Lattice vibrations - phonons

Transcription

1 Lattice vibratios - phoos So far, we have assumed that the ios are fixed at their equilibrium positios, ad we focussed o uderstadig the motio of the electros i the static periodic potetial created by the ios. But, of course, the ios are quatum objects that caot be at rest i well-defied positios this would violate Heiseberg s ucertaity priciple. So they must be movig! I the followig, we will assume that we are at temperatures very well below the meltig temperature of the crystal, so that the ios perform small oscillatios about their equilibrium positios. Of course, with icreasig T the amplitude of these vibratios icreases ad evetually becomes large eough that the ios start movig past oe aother ad the lattice order is progressively lost (the crystal starts to melt); our approximatios will ot be valid oce we reach such high temperatures. Dealig with small oscillatios is very ice (as you may remember from studyig Lagragia/Hamiltoia dyamics) because i the harmoic approximatio, i.e. where we expad potetials up to secod-order i these displacemets, such classical problems ca be solved exactly by mappig them to a sum of idepedet harmoic oscillators. Goig to the quatum solutio is the trivial, because it will simply be the sum of idepedet quatum harmoic oscillators. I this chapter we discuss this solutio ad its implicatios, ad for the time beig we will completely forget about the electros. We will retur back to them i the ext sectio. I assume you have already studied phoos i your udergraduate solid state physics course. It is very likely that there you cosidered some simple cases (chais with 1 atom ad atom bases) from which you iferred the geeral characteristics of the phoos of ay solid. Here we will geeralize this discussio to ay 3D crystal. This makes the discussio a bit more mathematical but it s really the same idea, so try to ot be blided by the math ad to see that the uderlyig physics is very simple, just harmoic motio. The geeral case of iterest to us is a 3D crystal with a total of N = N 1 N N 3 uit cells ad periodic boudary coditios. To be specific, let a i,i = 1,3 be the lattice vectors that defie the uit cell. For a simple cubic lattice, they would poit alog the 3 axes ad have equal legths, but of course much more complicated possibilities exist. I ay evet, we assume that the crystal s size is N 1 a 1 N a N 3 a 3 with periodic boudary coditios. As always, we will let N i i the ed. We idex uit cells with = ( 1,, 3 ) where each i = 1,N i, ad their locatios are, therefore, specified by R = 3 i=1 i a i. Let us assume that there are r atoms iside each uit cell, which we will idex with α = 1,r, ad let their masses be M α ad their equilibrium positios (with respect to the referece poit of the uit cell) be ρ α. These we will assume to be kow, i the followig. What iterests us is the motio about these (kow) equilibrium positios. The, the locatio of the α io of uit cell is R,α (t) = R + ρ α + u,α (t), where u,α (t) is its displacemet from its equilibrium positio. Util ow, we assumed that u,α (t) 0, i.e. ios were froze at their equilibrium positios. From ow o, we ll allow these quatities to be true variables, but their amplitudes are restricted to be much much smaller tha the separatio from their closest eighbors, so that we still have a well-defied crystal structure. For simplicity, we will solve this problem first i its classical versio, ad the quatize it ad get the quatum solutio. Also, I ll use the lagragia approach ad the switch to the Hamiltoia oe, before quatizig. Of course, oe ca do thigs i a differet order as well ad the fial aswer is the same, but I thik this is the simplest way to get to it. Usig the 3rN coordiates of the N r ios as the geeralized coordiates, the Lagragia for this correspodig classical problem is: L ios = N r =1α=1 M α R,α 1, α,α V α, α (1) where the secod term is the sum of iteractios betwee all pairs of ios. Note that eve though I said that we completely igore the electros, i fact V is screeed by them, i other words we would ot use the Coulomb form, but somethig with a shorter rage. Usually, i practice oe restricts it to earest-eighbor pairs oly, because of this. The formalism works for ay form, though, this is just a matter of how tedious the actual calculatios get. Agai, we assume this potetial to be kow. So far, this is exact. Here we make our harmoic approximatio, i.e. we switch to the variables u,α (t) ad expad the potetial to secod order i these small displacemets. This meas that we replace the iteractios by a sum of three terms. The first oe is the equilibrium cotributio, i.e. what we get whe we set u,α (t) = 0. This is just a overall costat eergy which is part of the cohesio eergy of the crystal. The first order term is the cotributio liear i u,α (t). This must vaish because we are expadig about equilibrium positios, therefore the potetial gradiets that appear i the coefficiets must all vaish. So oly the secod order correctio survives (i 1

