14-3 Landau Level
Landau level of conduction electron in the presence of magnetic fieldMasatsugu Sei SuzukiDepartment of Physics, SUNY at Binghamton(Date: May 30, 2014)We discuss the motion of electrons in an external magnetic field. First we consider theclassical motion of electron. The electrons undergo a circular motion in the plane perpendicularto the magnetic field, with the cyclotron angular frequency. Next we discuss the motion ofelectron in terms of quantum mechanics. The wavefunction depends on the choice of gaugessuch as Landau gauge, Coulomb gauge, and symmetric gauge. The energy of the electrons isquantized. Each quantized energy level is called the Landau level.Lev Davidovich Landau (January 22, 1908– April 1, 1968) was a prominent Soviet physicistwho made fundamental contributions to many areas of theoretical physics. His accomplishmentsinclude the independent co-discovery of the density matrix method in quantum mechanics(alongside John von Neumann), the quantum mechanical theory of diamagnetism, the theory ofsuperfluidity, the theory of second-order phase transitions, the Ginzburg–Landau theory ofsuperconductivity, the theory of Fermi liquid, the explanation of Landau damping in plasmaphysics, the Landau pole in quantum electrodynamics, the two-component theory of neutrinos,and Landau's equations for S matrix singularities. He received the 1962 Nobel Prize in Physicsfor his development of a mathematical theory of superfluidity that accounts for the properties ofliquid helium II at a temperature below 2.17 K.http://en.wikipedia.org/wiki/Lev Landau1
1.Cyclotron motion of electron in the presence of magnetic field (classical theory)yvr0RxFig.Cyclotron motion in classical mechanics.We consider the motion of the electron in the presence of a magnetic field B. The Lorentz forceis given byeF me r r Bc(Lorentz force)where me is the mass of electron, -e is the charge of electron (e 0), B is an external magneticfield along the z axis. The solution of this differential equation is as follows.r R r0 (cos( ct ), sin( ct ),0) ,v r0 c ( sin( ct ), cos( ct ),0) ,where the cyclotron angular frequency c is given by c eB me c2
The trajectory of an electron is a circle with a radius r0 centered at the position vector R R ( X , Y ,0) . Suppose that the relative co-ordinate from the center is given by the posion vector , ,0 . Then we haver ( x, y ) ( X , Y ,0) ,v c ( , ,0) ,where r0 cos( ct ) ,2. r0 sin( ct ) .Gauge transformation (classical theory)The Hamiltonian H [q -e (e 0)] is given byH 1q1e( p A) 2 q ( p A) 2 e ,2mec2mecWhere q -e (e 0) A is the vector potential and is the scalar potential. The electric field andthe magnetic field are expressed byB A,E 1 A.c tThe canonical momentum is defined byπ me v p qeA p A.ccThe gauge transformation is defined byA' A , ' 1 ,c twhere is an arbitrary function of the position vector r. The orbital angular momentum isdefined byL r p r (π eeA) r π r Acc3
sincep π (i)eAcSymmetric gaugeIn the presence of the magnetic field B (constant), we can choose the vector potential asex11A (B r) 022xeyez0B yz1( By , Bx,0)2where1 A (B r ) B2(ii)Landau gaugeIn the gauge transformation:A' A we choose 1Bxy . Then we have21B( y, x,0)2Thus the new vector potential A' is obtained asA' (0, Bx,0)(Landau gauge)3.Commutation relations in quantum mechanics (general gauge)We discuss the commutation relations in quantum mechanics. Since the gauge is notspecified, the discussion below is applicable for any gauge. We start with the quantummechanical operator,πˆ pˆ e ˆA.c4
