Physics 525, Condensed Matter Homework 3 Due Tuesday, 16 th October 2006

Transcription

1 Physics 55 Condensed Mtter Homework Due Tuesdy 6 th October 6 Jcob Lewis Bourjily Problem : Electron in Wek Sinusoidl Potentil Consider n electron moving in one-dimensionl periodic potentil Ur = cosπr/. We re to obtin the eigenenergies ǫ n q nd corresponding wvefunctions ψ nq r of the lowest two bnds treting the potentil perturbtively. Awy from the edge of the Brillouin zone there re no degenercies in the lowest energy bnd. Using this fct we re to compute ǫ q to order nd the corresponding wvefunctions to order. We begin by doing wht mounts to Fourier trnsforming ψr into momentum spce mking use of Bloch s theorem to write ψ q r = G c q G e iq Gr.. where G represents the reciprocl lttice which is in this one-dimensionl problem generted simply by b π. Following Ashcroft nd Mermin we will study the Schrödinger eqution in momentum spce: [ ] q G ǫ c q G + U G Gc q G =.. m G where U k re Fourier modes of the potentil. In our cse this is extremely esy to extrct: every kindergrtener should be ble to tke the Fourier trnsform of cosine; we find: U + = U = nd U i =± =... We re going to be interested in wve function concentrted well within the first Brillouin zone in the limit where there re no nerly degenerte bnds. Writing ǫ q = m q nd inserting our potentil U k into the Schrödinger eqution we see ǫ ǫ q cq = c q+b + c q b = c q = ǫ ǫ q c q+b + c q b... This expression does not reflect our interest in the first Brillouin zone: it is vlid for ll q. Indeed we see tht we cn itertively unfold the eqution to obtin n expnsion in terms of ǫ ǫ q : c q = ǫ ǫ q c q+b + c q b..5 = ǫ ǫ q ǫ ǫ q+b c q + c q+b + ǫ ǫ q b c q + c q b..6 [ ] = c q ǫ ǫ q ǫ ǫ q+b + ǫ ǫ q b + 8ǫ ǫ qǫ ǫ q+b ǫ ǫ q+b c q+b + c q+b + b b ;..7 &tc. Now becuse we re expnding in ǫ ǫ we my sfely drop the O q terms. Also notice tht in the curly brckets tht we hve the expressions ǫ ǫ q±b. Now becuse the bnds re non-degenerte nd we know the bnd energies re only ffected by terms leding in we cn re llowed to tke ǫ ǫ q±b ǫ q ǫ q±b to this order of pproximtion. With tht in mind we my divide both sides of the expression bove by c q /ǫ ǫ q obtining ǫ ǫ q = ǫ q ǫ q+b + ǫ q ǫ q b + O...8 Note dded in revision: the solution presented follows Ashcroft nd Mermin which is bsolutely horrendous. The entire first problem cn be done in couple of lines if you red the first few pges of Griffith s Quntum Mechnics chpter on time independent perturbtion theory first! Honestly believe me; lern non-degenerte perturbtion theory first nd see how to pply it in the degenerte cse nd the problem will be MUCH esier.

2 JACOB LEWIS BOURJAILY It is not ltogether delightful but this expression cn of course be quite drmticlly simplified. ǫ = ǫ q + m q + b q + b b q b + O = ǫ q + m + O q π ǫ = m q + m + O. q π..9 óπǫρ ǫδǫι πoιησαι Now we re sked to determine the corrected wve function to leding order in. To do this we strt by combining the right hnd sides of equtions..5 nd..7: ǫ ǫ q c q+b + c q b = c q ǫ ǫ q ǫ q ǫ q+b + ǫ q ǫ q b + O = c q+b + c q b = c q ǫ q ǫ q+b + ǫ q ǫ q b + O m = c q + O... q π Now remember tht eqution.. llows to write c q b in terms of c q nd c q b for exmple. Using this to rerrnge eqution.. we see m c q+b = c q b + c q + O q π = ǫ q ǫ q b c m q + c q b + c q + O q π m = c q q π ǫ q ǫ q b + O m = c q q + π + O ; q π π m c q+b = c q q + O... q π π We don t need to reproduce the bove steps for c q b : it comes for free once we hve c q+b : m c q b = c q+b + c q + O q π q π = c q m q π m c q b = c q q π Inserting this in the expnsion for ψ q r we find directly + O ; + q + O... π ψ q r = c q e iqr + c q+b e iqr e ibr + c q b e iqr e ibr + O = c q e iqr m [ + q e irb + + q e irb] + O ; q π π π ψ q r = c q e iqr + m cos q π πr iq πr π sin... óπǫρ ǫδǫι πoιησαι

