SOLID STATE PHYSICS PART III Magnetic Properties of Solids

1 SOLIDSTATEPHYSICSPART Dresselhaus1 Contents1 Reviewof Topicsin .. ,De nitionsand CommutationRelations.. Wave Mechanics.. and OrbitalAngularMomentum.. Schr odinger'sEquationfor Free Atoms.. for AngularMomentum.. 152 MagneticE ectsin Free ect.. neInteraction.. AngularMomenta .. 'sRule.. gurations.. {GordanCoe cients .. 263 Diamagnetismand Paramagnetismof .. MagneticInteractionsfor BoundElectrons.. BoundElectrons.. BoundElectrons.. ParamagneticIons.. CrystalFieldTheory.. OrbitalAngularMomentum.. 474 Paramagnetismand Diamagnetismof NearlyFree.}

2 Paramagnetism.. LandauDiamagnetism.. in 3D.. the MagneticEnergyLevels inkx.. the MagneticEnergyLevels Alongthe MagneticEnergyLevels .. for ConductionElectrons.. 615 Magneto-Oscillatoryand OtherE LandauLevel E ects.. LandauLevel Transitions.. Quantizationfor LargeQuantumNumbers.. 696 TheQuantumHallE ect(QHE) QuantumHallE ect.. 2D HallResistance.. ElectronGas.. ectof EdgeChannels.. the QuantizedHallE ect.. ect(FQHE).. 837 .. MolecularFieldModel .. Ferrimagnetism.. Energy.. 1178 .. Magnets.. ers.

3 :MagneticTape .. E ect.. Structures.. 1373 Chapter1 Reviewof Topicsin AngularMomentumReferences Sakurai,ModernQuantumMechanics, Chapter3. Schi ,QuantumMechanics, Chapter7. Shankar,Principlesof QuantumMechanics, Chapter12. Jackson,Classical Electrodynamics, 2nd Ed., thischapterwe reviewsometopicsin quantummechanicsthatwe willapplyto ourdiscussionof topicunderdiscussionhereis ,De nitionsand CommutationRelationsIn classicalmechanics,as in quantummechanics,angularmomentumis de nedby~L=~r ~p:( )Sincethe positionoperator~rand momentumoperator~pdo not commute, the componentsof the angularmomentumdo not note rstthatthe positionand momentumoperatorsdo not commute:(xpx pxx)f(~r) = hix@@xf(~r) hi@@x[xf(~r)] =i hf(~r):( )The resultof Eq.

4 Is conveniently writtenin termsof the commutatorde nedby [rj;pk] rjpk pkrjusingthe relation[rj;pk] =i h jk;( )where jkis a deltafunctionhavingthe valueunity ifj=k, and thatdi erent components of~rand~pcommute, and it is onlythe samecomponents of1~rand~pthatfail to we now applythesecommutationrelationsto the angularmomentumwe get:[Lx;Ly]=LxLy LyLx= (ypz zpy)(zpx xpz) (zpx xpz)(ypz zpy)= i hypx+i hxpy=i hLz:( )Similarly, the commutationrelationsfor the othercomponents are:[Ly;Lz] =i hLx[Lz;Lx] =i hLy:( )The commutationrelationsin and are conveniently summarizedin the symbolicstatement~L ~L=i h~L:( )Thesecommutationrelationsare basicto the propertiesof the angularmomentumin two components of the angularmomentumcommute, it is not possibleto nda representationthatsimultaneouslydiagonal izesmorethanone component of~L.

5 Thatis,thereis no wavefunctionthatis both an eigenfunctionofLxandLy, for if therewere, wecouldthenwriteLx =`x ( )andLy =`y :( ) and wouldthenimplythatLxLy =Lx`y =`x`y ( )and alsoLyLx =`y`x ( )so thatifLxandLycouldbe diagonalizedsimultaneously, thenLxandLywouldhaveto , we know thattheydo not commute;therefore, the otherhand,all threecomponents of angularmomentum,Lx,Ly, andLzcommutewithL2whereL2=L2x+L2y+L2z:( )For example[Lz;L2] =LzL2x L2xLz+LzL2y L2yLz=LxLzLx+i hLyLx LxLzLx+i hLxLy+LyLzLy i hLxLy LyLzLy i hLyLx= 0( )and similarlyfor [Lx;L2] = 0 and [Ly;L2] = 0. SinceLx,LyandLzdo not commute witheach other,it is convenient to selectone component ( ,Lz) as the component is convenient to introduceraisingand loweringoperatorsL =Lx iLy( )so thatwe can writeL2=L2z+12(L+L +L L+):( )FromEq.

