Problems in quantum condensed matter

Transcription

1 Gl Refael and Krll Shtengel June 4, 007 Book proposal Problems n quantum condensed matter From our experence as both teachers and students, the best way to learn subject matter s through problem-solvng. When we look for collectons of condensed matter physcs problems avalable to the advanced undergraduate or begnnng graduate student, however, the only book one readly fnds s Mhály and Martn s Sold State Physcs: Problems and Solutons []. Textbooks often have problems n the end of chapters, but these are mostly problems that demonstrate and practce the precedng analytcal technque ntroduced. We propose a book of quantum condensed matter physcs problems that wll fll ths gap and provde students wth a tool for practcng and sharpenng ther understandng of the vared and complex topcs of our feld. Ths book wll concentrate on modern condensed matter theory. It wll cater to advanced undergraduates, any level of graduate students, and engneerng professonals. The only prerequstes for effcacous use of the book would be havng taken electromagnetsm, quantum mechancs, and ntroductory sold-state Ashcroft and Mermn level classes. The problems wll range n dffculty from smple problems meant to acquant the student wth the physcal or expermental system, all the way to open-ended problems reflectng contemporary research. Intermedate problems wll requre some knowledge of varous technques, and detaled solutons wll allow students to learn the technques necessary. Emphass wll be put on the understandng of both the expermental detals and the analytcal technques. Needless to say that the most advanced problems wll be addressed to theorsts, but at least half of the problems wll be ntended to both theory- and experment-orented readers. We expect that each chapter wll consst of a bref ntroducton, lst of general references and revews, collecton of about ten problems, and solutons wth approprate specfc references. Table of contents A tentatve table of contents follows. In each topc we gve a lst of lkely themes for the problems, as well as the specfc expermental aspects we ntend to cover. Needless to say, snce condensed matter s an ntegrated subject, many problems wll pertan to more than just the subject matter of the chapter ther grouped under.. A revew of the theory of metals; second quantzaton; tght-bndng Hamltonans; Landau theory of Ferm lquds. Expermental topcs: transport and magneto-resstance measurements.. Introducton to mesoscopc physcs. Transport; Coulomb blockade; weak localzaton; shot-nose. 3. Quantum phase transtons. Dynamcal scalng; transverse-feld Isng model; Kosterltz-Thouless transton n D quantum systems. 4. Luttnger lqud and Bosonzaton. Sne-Gordon model, Kane-Fsher mpurty tunnelng, Landauer- Büttker formalsm. Expermental topcs: transport measurements; nanotubes. 5. Superconductvty. Ths chapter wll encompass several extended sub-topcs: a Low-temperature superconductvty; BCS wave-functon. b Hgh-temperature superconductvty; Cuprates phase-dagram phenomenology; Hubbard model. c Superconductor-metal-nsulator transtons. Dsorder effects n metals; phase-fluctuatons; Chakravarty- Schmd transton of Josephson junctons. d Expermental topcs: tunnelng, ARPES, STM, surface-acoustc waves.