3 PHYSICS 55: CONDENSED MATTER HOMEWORK b At the edge of the Brillouin zone there re degenercies for smll. We re to work perturbtively ner the zone edge to digonlize the single electron Hmiltonin within the two-stte Hilbert spce of the two nerly-degenerte lowest-energy free-electron momentum eigensttes of the sme crystl momentum. Then we re to dd the effects of the higher bnds perturbtively. We re to obtin the eigenenergies of the lowest two energy bnds to order nd the wve functions to order s before. We re to verify tht for smll these results mtch our work for prt when one moves fr enough wy from the edge of the Brillouin zone. We re to sketch the dispersions ǫ n q nd determine how smll must be for this perturbtion nlysis to be relible. We re going to proceed long lines similr to those encountered in prt. Specificlly let us strt by gin by equting the right hnd sides of equtions..5 nd..7 only this time we will not use the ssumption tht ll the eigenenergies re nondegenerte. ǫ ǫ q = ǫ ǫ q+b + ǫ ǫ q b + O..b. Now we re going to consider perturbing the system ner the Brgg plne t q = π ; this will men tht we cn consider the term ǫ ǫ q+b ζ where /ζ is t most liner in we will justify this nd give n explicit expression for ζ lter. Mnipulting eqution.b. we see tht ǫ ǫ q ǫ ǫq b = + ǫ ǫ q b + O ζ = ǫ ǫ ǫ q + ǫ q b + + ǫ qǫ q b ǫ q b =. ζ ζ This qudrtic is esily solved by clling upon kindergrten identities: ǫ = q ǫ + ǫq b + ± ǫ q ζ + ǫ q b + / ǫ q ζ ǫ q b + ǫ q b ; ζ ǫ = ǫ q + ǫ q b + ζ ± ǫ q ǫ q b + / +..b.5 ζ óπǫρ ǫδǫι πoιησαι To evlute the expression one cn simply insert the eqution itself itertively into ζ = ǫ ǫ q+b nd observe tht tht it lwys gives well-defined expression up to terms of order O. This bnd structure is shown in Figure. We should check tht this result mkes sense nd verify tht it grees with our previous work once we re fr enough wy from the Brgg plne. First notice tht t the Brgg plne where q = b q = π/ we hve ǫ q = π = ǫ π/ + 8ζ ± + / 6ζ = π m ± + 8ζ + O. Inserting this into definition of ζ s prescribed we obtin ǫ q = π = π m ± m π + O..b.6 The reson for being implicit here is tht the two cses we re interested ner nd fr from the Brgg plne give different results; but the implicit expression is lwys correct.

