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cha12 - Homework 4 due Problem 1 Ashcroft—Mermin 12.2(a...

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Unformatted text preview: Homework 4, due February 22, 1999 Problem 1. Ashcroft—Mermin 12.2 (a) We have so?) = "r7212? - M-l - 1'5 where we have taken the minimum energy to be zero and the minimum at the origin. This does not change the results. The area A(e, k2) is the area inside the curve given by km and Icy obeying 2 _ —1 2 —1 —1 2 e _ (M ) k +2(M )wykxky+(M )yyky+ h 9690”” 2 (M—1)mz [£me + 2 (M—1)yz kykz + (M—1)zz k2 Z We can write this in the form A(kw — kg? + 2B(k\$ — kg)(ky — k2) + C(ky — k3) = D with _ 71 _ 71 _ 71 A—(M )mm’B—(M )my,O—(M )3”! and (Rd—1).. (M‘lhz k3 = ‘sz ’ k3 = ‘sz and D = g — (M4)“ k3 — A(lcg)2 — 2Bkgk3 — C(k3)2 The area of the ellipse is A(e, k2) = 77D m which is linear in 6. Therefore we have m* _ ﬁaMech) _ 1 _ 27f 36 _ W Next we use (WI—1V1)“ = W17((M_1)M(M_I)yy ‘(M_1)W(M_1)W) (Mt. = det(MXAC — B2) from which the formula follows. (b) The density of states follows from 9(6) 2 ﬁg fd3k6(e — hgk’T-M-l .13) This can be rotated to principal axes 9(6) = a; fdgqae — @7qu -D*1 -q‘) where the matrix D is diagonal and det(D) = det(M) We now transfer to scaled coordinates and obtain 9(a) = 4—71; fdgsx/det(D)6(e _ 71:82) = comparing with g<e> = <m*>%Th;2 1: gives the required answer. det(D)#1/%g Problem 2. Ashcroft—Mermin 12.4 j’total = 2.771 = ZﬁglE—l E ﬁilE’ n n which gives [3—1 = 2/5771 TL From N _ pn _RTLH we ﬁnd and therefore 1 P RH _ PLRi H E —RH p _ P1‘ p1 This leads to the equations 2 7R2H2 : ,FIWRIEIH2 + pgﬁﬁgm R = R1 + R2 p277R2 H2 pﬁl "R? H2 pg "R; H2 1 p1 R1H + 1 p2 R2H "R1H2 —R1H P2+R2H2 —R2H p2 Combine these using complex arithmetic: p—iRH _ pl—iRlH + p2—iR2H ill-REHE _ pglnRElH2 pgnREH2 01' 1 — + + + p——z'RH _ p1+iR1H p2——iR2H Or 1 _ p1+p2+iH(R1~R2) _ ,0an — From this we get (91"102)2+1‘I2(RlJrR2)2 (P1+iR1 H) (P2+iR2 H) _ (Pr-131 H) (P2+iR2 H)(P1+P2 —iH(R1 +R2)) - _( +’R H)( +‘R H)( + —'H(R--R)) p—l—ZRH— M Z 1 (pfip:)22+H2p(lRli2R:)2 1 2 which has the correct denominator. (which is the enumerator of p ) is The real part of the enumerator P1P2(P1 + P2) — RiHR2H(P1 + P2) + H2031 + R2)(p1R2 + P231) which is equal to P1P2(P1 + P2) + 1112(0le + P23?) In the same way, the imaginary part gives —p1p2H(R1 + R2) + (p1 + p2)H(R2p1 + R1p2) + H3R1R2(R1 + R2) which is H(R2p§ + R1p§)+ H3R1R2(R1 + R2) and this gives the correct result for R. (c) From the equation for the Hall coefﬁcient we have Eli—moo R = % If the high ﬁeld Hall coeﬂicient has neff = 0 this means that Elim R = 00 and hence R1 —— R2 = 0, compensating hands. This —>OO gives P : P1P2(P1+p2)--H2R21(P1+P2) : p1p2--H2R2] (P1--P2)2 (P1+P2) Problem 3. Ashcroft—Mermin 12.6 Take a single band and consider a state with mm R,t = 0) = eik'RzﬂFﬁf = 0) if we can show the result for this wave function, it will also hold for a linear combination. H(F+§)=—f—mW%5-W%5+U(F+R’)+eE-(F+R)=H(ﬂ+eE-R since derivatives are invariant and the potential is periodic. Therefore .H(F+§)t .H(F)i _.e§.m woman = 6—1 n ¢(F+R,t = 0) = e-lTe z . eikﬁwm = 0) which gives ﬂ _ “77+ 1%, t) = e—[email protected]+iE~Re-i#¢(ﬁt = 0) where we could move and break up exponents since only the one with H depends on position7 the other two are just numbers. Hence we have W+ R‘. t) = eWﬁﬁch t) which is the required result. In this case we are able to derive the result of the semi-classical equation of motion directly from quantum mechanics! ...
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