2 Fgure : The AKLT state s constructed by consderng each spn- ste as consstng of two spn-/ parts that are symmetrzed. In each ste, one part forms a snglet sold lne to ts rght, and the other part forms a snglet to ts left. Parameterzng the spn-/ parts usng Schwnger Bosons enforces ther symmetrzaton. 6. Quantum magnetsm. Orgn and models; sgnatures of ground states; spn-wave theory. Expermental topcs: susceptblty measurements, neutron scatterng. 7. Quantum Hall effect. Phenomenology; composte bosons and fermons; edge-channels; zero-resstance states. Expermental topcs: transverse transport, sem-conductor heterostructures. 8. Low-temperature atomc physcs. Bose-Ensten condensaton; Gross-Ptaevsk equaton; Feshbach resonances; Superflud-nsulator transtons. Expermental topcs: tme-of-flght measurements; nose measurements. 9. Heavy-Fermon materals. Phenomenology; slave-boson methods; DMFT. Sample problems. The AKLT state of a -D spn- Hesenberg model Problem Consder the spn- Hesenberg chan, H = JŜ Ŝ+. Ths Hamltonan has very specal propertes: unlke ts half-nteger spn counterparts, t has a sngle ground state whch has an exctaton gap. It descrbes very well the compound NC H 8 N NO ClO 4, or NENP []. A state close to the ground state of the spn- Hesenberg model s the AKLT state, named after Affleck, Kennedy, Leb, and Tasak [3]. In terms of Schwnger-Boson operators Schwnger bosons wll be descrbed n the background materal, currently n footnote[4], t s gven by: AKLT = N [ â ˆb + ˆb â + ] 0. Ths state lterally consders each spn- ste as consstng of two spn-/ parts, where one of the parts forms a snglet to the rght, and the other to the left see Fg.. Ths maxmzes antferromagnetc correlatons wthout breakng any symmetry of the system, whch s forbdden by the Mermn-Wagner theorem [5].. Show that the state AKLT s a an egenstate and b the ground state of the AKLT Hamltonan: H AKLT = [ J Ŝ Ŝ+ + ] Ŝ 3 Ŝ+ 3 Hnt: what s the projecton operator for two spns to the s total = subspace?. Although AKLT s not an egenstate of the Hesenberg Hamltonan, Eq., we can use t to estmate the ground-state energy of the model. For ths purpose, calculate the expectaton value AKLT H AKLT. 4

3 Ths energy should be compared to the energy of a spn-/ snglet per bond. How dfferent are these energes? Soluton. The AKLT state s unque n the followng way: f we concentrate on each par of neghborng states, and gnore completely the rest of the chan, we see that at most the two stes have total spn. Ths s snce the basc element n the AKLT wave functon s the Schwnger boson snglet operator: â ˆb + ˆb â +. 5 Therefore the AKLT wave functon must be an exact ground state of operators projectng neghborng spns on the S total = subspace. We wll now show that the AKLT Hamltonan conssts of a sum of such operators. To construct such a projecton operator, we wrte an operator whch wll vansh for the S total = and S total = 0 states: P = S 6 4 total Stotal 6 where the normalzaton s obtaned by puttng S total = and Stotal = 3. Ths operator s ether 0, or n the case of S total =. By assgnng S total = S + S + we fnd: P = 6 4 S + S+ + S S + S + S + + S S + = 3 + S S S S +. But ths looks exactly lke one term n the AKLT Hamltonan. In fact: 7 H AKLT = P /3 8 Snce for each P we have P AKLT = 0, and P 0, AKLT s a ground state of H AKLT, wth: It turns out that ths state s also unque; for proof see Ref. [3]. H AKLT AKLT = L AKLT Our goal s to calculate AKLT H AKLT. Let us frst calculate the normalzaton N of the AKLT state: N AKLT AKLT = 0 â ˆb+ ˆb â + â ˆb + ˆb â Ths complcated product s smplfed n a smlar fashon to that of one-dmensonal partton functons, usng transfer matrces. Frst, we note that when we expand the product on the left, each term must be matched wth ts hermtan conjugate n the product to the rght, n order to produce a non-zero contrbuton to ths expectaton value. Let us make the followng defnton: f = â ˆb + f = ˆb â + Wth ths defnton we can wrte: N AKLT AKLT = τ 0 {τ =,} f τ f τ 0 The Schwnger-Boson SB vacuum can be wrtten as a product of the SB vacuum for each ste: 0 =

