PHY 140A Solid State Physics Solution to Homework 8 Xun Jia1 December 11 2006 1 Email jiaxun physics ucla edu Fall 2006 c Xun Jia December 11 2006 Physics 140A Problem 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 2 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 Solution k 1 a k k 2 y k x Figure 1 Reciprocal space of 2D simple square lattice a As in Fig 1 the reciprocal lattice of a simple square lattice in 2D is also a simple square lattice with spacing 2 a The 1st Brillouin zone is shown as in the shaded area Then at the corner of the zone k1 ex a ey a while at the midpoint of a side k2 ex a Therefore the energies corresponding to those two points are h 2 2 h 2 k1 2 2 2m 2m a2 h 2 k2 2 h 2 2 E2 2m 2m a2 E1 1 1 Fall 2006 c Xun Jia December 11 2006 Physics 140A obviously E1 2E2 b Similarly in 3D case the reciprocal lattice is simple cubic with spacing 2 a and at the corner of the first zone k1 ex a ey a ez a while at the midpoint of a side k2 ex a Therefore h 2 k1 2 h 2 2 3 2m 2m a2 h 2 k2 2 h 2 2 E2 2m 2m a2 E1 2 and hence E1 3E2 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 Problem 2 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 Plot roughly in the 111 direction the energies of all bands up to six times the lowest band energy at the zone boundary at k 2 a 12 21 12 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 v1 v2 v3 and w where the vi 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 Solution For fcc lattice with lattice a constant a the reciprocal lattice is bcc with lattice constant 4 a Along the 111 direction in the reduced zone scheme k w 1 1 1 a with w 1 1 From the general expression h 2 k G 2 2m h 2 kx Gx 2 ky Gy 2 kz Gz 2 2m k 2 3 Fall 2006 c Xun Jia December 11 2006 Physics 140A 6 5 2 0 0 2 0 0 4 1 1 1 1 1 1 3 2 1 1 1 1 1 1 1 0 0 0 0 1 2 0 8 0 4 0 0 0 4 0 8 1 2 w Figure 2 Band structure of fcc lattice in first zone along 111 direction in the lowest band G 0 then along 111 direction h 2 h 2 2 2 2 k k 3w 4 2m 2m a 2 h 2 at zone boundary w 1 we have 0 3 which gives the unit of energy 2m a in the plot as stated in the problem In general for G 2 v1 v2 v3 a with v1 v2 v3 integers for allowed bcc lattice points in reciprocal space we have h 2 k G 2 2m 2 2 2 h 2 2 2 2 w v1 w v2 w v3 2m a a a a a a k 5 in the unit of 0 this could be simplified as 1 k w 2v1 2 w 2v2 2 w 2v3 2 3 6 taking all possible combinations of integers v1 v2 and v3 such that G 2 v1 v2 v3 a forms a bcc structure remember in HW 1 we know this is equivalent to that the sum of vi are either all even or all odd we get all band structures as in Fig 2 The corresponding choices of vi are labelled next to the curves 3 Fall 2006 c Xun Jia December 11 2006 Physics 140A Remark In the solution above we take the reciprocal lattice vector G in the form G 2 2 2 v1 ex v2 ey v3 ez a a a 7 where integers vi 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 vi for first several energy bands are 000 111 1 1 1 111 1 1 1 200 and 2 00 An equivalent way of doing this problem is to consider G in the form of G v1 b1 v2 b2 v3 b3 2 2 2 v1 ex ey ez v2 ex ey ez v3 ex ey ez a a a 8 with bi the standard choices of the reciprocal vectors as in Kittel and the choices of vi are then all possible combinations of integers Now following a same calculation h 2 k G 2 2m 1 w 2 v1 v2 v3 2 w 2 v1 v2 v3 2 3 w 2 v1 v2 v3 2 k 9 This time corresponding to the lowest several bands since all combinations of vi are allowed now the choices are 000 111 1 1 1 100 1 00 110 and 1 1 0 A same figure as Fig 2 will then follow It is easy to verify the equivalence between above two ways of doing this problem Problem 3 Kronig Penny model a For the delta function potential and with P 1 find at k 0 the energy of the lowest energy band b For the same problem find the band gap at k a Solution a For the delta …