2 this approximatio) ad we have: L ios N r =1α=1 M α u,α 1 φ α i, α,α i,i where u αi,i = 1,3 are the cartesia coordiates of the displacemet u,α (t) ad φ α i αi = V α, α R αi R α i eq αi u αi u α i () are kow quatities, called atomic force costats. I fact, you should thik of φ as beig a 3rN 3rN matrix that cotais all the iformatio regardig the iteractios betwee ios (at this level of approximatio). Clearly, it has to satisfy some symmetry coditios. First, all its etries are real umbers, for obvious reasos. Secod, it is clearly a symmetric matrix: φ α i αi = φ αi α i To see the origi of the third symmetry requiremet, cosider the Euler-Lagrage (classical) equatios of motio for io (,α): M α ü αi = φ α i,α,i αi u α i If we displace all atoms equally, i.e. we chage all u α i u α i +δ i, the the force o io (,α) caot chage. From the equatio above, this meas that: α i αi = 0 α φ Similarly, we could rotate the crystal as a whole ad that would ot chage the forces, either. This gives a further set of geeral costraits (a bit of care is eeded here as the cartesia projectios of the force do chage, as described by a rotatio matrix). Fially, because we do have a uderlyig lattice structure, we must have: φ α i αi φ α i αi ( R R ) i.e. because the potetial depeds o the uit cells to which the ios belog oly through the distace betwee the uit cells, so do its derivatives. Of course, if there are additioal symmetries, for example very symmetric locatios of like ios iside the uit cell, the this matrix will reflect that as well. I ay evet, so we kow this matrix, ad the challege is to solve the equatios of motio writte above. We are lookig for ormal modes (because ay geeral solutio ca be writte as a liear combiatios of them); these are solutios where all atoms oscillate with the same frequecy. Moreover, because of the lattice structure, we expect solutios to reflect this periodicity (similarly located atoms should exhibit similar motio). So we try solutios of the form: u αi (t) = 1 Mα e αi e i q R iωt where e αi are the ew ukows. The factor i frot is usually pulled out, although of course it s a costat that could be icluded ito the e s. Before cotiuig, we use the periodic boudary coditios to fid the restrictios o the vector q. Expressig it i terms of the reciprocal lattice vectors b i,i = 1,3 (defied such that a i bj = πδ ij, this results i allowed mometa q = m i i N bi i where m i are itegers. Sice we oly have plae waves at discrete locatios, we fid that oly cosecutive N i values of m i lead to distict solutio, ad the they repeat itself. So, i fact, that are N distict values allowed for q, lyig iside the first Brilloui zoe. Hopefully this is ot a shock: by ow you should expect that aytime we deal with a periodic lattice, all quatities are periodic i (quasi)mometum ad therefore its values are restricted to the BZ. With this, the equatios of motio become: ω e αi = βj D α i αi ( q)e α i

3 where D α i αi ( q) = 1 Mα M β φ α i αi ( R )e i q R are 3r 3r matrices (kow, as well; they are called the dyamical matrix), oe for each allowed value of q. So the beauty is that usig the lattice s traslatioal ivariace, we separated the iitial set of 3rN coupled equatios, ito N distict sets of 3r equatios each; eedless to say, this is huge progress ad a ice example why oe should always take advatage of symmetries! Now, such homogeeous liear systems have o-trivial solutios oly for values ω which are eigevalues of the matrix D, while the e coefficiets are the correspodig eigevectors, which we choose to be orthoormal. Note that we kow that the matrix D is hermitia symmetric because φ is symmetric, so we re guarateed that the eigevalues are real. It turs out that they are also positive if the structure we chose truly is the equilibrium oe (i.e., total eergy is miimized, meaig that the matrix of secod derivatives is positive defied). So we re guarateed that ω are real umbers (we always choose the positive solutio this is a covetio, it makes o differece which solutio is chose so log as it s oly oe for each eigevalue). To summarize: for each of