Landau levels using Landau gauge

Transcription

1 Appendix A Landau levels using Landau gauge In chapter 1 we have given the eigenfunctions and energy eigenvalues of an electron in two dimension in presence of a transverse magnetic field. There we have used symmetric gauge to solve the problem. Here we sha.ll solve the same problem but in Landau gauge. This gauge was used by Landau who originally solved this problem. This is why the quantised energy levels of the electron in this problem are known as Landau levels. In Landau gauge A= B(O,x,O) (A.1) The time independent Schrodinger equation written explicitly in this gauge will be 1-1i 8 'l/; -i1i8?[ 0 + (-~-- ebxjc) 'l/;] = E'lj; (A.) ~n"t x uy Assuming periodic boundary condition in the y-direction the eigenfunctions will be of the form (A.3) 175

2 where k - y- 1fq (A.4) Ly with q = 0, ± 1, ± and Ly is the length of the sample in y direction. Inserting this form in the equation ( A.) we get the equation of an one dimensional simple harmonic oscillatar along the x-a.xis with its origin displaced from the origin of the co-ordinate sytem (x = 0). with a.nd ) d y 'f'n,ky ~m. :r c 1_(-1id <Pn,ky + (nk _ ebx),.~.. _ ) n d <Pn.),:y rn, d:r rnwc ( k l), + X - 'y <!Jn,ky eb Wc=- 7nC (A.5) (A.6) (A. 7) Then the eigenfunction <Pn,ky can be written as (A.S) where Hn is the nth Hermite polynomial and..\- = kyf. Also Lx is the dimension of the sample in x direction. Each wavefunction thus have a Gaussian in it centred at :r = X. The difference between the centre of two such Gaussians are gi\ en b_y ' y The energy eigen values a.re given by energy levels of a simple harmonic oscillator En= (n + 1/)nwc (A.9) 176

3 and they are independent of ky. These energy levels are known as Landau levels. Using eqs.(a.4 and A.8) one can see that the number of states in a given Landau level is (A.10) where A is the area. As all these states have same energy each Landau level is highly degenerate and the degeneracy is Nd. The lowest Landau levellll corresponds to n = 0 and the LLL wavefunction ( 4.8) '1/JLLL = 1 eikyye-(x-x)/1 (A.ll) ( -Ji LxLy)t [In eq. ( 4.8) Lx is set to 1 in the unit of magnetic length ( l) and denotes the full wavefunction] 177

4 Appendix B Expression of p 3, Pd and f3m in Hatree Fock treatment and the higher order contributions from the direct Coulomb energy B.l Spin-pseudospin stiffness in Hatree-Fock treatment Here we shall explicitly show the equality of the expressions of (pseudo )spin stiffness given in eq.(.93,.94) with those derived from Hatree Fock picture in Chapter 4 ( eq.4.38 and eq. 4.39). For that we should first note that the Exchange Coulomb term as given in eq.( 4.6) after explicitly written in terms of the Landau-gauge single particle wavefunctions turns out to be 178

5 (B.l) where \/a 1 a (q) is the Fourier transform of \/a 1 a (r 1 - r ). Here his they component of the wave vector and X = k/. In getting the final form of the equation we have used the identity (B.) It is known that (B.3) llsing this integral we get from eq. (B.l) we get (B.4) Where Ly is the c.imension of the sample in the y-direction. Now the spin stifrness coefficient pu 1 a is given by the ex ;nession pal a (B.5) 179

6 Applying eq. (B.4) we obtain for it pa1a = (-1---) J 47rl 7rl d(x _X )Ea1a (X1 - X) 1 ( ~) 3 z 4 j dqxld(kl- k)1va 1 a (qxl, (k1.,-- k)l)l (kl- k) -~ _(kll-kl) t _, :_ e e (B.6) We have explicitly kept the track of ls to show that pa1a f e 0 d has the dimension Here the discrete to continuum transformation ( 4.31) is used in several places. To get the final form of pa1a we will now change the wavevectors to their polar form by introducing variables k \Jq; + (h- k)\l tan- 1 ( qx ) kl- k (B.7) (B.S) Inserting these eqs. in the expression given in (B.6) finally we get (B.9) where for same layer and (B.10) (B.ll) for different layers. 180

