Solutions to Problems Fundamentals of Condensed Matter Physics

Transcription

1 Solutons to Problems Fundamentals of Condensed Matter Physcs Marvn L. Cohen Unversty of Calforna, Berkeley Steven G. Loue Unversty of Calforna, Berkeley c Cambrdge Unversty Press 016 1

2 Acknowledgement The authors thank Mr. Meng Wu, together wth Mr. Felpe H. da Jornada, Mr. Fangzhou Zhao and Mr. Tng Cao, for ther nvaluable help n preparng ths problem solutons manual.

3 Sec. I I.1. Crystal structure of MgB. (a) The unt cell and Wgner-Setz cell are dsplayed n Fg. 1. Top vew Sde vew B 1 B a Unt cell B Mg B 1 Unt cell Mg a 3 Mg Wgner-Setz cell B 1 y B a B B B 1 a 1 z x Wgner-Setz cell Mg Fgure 1: Unt cell and Wgner-Setz cell of MgB If a s the B-B dstance, then a 1 = a = a 3. We wll choose the followng conventon for the lattce vectors, ( ) 3 3 a 1 = a, a, 0 ( ) 3 3 a = a, a, 0 (1) n Cartesan coordnates. a 3 = (0, 0, c), (b) The unt cell volume s Ω prm = (a 1 a ) a 3 = 3 3 a c, and the Brlloun zone volume s Ω BZ = (π) 3 /Ω prm. The recprocal lattce vectors are gven by b = π Ω prm a a k ɛ k, where ɛ k s the Lev-Cvta symbol, or b 1 = π ( ) 3 Ω prm ac, 3 ac, 0 b = π ( 3 Ω prm ac, 3 ) ac, 0 () b 3 = π ( 0, 0, 3 ) 3 a, Ω prm n Cartesan coordnates. Note that other recprocal lattce vectors can be obtaned dependng on the orentaton of the real-space lattce vectors. The recprocal lattce vectors are dsplay n Fg. : Usng the conventon from our unt cell, the atoms are at τ Mg = a 3 /, τ B1 = a 1 /3 + a /3, τ B = a 1 /3 + a /3. 3

4 Top vew Sde vew b kz b3 Recprocal cell Brlluon zone b1 Recprocal cell Brlluon zone ky b kx Fgure : Brlloun zone (c) There are 4 pont symmetry elements n MgB that leave the crystal structure nvarant. There are 1 n-plane symmetry elements: 6 rotaton operatons, ncludng the dentty operaton: C6, C6,, 6C6 = I. There are also 6 n-plane mrror symmetres that go along the B atoms or between the B atoms, ncludng, for nstance, the mrror symmetry along the x = 0 and y = 0 planes, denoted by σx and σy, respectvely. One can also compose any of these 1 n-plane symmetry elements wth a mrror operaton along the z = 0 plane, σz. Ths gves the total of 4 symmetry elements. Note that ths system has nverson symmetry, whch s equvalent to σz C. (d) The rreducble part of the Brlloun zone s sketched below. It can be easly verfed that, f we apply all the 4 symmetry elements, we wll recover the full Brlloun zone. Top vew Sde vew ky Irr. BZ kz Irr BZ kx Fgure 3: Irreducble BZ I.. The GaN crystal. Ths problem s smlar to Problem I.1, except that a here s the length of the n-plane lattce vectors. The detals of the computaton wll be omtted. (a) We wll choose a dfferent orentaton for the unt cell vectors wth respect to problem I.1. The 4

