Quantum Physics III (8.06) Spring 2006 Solution Set 4

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1 Quantum Physics III 8.6 Spring 6 Solution Set 4 March 3, 6 1. Landau Levels: numerics 4 points When B = 1 tesla, then the energy spacing ω L is given by ω L = ceb mc = evcm3 1 5 ev/cm 511 kev = ev. In natural units the length l is 1 l = = c mω L mc = ω L c Notice that this length is the inverse of the geometric average of the two energy scales of the problem: the rest energy of the electron, and the cyclotron energy ω L. Finally, in cm, l = cm.. Transformation between basis vectors of different gauges 1 points a points Consider the difference A A = By, Bx, = x f, y f,. ev. The solution is easy to guess fx, y = Bxy. b 5 points Consider x in y basis. x = + To find x y we consider the following equalities dy x y y 1 x Y y = y x y = il x x y 3 where in the first equality we act Y to the right and in the second equality we act Y to the left. The right hand sides of equations and 3 give rise to a differential equation for x y which can be solved by x y = Ce i x l y with C a constant of integration. Setting 1 C = 1 we find from 1 πl 1 + x = πl 1 The normalization constant C can be fixed by requiring x x = δx x. e i x l y y. 4 1

2 Alternatively, one can solve this problem by comparing the commutation relation [X, Y ] = il to the canonical commutator of momentum p and coordinate x, [p, x] = i. We know from 8.5 that [p, x] = i implies p = i d/dx, which allows to translate all results of one particle QM in d=1 to the Landau level problem in d=. This is achieved by replacing p X, x Y, l, etc. After this identification, the relation we are seeking to establish in this problem is nothing but the familiar p = x e ipx/ x. c 5 points Consider the wavefunction of lowest energy eigenstate in the unprimed system: ψ x, y; y = 1 πl e iy x l 1 e y y l 4 = l e iy x l πl 1 4 Recall the formula for the delta function dk e iky y l k 5 π e ikx x dk = πδx x 6 Now we plug 5 into the right hand side of equation 1 of psets 4, exchange the order of integration and use 6 to compute the integrals: ixy ix dy y l RHS = e l l e e iy x l πl πl 1 4 = 1 ixy l e πl 1 4 = 1 ixy l e πl 1 4 = 1 e πl 1 4 = LHS. ix y l x x l e iky l k dk 1 π x x δ l k dk e iky y l π dy e iy e iky l k dk k x x l k 3. Landau levels in symmetric gauge 1 points a 3 points Given that the computation for the commutator [a, a ] is [v x, v y ] = i ω L m. 7 [a, a ] = m [v x + iv y, v x iv y ] ω L = i m [v x, v y ] ω L = i m i ω L ω L m = 1.

3 Now we check that H = ω L a a + 1 is indeed the correct Hamiltonian: ω L a a + 1 m = ω L v ω x + vy + i[v x, v y ] + 1 L where we have used again 7. b 4 points Since p = i x Also we observe that = m v we can write the the complex derivatives as z = 1 x i y = i p x ip y z = 1 x + i y = i p x + ip y 8 z = x + i y = A y ia x 9 l l l Bl Plugging 8, 9 this in the right hand side of the desired relation we get m a = v x + iv y ω L m 1 = q ω L m c A x + ia y + p x + ip y = l Ax ia y Bl + 1 p x + ip y = i z + l z l c 3 points Plugging the definition of a in the annihilation condition a ψ = we get z ψ = z 4l ψ. This can be viewed as an ordinary differential equation for ψ with z as independent variable and z as a parameter. The general solution is where fz is an arbitrary function of z. ψz, z = fze z z 4l d 4 points The basis of the ground state wave functions is ψ n = N n z n e z z 4l We compute the normalization of ψ n by first going to the polar coordinates r, φ. Then the integral over φ is easily computed. To take the integral over r we perform another change of 3

4 variables ρ = r : 1 = ψ n ψ n = N n = N n = πn n dzd zz z n e z z l π drdφr n+1 e r l dρρ n e ρ l = πn n 1 n n α n = πnn 1 n+1 n α n = πn nn!l n+1 e dρ αρ α= 1 l 1 α α= 1 l and therefore N n = πn! n+1 l n+ 1. Next we compute ψ n x + y ψ n by first noticing that x + y = z z, and second dividing by ψ n ψ n ψ n x + y ψ n = ψ n z z ψ n = ψ n z z ψ n ψ n ψ n [ n+1 = α n+1 = n + 1l. / 1 n α α n ] 1 α α= 1 l e 3 points The contour plots of the probability densities ψ n x, y for n =, 4, 1 are shown on the figures 1, and 3 respectively f 4 points The angular momentum L in complex notations is L = xp y yp x = i i z + z z z + z z z + z = z z z z. We do it for convenience, this way we don t need to keep trace of normalization constant, as it cancels out and we can use the insight we got when we computed ψ n ψ n 4

