Problem Set 9 - Solutions

Transcription

1 Problem Set 9 - Solutions PHY 576, Due Fridy, April 1, 013 t 3pm April 14, Squre lttice, free electron energies Show for simple squre lttice in two dimensions tht the kinetic energy of free electron t corner of the first zone is higher thn tht of n electron t the midpoint of side fce of the zone by fctor of. Let s do this step by step. The term first zone must be referring to the first Brillouin zone in k-spce. The reciprocl lttice of squre spce lttice is lso squre lttice. For rel spce i.e., crystl lttice of side, the reciprocl lttice points re seprted by steps of / in the k x nd k y directions. The Wigner-Seitz cells in reciprocl spce re squres lso. These hve boundries tht bisect the segments connecting reciprocl [ lttice points. The first Brillouin zone is therefore the squre, ] [, ]. A corner of the zone is k x, k y = ±, ±. A free electron with one of these four wvevectors hs kinetic energy E corner = h mk = h m. The fces of the wve zone hve one of k x, k y with mgnitude nd the other component zero, giving E fce = h m / = E corner /. b Wht is the corresponding fctor for simple cubic lttice in three dimensions? In three dimensions, the reciprocl lttice is cube nd the Wigner- Seitz cells of reciprocl spce re lso cubes, gin with sides /. The corners of the first Brillouin zone FBZ [, ] [, ] [, ] re k x, k y, k y = ±, ±, ± nd so electrons with pseudomomentum equls rel momentum for the free electron hve energy E corner = h m k x + ky + kz = 3 h m. At the fce of the FBZ, one of the k x,y,z hs mgnitude / while the other two re zero. This gives E fce = E corner /3 in d = 3. c Wht bering might the result of b hve on the conductivity of divlent metls? The result of b is tht for the cse of free electrons, electrons with 1

2 pseudomomentum t the fce of the FBZ cn hve n energy of 1/3 of tht of those in the corner. For divlent metl, the number of electrons is N, where N is the number of ions. This is the sme s the number of sttes in the first bnd. If the sttes in this bnd hd lower energies thn ny sttes in the second or third or higher bnds, then the simple cubic lttice of divlent ions would be n insultor. This is unlikely unless the lttice potentil Ur is extremely strong, though, s when U = 0, the lowest sttes in the second bnd hve the energy of sttes t the fce of the FBZ: the free second bnd sttes hve kinetic energy 1/3 of tht of the highest first bnd sttes. You hve better chnce of hving n insultor if your lttice is fcc, so tht the FBZ hs bcc shpe, which is more sphericl, i.e., closer to the filling of free electron sttes.. Free electron energies in reduced zone Consider the free electron energy bnds of n fcc crystl lttice in the pproximtion of n empty lttice i.e. potentil is ctully zero, but in the reduced zone scheme in which ll k s re trnsformed to lie in the first Brillouin zone for the fcc lttice. Plot roughly in the [111] direction the energies of ll bnds up to six times the lowest bnd energy t the zone boundry t k = / 1, 1, 1. Let this energy t the zone boundry be the unit of energy for your plot. This problem shows why bnd edges need not necessrily be t the zone center. Severl of the degenercies bnd crossings will be removed when ccount is tken of the crystl potentil. First note tht for fcc crystl lttice tht the reciprocl lttice is bcc nd the first Brilloiun zone FBZ is truncted octhedron. By your texts or stndrd construction, primitive lttice vectors in reciprocl spce re of the form G = m 1 b1 + m b + m 3 b3 for integer m i nd b1 = ˆX + Ŷ + Ẑ 1 b = + ˆX Ŷ + Ẑ b3 = + ˆX + Ŷ Ẑ 3 where ˆX, Ŷ, Ẑ re the dul vectors in reciprocl spce with ˆx ˆX = 1, ŷ ˆX = 0, etc. The [111] direction is given s the set of pseudomomentum vectors k whose energy we re to compute. The energy of ech bnd for free electrons is given by E G, k = h m k + G. 4 We re to compute this energy for ech G of smll enough mgnitude nd for k = u ˆX + Ŷ + Ẑ, 5

