PHY 140A: Solid State Physics. Solution to Homework #2

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1 PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, Emil:

2 Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice. (). Prove tht the reciprocl lttice primitive vectors (eqn.13 in Kittel) stisfy (π) 3 b 1 (b b 3 ) =. (Hint: Write b 1 (but not b or b 3 ) in terms of the i, nd use the orthogonlity reltions.14.) (b). Suppose primitive vectors c i re constructed from the b i in the sme mnner s the b i re constructed from the i. Prove tht these vectors re just the i themselves. (c). Prove tht the volume of Brvis lttice primitive cell is V =, where the i re the three primitive vectors. (In conjunction with the result of prt (), this estblishes tht the volume of the reciprocl lttice primitive cell is (π) 3 /V.) Two useful identities regrding vector multipliction re: (b c) = b (c ) = c ( b) (b c) = ( c)b ( b)c where, b, nd c re three vectors in three dimensionl spce. (1) (). write b 1 in terms of the i, nd use the orthogonlity reltions.14, we hve: b 1 (b b 3 ) = (π)( 3 ) (b b 3 ) = (π) [ 3 (b b 3 )] = (π) b = (π)3 ( 3 1 ) [] = = (π)(b b 3 ) ( 3 ) = (π) [( 3 b 3 )b ( 3 b )b 3 ] (π) 3 (b). Suppose primitive vectors c i re constructed from the b i in the sme mnner s the b i re constructed from the i, then c 1 stisfies: c 1 = (π)b b 3 b 1 (b b 3 ) = = ( 3 1 ) ( 1 ) = [( 3 1 ) ] 1 (π) 3 ( 3 1 ) ( 1 ) [] (π) 3 = [( 3 1 ) ] 1 [( 3 1 ) 1 ] = [( 3 ) 1 ] 1 1 = 1 () (3)

3 here we used the fct tht ( 3 1 ) 1 = ( 1 1 ) 3 = 0 due to eqution (1). Similrly, we cn proof tht: c = (π)b 3 b 1 b 1 (b b 3 ) = (4) c 3 = (π)b 1 b b 1 (b b 3 ) = 3 (5) Therefore, the reciprocl lttice for the reciprocl lttice is the originl lttice. Figure 1: A generl prllelepiped with edges of, b, nd c, where S = b. (c). As in Fig. 1, consider generl prllelepiped with edges of, b, nd c. The re of its bse prllelogrm spnned by vectors nd b is A = b sin θ = b, where θ is the ngle between nd b. Note tht S = b is perpendiculr to this bse, the height corresponding to this bse is thus h = c S S, which is just the projection of c onto the direction of S. therefore the volume of this prllelepiped is: V = Ah = c b b = c ( b) (6) b Thus, the volume of Brvis lttice primitive cell spnned by primitive vectors i is V =. Moreover, from prt (), the volume of the reciprocl lttice primitive cell is then (π) 3 /V. Problem # Interplnr seprteion.(problem.1 in Kittel.) Consider plne hkl in crystl lttice.

4 (). Prove tht the reciprocl lttice vector G = hb 1 + kb + lb 3 is perpendiculr to this plne. (b). Prove tht the distnce between two djcent prllel plnes of the lttice is d(hkl) = π/ G. (c). Show for simple cubic lttice tht d = /(h + k + l ). (). To prove tht the reciprocl lttice vector G = hb 1 + kb + lb 3 is perpendiculr to this plne, it suffices to show tht G is perpendiculr to two nonprllel vectors in this plne. For the plne (hkl), it intercepts xis 1,, nd 3 t rtio 1 h : 1 k : 1 l, thus the two vectors in the plne cn be chosen s ( 1 h 1 1 k ) nd ( 1 h 1 1 l 3). Obviously, they re not prllel to ech other. From direct clcultion: G ( 1 h 1 1 k ) = (hb 1 + kb + lb 3 ) ( 1 h 1 1 k ) = 0 (7) G ( 1 h 1 1 l 3) = (hb 1 + kb + lb 3 ) ( 1 h 1 1 l 3) = 0 (8) we know G is perpendiculr to those two vectors, nd hence perpendiculr to the plne. (b). Among the indices hkl, t lest one of them is nonzero, without lose generlity, h 0. Since, by definition, this fmily of plnes divide vector 1 into h prts with equl length, then the vector R 1 = 0 ends on one plne, nd R = 1 h 1 ends on n djcent plne. Therefore, the projection of R R 1 = 1 h 1 on to the direction perpendiculr to this fmily of plnes will be the distnce between two neighboring plnes. From prt (), G is perpendiculr to these plnes, hence: 1 d(hkl) = h G 1 G = 1 h 1 hb 1 + kb + lb 3 G = π (9) G (c). For simple cubic, the reciprocl lttice hs: b 1 = π x, b = π y, b 3 = π z (10) where x, y, nd z re unit vectors long x, y, nd z directions. Thus, nd hence: G = (hb 1 + kb + lb 3 ) = (π) (h + k + l ) (11) d(hkl) = (π) G = h + k + l (1) 3