the N allowed q, we fid 3r distict ormal modes with frequecies ω j ( q), j = 1,...,3r. The correspodig eigevectors e α (j) ( q) the defie the spatial profile of the correspodig solutio, i other words show i which directio the atoms of type α = 1,...,r iside each uit cell, are displaced whe this ormal mode is activated. Because the displacemet patters associated with the ormal modes are physical quatities, i other words must be real umbers, it is straightforward to check that we must have e (j) α ( q) = e (j) α ( q). The orthoormatio coditio also implies that we chose α e(j) α ( q) e (j ) α ( q) = δ j,j. The geeral solutio, the, is a liear combiatio of all these ormal modes: u α (t) = C j ( q) e (j) α ( q) ei q R iω j( q)t Mα N where C j ( q) are 3rN complex umbers which defie the amplitudes ad phases of the various ormal modes. Of course, to fid these we eed to be give 3rN iitial coditios. So this would ed the classical solutio. Before quatizig, let s rewrite the Lagragia i terms of ew geeralized coordiates ad speeds, correspodig to these ormal modes. I other words, we chage variables from the old u s to the ew Q s defied such that: u α (t) = Q j ( q,t) e (j) α ( q) ei q R Mα N Of course, we already kow that the solutios for the ew variables will be Q j ( q,t) = C j ( q)e iωj( q)t. The poit is that these are so simple, that oe ca easily guess that the Lagragia should look a lot simpler writte i terms of the Qs. After straightforward calculatios usig the equatios discussed above, oe fids ideed that: L ios = 1 Q j ( q,t) Q j ( q,t) ω j ( q) Q j ( q,t)q j ( q,t)] So yes, we rewrote the messy origial Lagragia as a sum over 3rN idepedet harmoic oscillators. Now let s tur this ito a Hamiltoia, ad the quatize it. For the first step, we replace the geeralized velocities with the geeralized mometa P j ( q,t) = Lios Q to fid: j( q,t) H ios = 1 Pj ( q,t)p j ( q,t)+ω j ( q) Q j ( q,t)q j ( q,t) ] Now we quatize by askig the mometa ad coordiates to be operators: Ĥ ios = 1 ˆPj ( q)ˆp j ( q)+ω j ( q) ˆQj ( q)ˆq j ( q)] where, as usual ˆQ j ( q), ˆP j ( q )] = i hδ q, q δ jj. Just like we do for ordiary quatum harmoic oscillators, it is coveiet to defie ladder operators for each mode: h ( ) hωj ˆQ j ( q) = b +b ( q) ( ) j, q ˆP j ( q) = i b ω j ( q) b j, q 3

4 where, as usual, b,b j, q ] = 0,b,b j, q ] = δ q, q δ jj are bosoic operators. I terms of these, we fially fid: Ĥ ios = ( hω j ( q) b b + 1 ) Here we have the zero poit motio eergy E GS = 1 hω j( q) which should also be icluded ito the cohesio eergy this cotributio is always there, ad it is precisely because eve i the groud-state, the ios do oscillate ad therefore cotribute somethig to the total eergy. Excited states are { } = b ] 0 with eergy! E ios ({ }) = hω j( q) ( + ) 1. Here, = 0,1,,... defies the eigestate of each ormal mode (oscillator). We are ow basically doe: we kow all eigevalues ad eigefuctios. Of course, i actual calculatios we eed to go back to the origial operators which were the displacemets of the ios out of equilibrium. Give the lik betwee us ad Qs, ad the Qs ad the bs, this is easily achieved: ˆ u α = ˆQ j ( q) e α (j) ( q) ei q R Mα N = h M α ω j ( q) e(j) α ( q) ei q R ) (b +b N j, q (3) ad you ca similarly fid the correspodig mometa. So we ca calculate ay desired matrix elemets of ay operators that ca be expressed i terms of the ios locatios ad mometa. Let see what phoos are. For simplicity, cosider the eigestate 0,...,0, =,0,...,0, i.e. the groud-state for all harmoic oscillators except which is i its th -excited state. The the eergy is E GS + hω j ( q) because we use the raisig operator b to go up times from the GS for this oscillator. Istead of thikig i these terms, we ca use the followig (totally equivalet, but more coveiet) laguage. We thik of the groud-state 0,0,...,0 = 0 as a vacuum for excitatios, ad say that state 0,...,0, =,0,...,0 has phoos of the mode j ad mometum q, where each such phoo costs a eergy hω. So the total eergy is agai E GS + hω j ( q) because of these preset phoos. I this laguage, the, b,b are phoo aihilatio ad creatio operators, because actig with b o a state will add to it a phoo of type j ad mometum q, etc. So phoos are bosoic particles of j = 1,...,3r distict types, ad such that if their mometum is q the their eergy is hω. If you thik about it, this is just like we thik about photos except those are the bosos we get whe we quatize the E&M fields ad there are oly types, correspodig to circularly left ad right polarized oes. But just like we say that photos are