where e 0, and  is a vector potential depending on the position operator r̂ . The commutationrelation is given by[ xˆ, ˆ x ] [ xˆ , pˆ x e ˆAx ] i 1̂ ,c[ yˆ , ˆ y ] [ yˆ , pˆ y e ˆAy ] i 1̂c[ xˆ, ˆ y ] [ xˆ , pˆ y e ˆAy ] 0c[ yˆ , ˆ x ] [ yˆ , pˆ x e ˆAx ] 0cwhere we use the relation[ xˆ, Aˆ x ] [ yˆ , Aˆ x ] 0 , [ xˆ, Aˆ y ] [ yˆ , Aˆ y ] 0We also get the commutation relation,e ˆeAx , pˆ y Aˆ y ]ccee [ pˆ x , Aˆ y ] [ pˆ y , Aˆ x ]ccˆe Ay e Aˆ x ic xˆic yˆ[ ˆ x , ˆ y ] [ pˆ x e Bzic[ ˆ x , ˆ y ] e Bzicorwhere5
Aˆ y xˆ Aˆ x Bz . yˆWe have the commutation relations,[ ˆ y , ˆ z ] e Bx ,icSuppose that B (0,0,B)[ ˆ x , ˆ y ] e B,ice By .icand[ ˆ z , ˆ x ] orBz B. Then we get[ ˆ z , ˆ x ] 0 .[ ˆ y , ˆ z ] 0 ,Note thate 2 B 2[ ˆ x , ˆ y ] i 2 , ic where the characteristic length l is defined by 2 4.c .eBHamiltonian (general gauge)The Hamiltonian Ĥ is given by11e22( pˆ A) 2 Hˆ ( ˆ x ˆ y ) .2me2mecWe define the creation and annihilation operators such thataˆ l( ˆ x i ˆ y ) ,2 aˆ ˆ x ( aˆ aˆ ) ,2l ˆ y l( ˆ x i ˆ y ) ,2 or ( aˆ aˆ ) ,i 2l6
[aˆ , aˆ ] l2l2l2 2ˆˆˆˆˆˆ[ i , i ] i[ , ] i( i) 1̂ ,xyxyxy2 2 2 2l2 ˆ x 2 ˆ y 2 2 2 2 2 2ˆˆˆˆˆˆˆˆ[(a a) (a a)] (aa aa) ( 2aˆ aˆ 1) .2222lllThus we have11 2Hˆ (aˆ aˆ ) c (aˆ aˆ )222melwhere 2 2 eB c 2c melme cmeeBWhen aˆ aˆ Nˆ , the Hamiltonian is described by1Hˆ c ( Nˆ ) .2We find the energy levels for the free electrons in a homogeneous magnetic field, which isknown as Landau levels. We mote that11Hˆ n c ( Nˆ ) n c ( n ) n221with the energy eigenvalue En (n ) c (n 0, 1, 2, 3, .).25.Guiding centers (general case)Here we derive the expression for the guiding centers in quantum mechanics (general gauge).The Hamiltonian Ĥ is given by1e1e1e122( pˆ A) 2 Hˆ ( ˆ x ˆ y ) [( pˆ x Ax ) 2 ( pˆ x Ay ) 2 ]2me2mec2mec2mecwhere e 0, and7
ecec ˆ x pˆ x Ax ,(i) ˆ y pˆ y Ay .The commutation relation (I)We have the commutation relations,222[ xˆ,2me Hˆ ] [ xˆ , ˆ x ˆ y ] [ xˆ , ˆ x ] 2i ˆ x(1)222[ yˆ ,2me Hˆ ] [ yˆ , ˆ x ˆ y ] [ yˆ , ˆ y ] 2i ˆ y(2)where[ xˆ, ˆ x ] i 1̂ ,[ yˆ , ˆ y ] i 1̂ ,[ xˆ , ˆ y ] 0 ,[ yˆ , ˆ x ] 0 ,[ ˆ x , ˆ y ] i(ii) 2,l2[ ˆ y , ˆ z ] 0 ,[ ˆ z , ˆ x ] 0The commutation relation (II):We also have the commutation relations,[ ˆ x ,2me Hˆ ] [ ˆ x , ˆ x ˆ y ] [ ˆ x , ˆ y ] ( 2i )222 2 ˆ yl22 222[ ˆ y ,2me Hˆ ] [ ˆ y , ˆ x ˆ y ] [ ˆ y , ˆ x ] 2i 2 ˆ x ,lFrom Eqs,(1)-(4), we have[ xˆ l2l 2 2 ˆ y ,2me Hˆ ] 2i ˆ x 2i 2 ˆ x 0 . l[ yˆ l2l2 2ˆˆˆ x ,2me H ] 2i y ( 2i ) 2 ˆ y 0 l8(3)
So we can define the guiding center as22lXˆ xˆ ˆ y , lYˆ yˆ ˆ x (general gauge)with [ Xˆ , Hˆ ] 0 and [Yˆ , Hˆ ] 0 . Thus we havedX 0dtdY 0,dt(constant motion).We note that22ll[ Xˆ , Yˆ ] [ xˆ ˆ y , yˆ ˆ x ] 22lll4 [ xˆ , ˆ x ] [ yˆ , ˆ y ] 2 [ ˆ x , ˆ y ] 242ll 2 i 1̂ 2 ( i 2 )1̂ il 21̂ lWe introduce the new operators as1 ˆbˆ ( X iYˆ ) ,2l1 ˆbˆ ( X iYˆ )2ll ˆ ˆ Xˆ (b b ) ,2l ˆ ˆ Yˆ (b b )2iorobeying12[bˆ, bˆ ] 2 [ Xˆ iYˆ , Xˆ iYˆ ] 2 i[ Xˆ , Yˆ ] 1̂2l2lThen we haveXˆ 2 Yˆ 2 l 2 ( 2bˆ bˆ 1̂)(i)Landau gauge9
ec ˆ x pˆ x Aˆ x pˆ x ,ec ˆ y pˆ y Aˆ x pˆ y eB xˆ pˆ y 2 xˆclThen the guiding center is given byl2l2l2 Xˆ xˆ ˆ y xˆ ( pˆ y 2 xˆ ) pˆ yl 22llYˆ yˆ ˆ x yˆ pˆ x (ii)Symmetric gauge ˆ x pˆ x Ax pˆ x eceB yˆ pˆ x 2 yˆ ,2c2leceB xˆ pˆ y 2 xˆ2c2l ˆ y pˆ y Ay pˆ y Then the guiding center is given byl2l21l2 Xˆ xˆ ˆ y xˆ ( pˆ y 2 xˆ ) xˆ pˆ y2l2 l2l21l2 ˆY yˆ ˆ x yˆ ( pˆ x 2 yˆ ) yˆ pˆ x2l2 6.Relative co-ordinate (general gauge)The guiding center is given by2lxˆ Xˆ ˆ y , 2lyˆ Yˆ ˆ x The operators for the relative co-ordinates are defined by10