5 Figure 1: ψ x, y is simply a gaussian which has a peak at. 5

6 Figure : We see, that for n = 4 the probability to detect the particle is concentrated in the circular area around th origin 6

7 Figure 3: The case of n = 1 is qualitatively similar to the n = 4 case, the radius of the ring grows roughly linearly with n, which, in agreement with part d 7

8 Let us compute the commutator of angular momentum L with raising operator a [L, a ] = i ] z [z z z z, + l z l = i ] z [z z, [ z z, l z ] l = i z + l z l Similarly = a. [L, a] = a. The commutator of angular momentum with the Hamiltonian is Acting with L on ψ n we obtain [L, H] =[L, ω L a a + 1 ] = ω L [L, a ]a + a [L, a] =. Lψ n = N n z z z z z n e z z 4l = N n nz n e z z 4l = nψ n. 4. Counting States in the lowest Landau Level in the symmetric gauge 6 points a 3 points The probability density is P = πr ψ n dr = πnnr n+1 e r l Taking derivative with respect to r and setting it to zero we get Therefore n + 1 r max l =. r max = l n + 1 b 3 points The largest value of n max allowed by the size of the material is determined from l nmax + 1 R. Therefore n max 1 R l = πr qb = Φ. hc Φ 8

9 5. Coherent state in the symmetric gauge 1 points Note: In this problem we call the eigenstate of a as φ in order to avoid confusion with the ground state eigenfunctions ψ n defined in problem 3. a 3 points The eigenstate equation simplifies to z aφ = i φ 1 l z φ = z z 4l φ. As we did in problem 3 we view this as an ordinary differential equation for φ z. The general solution is φ z, z = fze z z z z 4l 11 where fz is an arbitrary function of z. Here and below, one gets full credit for any choice of fz. b 3 points Multiplying the wave function by it s complex conjugate we have φ = fz e z z z z l = fz e x x +y y l. c 5 points This part count as extra bonus credit. Since the Hamiltonian is time-independent, the wave function at time t is given by φt = e iht φ. In order to find φt we act with e iht on both sides of 1 z e iht ae iht e iht φ = i e iht φ l which leads to z a tφt = i φt 1 l where at is the Heisenberg operator for a Since at satisfies the equation at = e iht ae iht. we find that da dt = i [a, H] = iω L a t = e iω Lt a. 13 9

10 Plugging 13 into 1 we get aφt = i z e iω Lt φt l = i z t φt l with z t = e iω Lt z φt satisfies exactly the same equation as φ with z replaced by z t compare with equation 1. Thus from 11 φt = fz exp z z t z z t 4l. 14 d 1 point Multiplying the answer to c by its complex conjugate we obtain φt = fz e x x t +y y t l 6. Off-diagonal conductance in two dimensions 1 points a 3 points Using the usual formula for the inverse of a matrix, we write 1 ρ ρ H σ = ρ +, ρ H ρ H ρ which gives us ρ σ = ρ + ρ H ρ H σ H = ρ +. ρ H b 3 points If ρ H is non vanishing, then lim σ = lim ρ ρ ρ ρ + ρ H =. Thus we can have both σ and ρ equal to in the presence of a non vanishing B field. c 4 points In two dimensions, the current density is j = I/l, that is, current per unit length rather than current per unit area, as in a three-dimensional sample. Therefore, the equation defining the Hall resistance, V y = R H I x, is the same information as E = ρ j. Notice that E y = V y /L and j x = I x /L. Therefore: ρ H = E y j x = E yl j x L = V y I x = R H. Thus the Hall resistance of the sample does not depend on its dimensions. Alternatively, we can give a more generic treatment NOT required for full credit: E x ρ ρ H j x =, E y ρ H ρ j y 1

11 therefore, we can write, V x V y = = = W L E x E y ρ W j x ρ H W j y ρ H Lj x + ρ Lj y ρ H W ρ L ρ H ρ L W I x I y, which is the matrix equation of the form V = RI. Clearly, when ρ =, the resistance matrix is independent of L/W. 11

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