emitted or absorbed whe a system (eg a electro) moves i betwee two eigestates, ad the photo is such that mometum ad eergy is coserved; just like that phoos are emitted or absorbed as the lattice switches from oe eigestate to aother (because of iteractios with somethig else, of course); ad the phoo(s) emitted or absorbed are such that total mometum ad eergy is coserved. That s it - but it s very coveiet to thik i this picture. Each of the 3r distict flavors of phoos is called a phoo mode. As already stated, they are characterized by their eergies hω j ( q) where j = 1,...,3r. These eergies deped o the mometum q of the phoo ad are periodic fuctio with BZ periodicity. Phoo modes are classified i a couple of ways. First, precisely 3 of these modes are so-called acoustic modes (correspodigly, we talk about acoustic phoos). By defiitio, a mode is acoustic if its eergy vaishes as q 0; i other words these modes are gapless. I fact, we ca be a bit more precise. It turs out that for these modes, ω j ( q) c j q as q 0, where c j is the speed of soud for that mode. Of the 3 modes, oe is logitudial, i.e. its polarizatio vectors e α (j) ( q) become parallel to q as q 0 (i other parts of the Brilloui zoe the polarizatios ca poit whichever way. We still call the mode logitudial ). If you look at the equatio for the displacemets ˆ u α, this meas that i this mode, for small q, all atoms are displaced parallel to q, hece the ame. As you guessed, there are also two trasverse modes, where the atoms are displaced perpedicular to q whe q 0. The speed of soud associated with these three modes is usually differet, although i highly symmetric systems the trasverse modes will have the same speed of soud. These modes are called acoustic precisely because they are activated whe soud propagates through the crystal. These acoustic phoos are also called Goldstoe modes. There is a Goldstoe theorem which states that everytime a system breaks a cotiuos symmetry, there is a gapless boso associated with that. The symmetry broke here is the (cotiuous) traslatioal symmetry, times 3 as we ca traslate i the x,y,z directios. The crystal oly has discrete traslatioal symmetry, ot cotiuous like free space, so there must be precisely 3 Goldstoe modes, ad these are the acoustic modes. The way to thik about this, is that if we traslate 4

5 all atoms by the same δ i, i = x,y,z, the although we have displaced all atoms, i reality we have t chaged their relative positios. So this should ot cost ay eergy, ad this is why these Goldstoe modes are gapless. Ideed, as q 0, oe ca show that the patter of displacemes ˆ u α i a state with oly acoustic phoos looks like a i-phase motio where all atoms are displaced i the same directio with a wavelike patter with a wavelegth λ 1/q. As q 0 this is ideed a overall traslatio ad will cost o eergy, hece the gapless modes. Note that if we have a sigle atom per uit cell, i.e. r = 1, the the 3 resultig modes are the three acoustic modes, ad that is all. For more complicated crystals with r, the remaiig 3(r 1) modes are called optical modes. These are gapped modes, ad i particular ω j ( q) remais fiite as q 0. Agai, i this limit we ca classify them as logitudial vs trasverse. So we ca also say that there are r 1 optical modes, each of which has oe logitudial ad two traverse types. These modes are called optical for the followig reaso. Oe ca show that i the limit q 0, the various atoms i the uit cell are displaced i ati-phase. To give a simple example, if we have a chai with two atoms per uit cell, the if its logitudial optical mode is excited, we fid that u 1 (t) u (t). I other words if oe (set of) atoms moves right, the other (set) moves left, ad viceversa. Remember that these are ios, ad they will geerally have differet ioic charges. So this ati-phase motio gives rise to oscillatig dipoles, ad those couple very strogly to light (oscillatig E&M fields). I other words, these modes ca be excited optically, hece their ame. 0.1 Phoos - whe is this cocept truly useful What we did so far is all exact, oce we accept the harmoic approximatio. So we have to worry what happes if that is ot a good approximatio ad we had to keep higher order terms i the potetial s expasio (for istace, because of goig to higher temperatures). We ca do that by buildig o what we already have, as we ll see soo, but the poit is that sice we chage the Hamiltoia the we obviously will chage the eigestates, ad the we have to woder whether the otio of phoos survives this chage (after all, we itroduced them precisely because we had a very specific structure for the eigeeergies, ie the sum of may idepedet harmoic oscillators). So let s add cubic correctios to the Hamiltoia - those would be the ext biggest cotributio. By the way, such higher order correctios are called aharmoic terms. The, the Hamiltoia ca be writte as: Ĥ ios = ( hω j ( q) b b + 1 ) + 1 φ(...)