4 hence Eq. conssts of products of the expectaton value of f-operators wth respect to the vacuum each ndvdual ste. For ste ths product s roughly: 0 f τ f τ f τ f τ 0 4 Snce τ and τ can each be or, let us organze ths product n a matrx. For: T, T τ, τ = T, T, T, we have: T = Now, t s easy to see that: wth τ 0 = τ L. But ths s just the trace: 0 b a b a 0 0 b b b b 0 0 a a a a 0 0 a b a b 0 N AKLT AKLT = N = = L {τ =,} =0 5 6 T τ, τ + 7 = T rt L. 8 The trace of a power of a matrx s most easly obtaned usng ts dagonal form. T s readly dagonalzed by the egenvectors:,,, 9 wth egenvalues: respectvely. Hence: 3 0 N = 3 L + L. Movng next to the evaluaton of AKLT Ŝ Ŝ+ AKLT, we begn by notng that snce the AKLT state s rotatonally nvarant, we should have: whch smplfes thngs dramatcally. AKLT Ŝ Ŝ+ AKLT = 3 AKLT Ŝz Ŝz + AKLT, Lookng Eq. 7 above, the nserton of Ŝz Ŝz + to the AKLT product modfes the transfer matrx T τ, τ + n locatons and +. At we have: 0 b T τ, τ + S τ, τ + = a Sz ˆ b a 0 0 b b Sz ˆ b b a a Sz ˆ a a 0 0 a b Sz ˆ a b 0 = 3 0 And, therefore: AKLT Ŝz Ŝz + AKLT = 3 L T τj, τ j+ S τ, τ S τ, τ + T τj, τ j+ j=+ T r 4 T L S 3 L {τ =,} j=0 = Now, we know that the trace on the RHS 4 s domnated by the egenvalue λ = 3 of the matrx T, whch corresponds to the egenvector, /. Therefore, neglectng the contrbuton of the other egenvalue of T due to the large length of the chan, we can wrte: Thus, we have: AKLT Ŝz Ŝz + AKLT = L 3 L, T AKLT J, S L = Ŝ Ŝ+ AKLT = 4 JL. 6 3 Note that a spn-/ snglet wth a Hesenberg Hamltonan would have the energy 3/4J per bond, hence the AKLT state exceeds ths energy by almost a factor of. On the other hand, a snglet between two spn- stes has energy J per bond, whch s nearly twce the energy per bond of the AKLT state. 4

5 . The Haldane gap Problem In 983, Duncan Haldane dscovered that a spn- Hesenberg chan, H = JŜ Ŝ+, 7 has an exctaton gap [6]. He had done so by developng a mappng of the Hesenberg Hamltonan to a nonlnear sgma model. Usng the AKLT state of the prevous problem, however, t s possble to demonstrate and evaluate the gap n a much smpler way.. For the purpose of fndng the Haldane gap, we are gong to use the sngle-mode approxmaton. Assumng that we know the ground state of H to be GS, we can guess that there wll be a band of excted states wth wave number k s gven by Âk GS = e kn  n GS, wth Ân a local operator on n ste n. Show that the expectaton value of the exctaton energy of  k GS s gven by: k = GS ]] [ k [H,, Âk GS 8 Assume that the followng condton apples:  k GS and  k GS have the same expectaton of the exctaton energy, and the same normalzaton - a consequence of tme reversal.. Gven that the lowest excted states of the Haldane spn- chan have total spn S total = as opposed to the ground state whch s a full snglet, wth S total = 0, construct approxmatons to these states usng the AKLT state of the prevous problem. How many such exctaton branches are there? 3. By fndng the approprate operator Ân, the states of and the dentty of, evaluate the Haldane gap. Here too, use the AKLT state as an approxmaton for the ground state. Answer. We begn by assumng that the excted states s approxmated by: A = GS  GS  GS. 9 The exctaton energy wth respect to the Hamltonan H s then: = A H A GS H GS. 30 Snce GS s assumed to be the ground state of H, H GS = E 0 GS we can wrte:  GS H H  GS GS ÂH  H GS = GS  GS =. 3 GS  GS In the second equalty we used the fact that the normalzaton of  GS and  GS, as well as ther energy expectaton values are the same. Addng the two expressons for, and dvdng by two, we arrve at: [ [ ]] = GS, H,  GS GS  GS 3 5