ˆ xˆ Xˆ l2 ˆ y , ˆ yˆ Yˆ l2 ˆ x Then the Hamiltonian is expressed by only the relative co-ordinates1 2 ˆ222Hˆ ( ˆ x ˆ y ) ( ˆ 2 )42me2melwhere[ ˆ, ˆ ] il 2 , [ Xˆ ,Yˆ ] il 27.Heisenberg's principle of uncertainty (general gauge)We consider the commutation relation of X̂ and Yˆ , l2l2l4l4 2ˆˆX ,Y [ xˆ ˆ y , yˆ ˆ x ] 2 [ ˆ x , ˆ y ] 2 ( i ) 2 il 2 since[ ˆ x , ˆ y ] i 2 2Using the Schwarz inequality, we get1 X Y l 2 ,2 x y 1 22 2In fact we have1 X Y 2 l 2 l 2 .2((Note))Schwartz inequality and B̂ are two Hermitian operators with the condition[ Aˆ , Bˆ ] iCˆ .Then we have a Heisenberg’s principle of uncertainty:11
A B 1C .2This principle is a direct consequence of the non-commutability between two observables.8.Translation operator derived from the commutation relationsWe have the commutation relationsi ˆ x[ xˆ , Hˆ ] me(1)i ˆ y[ yˆ , Hˆ ] me(2)We also have the commutation relations: 2ˆˆ ˆ y[ x , H ] ime l 2(3) 2 ˆ x ,[ ˆ y , Hˆ ] ime l 2(4)From these equations, we get[ ˆ x , Hˆ ] i 2 2 me ˆ i[ yˆ , Hˆ ] 2 [ yˆ , Hˆ ]y22me lme l i lor[ ˆ x yˆ , Hˆ ] 0l2We also have 2 2 me ˆ[ ˆ y , Hˆ ] i i[ xˆ , Hˆ ] 2 [ xˆ , Hˆ ]x22me lme l i l12
or[ ˆ y xˆ, Hˆ ] 02lThe pseudo-momentum for the translation operator as a generator can be defined by eBe( yˆ , xˆ ) πˆ B rˆKˆ πˆ 2 ( yˆ , xˆ ) πˆ lccor Kˆ πˆ 2 ( yˆ , xˆ ) ( ˆ x 2 yˆ , ˆ y 2 xˆ )lllThen the translation operator can be defined byiTˆ (a ) exp( Kˆ a ) Clearly the translation operator K̂ commutes with the Hamiltonian Ĥ . K̂ commutes with Ĥ .However, K̂ x and K̂ y do not commute;2 [ Kˆ x , Kˆ x ] i 2 .lThe translation operators do not commute,iTˆ (a )Tˆ (b) Tˆ (b)Tˆ (a ) exp[ 2 (a b) z ]l9.Translation operator (II) derived from the gauge transformation (Cohen Tannoudjiet al)We start with the definition of the translation operator Tˆ (a ) ; T Tˆ (a ) r T r Tˆ (a ) r a (r a )13
The Hamiltonian commutes with the translation operator.[ Hˆ , Tˆ (a )] 0whereiTˆ (a ) exp( pˆ a ) The Schrödinger equation for r is given byH ( r ) ( r ) 1e[ p A( r )]2 ( r ) E ( r )2 mecThe Schrödinger equation for r a ( r a ) is given by1e[ p A( r a)]2 T ( r ) E T ( r )2mecSuppose that the vector potential is given byAT ( r ) A( r a) 1111B ( r a) B r B a A( r ) B a2222orA AT 1B a AT 2(Gauge transformation)Then we have121212 r ( B a) a ( r B) a ( B r )As a result, we get the Schrödinger equation for T ' ( r ) as14
1e[ p A(r ))]2 T ' (r ) E T ' (r )2mecUnder such a gauge transformation, the wavefunction changes as T ' exp( ieiei ) T exp( ) exp( pˆ a ) c c When the vector a is in the x-y plane,ieirˆ ( B a )) exp( pˆ a) 2 cˆi e( B r )i exp( a ) exp( pˆ a ) 2cie( B r ) exp[ ( pˆ ) a] 2ci exp[ Kˆ a ] T ' exp( since the two operators in front of are commutable. The generator of the translation operatoris given byeeBKˆ pˆ B r pˆ ( yˆ , zˆ,0)2c2c(1)From the previouse discussion we geteeeeeKˆ πˆ B rˆ pˆ A B rˆ pˆ B rˆ B rˆ2cccccoreKˆ pˆ B rˆ2cwhich is the same as Eq.(1).((Note))Gauge transformationUnder the gauge transformation,15
' A' A ,1 c tthe wave function changes as ' exp( 10.ie ) , cAngular momentum (symmetric gauge)The angular momentum is defined byLˆ z xˆpˆ y yˆ pˆ x ,wherepˆ πˆ e ˆA,cec ˆ x pˆ x Aˆ x pˆ x yˆ ,2l 2ec ˆ y pˆ y Aˆ y pˆ y xˆ2l 2 l2l21l2Xˆ xˆ ˆ y xˆ ( pˆ y 2 xˆ ) xˆ pˆ y 2l2 l2l21l2Yˆ yˆ ˆ x yˆ ( pˆ x 2 yˆ ) yˆ pˆ x 2l2Then we havel2l2l2l21111Xˆ 2 Yˆ 2 ( xˆ pˆ y )( xˆ pˆ y ) ( yˆ pˆ x )( yˆ pˆ x )2222 241ll22 ( xˆ 2 yˆ 2 ) 2 ( pˆ x pˆ y ) ( xˆpˆ y yˆ pˆ x )4 41ll222 ( xˆ 2 yˆ 2 ) 2 ( pˆ x pˆ y ) Lˆ z4 16(1)