5 lattce vectors are gven by ( ) 1 3 a 1 = a, a, 0 ( ) 1 3 a = a, a, 0 a 3 = (0, 0, c), n Cartesan coordnates. Note that, n real space, the Wgner-Setz cell s a hexagon rotated 90 relatve to that n problem I.1. 3 The unt cell volume s Ω = (a 1 a ) a 3 = a c, and the recprocal lattce vectors are, b 1 = π ( 1, 1 ), 0 a 3 b = π ( ) 1 1,, 0 (4) a 3 (3) b 3 = π c (0, 0, 1), n Cartesan coordnates. The Brlloun zone volume s Ω BZ = sketched n Fg π3 3a. The Brlloun zone s c Top vew Sde vew b b 3 kz Brlluon zone ky b 1 kx Brlluon zone b Fgure 4: BZ (b) If we nspect the unt cell of GaN. whch s enclosed by sold lnes, we see that there are N atoms (1 spread throughout the corners, and 1 nsde), and there are Ga atoms nsde (1 spread along the edges of the unt cell, and 1 completely nsde the cell). So, there are 4 atoms n the unt cell. (c) A top vew of the structure s dsplayed n Fg. 5. There are 6 pont symmetry elements that do not nvolve glde or screw axes: 3 mrror planes along the atoms that form the hexagons, and 3 C 3 rotaton operatons (ncludng dentty), performed around the center of the hexagons. (d) Ga has 3 valence electrons, and N has 5 valence electrons. So, one cell has 16 electrons, or 8 flled bands. There are no half-flled bands. so GaN s an nsulator. So, t s not surprsng that ths materal has a bandgap. I.3. Born-Oppenhemer approxmaton for a molecule. Thnk about the smplest crude model that captures the physcs of ths problem. Imagne that we have two rgd ons wth mass M 5

6 and effectve charge +Q each, separated by a dstance x. Consder that you put a sngle effectve pont charge of mass m e and charge Q between the two ons, whch corresponds to the effectve charge created by the bond. At rest, we know that x = a. The electronc contrbuton to the energy can be wrtten as a sum over knetc energy and potental energy, and we assume that the knetc energy comes essentally from quantum confnement va the uncertanty prncple p x, E(x) = T (x) + U(x) (5) T (x) = p m e U(x) = Q x + Q( Q) = Q x/ x E(x) = m e x Q x m e x (6) Our model has a free parameter, Q. But we have a constrant: E (x = a) = 0 = Q = am e. So, [ ] E(x) = 1 m e 1 x xa. The electronc energy s the total energy at x = a, so E el m ea. The vbratonal energy can be obtaned f we realze that the total energy landscape as a functon of x can be approxmated by a quadratc curve near x = a. Ths gves rse to a harmonc potental, where the vbratonal modes are quantzed, E vb = ω. But snce we have a harmonc oscllator, ω = k/m wth k = E (x = a). Thus, E vb a. Mm e Fnally, the rotatonal energy can be calculated f we assume that the rotatonal moton s quantzed. We get that E rot = l(l+1) I wth I = M(a/). Ths gves E rot. Ma So, the rato of the electronc to vbratonal to rotatonal energy s gven by E el : E vb : E rot 1 : me M : me M. The consequence of ths fndng s that, snce M m e, we can separate the three degrees of freedom, and calculate the contrbuton to the vbratonal and rotatonal degrees of freedom as a perturbaton to the electronc one. Ths valdates the use of the Born-Oppenhemer approxmaton, whch treats the ons as fxed partcles when solvng for the electronc contrbuton. (7) (8) Fgure 5: Crystal structure of GaN 6