û αi û α 6 i û α i,, α,α,α i,i,i where φ(...) are kow coefficiets depedig o all those idexes, which are obtaied as 3rd order partial derivatives of the potetial. Of course, ext we rewrite the displacemet operators i terms of phoo creatio ad aihilatio operators like i Eq. (3) so that we do t mix laguages. The, the Hamiltoia ca be writte as: Ĥ ios = hω j ( q) ( b b + 1 ) j 1,j,j 3 q 1, q M j1,j,j3 ( q 1, q, q 3 ) (b j1, q 1 +b j1, q1 )(b +b j, q )(b j3, q 3 +b j3, q3 ) Clearly the three paratheses come from the expasio of the three displacemets, ad all the details of the iteractio are icluded ito the M coefficiets. Such terms are called aharmoic terms. I particular, they have to ecode coservatio laws which come from the symmetries of the system. Oe of this is coservatio of the (quasi)mometum because the lattice is ivariat to lattice traslatios. You might assume that this meas that q 1 + q + q 3 = 0, give that processes described i each parathesis lower the mometum of a state they act upo by their respective q, either by removig a phoo with that mometum or by creatig oe with its opposite mometum. That coservatio law would be true i free space, but ot for a lattice, i.e. ot for a discrete traslatioal group. Remember that because of this discretess, each q is cotraied to the first BZ so as to ot overcout how may distict allowed mometa we have, give that there is o differece betwee, say, a phoo with mometum q ad oe with q + G, where G is ay reciprocal lattice vector. Because of this, if you go through the exercise to calculate the M s, you ll fid that the coservatio law is M j1,j,j 3 ( q 1, q, q 3 ) G δ q1+ q + q 3, G, i.e. the total mometum exchaged ca be ay of the reciprocal lattice vectors icludig zero, of course. The way this works i practice is the followig. Suppose we keep q 1 ad q as free variables ad woder what is the allowed q 3 (with our covetio that all these vectors are restricted to the first BZ). The, if q 1 ad q are small eough, we ca fid a value q 3 = q 1 q that also lies iside the BZ, ad all is good, this looks like the usual mometum coservatio. However, if q 1 ad q are large ad poitig sort of i the same directio, the q 1 q will lie outside the BZ. I this case, there is precisely oe possible G such 5

6 that q 3 = G q 1 q is retured back to the first BZ ad this is the G that appears i the (quasi)mometum coservatio. By the way, processes where G 0 are called Umklapp processes. So let s ow cosider what sort of processes are described by this additioal term. We have three-phoo creatio or aihilatio processes, which go like b 1 b b 3 or b 1b b 3. Their cotributios tur out to be small because of eergy coservatio basically that makes them quite ulikely to happe (but they are ot forbidde!). The more likely processes, which cotribute more substatially, are of the type b 1 b b 3 or b 1 b b 3, i.e. either oe phoo is absorbed ad two are created i its place, or viceversa (you ca check that quasimometum is always coserved). Oe thig we deduce from this is that the umber of phoos is ot coserved, so i this sese they are ot like real particles, eg electros, which caot appear or disappear out of owhere. (Of course, this statemet is ot quite true. There are vacuum fluctuatios that ca create electro-positro pairs, but the we have the positro to deal with ad iclude i our models, but more importatly these are very high eergy processes which we are irrelevat at the low-eergy scale of iterest to us). Because of this o-coservatio, the chemical potetial for phoos is zero. Oe way to thik about it is that we itroduce a chemical potetial whe we wat to switch from a caoical descriptio (with a fixed umber of particles) to a grad-caoical oe (with a variable umber) H H µ ˆN. If the umber is variable o matter what, the there is o differece betwee the two formulatios, ad for that to be true we must have µ = 0. We ll see below the cosequeces of this. The more worrisome coclusio is that we caot actually thik of the state of our lattice i terms of it havig so may phoos of such ad such types, if the umber of phoos ad their types keep chagig. This should remid you of the discussio we had about quasiparticles, about whe it is meaigful to