yˆ )( pˆ x 2 yˆ ) ( pˆ y 2 xˆ )( pˆ y 2 xˆ )22l2l2l2l2 22 ( pˆ x pˆ y ) 4 ( xˆ 2 yˆ 2 ) 2 ( xˆpˆ y yˆ pˆ x )4ll2 22 ( pˆ x pˆ y ) 4 ( xˆ 2 yˆ 2 ) 2 Lˆ z4ll24 1 2ll2 ˆ222ˆˆˆˆ 4 [ ( x y ) 2 ( p x p y ) Lz ] l 4 ˆ x 2 ˆ y 2 ( pˆ x orl41 2l4l2 ˆ22222ˆˆˆˆˆˆ( ) (x y) (p p) Lzxyxy 2 2 4From Eqs.(1) and (2) we have l222Lˆ z ( ˆ x ˆ y ) 2 ( Xˆ 2 Yˆ 2 )2 2lorl42l 2 ˆ2222ˆˆˆˆX Y 2 ( x y ) Lz In this case, he Hamiltonian is rewitten as 2 ˆ122Hˆ ( ˆ x ˆ y ) ( Xˆ 2 Yˆ 2 ) Lz42me2m e lmel 2Note that ˆ e B Lˆ zLˆ zL2B zB mel 2me c where B e is the Bohr magneton.2me cIn general gauge, we have17(2)
ˆ x 2 ˆ y 2 2( 2aˆ aˆ 1)2landXˆ 2 Yˆ 2 l 2 ( 2bˆ bˆ 1̂)with[bˆ, bˆ ] 1̂The angular momentum in the symmetric gauge is given byl2 22Lˆ z ( ˆ x ˆ y ) 2 ( Xˆ 2 Yˆ 2 )2 2l22l ( 2aˆ aˆ 1) 2 l 2 ( 2bˆ bˆ 1̂)22 l2l ˆˆ ( aˆ aˆ b b)11.Schrödinger equation (Landau gauge)In the absence of an electric field, the H 1eB 212222Hˆ [ pˆ x ( pˆ y xˆ ) pˆ z ] [ pˆ x ( pˆ y 2 xˆ ) 2 pˆ z ]2mec2melwhen we use the Landau gauge and l 2 c (e 0). The guiding center is given byeB l2l2l2Xˆ xˆ ˆ y xˆ ( pˆ y 2 xˆ ) pˆ y . lWe note that this Hamiltonian Ĥ commutes with p̂ y and p̂ z .[ Hˆ , pˆ y ] 0and[ Hˆ , pˆ z ] 0Then n, k y , k z is a simultaneous eigenket of Ĥ , p̂ y and p̂ z .18
Hˆ n, k y , k z En n, k y , k zpˆ y n, k y , k z k y n, k y , k z ,andpˆ z n, k y , k z k z n, k y , k zWe also note thatl2l2ˆˆX n, k y , k z p y n, k y , k z k y n, k y , k z l 2 k y n, k y , k z In other words, n, k y , k z is the eigenket of X̂ with the eigenvalue X l 2 k yIn the position representation, we havey pˆ y n, k y , k z k y y n, k y , k z ,z pˆ y n , k y , k z k y y n , k y , k zor y n, k y , k z k y y n, k y , k z ,i y z n, k y , k z k z z n, k y , k z .i zThen the wave function has the form of ( x, y , z ) eik y y ik z z ( x) .whereX l 2 k y .We assume the periodic boundary condition along the y axis.19
( x, y L y , z ) ( x, y , z )oreik y L y 1orky X 2 nyl 2 Ly(ny: intergers)The Schrödinger equation for the wavefunction ( x, y, z ) is given by1 2 2 2[() ( x) () ] ( x, y, z ) E ( x, y, z )2me i xi y l 2i zwith c eB.me cThe Schrödinger equation for the wavefunction (x) is given byd212m E ( x) 4 ( x X ) 2 ( 2e k z 2 ) ( x)2dxl The energy eigenvalue is21 2kzE E (n, k z ) c (n ) 22me1l " ( x) [ 4 ( x X ) 2 12.1(2n 1)] ( x)l2Mathematica((Mathematica))20
Clear "Global " ; x : y : H1 : 12m— D , y e1 BcD , x &;— x &; z : — D , z &; Nest x, , 2 Nest y, , 2 Nest z, , 2 &;eq1 H1 x, y, z E1 x, y, z Simplify;rule1 Exp ky 2 kz 3 1 & ;eq2 eq1 . rule1 Simplify;mc eq3 eq2 . B FullSimplifye11 ky y kz z 2m 2 E1 m m 2 x2 2 2 ky m x — ky2 kz2 —2 x —2 x 13.We putLandau level and the wave function x,l 0 Xld d d 1 d ' ( x) ,dx dx d l d " ( x) Then we get a differential equation " ( ) [(2n 1) ( 0 ) 2 ] ( )The solution of this differential equation is n ( ) ( 2 n! )n 1 / 2e ( 0 ) 22H n ( 0 )or21d d d d 1 d 2 dx d dx d l 2 d 2
n ( x ) ( 2 n! )n 1 / 2e ( x X )22l 2Hn (x X)lThe coordinate X is the center of orbits. Suppose that the size of the system along the x axis is Lx.The coordinate X should satisfy the condition, 0 X Lx. Since the energy of the system isindependent of x0, this state is degenerate.0 X 2 k y LxorX 2k y 2 2 n y LxLyorny Lx Ly2 2Note that X 2 l 2LyThe area of each Landau state is A Ly X 2 l 2 .Thus the degeneracy is given by the number of allowed ky values for the system.g Lx Ly2 2 AABA 2c 2 0 2 02 2 eBwhere2 0 2 c 4.13563 10 7 Gauss cm2e22