4 JACOB LEWIS BOURJAILY E nergy Eigenvlu e Momentum q Figure. The second-order bnd structure for one-dimensionl system in wek sinusoidl potentil. Similrly we cn check tht eqution.b.5 gives the right nswer when we re fr enough wy from the Brgg plne. When we re fr from the Brgg plne then ǫ q ǫ q b so tht we my expnd ǫ = q ǫ + ǫq b + ± ǫ q ζ ǫ q b + / + ζ = = ǫ q + ǫ q b + ζ ± ǫ q ǫ q b ǫ q + ǫ q b + ζ ± ǫ q ǫ q b + ǫ q ǫ q b ǫ q ǫ q b = ǫ q + ǫ q b + ζ ± ǫ q ǫ + q b ǫ q ǫ q b ζ ǫ q ǫ q b ζ ǫ q ǫ q b + ζ + O. Tking the solution corresponding to the lower bnd ǫ q = ǫ q + ζ + ǫ q ǫ + + O q b ζ = ǫ q + ǫ q ǫ + q+b ǫ q ǫ + O q b / / + O nd this we recognize s eqution..8 which implies tht this formul.b.5 does indeed gree with our results from prt. óπǫρ ǫδǫι πoιησαι Our lst tsk is to determine the wve function for electrons t the Brgg plne to first order in. We will follow similr lines of thought to those trvelled in prt. Using the sme logic s there only this time being creful not to ignore degenercies we cn begin our work with the equtions c q+b + c q b = c q ǫ ǫ q+b + ǫ ǫ q b + O nd c q = ǫ ǫ q c q+b + c q b..b.7 The solutions corresponding to the respective ± sign the eqution.b.5 hve now switched this is simply becuse when we extrcted ǫ q ǫ q b from the squre root the signs one gin become rbitrrily ssigned.

5 PHYSICS 55: CONDENSED MATTER HOMEWORK 5 This system yields exctly our result in prt for the cse of c q+b : c q+b = c q ǫ ǫ q+b + ǫ ǫ q b c b q + O c q+b = c q ǫ ǫ q+b + ǫ ǫ q b ǫ ǫ q b + O c q+b = c q ǫ ǫ q+b + O c q+b = c q ǫ q ǫ q+b + O m = c q π q + π + O ; c q+b = c q q + O..b.8 q π π The story chnges however for c q b. It is not hrd to jump bit in the clcultion nd see c q b = c q ǫ ǫ q b + O..b.9 Now from our clcultion of the eigenenergies t the Brgg plne we know tht so we see ǫ π/ ǫ π/ b = ± m 6π q + π + O.b. c q b = c q + O ± m 6π q+ π = c q + O ± m 8π q+ π m q c q b = ±c q + 8 q π π + O..b. Putting ll this together we see ψ ± q= r = c π q e iqr ± e ibr m + q e ibr e ibr q π π so tht πr ψ + r c q e iqr e i m br cos + i q [ ] πr sin i e ibr ;.b. q π π nd ψ r c q e iqr ie i br sin πr + m q π q π [ ] πr cos e ibr..b. óπǫρ ǫδǫι πoιησαι

6 6 JACOB LEWIS BOURJAILY Figure. The first Brillouin zone dispersion for tight-binding model on twodimensionl squre lttice. Problem : Tight-Binding Model on Squre Lttice Consider tight-binding model on squre two-dimensionl squre lttice lttice spcing with on-site energy ǫ nd nerest-neighbour hopping mtrix element t: H = [ ] ǫ r r + t r r + ˆx + r r ˆx + r r + ŷ + r r ŷ. r We re to obtin the dispersion reltion for this model. Just for the ske of clering up nottion our Brvis lttice here will be generted by = nd = which hs the ssocited reciprocl lttice generted by b = π nd b = π. We will write ll moment in terms of the reciprocl lttice so q = q b + q b. Using Bloch s theorem it is quite esy to see tht the Hmiltonin of this system is given by Hψ = ǫ + t e i q + e i q + e i q + e i q ψ.. = ǫ + t e iπq + e iπq + e iπq + e iπq ψ.. = ǫ + t cosπq + cosπq ψ;.. ǫ q = ǫ + t cosπq + cosπq. This dispersion reltion is shown in the first Brillouin zone in Figure... b-d Let us sketch the Fermi surfce in the first Brillouin zone when the bnd is less thn nd more thn hlf-full ssuming prticle-like bnd t <. And we re to mke n ccurte drwing of the Fermi surfce for the cse of precisely hlf-filled bnd. When the Fermi surfce is very ner the bottom of the bnd energy then it is pproximtely circle: for q i we cn expnd the cosπq i s to see tht ǫq ǫ + t πt q + O q the solution to which is precisely circle. As the energy increses the Fermi surfce flttens out long the digonl directions becoming squre when the bnd is hlf-filled. When the bnd is more thn hlffilled the squre breks into four disjoint components which encircle the corners of the Brillouin zone. Expnding cosπq i bout q i shows tht when the bnd is nerly filled the Fermi surfce components do in fct become circles. These re shown in detil in Figure.