3 which gives vectors in reciprocl spce long the [111] direction with prefctor u chosen so tht k lies within the first Brillouin zone. By either drwing it out in 3D or by noting tht b 1 + b + b 3 = ˆX + Ŷ + Ẑ, so tht there is unit cell in reciprocl spce centered t u = in Eqn. 5, we cn see tht the rnge 1 u 1 gives k tht rech fces of the FBZ for Eqn. 5. Using G = 0 nd u = 1 gives the energy E 1 = 3 h m 6 for the ends of the [111] vector in the FBZ. Inserting the expression Eqn. 5 for ku into the free electron energy Eqn. 4 nd plotting vs. Eu, G/E 1 vs. u, i.e., Eu, G E 1 = [ u m1 + m + m 3 + u m 1 + m + m 3 + u m 1 + m + m 3 ] = 1 3 using the gnuplot file [ u m1 + m + m 3 + u m 1 + m + m 3 + u m 1 + m + m 3 ] 7 set yrnge [0:6] set xrnge [-1:1] fu,,b,c=u-*-+b+c**+u-*-b+c**+u-*+b-c**/3 plot fx,0,0,0,fx,1,0,0,fx,-1,0,0,fx,1,1,0, \ fx,-1,-1,0,fx,1,1,1,fx,-1,-1,-1 gives 6 5 fx,0,0,0 fx,1,0,0 fx,-1,0,0 fx,1,1,0 fx,-1,-1,0 fx,1,1,1 fx,-1,-1,

4 To see in prt how this figure comes bout, tke look t two primitive cells in reciprocl spce. The lower truncted octhedron is the FBZ nd the red line in the first FBZ is in the [111] direction. This line strts nd termintes on the center of hexgonl fces. The upper cell is the result of trnsltion by b 1 brely visible s fint line nd the blck line in its interior is k + G = k + b 1. Note tht the minimum distnce from the origin to this interior line in the second cell is not t either end or the center, s seen by the plot of the normlized k nd k + b1 included: Here is Mple code for mking this figure: reset: withplots: withgeom3d: defo := [45, 45]: r:= sqrt5: points1, -1, -1, -1: points, 1, 1, 1: dsegmentst, [s1, s]: pointo, 0, 0, 0: pointc1, -,, : pointc,, -, : pointc3,,, -: TrunctedOcthedronfbz, o, r: dsegmentg1, [o, c1]: dsegmentg, [o, c]: dsegmentg3, [o, c3]: trnsltionto1, fbz, g1: trnsltionto, fbz, g: trnsltionto3, fbz, g3: trnsltionst1, st, g1: trnsltionst, st, g: trnsltionst3, st, g3: qleft := drw[fbztrnsprency =.5, to1trnsprency =.5, stcolor = red, st1color = blck, g1], orienttion = defo: f0 := x -> x^: f1 := x -> 1/3*x-^+/3*x+^: qrite := plot[f0x, f1x], x = , y = 0.. 6: A := Arry1.. : A[1] := qleft: A[] := qrite: displya: 4

5 3. Kronig-Penney model The Kronig-Penney model is discussed in Chpter 7 of Kittel, nd we discussed it briefly in clss. The potentil is composed of brriers of width b nd height U 0 nd wells of width nd zero height. Assume the first brrier for positive x is between nd + b, nd tht the energy is E < U 0. Beginning with Schrodinger s eqution, prove tht inside the wells ψ 1 x = Ae iαx + Be iαx nd inside the brriers it is ψ x = Ce βx + De βx. Wht re α nd β? Well, Schrodinger s eqn. in one dimension is For regions of constnt U, h m ψ x + Uxψx = Eψx. h m ψ x = E Uψx. Solutions in generl for constnt U re just by ordinry differentil equtions, knowing tht there re two independent solutions nd checking tht exponentils work ψx = expx where h = E U. m For E U > 0, is imginry. For E U < 0, is rel. In ech cse, there re two roots for, since there is n in the eqution. So for the regions where U = 0, we hve ψx = Ae iαx + Be iαx, with rel α = iα is imginry, α = me/ h, while for the region where U = U 0 > E, ψx = Ce βx + De βx, with β = mu 0 E/ h. In the forbidden regions, U = U 0 > E, the wve function is exponentil sy, decying from either side of the brrier while in the llowed regions ψ is oscilltory. b Write down four equtions for the four coefficients A, B, C, D, using continuity of ψ nd its derivtives t x = 0 nd periodicity of ψ nd its derivtives t x = b. Find the mtrix Q such tht Q [A, B, C, D] = 0. We will use the Bloch theorem, so tht ψx + b + = e ikb+ ψx for some k. We will tke ψx = Ae iαx + Be iαx 0 x I 8 ψx = Ce βx + De βx b x 0 II 9 5