The value of g is the total magnetic flux. There is one state per a quantum magnetic flux 2 0 . 0 is the quantum fluxoid.E n,kz 2017.51512.5107.552.5-8Fig.14.-6-4-2246Energy dispersion relation for the Landau level.Creation and annihilation operator (Landau gauge)We define the creation and annihilation operators such thataˆ l( ˆ x i ˆ y ) ,2 aˆ l( ˆ x i ˆ y )2 with[aˆ , aˆ ] 1̂In the Landau gauge, we have ˆ x p̂ x , ˆ y ( xˆ Xˆ )2lThen we haveaˆ l[ pˆ x i 2 ( xˆ Xˆ )] ,l2 238kz
aˆ l[ pˆ x i 2 ( xˆ Xˆ )]l2 aˆ n, X n 1 n 1, Xaˆ n, X n n 1, X ,(i)Wavefunction of the ground stateaˆ n 0, X 0x aˆ n 0, X i1 (x X )[l ] x n 0, X 0l2 xThe normalized wavefunction can be solved by using Mathematicax n 0, X 1 lexp[ ( x X )2]2l 2(Gaussian)In general we have N1x aˆn 0, XN!1( x X )2 x X N( l ) exp[ ]l2l 2 x2 N N! lx N , X ( i ) N x X( x X )2HN () exp[ ]l2l 22 N N! l1withaˆ l [ pˆ x i 2 ( xˆ Xˆ )]l2 For simplicity we put the factor ( i) N .((Mathematica-1))Wavefunction of the ground state24
Clear "Global " ; x XL1 D , x An : L12&;eq1 An x ;eq2 DSolve eq1 0, x , x Simplify x x x 2 X 2 L12C 1 1 x x . eq2 1 . C 1 C1 ;eq3 Integrate 1 x 2 , x, , Simplify , L1 0 &;eq4 Solve eq3 1, C1 ; 0 x : 1 x . eq4 2 Simplify; 0 x x X 22 L12L1 1 4((Mathematica-II))Wavefunction of excited stateWe calculate the two functions by using the Mathematica. ( x, N ) ( x, N ) 12 N N! l( l12 N N! l( x X )2 x X N) exp[ ] l2l 2 xHN (x X( x X )2) exp[ ]l2l 2where H N (x) is a Hermite polynomial.15.Schrödinger equation (symmetric gauge)In the absence of an electric field, the H1 2Hˆ [( pˆ x 2 yˆ ) 2 ( pˆ y 2 xˆ ) 2 pˆ z ]2me2l2l25
when we use the symmetric gauge and l 2 c (e 0). Ĥ can be rewritten aseB 2 2 Lˆ z122222ˆˆ Hˆ ( pˆ x pˆ y pˆ z ) (xy)2 me8me l 42me l 2 Lˆ112222 ( pˆ x pˆ y pˆ z ) me c ( xˆ 2 yˆ 2 ) B B z 2 me8where L̂z is the orbital angular momentum.e B 2 B B22mel2me c B is the Bohr magneton and is defined by B e 2me cThe third term of Ĥ is the Zeeman energy. The orbital magnetic moment isμˆ L BLˆ. So the Zeeman energy is given by μˆ L B ( BLˆLˆ) B B B z . We note that[ Hˆ , Lˆ z ] 0 ,[ Hˆ , pˆ z ] 0 ,[ Lˆ z , pˆ z ] 0Therefore, there is a simultaneous eigenket of Ĥ , L̂z , and p̂ z .Hˆ E ,Lˆ z m ,pˆ z k z 26
((Note))We note that2me l 4 ˆ ˆ22ˆˆ(X Y ) (H Lz )2 me l 2Thus this operator is related to the square of radius the guiding center commutes with Ĥ and L̂z .( Xˆ 2 Yˆ 2 ) 2me l 4 ˆ ˆ(H Lz ) 2me l 2 2me l 4m 2(E ) me l 2 2 2me l 4( E m c ) 21Suppose that E (n ) c , we have22mel 42mel 41 c (n m )(E m) c22 21 2l 2 (n m )2or ( Xˆ 2 Yˆ 2 ) 2l 2 (n m 1) 0wherem n, n-1, n-2, ,- .16.Mathematica: Proof of [ Hˆ , Lˆ z ] 0 .27
Clear "Global " ;px : — H1 : 12m12D , x &; py : — D , y &; Nest px, , 2 Nest py, , 2 m 2 x 2 y 2 &;Lz : x py y px &;Lz x, y x — 0,1 x, y y — 1,0 x, y H1 Lz x, y Lz H1 x, y Simplify0H1 Lz x, y Simplify1 — m 2 x x2 y2 2 0,1 x, y 2mx —2 0,3 x, y m 2 x2 y 2 1,0 x, y m 2 y3 2 1,0 x, y y —2 1,2 x, y x —2 2,1 x, y y —2 3,0 x, y Lz H1 x, y Simplify1 — m 2 x x2 y2 2 0,1 x, y 2mx —2 0,3 x, y m 2 x2 y 2 1,0 x, y m 2 y3 2 1,0 x, y y —2 1,2 x, y x —2 2,1 x, y y —2 3,0 x, y 17.Cylindrical co-ordinates (symmetric gauge)The Schrödinger equation is given by 1 21 2 2 1 2[( ) 2 2 2 ] i BB me c 2 E 2me 8 z 28
where , , z Lˆ z ( , , z ) m ( , , z ) ,i , , z pˆ z ( , , z ) k z ( , , z )i z ( , , z ) e im ( , , z ) e ik zzThen the wavefunction has the form of , , z ( , , z ) e ik z e im ( )zThen we have " ( ) 1 ' ( ) f ( ) ( ) 0or2 2 m 2 "( ) ' ( ) ( 2 4 2 ) ( ) 0 4l l1with2me E 2 B Bme 2 cm22 kzf ( ) mme2222 4 22 2 22meme 2 cm2( )() EkBmzB 22me4 2 2 2 2 m 2 l 2 4l 4 222wherel2 c eB1 c c (E eB,me c B e ,2me c1 2 2 2 2 m(E k z B Bm ) kz ) 2 me2me2 c29