6 . The next step s to guess what the excted states look lke. An excted state would presumably have a structure of a quas-partcle, delocalzed at the equvalent of a k momentum state. Startng wth the AKLT state, t seems plausble that the role of the quaspartcle would be played by a trplet replacng the snglet operator â ˆb + ˆb â +. Ths would ndeed yeld a spn- state. The trplet â ˆb + + ˆb â +, â â +, ˆb ˆb +. Thus we fnd three branches, or bands operator can be one of three: of exctatons: k, + = N k+ e â kn nâ n+ â mˆb m+ ˆb mâ m+ 0, n m n k, 0 = N k0 e â kn nˆb n+ + ˆb nâ n+ â mˆb m+ ˆb mâ m+ 0, n m n k, = e ˆb kn nˆb n+ â mˆb m+ ˆb mâ m+ 0. N k n Due to rotatonal symmetry, these three states wll gve rse to states wth the same energy. They represent the m z =, 0, states of the the trplet of exctatons. 3. Because the three states n Eq. 33 are degenerate, t suffces us to consder one of them. It s most convenent to take the m z = 0 trplet state, snce t can be produced usng the applcaton of Sz ˆ. Also, snce we are after the lowest energy state, we need to choose k = 0 or π. m n Let us defne the states j wth a localzed m z = 0 trplet: j = N â jˆb j+ + ˆb jâ j+ j 33 â ˆb + ˆb â + 0, 34 N s the same normalzaton as for the AKLT state. Our guess for the lowest excted state s thus: A = j j j, 35 correspondng to the state π, 0 n Eq. 33. In prncple, we need to check whether the j staggerng term gves a lower energy than just havng a constant 0, 0. As t turns out, t does. We leave t to the reader to verfy that the energy for a non-staggered trplet state s hgher. The next step s to construct ths state, and fnd the approprate Â. We frst note the followng commutator: [Ŝz, â jˆb j+ ˆb ] jâ j+ = δ,j δ, j + â jˆb j+ + ˆb jâ j+ 36 Thus: And we have: whch s also Hermtan. j j = j Ŝj z AKLT 37 j j  = j j Ŝ z j 38 Note that usng the spn operators we can construct all three branches of trplet exctatons: k, ± n k, 0 n e kn S ± n AKLT, e kn S z n AKLT. 39 In antferromagnetc spn systems t s often the case that the operator used for the sngle-mode approxmaton are ndeed the spn operators. Snce the full spn operator s a vector, t produces a spn- state out of snglets, fulfllng the expectaton out of the lowest excted states over a snglet 6

7 ground state. Note that formula 39 fals at k = 0, snce the AKLT state s an exact snglet state. In ths case the approprate state requres e kn n - an nterestng fact to show. [ ]] Now we need to evaluate the commutator [Â, H, Â. We have: [ [ ] H, Â = = 4 Ŝ+ Ŝ + + Ŝ ] Ŝ+ [Ŝ, Ŝz + [Ŝ+, Ŝz Ŝ Ŝ ] Ŝ + + [Ŝ, Ŝz, ] j Ŝj z j ] [Ŝ+ ] Ŝ+ +, Ŝz where another / was added to offset the double countng from the Ŝz sum. Usng ] and [Ŝ+, Ŝz = Ŝ+, we get: = Ŝ+ Ŝ + + Ŝ Ŝ ] [Ŝ, Ŝz = Ŝ, 4 And repeatng the same steps gve: [ ]] [Â, H, Â = Ŝ+ Ŝ + + Ŝ Ŝ By usng rotatonal symmetry agan we fnd: [ ]] AKLT [Â, H, Â AKLT = AKLT Ŝz Ŝz + AKLT = 8 L The remanng part s the evaluaton of the denomnator, AKLT ÂÂ AKLT. For ths purpose we need to know the z-z correlaton functon n the AKLT state. Followng the methods of problem., we have: AKLT Ŝz Ŝz j AKLT = 3 L, T L j, S T j S The matrx S swtches the domnatng egenvector of T,,, nto the sub-domnant egenvector: 0 S = = 45 0 whch has egenvalue under T, λ =. Note that also: wthout the extra mnus sgn. Hence:, S T j S 44, S =, 46 = 4 j 47 and thus: 4 AKLT Ŝz Ŝz j AKLT = 3 j + j. 48 In order to calculate the denomnator of Eq. 3 we also need: AKLT Ŝz AKLT = Ŝ 3 AKLT AKLT =