7 I.4. Hartree-Fock approxmaton. Let us frst derve some useful denttes related to Slater determnants. Let s defne the antsymmetrzaton operator A as  = 1 ( 1) P ˆP, (9) N! P where N s the number of partcles and occuped orbtals, P s a permutaton, and ˆP s the operator assocated to the permutaton P. If the overall permutaton contans an odd number of parwse permutaton, ( 1) P = 1, otherwse the permutaton s even and ( 1) P = 1. The dentty permutaton s even. For nstance, consder the followng permutatons and ther sgns: P = (1,, 3, 4,..., N) ( 1) P = 1 P = (, 1, 3, 4,..., N) ( 1) P = 1 P = (, 3, 1, 4,..., N) ( 1) P = 1 (10) We denote the tral Hartree-Fock wavefuncton n a shorthand notaton, Ψ HF (1,,..., N) = N!  φ 1 (1)φ ()... φ N (N), (11) where φ s a sngle-partcle orbtals, the subscrpt n φ () denotes the orbtal ndex, and the argument denotes the space-tme coordnate. We take a spnless wavefuncton for the sake of smplcty, although the proof here can be trvally extended to the case wth spn. We defne the permutaton operator ˆP actng on Ψ by permutng the orbtals ndces, ˆP φ 1 (1)φ ()... φ N (N) = φ P1 (1)φ P ()... φ PN (N) (1) The followng propertes can be easly proved for Â,  = Â,  =  (13) In addton,  commutes wth any ] operator that preserves partcle ndstngushablty, such as the crystal Hamltonan, [Â, Hxtal = 0. In partcular, we can show that: We can proof a couple of useful propertes nvolvng Â: (a) Ψ HF Ψ HF = 1 (b) Ψ HF ĥ() Ψ HF = φ ĥ(1) φ (c) Ψ HF ĝ(, ) Ψ HF = [ φ φ ĝ φ φ φ φ ĝ φ φ ] We wll prove relaton 3 n the followng: Â, ) = 0 (14) 1,..., M Ψ HF ĝ(, ) Ψ HF = N! Âφ 1(1)... φ N (N) ĝ(, ) Âφ 1(1)... φ N (N) = N! φ 1 (1)... φ N (N) ĝ(, ) Âφ 1(1)... φ N (N) = ( 1) P φ 1 (1)... φ N (N) ĝ(, ) φ P1 (1)... φ PN (N) P (15) 7

8 We wll now partcularze to the case where = 1 and =, Snce ĝ(1, ) doesn t depend on any coordnate larger than and the orbtals are orthogonal to each other, we can separate the orbtals nto two sets, one wth ndces 1 or, where ĝ(1, ) wll act upon, and another set wth ndces larger than. There can be no cross-terms between orbtals because they are orthogonal. Ths way, we get: Ψ HF ĝ(1, ) Ψ HF = P = P = P ( 1) P φ 1 (1)... φ N (N) ĝ(1, ) φ P1 (1)... φ PN (N) ( 1) P φ 1 (1)φ () ĝ(1, ) φ P1 (1)φ P () ( 1) P φ 1 (1)φ () ĝ(1, ) φ P1 (1)φ P () (16) φ 3 (3) φ P3 (3) φ N (N) φ PN (N) = P ( 1) P φ 1 (1)φ () ĝ(1, ) φ P1 (1)φ P () δ 3 P3 δ N PN Now, there are only two permutatons such that P = for >, namely: So, P = (1,, 3, 4,..., N) ( 1) P = 1 P = (, 1, 3, 4,..., N) ( 1) P = 1 Ψ HF ĝ(1, ) Ψ HF = φ 1 (1)φ () ĝ(1, ) φ 1 (1)φ () φ 1 (1)φ () ĝ(1, ) φ (1)φ 1 (). = φ 1 φ ĝ φ 1 φ φ 1 φ ĝ φ φ 1, (17) (18) where we assume that the coordnate ndces (e.g., (1) and ()) follow the same order n the bras, kets, and n the operators. A generalzaton for other orbtal ndces gve relaton 3, as desred. Relatons 1 and are easer cases that can be proved n the same fashon, but n those cases only the dentty permutaton yelds a non-zero contrbuton to the matrx element. Now, we can wrte the crystal Hamltonan as H xtal = ĥ() + ĝ(, ), (19) where, n Rydberg atomc unts, ĥ() = 1 m + V ext (r ) ĝ(, ) = 1 e. = 1 (0) r r v(, ) In order to fnd the Hartree-Fock equatons, we wsh to mnmze the expectaton value of the total energy wth respect to the sngle-partcle orbtals that make up the Hartree-Fock wavefuncton, wth the constrant that the orbtals are orthonormal, whch can be ntroduced va a Lagrange multpler, L = Ψ HF H xtal Ψ HF λ ( Ψ HF Ψ HF 1) (1) 8