((Matheamtica))30
Clear "Global " ; Clear B ; r1 z2 2 ;ux Cos , Sin , 0 ; uy Sin , Cos , 0 ;1 , 0, z ; x ux.r;uz 0, 0, 1 ; r , 0, z ; ur r1y uy.r;z uz.r;P : — Grad , , , z , "Cylindrical" &;Px ux.P & Simplify; Py uy.P & Simplify;Pz uz.P & Simplify;L : — Cross r, Grad , , , z , "Cylindrical" &;Lx : ux.L & Simplify; Ly : uy.L & Simplify;Lz : uz.L & Simplify;Lz , , z — 0,1,0 , , z Pz , , z — 0,0,1 , , z H1 12 me Nest Px, , , z , 2 Nest Py, , , z , 2 Nest Pz, , , z , 2 B B—18me c2 2 , , z Lz , , z FullSimplify31
1 me2 4 c 2 , , z 8 me 24 2 —2 0,0,2 , , z 2 B me B 2 0,1,0 , , z —2 0,2,0 , , z 1,0,0 , , z 2,0,0 , , z rule1 Exp m 2 Exp kz 3 1 & ;eq1 H1 E1 , , z . rule1 Simplify1 kz z m 8 me 2 8 E1 me 2 8 B m me B 2 me2 4 c2 4 m 2 —2 4 kz2 2 —2 4 —2 14 2 —2 8 E1 me 2 8 B m me B 2 me2 4 c2 4 m 2 —2 4 kz2 2 —2 4 —2 Expandkz2 m 2 2 E1 me 2—22 B m me B me2 2 c 2 22 —4—We use a new variable, 22l 2.(Landau and Lifshitz, Quantum Mechanics). Then we havem2222 "( ) '( ) [ ( )] ( ) 0l2l2l22l 2 2 l 2or m2 " ( ) ' ( ) ( ) ( ) 044 where32
1 '( ) 1 1 1 ' ( ) ' ( ) 2 ' ( )2 d ll ( )d d ( ) 2l l 2 ( ) 4l 4 [ ' ( ) " ( )]l "( ) l2[ ' ( ) " ( )]l4 12 2 ' ( ) 2 " ( ) llWe solve this differential equation using by the series expansion.(i)In the limit of ,14 " ( ) ( ) 0The solution of ( ) is approximately given by12 ( ) exp( )(ii)In the limit of 0 " ( ) ' ( ) m2 ( ) 04 The asymptotic form of ( ) is given by ( ) m / 2So that we seek a solution in the form33
( ) exp( ) m / 2 u( )2The differential equation for u( ) is given by12 u" ( ) (1 m )u' ( ) (1 m 2 )u( ) 0 .18.Solving of the differential equationFor convenience, we use a new variable x instead of .1xu" ( x ) (1 m x )u' ( x ) (1 m 2 )u( x ) 02We assume that u(x) can be expressed by u( x ) x p C (k ) x kk 0We get the coefficients for the first several termsp ( p m )C (0) 0 ,1 (1 m 2 p 2 )C (0) (1 p )(1 m p )C (1) 0 ,21 (3 m 2 p 2 )C (1) ( 2 p )( 2 m p )C ( 2) 02From the first equation, we get p 0 or p m . We need to choose p 0, so that (x) is finiteat x 0. Then we get the recursion relationC ( k 1) (1 m 2k 2 )2(1 k )(1 m k )C (k ) .Suppose that34
(1 m 2nr 2 ) 0 .or 1 m nr .2Then we have C (nr 1) C (nr 2) . 0 . u(x) is a polynomial of x with the order of nr ,nrun ( x) C (k ) x k C (0) C (1) x C (2) x 2 . C (nr ) x n rk 0with the recursion formulaC (k 1) {(1 m 2k (1 m 2nr )}C (k )2(1 k )(1 m k )(k nr )C (k )(k 1)(k m 1)Note that u (x) satisfies the differential equation,xu" ( x) (1 m x)u ' ( x) nr u ( x) 0 .The solution is given by the generalized Laguerrel polynomialmu ( x) Lnr ( x)Note that in the confluent hypergeometric representation (Arfken), we have the relationmLnr ( x) (nr m )!M ( nr , m 1, x)nr ! m !Then we get the solution of the above differential equationx2 ( x) A exp( ) x m / 2 Lnm ( x)rwithx 22l 235
The normalized wavefunction is ( x) nr !1 im ik z z 2 2 e eexp( 2 ) (nr m )!4l 2l 2 lThe energy eigenvalue is obtained as1 m 2 2 m ( En k z ) nr 2me22 2 c1orEnr c (nr 1 m m 2 2) kz222meIt is useful to introduce a new quantum number n as1n nr (m m )2for nr 0, 1, 2, 3, .When m 0,1n nr ( m m ) nr m2when m 0,n nrThe standard form of the Landau spectrum is1 2 2E n c ( n ) kz2 2meFinally we get the expression for the wavefunction36m /2mnrL ( 22l 2)
