Kai Sun. University of Michigan, Ann Arbor

Transcription

1 Ki Sun University of Michign, Ann Arbor

2 How to see toms in solid? For conductors, we cn utilize scnning tunneling microscope (STM) to see toms (Nobel Prize in Physics in 1986) Limittions: (1) conductors only nd (2) surfce only Imges from wikipedi

3 How do we know there re toms or molecules? Now we hve technique to see tom directly (Scnning tunneling microscope, developed in the 80s), but people knows the eistences of toms long before we cn see them. Q: How did people know there re toms before we cn relly see them? A1: For liquid/gs, from the study on Brownin motion (Einstein 1905) A2: For solids, from the X-ry crystllogrphy Brownin motion: rndom moving of prticles suspended in fluid

4 X-ry scttering A single crystl Polycrystl: mny smll pieces of crystls

5 Sctterings nd Diffrction

6 Crystl Plnes Crystl plnes: we cn consider 3D crystl s lyers of 2D plnes. These 2D plnes re crystl plnes Ø It is geometry concept, insted of mechnicl one. Sometimes, these 2D plnes re wekly coupled (grphene) For other cses, the 2D plnes re coupled very strongly.

7 Crystl Plnes

8 Inde System for Crystl Plnes Ø Define three es 1,2 nd 3 using the three lttice vectors #, $ nd % The three lttice vectors could be primitive or conventionl lttice vectors Ø Find the intercepts on the es in terms of lttice constnts #, $ nd % : We find three numbers We only consider the cse tht they re rtionl number i.e. p # /q #, p $ /q $ nd p % /q % (for the figure: 3, 2, 2) Ø Tke the reciprocls of these numbers Ø q # /p #, q $ /p $ nd q % /p % (for the figure: 1/3,1/2,1/2) Ø Find the lest common multiple of p #, p $ nd p % : n (for the figure: 6) Ø So, we cn define three integers h = n -., k = n - 2 nd l = n - 4 /. / 2 / 4 Ø We use (hkl) to mrk this crystl plne

9 Inde System for Crystl Plnes Ø If one of the integer is negtive, we put br on top of it (hk7l) Ø We use prentheses (hkl) to mrk crystl plnes Ø We use squre brket hkl to mrk directions in crystl:v = h # + k $ + l % Ø In cubic crystls, the direction hkl is perpendiculr to the plne (hkl), but this is not true for generic lttices

10 The Brgg lw: mirror reflection by crystl plnes Constructive interference Destructive interference Ø The pth difference: 2d sin θ Ø Constructive interference: 2d sin θ = n λ Ø Intensity peks: θ = rcsin F G $H θ: Brgg ngle The direction of the bem is chnged by 2 θ

11 The Brgg lw: mirror reflection by crystl plnes Brgg s lw: 2d sin θ = n λ. So sin θ = n λ/2d Ø sin θ 1, so G 1. In other words: λ 2 d $H Ø d~size of n tom (bout 1A = 10 U#V m), so λ~1a Ø Visible light, λ~ A, too lrge for Brg sctterings Ø X-rys, electrons, or neutrons Ø Prticles (v c): E = /2, so p = 2 E m $[ Ø Prticles re wves (QM) k = p/ħ λ = 2π k = 2πħ p = h 2 E m

12 Fourier Anlysis: very powerful tool for periodic functions A 1D emple: For 1D function with periodicity, n + = n(), we cn lwys write it s the sum of cos nd sin functions: n = n V + [C / cos with p being n positive integer. /fv It is esy to check tht n + = n : n + = n V + {C / cos /fv = n V + C / cos /fv = n V + C / cos /fv + + S / sin + + S / sin = n() + S / sin + S / sin ] + } +

13 Fourier Anlysis Another wy to write down the Fourier series: with p being n integer. n = n / ep i / Ø Equivlent to the cos nd sin formul shown on the previous pge (e n o = cos + i sin ) Ø If n is rel function [n = n ], n / = n U/ n = n / ep i / = n V + n / cos /fv + i n / sin = n V + (n / + n U/ ) cos i /fv = n V + C / cos /fv = n V + n / ep i /fv + n U/ cos i + n U/ ep i + i(n / n U/ ) sin + S / sin i n U/ sin