thik i terms of them occupyig a Fermi sea. It is precisely the same here: if the aharmoic terms are so small that this scatterig ad chagig of the phoos is very very slow, the we ca igore it ad it is perfectly useful ad fie to thik i terms of phoos. If the aharmoric terms are large ad these processes happe o a time-scale similar to whatever process is of iterest to us, the a descriptio i terms of phoos is meaigless ad we have to do better. The good ews is that for all reasoable materials at reasoable temperatures (i.e, well below the meltig poit) the phoo lifetimes are log eough that it is perfectly ok to thik i terms of them as excitatios with (essetially) ifiite lifetimes. 0. The lattice cotributio to various properties of the solid Now we ca discuss the cotributio of the ios to properties of the solid. I will oly do so briefly because this should have bee discussed i solid state/stat mech courses, so I ll just remid you how this works. Like for ay bosos with µ = 0, i thermal equilibrium the average umber of phoos of type j ad mometum q is give by the Bose-Eistei distributio: b b 1 = e β hωj( q) 1 This tells us that at low temperatures, which here meas k b T hω j ( q), there are basically o phoos of that type, whereas at high temperatures their umber icreases liearly with temperature, b b kbt hω j( q). Note that what is high ad what is low temperature ow depeds very much o the type of phoos: oftetime temperatures of iterest are high for acoustic phoos but low for optical phoos, as the latter have much higher eergies tha the former. You might also worry what happes with the q = 0 acoustic phoos, of which it seems we have a ifiite umber o matter what. It turs out they always fall out of calculatios, as I ll show i you i the ext example. So we could just remove them from cosideratio, if that makes you feel better. Physically, this is because they describe actual lattice traslatios, as we discussed, ad those caot cotribute to ay physical property. The lattice cotributio to the eergy of the system is, the: E ios = 1 hω j ( q) e β hωj( q) ] Oce we kow the phoo eergies, this ca be calculated. Note that at temperatures that are high for all phoo modes, we fid this eergy to be 3rNk B T. This is the Dulog-Petit law which follows from the equipartitio theorem for a classical system. Of course, at high eough T a classical descriptio of our quatum system must become valid. I fact, the measurable quatity is the specific heat, i.e. how much eergy we must give to the system to chage its temperature, eg at costat volume. The cotributio from the lattice (remember that there is a separate oe, due 6

7 to the electros) is: C V = E ios T This is a rather messy expressio, but it ca be calculated. Of course, i practice we take N, i.e. the sums over allowed mometa are replaced by itegrals. Oe ca also make various approximatios (like Debye or Eistei approximatios) for the phoo dispersios hω j ( q) to further simplify thigs so that those itegrals ca be evaluated easily. You should have see all this discussed before. If ot, please look at ay stadard textbook it s basically just a matter of doig some itegrals. At this poit we could stop ad move o, but I d like to show you aother way to calculate phoo spectra, where we do the calculatio at a quatum level all the way. Of course, it gives precisely the same aswers as the classical calculatio that was the quatized, that we discussed here. This will give us a chace to practice equatio-of-motio techiques for diagoalizig bosoic Hamiltoias, ad a explicit example of acoustic ad optical modes, which should better illustrate the geeral stuff I talked about so far. We will chose a example that is simple eough that we ca do it fast, but geeral eough that you should be able to geeralize it to basically ay other case of iterest. 0.3 Calculatig the logitudial phoos of a chai with a two-site basis Let s cosider a chai of alteratig ios with masses m ad M, respectively. Let the lattice costat be a. The we ca choose the equilibrum positio of ios of mass m to be at x = a, = 1,N where N is the umber of uit cells (ad surprise surprise, we ll use PBC agai). The ios of mass M are located X = a+δ, where δ is a kow distace. I the simplest case we would have δ = a/ so that each io is halfway betwee two ios of the other type, but i geeral that does t have to be true. Let u,u be the displacemets out of equilibrium of these ios. We ll oly cosider displacemets alog the chai, so oly the logitudial modes are beig discussed. If we make the harmoic approximatio ad