( x) 19.nr ! 2 2 1 eim eik z zexp( 2 ) (nr m )!4l 2l 2 l1eim eik z z2 lm /2mLnr ( 22l 2)(m m )m /2]! 2 2 2m2 exp( 2 ) ( 2)L 1n ( m m ) 2l(m m )4l 2l 2[n ]!2[n The quantum numbers n and mWe get the relation from the above discussion,1n nr (m m )2for nr 0, 1, 2, 3, .where n 0, 1, 2, 3, . From these conditions we have the values of (n, m) as followsFor nr 0,n 0,n 1,n 2,n 3,m 0, -1, -2, .,- m 1,m 2,m 3,For nr 1,n 1,n 2,n 3,n 4,m 0, -1, -2, .,- m 1,m 2,m 3,For nr 2,n 2,n 3,n 4,n 5,m 0, -1, -2,., - m 1,m 2,m 3,For nr 3n 3,n 4,n 5,n 6,m 0,-1, -2, . - m 1,m 2,m 3,37
In other words,(i)(ii)(ii)(ii)The ground state:n 0, m 0, -1, -2,. - The first excited state:n 1 m 1, 0, -1, -2, ., - The second excited state:n 2 m 2, 1, 0, -1, -2, ., - The third excited state:n 3 m 3, 2, 1, 0, -1, -2,., - (the degeneracy g0 )(the degeneracy g1 ).(the degeneracy g2 ).(the degeneracy g2 ).n43n m-121-4Fig.-3-2-1001234mm-n plane. n m 1 . n is a positive integer. n, m ( x) 1 im ik z ze e2 l(m m )m /2]! 2 2 2m2 L 1exp( 2 ) ( 2)n ( m m ) 2l(m m )4l 2l 2[n ]!2[n The probability density is given byPn ,m ( ) 2 n ,m ( )238
where P ( ) d 10We make a plot of Pn ,m ( ) by using the Mathematica (Plot3D) for several cases(i)n 0, m 0, -1, -239
(ii)n 1 m 1, 0, -1, -240
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(iii)n 2 m 2, 1, 0, -1, -243
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(iv)n 3 m 0, 1, 2, 346
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20.Mathematica: solving the differential equation by series expansion49
Clear "Global " ;eq1 x f'' x f' x x4 a24xf x ;rule1 f Exp 2 a 2 u & ;a m ;eq11 eq1 . rule1 Simplify1 x 2 a 2 x2 1 a 2 u x 2 1 a x u x x u x eq12 1 a 2 u x 2 1 a x u x x u x 2 Expand Simplify 1 1 a 2 u x 1 a x u x x u x 2rule2 u p C k k & ;10k 0eq21 eq12 . rule2 Expand;eq22 eq21 x1 p Simplify;list1 Table n, Coefficient eq22, x, n , n, 0, 4 Simplify;list1 TableForm01234p a p C 0 12 1 a 2 p 2 C 0 1 p 1 a p C 1 12 3 a 2 p 2 C 1 2 p 2 a p C 2 12 5 a 2 p 2 C 2 3 p 3 a p C 3 12 7 a 2 p 2 C 3 4 p 4 a p C 4 list2 list1 . p 0 ; list2 TableForm 12 1 a 2 C 0 1 a C 1 001 12 3 a 2 C 1 2 2 a C 2 234 12 5 a 2 C 2 3 3 a C 3 12 7 a 2 C 3 4 4 a C 4 50
Determination of recursion formularule3 u C n n & ;k 3n k 3eq3 eq12 . rule3 Expand;eq31 eq3 x4 k Simplify;list3 Table n, Coefficient eq31, x, n , n, 3, 5 Simplify;list3 TableForm345 12 1 a 2 k 2 C 1 k k a k C k 12 1 a 2 k 2 C k 1 k 1 a k C 1 k 12 3 a 2 k 2 C 1 k 2 k 2 a k C 2 k eq4 list3 2, 2 1 1 a 2 k 2 C k 1 k 1 a k C 1 k 2eq5 Solve eq4 0, C k 1 C 1 k 1 a 2 k 2 C k 2 1 k 1 a k DSolve eq12 0, u x , x Simplify u x C 1 HypergeometricU C 2 LaguerreL 1 1 a 2 , 1 a, x 21 a , a, x 2 2Solving differential equation:seq1 x F '' x 1 Abs m x F ' x nr F x 0;DSolve seq1, F x , x F x C 1 HypergeometricU nr, 1 Abs m , x C 2 LaguerreL nr, Abs m , x 21.Wavefunction for the Coulomb gaugeHere we use the Coulomb gauge such that A 0 . Then we get1e1e2eHˆ ( pˆ A) 2 [ pˆ 2 2 A2 ( pˆ A A pˆ )]2mc2mccwhere51
pˆ A A pˆ pˆ x Ax pˆ y Ay pˆ z Az Ax pˆ x Ay pˆ y Az pˆ z [ pˆ x , Ax ] [ pˆ y , Ay ] [ pˆ z , Az ] 2 A pˆ A 2 A pˆiThen we have1e2 2 e 2ˆˆH [ p 2 A ( A 2 A pˆ )]2mcc i21ee 2e ( pˆ 2 2 A 2 A A pˆ )2mciccSince A 0 , we have1 2 e 2 2 2e1 2 e 2 B 2 2 eBˆˆˆH (p 2 A A p) pˆ xˆ xˆpˆ y2mcc2m2mc 2mcwherel2 c ,eB c 2 eB ,2mlmcm c 2e2 B 2mc 21 2 e2 B 2 21 2 m c 2Hˆ pˆ xˆ c xˆpˆ y pˆ xˆ c xˆpˆ y22m2mc2m22The first and second terms of this Hamiltonian are that of the simple harmonics along the x axis. We note that this Hamiltonian Ĥ commutes with p̂ y and p̂ z .[ Hˆ , pˆ y ] 0and[ Hˆ , pˆ z ] 0Then n, k y , k z is a simultaneous eigenket of Ĥ , p̂ y and p̂ z .Hˆ n, k y , k z En n, k y , k zand52