7 PHYSICS 55: CONDENSED MATTER HOMEWORK Figure. Severl Fermi surfces observerd for tight-binding squre lttice model. Drk colouring indictes lower energy Fermi surfces re included for the bnd both more thn nd less thn hlf filled. The hlf-filled Fermi surfce is the clerly visible squre in the plot Figure. Fermi surfces in the tight-binding squre-lttice model with t > next-tonerest-neighbour couplings for vrious vlues of t / t. From left to right: t / t = / t / t = / nd t / t = 7/. Notice the shrp trnsition t t / t = /. e-f We re to dd mtrix element t for hopping between next-to-nerest-neighbour sites nd sketch how the Fermi surfce of the hlf-filled bnd chnges for t > nd t <. It is simple enough to write down the new dispersion reltion coming from the the Hmiltonin similr to prt bove. Following tht nlysis we find Hψ = ǫ + t cosπq + cosπq + t e i q + + e i q + e i q + e i q + ψ = ǫ + t cosπq + cosπq + t e iπq+q + e iπq+q + e iπq q + e iπq q ψ = ǫ + t cosπq + cosπq + t cosπq + q + cosπq q ψ; ǫ q = ǫ + t cosπq + cosπq + t cosπq + q + cosπq q..f.5 This modifiction cn hve rther drstic effect on the Fermi surfce especilly if t / t cn be s lrge s round ±. Using eqution.f.5 we hve little difficulty plotting Fermi surfces for vrious vlues of t /t. In Figure we show three qulittively different Fermi surfces for t > for different vlues of t / t nd in Figure 5 we show these for t <.

8 8 JACOB LEWIS BOURJAILY Figure 5. Fermi surfces in the tight-binding squre-lttice model with t < nextto-nerest-neighbour couplings for vrious vlues of t / t. From left to right: t / t = / t / t = / nd t / t = 9/. Notice the shrp trnsition t t / t = /. Problem : Bnd Structure of Grphene nd Nnotubes Recll the honeycomb lttice used to describe grphene in homework. We re to consider tightbinding model with single level per site on two-dimensionl honeycomb lttice with only nerestneighbour hopping with on-site energy ǫ nd nerest neighbour hopping mtrix element t. -b We re to find the energy bnds of this model nd determine t wht moment the two bnds re degenerte. Just to get our berings let us recll the Brvis nd reciprocl lttices of the honeycomb lttice: R = with = nd =.. Q = b b with b = π nd b = π... In this model the wve function on ech Brvis cell contins two linerly independent prts coming from the two toms in ech cell; let s cll them toms A nd B. The Hmiltonin of the system cn be described by mtrix the digonl prts coming from the on-site energy ǫ nd the off-digonl prts describing the hopping mtrix elements. The two off-digonl entries re Hermitin conjugtes of ech other: one describe hopping from A B nd the other describes hopping from B A. Becuse the two processes re conjugte it is sufficient to describe one. Let q = q b +q b where q nd q re not required to be integers. Although it will be very quickly brushed wy let us sy tht the vector v AB connects the tom t site A to tht t site B. The hopping or off-digonl prt of the Hmiltonin is given by H A B = te i q vab + e i q + e i q.. t + e iπq + e iπq q.. = t + e iπq q cosπq...5 Let us briefly observe tht if H AB were represented s re iθ then the solution to the eigenvlue eqution is ǫ ǫ re iθ re iθ ǫ ǫ = = ǫ = ǫ ± r...6 Using this nd the work bove we cn directly write down the dispersion reltion: ǫ q = ǫ ± t + cosπ q q cosπq + cos πq...7 óπǫρ ǫδǫι πoιησαι The proportionlity is used to ignore phse fctor which will not ffect our nlysis of energy eigenvlues.