if we restrict the iteractios to oly those of earest-eighbor pairs, the resultig Hamiltoia is: Ĥ = ] P M + p + m k1 (U u ) + k (U u +1 ) I should put hats o all the P,p,u,U because they re operators, but it s too bothersome, so bear with me. I the secod term, I allowed for differet iteractios betwee ios i the same uit cells, which are δ apart, ad ios from eighborig cells, which are a δ apart. All the details are iside the elastic costats k 1,k > 0. Of course, they must be positive because we expaded about a equilibrium state. At this poit we could go back to the classical equatios of motio ad follow the geeral recipe we discussed ad solve the problem - I bet you ve doe this i a previous course, but maybe you should practice just to make sure you ca do it! After all, it s just solvig a classical problem ad it should t give you ay trouble. As I said, we ll do this quatum mechaically all the way. First step is to rewrite Ĥ = Ĥ1 +Ĥ +Ĥ1, where: Ĥ 1 = P M + (k 1 +k )U ] ] oly ivolve type 1 of ios, Ĥ = p M + (k 1 +k )u ] oly ivolve type, ad Ĥ 1 = k 1 u U +k U u +1 ] ivolve both. If we did t have the last part, the problem would be trivial because we are summig over idepedet Harmoic oscillators. Let Ω = (k 1 +k )/M ad ω = (k 1 +k )/m be the frequecy of these oscillatios for the two types of ios. Defiig the usual raisig ad lowerig operators: Û = h ( B +B MΩ ) hmω ( ) ˆP = i B B 7

8 ad we fid right away that: û = h ( b +b mω ) Ĥ 1 = hω B B + 1 ] = q hmω ( ) ˆp = i b b hω B qb q + 1 ] ad Ĥ = hω b b + 1 ] = q hω b qb q + 1 ] where for the secod set of sums we Fourier trasformed to mometum space: b q = e iq(a) N b B q = e iq(a+δ) N B Note that I used the actual equilibrium positios i the Fourier expoetials. I could remove the δ from the secod defiitio that is just a overall chage i the phase ad should have o effect o the fial results (you might wat to checkthatstatemet). Ofcourse,thePBCrestricttheallowedq tobeoftheformmπ/n,wherem = N/+1,...,N/ places them iside the first BZ. Note that if we could igore Ĥ1, the we would be doe: both these terms are diagoal, ad the aswer would be two optical phoos with frequecies ω,ω, i.e. two Eistei modes. The reaso the acoustic mode is missig is because if we igore Ĥ1 the, i reality, we igore the iteractios betwee atoms, as here each oe is sittig i its ow harmoic well. No surprise that soud could ot propagate i such a system, because that requires atoms to push each other aroud to create the propagatig wave. But i ay evet, this gives you a idea of how Eistei models ca arise (if you do t remember what those are, I ll remid you at the ed of this sectio). So let s see what s the effect of the iteractio terms betwee ios. First, we have to do a bit of work to write Ĥ1 i terms of these operators. The ed result is: Ĥ = q ] hωb qb q + hωb qb q f(q)(b q +b q)(b q +B q) h where f(q) = (k mmωω 1e iqδ + k e iq(a δ) ). By the way, why are there o Umklapp processes here? Also, I will igore the costat terms which are absorbed ito the cohesio eergy. Now, remember the equatio of motio idea. We re aimig to put this i its diagoal form: Ĥ = q hω j (q)b j,q b j,q j=1 The questio is what are the ew eergies ad the ew operators b 1,q,b,q ad how are they related to the old b q,b q. For the diagoal form: hω j (q)b j,q = b j,q,ĥ] (the ew operators are also bosoic). Now, the most geeral lik betwee the ew ad old operators is: b j,q = αb q +βb q +γb q +δb q where, of course, the coefficiets will deped o j,q but I wo t write that. You might thik that the ew b could deped o all the old b,b,b,b, ot just these four. But mometum coservatio comes to our rescue: oly these 4 guaratee that a mometum q is removed from the system whe b j,q acts o it. So the equatio above (called the equatio of motio) becomes, i terms of the old operators: ( ) hω j (q) αb q +βb q +γb q +δb q = αb q +βb q +γb q +δb q,ĥ] 8

9 ad the problem is reduced to fidig the coefficiets ad eergies which satisfy this. We could proceed with this, or we ca further simplify as follows. For istace, if I wat to pick oly the α term o the lfs, I should take aother commutator with b q. I other words: hω j (q)α = αb q +βb q +γb q +δb q,ĥ],b q] = αb q,ĥ],b q]+βb q,ĥ],b q]+γb q,ĥ],b q]+δb q,ĥ],b q] The double commutators are simple-eough to calculate umbers, so this is the first of 4 liear equatios that will give us a eigevalue problem. I hope this remids you a bit of what we did i the HFA (except there it was a approximatio; this is exact because the Hamiltoia is quadratic). Here is what I get after doig the