PHY 140A: Solid State Physics. Solution to Homework #8

Transcription

1 HY 140A: Solid State hysics Solution to Homework #8 Xun Jia 1 December 11, jiaxun@physics.ucla.edu

2 roblem #1 Sqrare lattice, free electron energies. (a). Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first zone is higher than that of an electron at midpoint of a side face of the zone by a factor of. (b). What is corresponding factor for a simple cubic lattice (three dimensions)? (c). What bearing might the result of (b) have on the conductivity of divalent metals? k 1 /a k k y k x Figure 1: Reciprocal space of D simple square lattice. (a). As in Fig. 1, the reciprocal lattice of a simple square lattice in D is also a simple square lattice with spacing π/a. The 1st Brillouin zone is shown as in the shaded area. Then at the corner of the zone, k 1 = e x π/a + e y π/a, while at the midpoint of a side k = e x π/a. Therefore, the energies corresponding to those two points are: E 1 k 1 E k 1 π a π a (1)

3 obviously, E 1 = E. (b). Similarly, in 3D case, the reciprocal lattice is simple cubic with spacing π/a, and at the corner of the first zone k 1 = e x π/a + e y π/a + e z π/a, while at the midpoint of a side k = e x π/a. Therefore: and hence E 1 = 3E. E 1 k 1 E k 3π a π a () (c). In the divalent case, when the band gap at the midpoint of a face is smaller than the the difference between the kinetic energy at this point and that at the corner, some electrons will go into the second band, resulting in both empty states and filled states in both bands. Whenever this happens, the material would be a conductor rather than an insulator. Whether this condition is satisfied depends on the details of the material, however, since the kinetic energy difference is so large, by a factor of 3 from part (b), it is likely for this condition be satisfied in divalent materials. roblem # Free electron energies in the reduced zone scheme. Consider the free electron energy bands of an fcc crystal lattice in the approximation of an empty lattice, but in the reduced zone scheme in which all k s are transformed to lie in the first Brillouin zone. lot roughly in the [111] direction the energies of all bands up to six times the lowest band energy at the zone boundary at k = (π/a)( 1, 1, 1 ). Let this be the unit of energy. This problem shows why band edges need not necessarily be at the zone center. Several of the degeneracies (band crossings) will be removed when account is taken of the crystal potential. Write the energy as a function of v 1, v, v 3, and w, where the v i are the integer coefficients appearing in the expression for the general reciprocal lattice vector and w is a number between 0 and 1 representing the length of k. For fcc lattice with lattice a constant a, the reciprocal lattice is bcc with lattice constant 4π/a. Along the [111] direction, in the reduced zone scheme, k = πw(1, 1, 1)/a, with w [ 1, 1]. From the general expression: ɛ(k) (k + G) [(k x + G x ) + (k y + G y ) + (k z + G z ) ] (3)

4 [,0,0] [1,1,-1] [-1,-1,1] [-,0,0] 1 [1,1,1] [-1,-1,-1] 0 [0,0,0] w Figure : Band structure of fcc lattice in first zone along [111] direction. in the lowest band, G = 0, then along [111] direction: ( ɛ(k) k π ) 3w a (4) ( ) at zone boundary w = ±1, we have ɛ 0 = 3 h π, which gives the unit of energy a in the plot as stated in the problem. In general, for G = π(v 1, v, v 3 )/a with v 1, v, v 3 integers for allowed bcc lattice points in reciprocal space, we have: ɛ(k) (k + G) [ (π a w + π a v 1 ) + ( π a w + π ) ( π a v + a w + π ) ] a v 3 (5) in the unit of ɛ 0, this could be simplified as: ɛ(k) = 1 3 [(w + v 1) + (w + v ) + (w + v 3 ) ] (6) taking all possible combinations of integers v 1, v and v 3, such that G = π(v 1, v, v 3 )/a forms a bcc structure (remember in HW#1, we know this is equivalent to that the sum of v i are either all even, or all odd,) we get all band structures as in Fig.. The corresponding choices of v i are labelled next to the curves. 3