9 PHYSICS 55: CONDENSED MATTER HOMEWORK Figure 6. Energy bnds clculted for grphene: the bnds re shown seprtely on the right nd left nd shown together in the middle. Notice there re precisely six degenercies locted in the firs Brillouin zone. We see tht degenercy implies tht the bove discriminnt vnishes. This will be the cse if e iπq q cosπq =...8 At first glnce there re two possibilities we my try: first we know tht e iπq q R so it must be ±. If e iπq q = then q q = n for n Z; the eqution bove requires cosπq = which mens tht q = or. This gives us n infinite clss of degenerte solutions nd by dding nd subtrcting reciprocl lttice vectors we find six essentilly equivlent degenercies t the corners of the first Brillouin zone:..9 where the components refer to the vlues of q q in q = q b + q b. Now becuse dding nd subtrcting lttice vectors brought us into the other condition for degenercy with q q n odd integer we know tht ll of the degenercies hve been ccounted for. The energy bnds re plotted in Figure 6 where the six degenerte points re clerly visible. c-d Describe nd sketch the topology for vrious Fermi surfces tht cn occur s the filling of bnds is vried. We should lso describe the Fermi surfce when the lower bnd is completely filled. For very low ǫ F the fermi surfce is circle inscribed within the bowl seen in Figure 6. As the energy increses the Fermi surfce ppers more nd more hexgonl until finlly it breks into six rcs one bout ech of the corners of the first Brillouin zone. These six regions shrink s ǫ F ǫ when they vnish. This is shown in Figure 7. As ǫ F grows bove ǫ the Fermi surfce lies on the upper bnd nd progresses in reverse of the lower-bnd: for low energies bove ǫ the Fermi surfce is composed of six distinct circulr components which grow until they become nerly hexgonl; t high energies the Fermi surfce gin pproches single circulr section. This progression is lso shown in Figure 7. In the cse when the lower bnd is completely full the Fermi surfce is the union of the six distinct Dirc points of course only two of them re inequivlent.

10 JACOB LEWIS BOURJAILY Figure 7. Contours indicting the Fermi surfces for vrious ǫ F in the lower left nd upper right bnds for the grphene tight-binding model. In both plots the energy is lower in the drker region regions in white on the left plot mtch to the regions in blck on the right plot. e If we were to compctify sheet of grphene in one direction the result would be crbon nnotube. Depending on which Brvis lttice vector is tken s the compctifying direction there my or my not be bnd gp if there is gp then the nnotube is n insultor; if the two bnds re degenerte then the tube is metllic. We re to determine which types of crbon nnotubes will be metllic nd which would be insulting. The compctifiction of grphene into crbon nnotube cn be described s tking quotient of the Brvis lttice by one of the lttice vectors written R/ r where r = r + r where in this cse r r Z. Tht is to sy trvelling in the direction r brings you round the nnotube nd bck to where you strted: the toms t sites relted by R + r re not merely relted to those t R but re ctully the sme toms. This mens tht there is no phse fctor for trvelling ny multiple of times long r. Precisely this requires tht e i q r = q..e. This plces strong constrint on the llowed q s indeed it breks our continuous bnd of llowed vlues to discrete set. Whether or not the nnotube will be metl or n insultor is completely determined by whether or not this discrete subset of llowed moment include the Dirc points explored bove. The condition e i q r = e iπqr+qr = is tht q r + q r Z. Recll tht the six Dirc points were locted t q with components with respect to b b bsis given in eqution..9. There re two types of points to check: the first two Dirc points listed in eqution..9 will be present in the nnotube iff r.e. r Z = r r Z;.e. the second type of point will be present in the nnotube iff r + r Z = r + r Z..e. It is not hrd to show tht these two conditions re in fct equivlent 5. Therefore crbon nnotube will be metl in this model only if r r = l for some l Z..e. óπǫρ ǫδǫι δǫιξαι 5 Sy r r = l then r + r = l + r + r = l + r Z; nd conversely sy r + r = m then r r = m r r = m r Z. QED