commutators: hω j (q)α = hω j (q)β = hω j (q)γ = hω j (q)δ = hωα f(q)(β δ) hωβ f( q)(α γ) hωγ f(q)(β δ) hωδ f( q)(α γ) Clearly, the desired ω j (q) are the eigevalues, ad (α,β,γ,δ) is the correspodig eigevector, for the matrix o the rhs. This should puzzle us, because it seems to suggest 4 differet solutios, ot like we expect. But remember that eve i the other formulatio we got twice as may solutios as expected, but we discarded the egative ω values. We have exactly the same deal here: we keep oly the positive eigevalues. Addig ad subtractig the first ad third equatio, we fid hω j (q)(α γ) = hω(α + γ) ad hω j (q)(α + γ) = hω(α γ) f(q)(β γ), which combie ito: ( hωj (q)) ( hω) ] (α γ)+f(q) hω(β γ) = 0. Same tricks for the other equatios lead to: f( q) hω(α γ)+ ( hω j (q)) ( hω) ] (β γ) = 0. The phoo eergies are ow easily recovered from the resultig quadratic equatio. Discardig the egative values, we fid: 1 ω ± (q) = ω +Ω ± (ω +Ω ) 16ω0 4 qa ] si where ω0 4 = k1k mm is also a frequecy. Let s look at the solutio first. For q 0 we fid ω (q) ω 0 a ω +Ωq, so this is the acoustic mode. It icreases mootoicaly ad reaches its maximum value at the Brilloui zoe edges q = ± π a. If you calculate the eigevectors you ca check that the ios move i phase, for q 0. The other mode is the optical mode, sice ω ± (0) = ω +Ω is fiite. This eergy decreases with icreasig q, but it is easy to verify that it is above the acoustic mode for all eergies. So the phoo spectrum is like that sketched o the left, i the ext figure. At this poit, you should be curious to verify that if you set m = M,k 1 = k, i.e. make all ios idetical ad equally spaced, you recover the spectrum for a chai with a oe-io basis, which should oly have the acoustic mode! Does that work out? O the right of the figure, I have sketched for you the ideas behid the Debye ad the Eistei approximatio. The latter is straightforward, we simply take the optical phoo frequecy to be idepedet of q. That simplies calculatios tremedously, of course. It works well because usually those bads are fairly arrow, i.e. early costat. The Debye approximatio is for acoustic modes, ad the idea is to replace the true dispersio with its log wavelegth expressio c s q, where c s is the speed of soud for that mode. Agai, this simplifies calculatios quite a bit. The logic behid this is that if the temperature is low eough that oly phoos with small q are excited i large umbers, the we oly have to worry about describig their eergies well ad this is what the approximatio does. If we re at temperatures well above the highest value, the this mode basically behaves like a classical harmoic oscillator ad we get the right results (equipartitio theorem ad all that) agai. If T is comparable to the phoo eergies at the top of the bad, the we better use the correct expressio if we wat to be accurate. 9

10 ω(q) ω (q) optical Eistei model acoustic Debye model π/a π/ a q kd k D q Figure 1: Left: eergy of the acoustic ad optical logitudial phoos, ω ± (q), for the chai with two atoms i the uit cell. Right: simplified Debye model for the acoustic phoos, assumig liear dispersio at all q but i a spherical Brilloui zoe; ad Eistei model for the optical phoos, assumig the same eergy for all q. I 3D, the Debye approximatio has a secod compoet. Remember that there the BZ has some o-spherical shape, eg it ca be a cube. This leads to a problem, because hω( q) = c s q is oly useful if we ca itegrate i spherical coordiates, but we ca t do that i a cubic (let s say) domai. So the oe deforms the BZ to have the same volume but a spherical shape, so that the itegral ca be doe easily (at low-t the phoos ear the BZ edges are ot excited so they do t cotribute to quatities of iterest, so it does t matter how accurately or badly we model them). We eed to keep the same volume because remember that the BZ volume is such that precisely N distict q are allowed, ad they have a desity that is set by the lattice costat ad system size. To get the correct umber of solutios thus requires us to keep the same overall volume. Doig this allows oe to defie a Debye vector q D, which is the radius of this spherical BZ. For istace, if the true Brilloui zoe is cubic ad of size π a π a π a, the we must have 4πq3 D 3 = ( ) π 3. a Oe the also talks about a Debye frequecy, which is hωd = c s q D, i.e. the maximum value allowed for these phoos i this approximatio. This is a good approximatio if k B T hω D or if k B T hω D. Now we ca go back ad discuss the ifluece of this lattice motio o the electroic behavior. 10