8 Now: AKLT ÂÂ AKLT = = L 3 + j> 4 3 j + j AKLT Ŝz Ŝz j AKLT j = L = L 3 50 Fnally, we can evaluate the gap: J Ths should be compared to the numercal result: [7] 8 9 = J = 0.J. 5 9 = 0.35J. 5.3 Graphene: Tght-bndng model on a honeycomb lattce problem: Solvng the tght-bndng model on the honeycomb lattce s the startng pont for understandng the electronc structure of graphene - a sngle sheet of graphte. Ever snce the success n producng sngle-layer Graphene sheets, the system has become a central topc of nvestgaton n Condensed Matter Physcs and nanotechnology. The tght-bndng Hamltonan s defned, n the second-quantzed notaton, as Ĥ TB = µ c c t c c j + c j c,j 53 where the summaton n the second term s performed over all pars of nearest neghbor stes, j. The spn ndeces are omtted for clarty; the Hamltonan s spn-ndependent. Ths s a sngle-partcle Hamltonan, so t can be convenently wrtten n the bass of Wanner states R : Ĥ TB = µ R R t R R + δ 54 R R where δ s are the vectors connectng each ste to one of ts neghbors. Wanner states are the orthonormal states whch are localzed n the vcnty of a gven ste: R c 0. For the purpose of ths problem, we concentrate on the case of a two-dmensonal honeycomb lattce.. The prmtve cell of a honeycomb lattce contans two stes e.g., a par of stes connected by a vertcal bond, as depcted n Fg.. These two stes belong to two dfferent sublattces, e.g., the bottom ste only connects to top stes n the same and other prmtve cells of the honeycomb lattce. Introduce two sets of Wanner functons, r R A and r R B descrbng the electrons localzed around the stes of each sublattce. Rewrte Equaton 54 n terms of R A s and R B s. Ths should allow you to specfy the sets of δ s for each sublattce.. Dagonalze ths Hamltonan and fnd the energy egenvalues. How many bands are there? 3. What s the Brlloun zone for ths lattce? Plot the energy spectrum as a functon of k along two drectons of hgh symmetry. Soluton: δ 8

9 A B a Fgure : The honeycomb lattce wth the two stes, A and B formng the bass. Two prmtve vectors d, are also shown.. Let us start by smply re-labelng the states: and R A R + v A R, A 55 R B R + v B R, B 56 where we choose R to pont to the mddle of the vertcal edge connectng the two atoms of one bass. In ths case, v A = v B = a /, 0. In fact, the partcular choce of whch pont n the bass s specfed by R, and what are the correspondng v, s mmateral; as you wll see, v, do not enter any the followng. What s mportant, s to keep track of the two sublattces because the sets of vectors δ connectng each atom to ts neghbors s dfferent for these sublattces. Wth ths labelng of the states, the Hamltonan becomes Ĥ TB = µ R { R, A R, A + R, B R, B } t R { R, B R, A + R + d, B R, A + R d, B R, A + R, A R, B + R, A R + d, B + R, A R d, B } 57 Here the vectors d, = a 3 /, ± 3/ = d 0 /, ± 3/ are the prmtve vectors of the underlyng trangular also sometmes referred to as hexagonal not to be confused wth honeycomb! Bravas lattce wth the lattce constant d 0 = a 3. The second lne of Eq. 57 descrbes hops from a gven A-ste to all three of ts surroundng B-stes, one wthn the same bass and two n the neghborng ones. The thrd lne descrbes hops n the opposte drecton. Convnce yourself that all possble hops are ncluded n ths Hamltonan. Wrtten ths way, the Hamltonan s explctly Hermtan, as of course t should be.. The next step s to perform the Fourer transform. Wrte R, A = e k R k, A 58 N whch can be nverted as k, A = N R exp k R R, A, wth the dentcal expressons for k, B and R, B. k 9