6 Know thatL+andL are non-hermitianoperators,becauseLxandLyare both Hermitianoperatorsand have real withL2, we have the commutationrelations:[L2;L+] = 0and[L2;L ] = 0:( )Furthermore,L+does not commute withLzbut rather[Lz;L+] = [Lz;(Lx+iLy)] =i h(Ly iLx) = hL+( )and likewise[Lz;L ] = hL :( )By the sameprocedurewe obtain[L+;L ] = [(Lx+iLy);(Lx iLy)] = 2 hLz( )andLx=12(L++L )Ly= i2(L+ L )( ) will now use the ndthe eigenvaluesof the us choose as our representation,one thatdiagonalizesbothLzandL2andwe will use quantumnumbersmand`to designatethe ,example,if is an eigenfunctionofLzwe can writeLz =m h ( )wherethe eigenvalueofLzism h.

7 FromEq. can thenwritethe matrixelement ofLzin thejm`irepresentationashm`jLzjm`i=m h( )wheremis a dimensionlessreal integerdenotingthe magnitudeofLz, while`is the maxi-mum valueofmand hhas the dimensionof , we can thinkofmas tellinghow many unitsof angularmomentumthereare onlydiagonalmatrixelements, the above relationcan be writtenmoregenerallyashm0`0jLzjm`i=m h mm0 ``0:( )We will now computethe matrixelements ofL+,L , andL2in thejm` commutationrelation[L2;L+] = 0( )we obtainhm0`0jL2L+ L+L2jm00`00i= 0( )for alljm` diagonalin thejm`irepresentationby hypothesisand thereforeis speci edby somefunctionof`, the quantumnumber associatedwithL2.

8 We thus obtainL2jm00`00i= (L2)`00jm00`00i( )wherewe have writtenthe eigenvalueof the operatorL2usingthe notation(L2)`00. We willnow show that(L2)`00= h2`00(`00+ 1):( ) write[(L2)`0 (L2)`00]hm0`0jL+jm00`00i= 0:( )For Eq. be satis ed,we see thatthe matrixelements ofL+must vanishunless`0=`00which impliesthatL+must be diagonalin`.Thecommutationrelation[Lz;L+ ] = hL+given by Eq. `jLzL+ L+Lzjm00`i= hhm0`jL+jm00`i( )Now exploitingthe eigenvaluerelationin Eq. `i=m00 hjm00`iwe get:(m0 m00) hhm0`jL+jm00`i= hhm0`jL+jm00`i( )so that(m0 m00 1) hhm0`jL+jm00`i= 0 which is conveniently expressedas hhm0`0jL+jm00`00i= `0`00 m0;m00+1 m0 h( )where m0is a dimensionlessnumber.

9 Thus, not onlyare the matrixelements ofL+in thejm`irepresentationdiagonalin`, but theyare non-vanishingonlyon o -diagonalpositionsofm(thatis wherethe indexofm0exceedsthe indexm00by unity). Thematrixelement forL+furthermoreimpliesthe matrixelement ofL sincethesematrixelements are relatedby the Hermitiantransposehm0`0jL jm00`00i= `0`00 m0;m00 1 m0 h:( )We can evaluate m0explicitlyby usingthe commutationrelation[L+;L ] = 2 hLzfromEq. ,so thatin takingmatrixelements of [L+;L ], we needonlyconsiderwhollydiagonalmatrixelem ents ofhm`j[L+;L ]jm`i= 2 hm h= 2m h2( )becauseLzis diagonalin bothmand`.

10 Sincethe matrixelement of a productof twooperatorsO1andO2in generalcan be writtenashnjO1O2jn0i= n00hnjO1jn00ihn00jO2jn0i( )4Eq. sumover all possiblem00and` +is diagonalin`and has onlyone non-vanishingmatrixelement inm, thenEq. the termshm`jL+jm 1;`ihm 1;`jL jm;`i hm;`jL jm+ 1;`ihm+ 1;`jL+jm;`i=j m 1j2 h2 j mj2 h2= 2m h2( )which has a solutionj mj2= m(m+ 1)( )so that (m 1)m+m(m+ 1) = 2mis satis can add any constantCtothe solutionof Eq. obtainanotherequallyvalid solutionj mj2=C m(m+ 1):( )Sincej mj2is positive, de nite,we requirethatC be chosento guarantee be restrictedto the rangefrom `to +`and C =`(`+ 1) orj mj2=`(`+ 1) m(m+ 1):( )Thisthenmeansthatthe raisingoperatoracts on the statejm;`ito producea statejm+1;`iL+jm;`i= h m+1jm+ 1;`i:( )Startingwiththe lowest statem= `, the raisingoperatorL+producesa physicalstateuntilm=`is reached at which timeL+j`;`iproducesa non-existent or null possiblevaluesof m, we can evaluatethe matrixelement ofL2in them;`representation:hm;`jL2jm;`i=hm;`j1 2(L+L +L L+) +L2zjm.