6 Continuity t x = 0 gives or A + B = C + D A + B C D = 0. The slope t x = 0 from the [0, ] intervl is iαa B nd from the [ b, 0] intervl is βc βd, so setting these equl gives iαa iαb βc + βd = 0. It gets more complicted mtching t x =, but it is not too bd. Using form I, ψ = Ae iα + Be iα, while using Bloch nd II gives so tht ψ = e ikb+ ψ b = e ikb+ Ce βb + De βb e iα A + e iα B e ikb++βb C e ikb+ βb D = 0. A similr substitution for the derivtive gives iαe iα A iαe iα B βbe ikb++βb C + βbe ikb+ βb D = 0. Rewriting this in mtrix form gives A 0 Q B C = 0 0 D 0 where Q = iα iα β β e iα e iα e ikb++βb e ikb+ βb iαe iα iαe iα βe ikb++βb βe ikb+ βb. c By setting the determinnt of Q = 0 nd tking the limit of the delt function potentil, one cn show tht P/αsinα + cosα = cosk You don t hve to show this. For the delt-function potentil nd with P << 1, find t k = 0 the energy of the lowest energy bnd. For P 1, P α sinα 1, since sinα/α < 1. For k = 0, 6

7 cosk = 1 nd so cosα = 1 P sinα α 1, so α 1 nd sinα/α 1, with 1 cosα α P giving P α nd since E is the kinetic energy where U = 0 E = h α m h P m. d For the sme problem find the bnd gp t k = /. Now we hve to solve P sinα α for α / = k. Writing α = + δ, P sinα α + cosα = 1, + cosα = P sinδ + δ For smll δ, this gives to lowest order in P, cosδ = 1. P δ δ 1 = 1 10 P δ = δ 11 which hs two roots which re, to lowest order in P or δ, δ = 0, P. These re two solutions for α t k = /. This gives bnd gp of = E α = + P E α = [ = h m + P ] = h P m = h P m 15 7

8 4. Squre lttice Consider squre lttice in two dimensions with the crystl potentil: Ux, y = 4U cosx/cosy/. 16 Apply the centrl eqution to find pproximtely the energy gp t the corner point /, / of the Brillouin zone. In will suffice to solve determinntl eqution. As described in Problem 1 of this set nd s you know by now nywy, the squre crystl lttice hs reciprocl lttice tht is lso squre. The first Brillouin zone is squre of side /. To pply the centrl eqution, we should know the components of the Fourier series for Ux, y. These cn be found directly by integrtion of Ux, ye i k x,y or by noting tht Ux, y = 4U cosx/ cosy/ 17 = U [ cosx/] [ cosy/] 18 = U e x/ + e x/ e y/ + e y/ 19 = U e x/+y/ + e x/ y/ + e x/+y/ + e x/ y/ 0 = U e i G ++ x + e i G + x + e i G + x + e i G x 1 = G C G e i G x where G ±± = ± ˆX ± Ŷ nd C G = U for G = G ±± nd C G = 0 for ll other G. The centrl eqution cn be written s h k m E C k + U GC k G = 0. 3 G The pproximtion tht is mde here is tht only the Fourier coefficients C k for nerly degenerte sttes tht re directly connected by U G re of importnce. Solving the full centrl eqution will mix in other coefficients, but only t higher order in U. So tking the pseudomomentum k = /, /, there is only degenerte stte tht is seprted by ny vector G ±±. This is k = /, / = k G++. Then the centrl eqution sets two constrints: h k m E h k m E C k + U G ++ C k G ++ = 0 4 C k + U G C k G =

9 Noting k = k nd setting h k m = λ, C k = C nd C k = C, λ EC UC = 0 6 UC + λ EC = 0 7 nd noting tht this eqution is solvble only for zero determinnt of the corresponding mtrix, λ E U U λ E = 0, gives λ E U = 0 finlly giving the energies for sttes with pseudomomentum k = /, / by the reltion E = λ ± U so tht the bnd gp is λ + U λ U = U. 9