pˆ y n , k y , k z k y n, k y , k z ,andpˆ z n, k y , k z k z n, k y , k zThus the wave function is described by the form, ( x , y , z ) n ( x ) eik y y ik z z REFERENCESD, Yoshioka, The Quantum Hall Effect (Springer, 1998).S.F. Ezawa, Quantum Hall Effect Field Theoretical Approach and Related Topics, second edition(World Scientific, 2008).Schaum's Quantum MechanicsAPPENDIXA.1Hamiltonian1eHˆ ( pˆ Aˆ ) 2 e ,2mcwhere e 0. The canonical momentum is defined byπˆ mvˆ pˆ e ˆA.cA.2Guiding center and relative co-ordinate(a)General case[ xˆ, ˆ x ] i 1̂ ,[ yˆ , ˆ y ] i 1̂ ,[ xˆ , ˆ y ] 0 ,[ yˆ , ˆ x ] 0 ,53
[ ˆ x , ˆ y ] i 2,l2[ ˆ y , ˆ z ] 0 ,[ ˆ z , ˆ x ] 0where B is the magnetic field. B is directed along the z axis.l2 (b)c .eB c eBme cLandau gauge22lxˆ Xˆ ˆ y , ˆ xˆ Xˆ [ ˆ, ˆ ] il 2[ Xˆ , Yˆ ] il 2lyˆ Yˆ ˆ x l2 ˆ y , 2l2 ˆ yˆ Yˆ ˆ x2ll[ ˆ, Xˆ ] [ ˆ y , xˆ ˆ y ] 0 2ll2[ ˆ , Xˆ ] [ ˆ x , xˆ ˆ y ] 0 , 22ll[ ˆ, Yˆ ] [ ˆ y , yˆ ˆ x ] 0 2ll2ˆˆˆˆ[ , Y ] [ x , y ˆ x ] 0
such as Landau gauge, Coulomb gauge, and symmetric gauge. The energy of the electrons is quantized. Each quantized energy level is called the Landau level. _ Lev Davidovich Landau (January 22, 1908- April 1, 1968) was a prominent Soviet physicist who made fundamental contributions to many areas of theoretical physics. His accomplishments
Landau-forbidden phase transitions between Landau-allowed phases Naive expectation: Breakdown of LGW paradigm at QCP when one of the two proximate phases has non-Landau order. Very interesting that LGW can also break down at critical point between two Landau-allowed phases. LRO, no LRE LRO', no LRE Eg: AF - VBS in spin-1/2 square lattice
Landau Levels within Landau Levels So Fractional Quantized Hall States are understood as filled Landau Levels of particles in a fictitious magnetic field, within a partially-full Landau Level of the true magnetic field. The fictitious magnetic field is an emergent phenomenon, an effective interaction that emerges from
Landau Forte Academy QEMS is a member of the family of Academies operated by Landau Forte Charitable Trust. This is an educational charitable trust founded by Martin . Landau and the Forte family over twenty years ago. The Trust expects each of the Academies to provide a Landau . Forte style of education which is distinctive, high quality and
of the Landau-level coincidence in the quantum Hall effect regime. FIG. 1. Color online Landau fan diagram for a two-subband system with a subband energy separation of 14.2 meV. We show only the lowest six Landau levels and they are spin splitted. The encircled region marks a p
Dec 16, 2015 · The Landau damping closures have similar impact on the ELM size as flux-limited heat flux Nonlinear simulation shows that the energy loss of am ELM are similar with Landau damping closure or flux-limited heat flux in 6-field Landau-fluid simulations. 1. C.H. Ma,
Landau, Frances . Landau, Jacob, 1917-2001 (lcnaf) Lev-Landau, Samual, 1895-1979 (lcnaf) Shahn, Ben, 1898-1969 (lcnaf) Organizations . Académie Julian (lcnaf
Landau Damping In Plasma Main ingredients of Landau damping: wave particle collisionlessinteraction. Here this is the electric field energy transfer: the wave the (few) resonant particles The result is the exponential decay of a small perturbation. Landau damping is a fundamental mechanism in plasma physics.
A Pre-Revolution Time Line Directions: Using the list in the box, fill in the events and laws that led up to the American Revolution. Write the event or law below each year. You may need to do some online research to complete this exercise. Boston Tea Party, Stamp Act Congress, Intolerable Acts, The French and Indian