14 Rel function n = n / ep i Ø If n is rel function [n = n ], n / = n U/ / n = n / ep i Compre n nd n, it is esy to see tht if n = n, we must hve n / = n U/ n = n V + (n / + n U/ ) cos i + i(n / n U/ ) sin /fv = n V + (n / + n / ) cos i /fv = n V + 2Re n / cos i /fv = n V + C / cos /fv So: C / = 2Re(n / ) nd S / = 2Im(n / ) / = n U/ / + i(n / n / ) sin 2Im(n /) sin + S / sin ep i

15 Reciprocl lttice n = n / ep i / Periodic function: lttice $t / u : the reciprocl lttice

16 Proof: Inversion of Fourier Series n = n / ep i / u n / = U# v d n ep i u V r.h.s. = U# v d n ep i V u = U# v d n /w ep i ep i V /w u = U# n /w v d /w Ø If p w u p, d ep i $t /{ U/ V Ø If p w u = p, d ep i $t /{ U/ V u u V r.h.s. = U# n /w /w = ep i w p u {ep i $t /{ U/ $t / { U/ n u u u = d 1 V V = d ep 0 δ /,/w = n /w /w 1}=0 = δ /,/ { = n / = l. h. s.

17 Higher dimensions n = n / ep i / u n / = U# v d n ep i V n r = n ep ig r n = V U# v dv n r ep G r n r is 3D periodic function: n r = n r + T, Ø T is the lttice vector: T = u # # + u $ $ + u % % u #, u $ nd u % re three rbitrry integers #, $ nd % re the lttice vectors, Ø G is the reciprocl lttice vector: G = v # b # + v $ b $ + v % b % v #, v $ nd v % re three integers b # = 2π u 2 u 4 u. (u 2 u 4 ), b $ = 2π u 4 u. u. (u 2 u 4 ) nd b % = 2π u. u 2 u. (u 2 u 4 ) Ø Another wy to define b #, b $ nd b % : n b = 2πδ n, If i j, then n b If i = j, then n b n = 2π For orthorhombic nd cubic lttices, b n = 2π/ n, but in generl this is not ture.

18 Brvis Lttice nd Reciprocl lttice T = u # # + u $ $ + u % % G = v # b # + v $ b $ + v % b % $ b # # Rel spce b $ k-spce/ momentum spce Ø Reciprocl vectors plys the role of wve-vector (momentum) in Fourier trnsformtions. Ø Three primitive lttice vectors #, $ nd % defines Brvis lttice with lttice sites locted t T = u # # + u $ $ + u % % (lttice vectors) Ø These primitive lttice vectors lso give us three reciprocl lttice vectors b #, b $ nd b % Ø Using b #, b $ nd b % (s primitive lttice vectors ), we cn define lttice, which is clled the reciprocl lttice: G = v # b # + v $ b $ + v % b % (reciprocl lttice vectors). Ø The primitive (conventionl) unit cell in the reciprocl lttice is known the Brillouin zone. Dulity: A Brvis lttice nd its reciprocl lttice re dul to ech other (1) If we know one, we know the other. (2) X-ry mesures the reciprocl lttice, nd thus we cn obtin the Brvis lttice

19 Diffrction Conditions Ø Number of prticles: F(k ) $ Ø k = k w k Scttering mplitude: F = dv n r ep[i(k k ) r ] = dv n r ep i k r Scttering strength t r Pth difference

20 Diffrction Conditions Scttering mplitude: F = dv n r ep[i(k k ) r ] = dv n r ep i k r Scttering strength t r Pth difference Ø n r is periodic function: n r = n r + T with T = u # # + u $ $ + u % % So we know: n r = n ep ig r Using n, we cn rewrite F s: F = v dv n ep ig r Ø If k = G, F = dv n = n V ep i k r = v dv n Ø If k differs from G slightly, F will be very tiny Ø So there is pek, whenever k being lttice vector. ep i(g k) r

21 Diffrction Conditions k = k w k = G For elstic sctterings ( k w = k ) k w = k + G = k So (k + G ) (k + G ) = k k 2k G + G $ = 0 If G is reciprocl lttice, so is G 2k G G $ = 0 So, the diffrction condition: Q: Is this the sme s the Brgg s lw (2d sin θ = n λ)? A: Yes 2k G = G $