5 Problem #3 (). Show tht the density of lttice points (per unit re) in lttice plne is d/v, where v is the primitive cell volume nd d the spcing between neighboring plnes in the fmily to which the given plne belongs. (b). Prove tht the lttice plnes with the gretest density of points re the {111} plnes in fce-centered cubic Brvis lttice, nd the {110} plnes in bodycentered cubic Brvis lttice. (This is most esily done by exploiting the reltion between fmilies of lttice plnes nd reciprocl lttice vectors.) (). Within ech plne, the lttice is -d lttice. Let the re of the primitive cell for this -d lttice be A, then since v is volume of primitive cell for 3-d lttice, it follows tht A d = v. Since every primitive cell contins exctly one lttice points, the number density of lttice points on this plne ρ is: ρ = 1/A = d/v (13) (b). Once the type of lttice is given, the volume of primitive cell v is determined. Thus from prt (), the lrger the interplne distnce d is, the greter the density of points ρ is. From Eqn. (9), the lrgest d corresponds to the fmily of plnes for which the length of G, reciprocl lttice vector perpendiculr to them, is minimized. Thus, if we could find out G 0 in reciprocl spce with lest length, then the rel spce lttice plnes perpendiculr to this G 0 will hve lrgest density of lttice points. For fcc, choose: 1 = (y + z), = (z + x), 3 = (x + y) (14) then the reciprocl lttice hs: b 1 = π (y + z x), b = π (z + x y), b 3 = π (x + y z) (15) Above equtions indicte tht the reciprocl lttice of fcc is bcc. Since the shortest vector in bcc is the one from the corner of the cube to the body center, without lose generlity, tke: G 0 = b 1 = π (y + z x) (16) then in the rel spce lttice, the lttice plnes perpendiculr to this G 0 belong to fmily of {111}, i.e. the lttice plnes with the gretest density of points re the {111} plnes in fce-centered cubic Brvis lttice. 4

6 For bcc, choose: 1 = (y + z x), = (z + x y), 3 = (x + y z) (17) then the reciprocl lttice hs: b 1 = π (y + z), b = π (z + x), b 3 = π (x + y) (18) Above equtions indicte tht the reciprocl lttice of bcc is fcc. Since the shortest vector in fcc is the one from the corner of the cube to the fce center, without lose generlity, tke: G 0 = b 1 = π (y + z) (19) then in the rel spce lttice, the lttice plnes perpendiculr to this G 0 belong to fmily of {110}, i.e. the lttice plnes with the gretest density of points re the {110} plnes in body-centered cubic Brvis lttice. Remrk: You cn proof these in mthemticl wy, sy, write explicitly the expression of G nd minimize it. Problem #4 Hexgonl spce lttice.(problem. in Kittel.) The primitive trnsltion vectors of the hexgonl spce lttice my be tken s: 1 = x + y, = x + y, 3 = cz (0) (). Show tht the volume of the primitive cell is c/. (b). Show tht the primitive trnsltions of the reciprocl lttice re: b 1 = π x + π y, b = π x + π y, b 3 = π c z (1) so tht the lttice is its own reciprocl, but with rottion of xes. (c). Describe nd sketch the first Brillouin zone of the hexgonl spce lttice. (). By direct clcultion, the volume of the primitive cell is: V = = ( x + y) [( x + y) (π c z)] = c () 5

7 (b). From Eqn. (.13) in Kittel, we cn obtin tht: b 1 = (π) 3 = π x + π y b = (π) 3 1 = π x + π y (3) b 3 = (π) 1 = π c z 3 nd b 3 re both long z xis; in the xy plne, the lttice nd the reciprocl lttice re s in Fig., obviously, the reciprocl lttice is lso hexgonl lttice, but with rottion bout z xis by n ngle of 30. () (b) Figure : () xy plne of hexgonl lttice. (b)xy plne of the reciprocl lttice for hexgonl lttice. Shded re indictes the Brillouin zone. (c). The first Brillion zone of the hexgonl lttice is lso hexgonl structure. The cross section of the Brillouin zone in xy plne is illustrted by the shded re in Fig. (b). Sweep this shded re long z xis from z = π c to z = π c, we get the whole first Brillion zone. 6

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