11 PHYSICS 55: CONDENSED MATTER HOMEWORK Problem : Thermodynmics Ner Dirc Point Let us return our ttention to the tight-binding model of grphene from problem. We my for the ske of convenience tke ǫ =. The points where two bnds become degenerte re clled Dirc points. We re to determine the low-temperture behviour of the specific het nd mgnetic spin susceptibility ner the Dirc point for grphene. The first step in our nlysis will be to expnd the dispersion reltion..7 found in problem bove ner the Dirc points. Now becuse the six Dirc points re obviously trnsltionlly relted it is sufficient to consider just one for the moment. Let us expnd eqution..7 bout the Dirc point / / 6 : ǫ = ±t + cos π + πδq cos = ±t + + 9π δq π + πδq = ±t + + 9π δq +... = ±t + + 9π δq +... = ±t + + = ±t 6π δq +... / ; π + cos + πδq +... πδq π δq / + πδq π δq πδq π δq πδq π δq +... πδq π δq πδq + π δq + 9π δq πδq π δq πδq π δq + π δq + π δq +... / / / +... / ǫ ±tπ 6δq. This llows us to compute the density of sttes bout single Dirc point is given by d q gǫ = π δ π 6δq dǫ = dǫ δq dǫ tπ 6 ; g one Dirc point ǫ = ǫ 6t π = g tot ǫ = ǫ t π. With the density of sttes we my compute the totl energy 7 u = u + = u + t π dǫ gǫfǫǫ ǫ dǫ e ǫ/kbt + = u + ζ t π k B T + OT where ζn is the Riemnn zet function. Therefore we see tht c v = 9ζk B t π T + OT We will expnd in δq so tht δq = δq. 7 Using Mthemtic for the integrls.

12 JACOB LEWIS BOURJAILY To find the mgnetic susceptibility we will begin by referring to the textbook or clss notes wherein it is found tht the totl mgnetiztion in the Puli model is given by M = µ H dǫ g ǫfǫ. Evluting this integrl directly we see.. χ = M H = µ k BT log t π + OT...5 b Consider doping grphene so tht ǫ F is just bove the Dirc point but by n mount much less tht of T; we re to gin describe the low-temperture pproximtions of the specific het nd mgnetic spin susceptibility. I m pretty sure tht the picture we re supposed to envision is tht we re some ǫ F seprted from the Dirc point yet close enough to it tht gǫ cn be still viewed s liner function of ǫ. If this is the pproprite then we cn tke gǫ = ǫ ǫ F t π nd fǫ = e ǫ ǫ f/k BT + nd integrte bove the Fermi surfce 8. Using computer lgebr pckge we find u = u + t π ǫ F ǫǫ ǫ F dǫ e ǫ ǫf/kbt + = ǫ FkB t π T + ζk B t π T + OT. It is comforting tht this reproduces our erlier result for vnishing ǫ F. This llows us to directly conclude tht c v = ǫ FkB 6t π T + 9ζk B t π T + OT..b. Now to find the mgnetic susceptibility we perform the sme steps s before nd see M = µ H dǫ gǫfǫ = µ H t π k B T log + ǫ F.b. nd so χ = µ t π k B T log + e F..b. 8 There re lterntive wys of looking t this.