7 B. Coefficient of the capacitance term in Hatree-Fock tr~eatment Here -vve have explicitly establish the equality between 3 in.90 and f3m in the equation The exchange contribution to,8m is given by (B1) Using eq.(b.l) and (4.31) r.h.s can be rewritten as (B.13) In the last step we have used the dimensionless polar form of the wave vector defined through eq. (B.8). lt is also clear that the dimension of this term is The contribution to this coefficient from the direct Coulomb energy can also be derived in a similar manner. For that let us first notic ~ ll0j(j x1xx1x,.i drl dr va 10 (rl - r)d>);] (id<p~>:, (r )o\jrd<px (r) dr1 dr lia 1a, ( r1 - r [ e- ( (xj ~:i"z+ :x ~:? ; l ) J iLf y 1 J dqdrl dr vala ( q'"')eiif-u'j--i"1 ( 7r ) 1r L~P 181

8 Applying eq. (B.4) we obtain for it P a1a = (-1- -) J 4x 1 x [ d(x -X )Eala (Xl- X) 1 ( ~) 3 [ 4 j dqxld(kl- k)lva 1 a (qxl,(kl- k)l)l (k1- k) -~ _(kr!-k!)!...:... :_ e e (B.6) We ha\ e explicitly kept the track of ls to show that pa 1 a f e 0 d has the dimension Here the discrete to continuum transformation ( 4.31) is used in several places. To get the final form of pa 1 a their polar form by introducing variables we will now change the wavevectors to k I J q; + ( k1 - k) 11 tan- 1 ( qx ) kl- k (B.7) (B.S) Inserting these eqs. in the expression given in (B.6) finally we get (B.9) where (B.10) for same layer and (B.ll) for different layers. 180

9 B. Coefficient of the capacitance te~rm in Hatree-Fock treatment Here vye have explicitly e~;tablish the equation The exchange contribution to f3m is given by the equality between,b in.90 and f3m in (B.l) Using eq.(b.l) and (4.31) r.h.s can be rewritten as (B.l3) In the last step we have used the dimensionless polar form of the \vave vector defined through eq. (B.8). It is also clear that the dimension of this term is The contrib.ution to this coefficient from the direct Coulomb energy can also be derived in a simtlar manner. For that let us first notice 181

10 (B.l4) Using the relation (B.15) we finally get (B.16) Now the contribution to the coefficient of the capacitance term from the direct Coulomb energy or Hatree energy is given by fjh = _1_ " 87rl L..t X1-X (Ds- Dd) (B.17) Substituting the expression for (Ds - Dd) from eq. (B.16) and also eqs. ( 4.31 and B.8) we finally obtain fjh = 3~3[4 J dq J d(xl- X)eiqx(Xl-X)8(qy)V-(q) (B.18) In the last line again after making the k a dimensionless quantity we have absorbed the overalll 3 in the unit ( ; 1 ~). Here we have also used the identity r)o Jo kdke_k = 1 (B.l9) to write both fjh and f3ex in the same fashion. Finally adding up eqs. (B.l3) and (B.18) we get the desired equality (B.0) 18

11 B.3 Higher order contributions from the direct Coulomb energy In deriving the full C P3 energy funtional ( 4.40) in chapter 4 we have neglected the higher ( 81) order contributions coming from the direct Coulomb energy ( 4.). The reason is the follmving. From the equations (B.l6 and 4.) it can be easily checked that such a term apa.rt from a constant multip.1icative factor will look like Thus such a term i:o associated with the zero wavevector (..\ == oo) component of the Coulomb interaction and nonvanishing because of the long range nature of the Coulomb interaction. However, for a finite size ~ample this contribution can be neglected. Hence we have not included contribution from it in our final derivation of the energy functional ( 4.40) 183