22 Diffrction Conditions 2k G = G $ v (hkl) plnes re perpendiculr to G = h b # + k b $ + l b % Proof: three intercepts re p # /q # #, p $ /q $ $, p % /q % % And h = n -., k = n - 2 nd l = n - 4 /. / 2 / 4 It is esy to show tht G is perpendiculr to lines 12, 23 nd 31 So G is perpendiculr to the plne. p # G q # p $ # q $ = 2π h p # k p $ = 2π n n = 0 $ q # q $ v Seprtion between two neighboring lyers for (hkl) plnes: d hkl = G G p # q # # = 2π h p # q # G = 2πn/ G From figure on the previous slides: G = 2 k sin θ, So 2πn/d hkl = 2 k sin θ. 2 sin θ d hkl = 2πn = n λ k

23 Lue Equtions k = G Using the following two conditions: k = G = v # b # + v $ b $ + v % b % nd n b = 2πδ n, it is esy to note tht n k = 2πv n where i = 1,2 nd 3.

24 Structure fctor Scttering mplitude: F = dv n r ep i k r When the diffrction condition is stisfied k = G, one cn prove tht every unit cell contributes the sme mount to the integrl: v This is becuse ## dv n r ep i k r = v dv n r ep i k r #$ v #$ dv n r ep i k r = v dv n r + T ep i k r + T ## = v dv n r ep i k r ## Here we used the fct tht ep ig T = 1. Therefore, F = N v dv n r ep i k r = NS where N is the number of unit cells nd S is clled the structure fctor.

25 Atomic structure fctor S = v dv n r ep ig r Ø Similr to the scttering mplitude F, but the integrl is limited to 1 unit cell (F integrtes over the whole spce): S = F /N Ø Wvevector must be reciprocl lttice vector (must stisfy the diffrction condition) š If single cell contins s toms (j = 1,2,, s), n r = # n (r r ) where r is the loction of ech tom nd n (r ) is the contribution from one tom. Therefore: š S = v dv n (r r ) ep ig r š # = v dv n (r r ) ep[ ig (r r )] ep ig r š # š = ep ig r # v dvn (ρ ) ep ig ρ = ep ig r # f

26 Atomic structure fctor where the structure fctor is: š F = NS S = ep ig r # where the tomic structure fctor is: f f = v dvn (ρ ) ep ig ρ For sme type of tom, they shre the sme the tomic structure fctor! Using bcc nd fcc s n emple

27 bcc In one conventionl cell, there re two identicl toms locted t (0,0,0) nd ( # $, # $, # $ ) $ S = ep ig r # f = f {1 + ep ig If G = v # b # + v $ b $ + v % b %, we hve $ S = ep ig r # Ø If v # + v $ + v % = even, S =2 f Ø If v # + v $ + v % = odd, S =0 1 2 # $ % } f = f {1 + ep[ i π v # + v $ + v % ]} Nïve diffrction condition: pek if k = G = v # b # + v $ b $ + v % b % Not quite true for bcc (conventionl cell) Hlf of k (v # + v $ + v % = odd) hs no pek. Insted, the mplitude is 0!

28 fcc In one conventionl cell, there re four identicl toms locted t (0,0,0), ( # $, # $, 0), (# $, 0, # $ ) nd (0, # $, # $ ) S = ep ig r # f 1 = f{1 + ep ig 2 # $ + ep ig If G = v # b # + v $ b $ + v % b %, we hve $ S = ep ig r # For v #, v $ nd v % Ø All odd or ll even: S =4 f Ø 1 odd nd 2 even, S =0 Ø 2 odd nd 1 even, S =0 1 2 # % + ep ig 1 2 $ % } f = f{1 + ep i π v # + v $ + ep i π v # + v % + ep[ i π v $ + v % ]} Nïve diffrction condition: pek if k = G = v # b # + v $ b $ + v % b % Not true for fcc (conventionl cell) Some of k (1 odd 2 even or 2 odd 1 even) hs no pek. Insted, the mplitude is 0!

29 Atomic structure fctor of n isotropic tom f = v dvn (ρ ) ep ig ρ Ø Assume the tom is isotropic (pproimtion) Ø n (ρ ) is the density of electrons f = 2π v ρ $ dρ sin α dα n ρ ep igρ cos α = 2π v ρ $ dρn ρ ep igρ ep igρ i Gρ = 4π v dρn ρ sin Gρ Gρ ρ $ For very smll Gρ, ª ª 1 where Z is the number of tomic electrons f = 4π v dρn ρ ρ $ = Z

30 Single crystls vs polycrystls A single crystl Polycrystl: mny smll pieces of crystls