U = T ) U 5 5U 3=2 3=2 C = = T = T . (89) T 2 2T This gives Z Z T 0 T c (T ) 5 5 5U 0 01=2 0 3=2 S = dT = T dT = T = (90) 0 T 2 3 3T 0 0 which leads to 2 =UTSN = U (91) 3 The pressure belowT is 0 3=2 5U m 5=2 p = = = 0:08 (k T ) (92) B 3 V 3V h which is independent of V. This determines the phase boundary p =p (v ) (93) c c c V with speci…c volume v = at the transition: hNi 2 h 5=3 p = 1:59 v : (94) c m IV. SECOND QUANTIZATION A. The harmonic oscillator: raising and lowering operators Lets …rst reanalyze the harmonic oscillator with potential 2 m 2 V (x) = x (95) 2 16where is the frequency of the oscillator. One of the numerous approaches we use to solve this problem is based on the following representation of the momentum and position operators: r  y xb = ba +ba 2m r  m y pb = i ba ba : (96) 2 From the canonical commutation relation x; b pb =i (97) follows   y ba;ba = 1   y y ba;ba = ba ;ba = 0: (98) Inverting the above expression yields r   m i ba = xb + pb 2 m r   m i y ba = xb pb (99) 2 m y demonstrating thatba is indeed the operator adjoined toba. We also de…ned the operator y b N =baba (100) which is Hermitian and thus represents a physical observable. It holds    m i i b N = xb pb xb + pb 2 m m m 1 i 2 2 = xb + pb p;b xb 2 2m 2   2 2 1 pb m 1 2 = + xb : (101) 2m 2 2 We therefore obtain   1 b b H = N + : (102) 2  1 b SincetheeigenvaluesofH aregivenasE = n + weconcludethattheeigenvaluesof n 2 b the operatorN are the integersn that determine the eigenstates of the harmonic oscillator. b Njni =njni: (103) 17  y Using the above commutation relation ba;ba = 1 we were able to show that p bajni = njn 1i p y ba jni = n + 1jn + 1i (104) y The operatorba andba raise and lower the quantum number (i.e. the number of quanta). For these reasons, these operators are called creation and annihilation operators. B. second quantization of noninteracting bosons While the above results were derived for the special case of the harmonic oscillator there is a similarity between the result   1 E = n + (105) n 2 for the oscillator and our expression X E = " n (106) fn g p p p p for the energy of a many body system, consisting of non-interacting indistinguishable par- ticles. While n in case of the oscillator is the quantum number label, we may alternatively argue that it is the number of oscillator quanta in the oscillator. Similarly we can consider the many body system as a collection of a set of harmonic oscillators labelled by the single particle quantum number p (more generally by p and the spin). The state of the many body system was characterized by the setfng of occupation numbers of the states (the p number of particles in this single particle state). We the generalize the wave functionjni to the many body case jfngi =jn ;n ;:::;n ;:::i (107) p 1 2 p and introduce operators p ba jn ;n ;:::;n ;:::i = n jn ;n ;:::;n 1;:::i p 1 2 p p 1 2 p p y ba jn ;n ;:::;n ;:::i = n + 1jn ;n ;:::;n + 1;:::i (108) 1 2 p p 1 2 p p That obey h i y 0 ba ;ba = : (109) 0 p p;p p 180 0 It is obvious that these operators commute if p =6 p. For p =p follows p y ba ba jn ;n ;:::;n ;:::i = n + 1ba jn ;n ;:::;n + 1;:::i p 1 2 p p p 1 2 p p = (n + 1)jn ;n ;:::;n ;:::i (110) p 1 2 p and p y y ba ba jn ;n ;:::;n ;:::i = n ba jn ;n ;:::;n 1;:::i p 1 2 p p 1 2 p p p = n jn ;n ;:::;n ;:::i (111) p 1 2 p y y which givesba ba ba ba = 1. Thus the commutation relation follow even if the operators p p p p are not linear combinations of position and momentum. It also follows y nb =ba ba (112) p p p for the operator of the number of particles with single particle quantum number p. The P y b total number operator isN = ba ba . Similarly, the Hamiltonian in this representation is p p p given as X y b H = " ba ba (113) p p p p which gives the correct matrix elements. We generalize the problem and analyze a many body system of particles with single particle Hamiltonian 2 pb b h = +U (br) (114) 2m which is characterized by the single particle eigenstates b hji =" ji: (115) isthelabelofthesingleparticlequantumnumber. Wecanthenintroducetheoccupation number representation with jn ;n ;:::;n ;:::i (116) 1 2 h i y 0 andcorrespondingcreationanddestructionoperators ba ;ba = . Wecanthenperform 0 ; a unitary transformation among the states P P j i = U j i = j ih j i (117) 19The statesj i are in general not the eigenstates of the single particle Hamiltonian (they only are if U =h j i =  ). We can nevertheless introduce creation and destruction operators of these states, that are most naturally de…ned as: P ba = h j iba (118) and the corresponding adjoined equation P y  y ba = h j i ba : (119) This transformation preserves the commutation relation (see below for an example). We can for example chose the basis as the eigenbasis of the potential. Then holds in second quantization X y b U = h jU (r)j ia a (120) and we can transform the result as X 0 y b 0 U = h j ih jU (r)j ih j iba ba 0 ; ; X 0 y = h jU (r)j iba ba 0 (121) 0 ; R 0 3 It holds of courseh jU (r)j i = d r (r)U (r) 0 (r). In particular, we can chosej i =jri such thath j i =hrj i =  (r). In this case we b use the notationba = (r) and our unitary transformations are r P b (r) =  (r)ba y P  y b (r) =  (r)ba (122) The commutation relation is then  0 ; h i h i P y 0  0 b b (r); (r ) =  (r) (r ) ba ;ba 0 0 0 ; P P  0 0 =  (r) (r ) = hrj ih jri 0 0 = hrjri = (rr ) (123) and it follows Z y 3 b b b U = d rU (r) (r) (r) (124) 20

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