cha12 - Homework 4, due February 22, 1999 Problem 1....

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Unformatted text preview: Homework 4, due February 22, 1999 Problem 1. Ashcroft—Mermin 12.2 (a) We have so?) = @7212? - M-l - I? 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 g6 =(M—1)m 1936+ 2(M—1)my kmky +(M—1)yy 123+ 2 (M4)“ Icme + 2(M—1)yz kykz +(M—1)Zz kg We can write this in the form A(kw — W + 230% — text, — k2) + Cosy — k3) = D with _ 71 _ 71 _ 71 and (Rd—1).. (M‘ILZ k3 = — (Me)... kz v ’“3 = _ (M-l)... k2 and D = — (M4)“ kg — A(kg)2 — 23k:ng — C(kg)2 The area of the ellipse is A(e, k2) = 77D m which is linear in 6. Therefore we have m* _ fiaMech) _ 1 _ 2w Be _ W E Next we use (WI—1V1)“ = W17((M_1)M(M_I)yy ‘(M_1)W(M_1)W) (MLZ = det(MxAc — B2) from which the formula follows. (b) The density of states follows from 9(6) 2 fig fd3k6(e — gym—1 .1?) This can be rotated to principal axes 9(6) = so: fd3q6(e — @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 me = <m*>%Th;2 gives the required answer. det(D)#1/%g Problem 2. Ashcroft—Mermin 12.4 j’total = = E fiilE’ n n which gives [3—1 = 2/5771 TL From N _ pn p” _ RnH p” we find ~71 _ 1 pm RnH p” — pn+RnH pn and therefore 1 P RH _ PLREHE —RH p _ 1 p1 R1H + 1 p2 R2H '01—'RiH2 —R1H p1 P2+R2H2 —R2H p2 This leads to the equations pLR2H2 : p§~R1§1H2 + pgfifigm R 2 R1 + R2 p277R2 H2 pfil "R? H2 pg "R; H2 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 p + iRH = which has the correct denominator. (which is the enumerator of p ) is (91"102)2+1‘I2(RlJrR2)2 (P1+iR1 H) (P2+iR2 H) _ (Pr-131 H) (P2+iR2 H)(P1+P2 —iH(Rl +R2)) (91+iR1 H)(P2 +iR2H) (Pl +P2 —iH(R1 "32» (P1+P2)2+H2(R1”R2)2 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 lim R = fl If the high field Hall coeflicient has neff = 0 this means H—>oo R1+R2 that filim R = 00 and hence R1 —— R2 = 0, compensating hands. This —>OO gives _ P1P2(P1+p2)--H2R21(P1+P2) _ p1p2--H2R2] p _ (P1--P2)2 _ (P1+P2) Problem 3. Ashcroft—Mermin 12.6 Take a single band and consider a state with «M 1%,]: = 0) = eik'RzflFfif = 0) if we can show the result for this wave function, it will also hold for a linear combination. H(F+§)=—%fi§3-m+U(F+R)+eE-(F+R)=H(F)+eE-R’ since derivatives are invariant and the potential is periodic. Therefore .H(F+§)t .H(F)i _.e§.m wow 1%, t) = e-1 5 W4 Rh: 2 0) = e-lTe ZTelk'Rwfi’J = 0) which gives fl _ “77+ 1%, t) = e—i@+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) = er“; We: 0 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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This note was uploaded on 06/07/2010 for the course MECH 1122 taught by Professor Dont during the Spring '10 term at A.T. Still University.

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cha12 - Homework 4, due February 22, 1999 Problem 1....

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