5 Remark: In the solution above, we take the reciprocal lattice vector G in the form: G = π a v 1e x + π a v e y + π a v 3e z (7) where integers v i are chosen such that the sum of them is either odd or even, to ensure that the reciprocal space lattice is bcc, as it should be for a fcc lattice in real space. After calculation, we end up with the expression of energy in the form of Eqn. (6), and the choices of v i for first several energy bands are [000], [111], [ 1 1 1], [11 1], [ 1 11], [00] and [ 00]. An equivalent way of doing this problem is to consider G in the form of: G = v 1 b 1 + v b + v 3 b 3 = π a v 1[ e x + e y + e z ] + π a v [e x e y + e z ] + π a v 3[e x + e y e z ] (8) with b i the standard choices of the reciprocal vectors as in Kittel, and the choices of v i are then all possible combinations of integers. Now following a same calculation: ɛ(k) (k + G) = 1 3 [(w + ( v 1 + v + v 3 )) + (w + (v 1 v + v 3 )) + (w + (v 1 + v v 3 )) ] (9) This time, corresponding to the lowest several bands, since all combinations of v i are allowed now, the choices are [000], [111], [ 1 1 1], [100], [ 100], [110], and [ 1 10]. A same figure as Fig. will then follow. It is easy to verify the equivalence between above two ways of doing this problem. roblem #3 Kronig-enny model. (a). For the delta-function potential and with 1, find at k = 0 the energy of the lowest energy band. (b). For the same problem find the band gap at k = π/a. (a). For the delta-function potential Kronig-enny model, we derived the equation relating K and k in Kittel: sin Ka + cos Ka = cos ka (10) Ka 4

6 When k = 0, cos ka = 1. For the lowest energy band, we are looking for the K value closest to zero solving above equation. Since when = 0, the solution is obviously K = 0, it is reasonable to expect that when 1, K will be very close to zero, namely Ka 1. Expand Eqn. 10 for small Ka, we have: Ka [Ka + o(ka)3 ] + [1 1 (Ka) + o(ka) 4 ] = 1 (11) we then have, to the leading order: therefore the energy is: K a (1) ɛ K h ma (13) (b). When k = π/a, cos ka = 1. Similarly, when = 0, K = π/a will solve Eqn. 10 exactly, then we expect when 1, the solutions will be close to π/a. Expand Eqn. 10 around K = π/a, we have: 1 = sin[π + (Ka π)] + cos[π + (Ka π)] π + (Ka π) = π [1 1 π (Ka π) + ][ (Ka π) ] + [ (Ka π) + ] = 1 ( π (Ka π) + π + 1 ) (Ka π) + (14) so to the leading order, the two solutions K: K 1 π/a K π a + therefore, the energy gap at k = π is: roblem #4 a( π + π ) π a + πa (15) = ɛ(k ) ɛ(k 1 ) [K K1] [ (π a + ) ] ( π ) (16) h πa a ma Square lattice with a weak periodic potential. Consider a square lattice in two dimensions with crystal potential: U(x, y) = 4U cos(πx/a) cos(πy/a) 5

7 Apply the central equation to find approximately the energy gap at the corner point (π/a, π/a) of the Brillouin zone. It will suffice to solve a determinantal equation The potential U(x, y) is periodic in both x and y direction with period a, and the fourier transform of the potential with G = (π/a, π/a) is: U G = 1 a a/ dx a/ dyu(x, y)e i( π a x+ π a y) a/ = 4U a = U a/ a/ a/ dx a/ a/ dy cos( π a x) cos(π a y)ei( π a x+ π a y) (17) this is also the fourier component of G (You can check the calculation exactly, indeed, this is so since the potential U(x, y) is real). Then at k = (π/a, π/a) = G/, from the central equation: ( ) 1 (λ k ɛ)c G UC ( 1 ) G = 0 ( ) 1 UC G + (λ k ɛ)c ( 1 ) (18) G = 0 where λ k = h k /. To have nontrivial solutions of C ( 1 G) and C ( 1G), it must be: ( ) λk ɛ U det = 0 (19) U λ k ɛ which gives the energy: therefore, the energy gap is: roblem #5 Study for the final. Good luck! ɛ 1, = λ k ± U (0) = ɛ 1 ɛ = U (1) 6