PHY4905: Nearly-Free Electron Model (NFE)

Transcription

1 PHY4905: Nearly-Free Electro Model (NFE) D. L. Maslov Departmet of Physics, Uiversity of Florida (Dated: Jauary 12, 2011) 1

2 I. REMINDER: QUANTUM MECHANICAL PERTURBATION THEORY A. No-degeerate eigestates This is a quic review of how the perturbatio theory is applied to a system with odegeerate eigevalues. Cosider a Schroediger equatio (H 0 + U)ψ = Eψ, (1.1) where U is a small perturbatio. Suppose that H 0 has a discrete spectrum ad oe of its eigefuctios are degeerate H 0 ψ = E ψ. (1.2) Suppose we wat to fid correctios to the eergy ad eigefuctio of state : E = E + E (1), (1.3) ψ = ψ + ψ (1), (1.4) where E (1) ad ψ (1) are of the first order i U. Now, substitute Eqs. (1.3,1.4) ito (1.1), tae ito accout Eq. (1.2), ad eglect all terms which are order of U 2 : =E 0 ψ (H 0 + U) ( ) ( )( ) ψ + ψ (1) = E + E (1) ψ + ψ (1) H 0 ψ +H }{{} 0 ψ (1) + Uψ + Uψ (1) }{{} order U 2 = E ψ + E (1) ψ + E ψ (1) + E (1) ψ (1) }{{} order U 2 H 0 ψ (1) + Uψ = E (1) ψ + E ψ (1) (1.5) Sice the eigefuctios of the uperturbed problem form a complete basis, ay fuctio, icludig the correctio to the eigefuctio do the perturbatio, ψ (1), ca be expaded over this basis as ψ (1) = m c m ψ m. (1.6) The m = term is already icluded i the defiitio of the first-order correctio, Eq. (1.4). Coefficiets c m, which determie the admixtures of the other states to the uperturbed state, arise oly because of the perturbatio ad, therefore, are small as log as U is small. Subsititig Eq. (1.6) ito Eq. (1.5) we obtai c m E m ψ m + Uψ = E (1) ψ + E m 2 m c m ψ m. (1.7)

3 [ Now we multiply this equatio by accout that [ ψ l ψ ] ψ m = δ l,m. This yields ] ad itegrate over the coordiate, taig ito E (1) = U, where U lm = [ ψ l ] Uψ m. Thus the first-order correctio to the eergy of state umber is just a expectatio value of the perturbatio potetial i this state. I order to fid the correctio to the eigefuctio, [ ] multiply Eq. (1.7) by with. This yields or ad ψ ψ (1) c E + U = E c c = = Liewise, oe ca fid secod-order correctios U E E U ψ E E (1.8) E = E + E (1) + E (2) (1.9) ψ = ψ + ψ (1) + ψ (2). (1.10) Substitutig these expasios agai ito Eq. (1.1), taig ito accout (1.2) ad (1.5), ad eglectig terms of order U 3 (but eepig terms of order U 2 ), we fid H 0 ψ (2) + Uψ (1) = E (2) ψ + E (1) ψ (1) + E ψ (2). (1.11) The secod-order correctio to the eigefuctio ca also be expaded over the set of ψ ψ (2) = m d m ψ m. Substitutig this expasio ito (1.11) ad usig (1.8), we obtai m d m E m ψ m +U m U m E E m ψ m = E (2) ψ +U m U m E E m ψ m +E m d m ψ m. 3

4 [ Multiplyig this equatio by ψ ] ad itegratig, we obtai E (2) If U is a real fuctio, U m = U m. Therefore, E (2) = U m U m. m E E m = m U m 2. (1.12) E m E Notice that both the 1st order correctio to the eigefuctio, Eq. (1.8) ad the 2d order correctio to the eigeeergy, Eq. (1.12), diverge if the at least two of the states are degeerate, i.e., if E m = E. I this case, the perturbatio theory costructed i this sectio diverges, ad oe has to select the eigefuctios of the zeroth order i a differet way. This is doe i the ext sectio. B. Degeerate eigestates Now suppose that some eigestate of a o-perturbed Hamiltoia, H 0, is doubly degeerate, that is H 0 ψ 1 = E 0 ψ 1 (1.13) H 0 ψ 2 = E 0 ψ 2. (1.14) The a arbitrary liear combiatio of the two eigefuctios ψ = c 1 ψ 1 + c 2 ψ 2 is also a eigefuctio. Choosig this combiatio as the zeroth order approximatio, we will fid the first-order correctio to the eigeeergy E = E 0 + δe from (H 0 + U) (c 1 ψ 1 + c 2 ψ 2 ) = (E 0 + δe) (c 1 ψ 1 + c 2 ψ 2 ) c 1 E 0 ψ 1 + c 2 E 0 ψ 2 + c 1 Uψ 1 + c 2 Uψ 2 = c 1 E 0 ψ 1 + c 2 E 0 ψ 2 + c 1 δeψ 1 + c 2 δeψ 2 c 1 Uψ 1 + c 2 Uψ 2 = c 1 δeψ 1 + c 2 δeψ 2 Multiplyig the last equatio first by ψ2 ad itegratig, ad the by ψ1 ad itegratig, we obtai a liear 2 2 system (usig U 21 = U12) c 1 U12 + c 2 U 22 = c 2 δe c 1 U 11 + c 2 U 12 = c 1 δe. 4

