Solution manual - Chapter 12 Jens Zamanian March 27, 2014 Chapter 12 12.3 When the energy is in the form Ek = E0 + ~2 (k 2 1 k0 ) M (k k0 ) , (1) where M is independent of k we get v(k) = 1 ~2 1 ⇥ rk E k = M ~ 2 ~ 1 (k k0 ) + (k k0 )M 1 ⇤ . (2) However, Mij1 = @ 2 ✏/(@ki @kj ) is symmetric Mij1 = Mji1 and we hence have v(k) = ~M The time derivative of this is then 1 (k k0 ). (3) ˙ ˙ Mv(k) = ~k, (4) where we have also multiplied by M. Finally we use the semiclassical equation of motion ~k˙ = eE and we get Mv˙ = eE. (5) In order to derive the conductivity we assume the simplest model for collisions which is done by adding the term Mv/⌧ , where ⌧ is the relaxation time. We then get Mv˙ = eE Mv . ⌧ (6) If we apply an electric field and wait a bit until the system is in a steady state we will have v˙ = 0 and we get that the average velocity will then satisfy the equation above with the left hand side set to zero. Solving for the average velocity we then get v= ⌧ eM 1 E. (7) This velocity give rise to a current j= Comparing this with Ohms law j = env = ⌧ e2 M 1 E. (8) E, we see that = n⌧ e2 M 1 . Note that this is the same result as in the Drude model with the change 1/m ! M 1 (9) 1 . 12.4 a) If we have several partially filled bands, each band will have a contribution ✓ ◆ ⇢n Rn H ⇢n = . Rn H ⇢n (10) For an applied electric field E the current in each band will then be jn = ⇢ n 1 E (11) and we hence have the total current j= X n jn = X X ⇢n 1 E = n ⇢n 1 n ! E. (12) Inverting the matrix within the parenthesis we get X E = ⇢j = ⇢n 1 n and the total resistivity hence is ⇢= X ⇢n n 1 ! ! 1 j. (13) 1 . b) The inverse of the resistivity tensor for each band is given by ✓ ◆ 1 ⇢n Rn H ⇢n 1 = 2 . Rn H ⇢n ⇢n + Rn2 H 2 For two bands we hence have the inverse of the total resistivity tensor given by ✓ ◆ ✓ ◆ 1 1 ⇢1 R1 H ⇢2 R2 H 1 ⇢ = 2 + 2 . R1 H ⇢1 R2 H ⇢2 ⇢1 + R12 H 2 ⇢2 + R22 H 2 (14) (15) (16) Inverting this (which is most easily done with a computer, or some smart applications of formulas for matrices) yields ⇢11 = ⇢22 ⇢12 = ⇢21 = = ⇢1 ⇢2 (⇢1 + ⇢2 ) + (⇢1 R22 + ⇢2 R12 )H 2 (⇢1 + ⇢2 )2 + (R1 + R2 )2 H 2 ⇢1 R22 + ⇢2 R12 + R1 R2 (R1 + R2 )H 2 H. (⇢1 + ⇢2 )2 + (R1 + R2 )2 H 2 (17) (18) Comparing this with the general form (10) gives the desired results that ⇢ = R = ⇢1 ⇢2 (⇢1 + ⇢2 ) + (⇢1 R22 + ⇢2 R12 )H 2 (⇢1 + ⇢2 )2 + (R1 + R2 )2 H 2 ⇢1 R22 + ⇢2 R12 + R1 R2 (R1 + R2 )H 2 . (⇢1 + ⇢2 )2 + (R1 + R2 )2 H 2 (19) (20) c) In the high field limit H ! 1 the above result yields (keep only the terms proportional to H in the numerator and denominator) R1 = R1 R2 . R 1 + R2 2 (21) This should be compared with 1 . ne↵ ec R1 = (22) In the limit of ne↵ ! 0 we see that R1 ! 1 and we must have R1 + R2 ! 0, which is possible if one of the bands consists of electrons and the other of holes (so that R1 = R2 ), i.e. the two bands are compensating. In this case we have for the resistivity ⇢1 ⇢2 (⇢1 + ⇢2 ) + (⇢1 R22 + ⇢2 R12 )H 2 ⇢1 ⇢2 (⇢1 + ⇢2 ) + (⇢1 R22 + ⇢2 R12 )H 2 (⇢1 R22 + ⇢2 R12 )H 2 ! ! 2 2 2 2 (⇢1 + ⇢2 ) + (R1 + R2 ) H (⇢1 + ⇢2 ) (⇢1 + ⇢2 )2 (23) and we have a resistivity which depends on the magnetic field (and which goes to infinity in the strong field limit). ⇢= Bloch Oscilations i) Assume we have a one dimensional band with energies Ek = E0 cos(ak). (24) From the semi-classical theory we then have the evolution equations 1 dEk ~ dk ~k˙ = eE, r˙ = v(k) = (25) (26) where E is an applied electric field. Solving the second equation gives k(t) = k(0) eEt , ~ (27) where we will set k(0) = 0. Furthermore, from (24) and (26) we have ✓ ◆ aE0 aE0 eEt r˙ = sin(ak) = sin , ~ ~ ~ sin(x). Integrating this we get ◆ ✓ aE0 eEt r(t) = r0 + cos . eE ~ (28) where we have used that sin( x) = (29) By setting r0 = 0 we get the required equation. ii) If we now introduce damping in the evolution equation for the crystal momentum we get r˙ = ~k˙ = 1 dEk = v(k) ~ dk mv(k) eE . ⌧ (30) (31) Inserting the velocity v(k) = (aE0 /~) sin(ak) into these equations we get r˙ = ~k˙ = aE0 sin(ak) ~ maE0 sin(ak). eE ⌧~ (32) (33) Now we define the dimensionless variables x= r , a y = ak, and 3 s= t , ⌧ (34) and the equations then obtain the form dx ds dy ds = = E0 ⌧ sin y ~ aeE⌧ ma2 E0 sin y. ~ ~2 (35) (36) Inserting the values given in the problem (and working with three significant figures) we get dx ds dy ds = 15.2 sin y = 0.304 (37) 0.525 sin y. (38) The numerical solutions of this is shown in Fig. 1 where x and u= v(k) E0 ⌧ = sin(ak) = 15.2 sin(y), a/⌧ ~ (39) are plotted as functions of s. It is seen that the e↵ect of the damping is to introduce a steady state Figure 1: The position x = r/a (red) and u = v/(a/⌧ ) (green) as functions of the time s = t/⌧ . The electron approaches a linear motion with constant velocity in the direction opposite to the applied field. velocity u = 8.7 under the influence of the accelerating electric field and the collisional damping. An idea for an analytic solution is to start with the observation that the equation for the crystal momentum is a separable di↵erential equation and write  dk maE0 ~dk = eE + sin(ak) dt ) = dt. (40) ⌧~ eE + (maE0 /(⌧ ~) sin(ak) Now integrating both sides of the equation yields an algebraic equation for k(t) which may in principle be solved. Let me know if you can do this! 4