9 We set δl = 0 and obtan, δ φ ĥ φ + 1 { N } [ φ φ ˆv φ φ φ φ ˆv φ φ ] δ φ φ = 0 () Note that we don t have to mpose orthogonalty constrant because, for small perturbatons n the orbtals, the change n the orthogonalty wll be quadratc n δφ. Then, δφ ĥ φ λ δφ φ + 1 [ δφ φ ˆv φ φ δφ φ ˆv φ φ ] + 1 [ φ δφ ˆv φ φ φ δφ ˆv φ φ ] + h.c. = 0 (3) Note that the double sums can be combned nto a sngle sum, snce ˆv(1, ) = ˆv(, 1). We seek crtcal solutons for any orbtals. Because the lnear ndependence of the orbtals, we only need to solve the lnear system for the bras. For a gven orbtal wth ndex, we obtan ĥ φ + [ φ () ˆv φ (1)φ () φ () ˆv φ (1)φ () ] λ φ = 0. (4) Ths equaton can be dentfed as the Hartree-Fock equaton, where we defne the Fock operator as, ĥ φ + F φ = ε φ, (5) ˆF φ. = [ φ () ˆv φ (1)φ () φ () ˆv φ (1)φ () ]. (6) In real space, the Hartree-Fock equaton reads, 1 m φ (x) + V ext (x)φ (x) e [ d 3 x φ (x ) φ (x) x x =1 d 3 x φ (x )φ (x ] )φ (x) x x = ε φ (x). The total energy of the system s not the sum of the orbtal egenvalues ε, but rather (7) E = Ψ HF H xtal Ψ HF = φ ĥ φ + 1 [ φ () ˆv φ (1)φ () φ () ˆv φ (1)φ () ] = = φ ĥ φ + 1 ε 1 φ ˆF φ ˆF φ φ (8) For the second part of the queston, we assume that we promote one electron from an occuped orbtal a to an unoccuped orbtal b,.e., we assume that sngle-partcle orbtals reman frozen, and 9

10 that orbtal a, whch was prevously occuped, becomes unoccuped, and that orbtal b, whch as prevously unoccuped, s now occuped. We need to assume that sngle-partcle orbtals are frozen: f we solve the HF equatons self-consstently, we always arrve at the ground-state orbtals). The exctaton energy E s the energy dfference from the ground-state energy, E 0, to the excted state energy, E a b. The energy dfference s composed of two parts: E 1, whch depends on ĥ(1), and E, whch depends on ĝ(1, ). It s clear that E 1 = φ b ĥ φ b φ a ĥ φ a. The change E s a lttle bt harder to compute as the whole Fock operator changes as an electron gets promoted, E = 1 1 [ φ b φ ˆv φ b φ φ b φ ˆv φ φ b ] [ φ a φ ˆv φ a φ φ a φ ˆv φ φ a ] (9) The exctaton energy, wthn the Hartree-Fock approxmaton, s therefore E a b = E 1 + E = (ε b ε a ) 1 [ φ b φ ˆv φ b φ φ b φ ˆv φ φ b ] + 1 [ φ a φ ˆv φ a φ φ a φ ˆv φ φ a ]. (30) Note that, wthn the HF approxmaton, the exctaton energy s not ust the dfference between the egenvalues, but t also contans a correcton term, whch can be understood as an approxmaton to the excton bndng energy. I.5. Born-von Karman boundary condton. (a) See Fg. 6. E(k) E(k) E(k) -3π/a-π/a -π/a π/a π/a 3π/a -3π/a-π/a-π/a π/a π/a 3π/a -3π/a-π/a-π/a (a) (b) (c) π/a π/a 3π/a Fgure 6: Smple bandstructure dagrams for a one dmensonal perodc sold n the lmt of V (r) 0 expressed n the extended zone (a), repeated zone (b), and reduced zone (c) scheme. (b) k 1 = Ω (π) 3 I.6. Energy bands of elemental solds. d 3 k = ΩΩ BZ (π) 3 = Ω Ω c = N (31) 10