5 A o-zero solutio exists, if the determiat is equal to zero. Solvig the resultig quadratic equatio, we obtai ad δe = U 11 + U 22 2 ± E = E 0 + U 11 + U 22 2 (U 11 U 22 ) 2 ± 4 (U 11 U 22 ) U 12 2 (1.15) + U 12 2 (1.16) Therefore, a perturbatio ot oly shifts but also splits the degeerate eergy levels. C. Eergy spectrum of a early-free electro model i 1D The eigestates of free problem i 1D with eergies ψ (x) = eix L E 0 = h2 2 2m are doubly degeerate because states with ad have the same eergy. This is true for ay, but a periodic perturbatio has o-zero matrix elemets oly betwee particular states. To see this, we expad a periodic potetial ito the Fourier series U (x) = m e 2πim x a um. Obviously, U (x + a) = m u m = m } e2πim {{} e 2πim x au m = m e2πim x au m = U (x). =1 Calculate the matrix elemets betwee states with ad M = 1 L = m x+a e2πim a e i x U (x) e ix = 1 U m e i x e 2πim x a e ix L/2 L u m L/2 L eiqx, m L/2 where q = 2πm/a. I assume here that the system is a fiite segmet of legth L. Now, L/2 L eiqx = eiql/2 e iql/2 iql = si (ql/2), ql/2 5

6 where I used that si (x) = (e ix e ix )/2i. Fuctio si (x)/x is plotted i Fig. 1. Its first two zeroes are located at x = ±π or, comig bac to the origial variables, at q = ±2π/L. As L, the witdh of the cetral pea shris to zero which meas that the fuctio is o-zero oly at q = 0, where its value is equal to 1. Therefore, the limit of L 0 of this fuctio is just a Kroecer symbol (discrete δ fuctio) lim L L/2 L eiqx = δ q,0, ad the matrix elemet is o-zero oly if the iitial ad fial states are related by the reciprocal lattice waveumber = + 2πm/a. This relatio is called Bragg s law amed after a family team of British scietists, Sir W. H. Bragg ad his so Sir W. L. Bragg, who discovered diffractio of X-rays from crystals i Bragg s law is relevat whe a wavy of ay type (electromagetic, soud, quatummechaical, etc.) is scattered off a periodic potetial. O the other had, the iitial ad fial states must have the same eergy h 2 2 2m = h2 2 2m 2 = 2 ( + 2πm/a) 2 = 2 = πm/a. Sice m rus from to +, we might as well flip the sig of m i the last equatio = πm/a. Therefore, the periodic potetial has the largest effect o the states with wavevectors = ±π/a, ±2π/a... Near each of this poits, the eergy spectrum of the free problem splits as described by equatio (1.15), where the matrix elemets are those of the periodic potetial. As a example, let s cosider the first pair of degeerate states, state 1 with wavevector = π/a ad state 2 with wavector = π/a I Eq. (1.15), 6

7 K15 K10 K x K0.2 FIG. 1: Fuctio si(x)/x. U 11 = U 12 = ( ) e iπx/a U (x)e iπx/a = L L U (x) = u 0 = U 22 ( ) e iπx/a U (x)e iπx/a = u 1 = u 1 L I the last lie, we used the followig property of Fourier coefficiets of real fuctios, U (x) = e 2πimx/a u m m }{{} = m relabelig the summatio idex m m = U(x) = m e 2πimx/a u m u m = u m. e 2πimx/a u m Therefore, right at = ±π/a the eergy is E = h2 π 2 2ma 2 + u 0 ± u 1. Costat u 0 ca be be dropped because this is just a shift of eergy. The eergy gap is ( h 2 ) ( π 2 h 2 ) E G = 2ma + u π u 1 2ma + u 2 0 u 1 = 2 u 1. Liewise, the eergy gap at = ±2π/a is give by E G = 2 u 2. The magitude of the gap decreases with its order because the Fourier harmoics fall off with idex m. 7

8 FIG. 2: Eergy levels from Eq. (1.19) as fuctio of ear = π/a. E is i uits of h 2 π 2 /ma 2, is i uits of π/a. u 12 is 0.2 i this uits. Red: plus sig; blue: mius sig; dashed: free electro spectrum h 2 2 /2m. If we wat to ow the behavior of spectrum at ay poit, we eed to cosider almost but ot completely degeerate eergy levels, that is, istead of equatios (1.12,1.14), we eed to write H 0 ψ 1 = E 1 ψ 1 (1.17) H 0 ψ 2 = E 2 ψ 2, (1.18) where E 1 E 2. A simple geeralizatio of the derivatio i the previous sectio leads to to the followig result for the perturbed eergy E = E 1 + E 2 + U 11 + U 22 2 ± (E 1 E 2 + U 11 U 12 ) U Let us cosider a viciity of the = π/a poit. The degeerate poit is at = 2π/a. Choose the periodic potetial such that U 11 = U 22 = u 0 = 0. The, ear = π/a ( ( E = h2 1 2m 2 + 2π ) ) ( 2 2 ( ) ) 2π 2 2 ( ) 2 a 2m ± + 2 a 4 h 2 u 1 2 (1.19) The resultig spectrum ear = π/a is show i Fig. 2. 8