10 Now, the Hamltonan becomes Ĥ TB = µ k { k, A k, A + k, B k, B } t k { k, B k, A + e k d + e k d + k, A k, B + e k d + e k d }. 59 In ths form, the Hamltonan s almost dagonal ts terms do not mx states wth dfferent wave vectors k. However, t stll mxes k, A and k, B n other words, we just block-dagonalzed the Hamltonan. The remanng task s to dagonalze these blocks, whch n matrx notaton.e., n k, A, k, B bass look lke ths: µ t + e Ĥ k = k d + e k d t e k d + e k d µ The egenvalue equaton s or µ E k = t + e k d + e k d + e k d + e k d = t [3 + cos k d + cos k d + cos k d + d ] 6 E k = µ ± t 3 + cos k d + cos k d + cos k d + d = µ ± t cos k xd 0 cos k yd 0 + cos k y d 0 6 Ths s the desred dsperson relaton. Clearly s has two bands, correspondng to the choce of sgn n the last expresson. 3. The recprocal lattce s another trangular lattce, rotated by π/6, wth the prmtve vectors b = π, = π 3, d 0 3 3a b = π d 0, = π 3 3a 3, The frst Brlloun zone s a hexagon wth two most symmetrc drectons beng that pontng from ts center to one of ts corners e.g., along the k x axs and the other one beng that pontng to the mddle of one of ts sdes e.g., along the k y axs see Fgure 3 below. Allowng k = k x ê x and k = k y ê y, we obtan E kx = µ ± t cos k xd 0 + cos k x d E ky = µ ± t cos k yd Whle n the k y -drecton, the gap between the bands, t cos 3 k yd 0 always remans postve, n the k x -drecton t vanshes at the corner of the Brlloun zone,.e. at k x = 4π/3d 0. Snce the Brlloun zone s a hexagon, t has sx corners, but they can be broken nto two set of three so that wthn each set the corners are dentfable wth each other: they are related to one another by the recprocal lattce vectors b,. Therefore, there are two dstnct ponts n the frst Brlloun zone where the band gap vanshes lnearly these are known as Drac ponts and are the man feature of the electronc spectrum n graphene. 0

11 k y k x 4π b b 3d 4π 0 3d 0 Fgure 3: The frst Brlloun zone for graphene. b and b are two prmtve vectors whose choce s not unque for the recprocal trangular lattce. The corners of the Brlloun zone that are related by the recprocal lattce translatons and therefore dentfable wth each other are color-coded red and yellow. E k E k 4 4 k xd k y d 0 Fgure 4: The energy spectrum for k n x and y drectons n the frst Brlloun zone of graphene. References [] László Mhály and Mchael C. Martn. Sold State Physcs : Problems and Solutons. John Wley, New York, 996. [] Partha P. Mtra, Bertrand I. Halpern, and Ian Affleck. Temperature dependence of the electron-spnresonance spectrum of the chan-end s=/ modes n an s= antferromagnetc chan. Phys. Rev. B, 450: , Mar 99. [3] I. Affleck, T. Kennedy, E. H. Leb, and H. Tasak. Rgorous results on valence-bond ground states n antferromagnets. Phys. Rev. Lett., 59:799, 987. [4] The Schwnger-Boson parametrzaton of spns uses two flavors of Bosons wth annhlaton operators, â and ˆb, wth usual commutaton relatons: [â, â ] = [ˆb, ˆb ] =. The â Bosons represent a spn-/ part pontng up, and ˆb Bosons a spn-/ part pontng down. The spn operators become: Ŝ total = â â + ˆb ˆb, Ŝz = â â ˆb ˆb and Ŝ+ = â ˆb, Ŝ = ˆb â. [5] N. D. Mermn and H. Wagner. Absence of ferromagnetsm or antferromagnetsm n one- or twodmensonal sotropc hesenberg models. Phys. Rev. Lett., 7:33 36, Nov 966. [6] F. D. M. Haldane. Nonlnear feld theory of large-spn hesenberg antferromagnets: Semclasscally quantzed soltons of the one-dmensonal easy-axs nel state. Phys. Rev. Lett., 50:53, 983. [7] A. Auerbach. Interactng electrons and quantum magnetsm. Sprnger Verlag, New York, 994.