Overflow from last lecture: Ewald construction and Brillouin zones Structure factor

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Edmund Simmons
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1 Lecture 5: Overflow from lat lecture: Ewald contruction and Brillouin zone Structure factor Review Conider direct lattice defined by vector R = u 1 a 1 + u 2 a 2 + u 3 a 3 where u 1, u 2, u 3 are integer and a 1, a 2, a 3 are primitive tranlation vector The reciprocal lattice i defined by (primitive) vector b 1, b 2, b 3 defined in the following way: b 1 = 2π a 2 a 3 a 1 a 2 a 3 b 2 = 2π a 3 a 1 a 1 a 2 a 3 b 3 = 2π a 1 a 2 a 1 a 2 a 3 The reciprocal lattice i alo a lattice, with all point acceed by reciprocal lattice vector G = ν 1 b 1 + ν 2 b 2 + ν 3 b 3 Where ν 1, ν 2, ν 3 are integer Another important property of the reciprocal lattice i that R G = 2πn where n i an integer. One important application of the reciprocal lattice i diffraction a type of experiment ued to determine the repeating tructure of a crytal by hining x-ray, neutron, or electron onto the crytal and invetigating how much the beam i deflected. In previou coure, you might have een the Bragg forula for diffraction: 2dinθ = nλ Where n i an integer, λ i the wavelength of light (or debroglie wavelength of particle), and d i the pacing between identical crytal plane. Any crytal can be ubdivided into plane (containing atom) in everal different way. There i another way to expre thi ame formula uing the reciprocal lattice. If a plane wave with wavevector k i incident on the crytal (~e ik r )and the outgoing wave ha wavevector k (~e ik r ), a diffraction peak will be een only if k k = G We can alo ue the reciprocal lattice to find pacting between identical lattice plane more eaily. If we have a plane decribed by miller indice (hkl) It turn out that d(hkl) = 2π/ G where G = hb 1 + kb 2 + lb 3. It alo turn out that G hkl i normal to the plane decribed by the (hkl) miller index Ewald contruction

$The Ewald contruction i a way of viualizing a diffraction experiment, and alo illutrating that reciprocal pace i an empty pace. 1. Draw reciprocal lattice 2.$

2 The Ewald contruction i a way of viualizing a diffraction experiment, and alo illutrating that reciprocal pace i an empty pace. 1. Draw reciprocal lattice 2. Pick a point at the origin and draw the wavevector of the incident beam, k, the outgoing beam, k, and their difference, Δk 3. Remember, if Δk = G (a reciprocal lattice vector), we will get contructed interference and a finite diffraction ignal 4. The vector k and k define a phere, the Ewald phere. If thi phere interect another point on the reciprocal lattice, the laue condition (tated in tep 3) will apply. However, thi i very difficult to accomplih jut by chance becaue the reciprocal lattice i mainly empty pace. 5. There are three way to get around thi difficulty Note: in the image above the difference between k and k i called K, but in the ret of thee note it i called G Method 1: rotate the crytal (The direct lattice i defined relative to the crytal, o when we rotate the crytal, we rotate the direct lattice, and hence we rotate the reciprocal lattice)

$Method 3: powder diffraction.$

3 Method 2: ue polychromatic x-ray (e.g. white light) o that many value of k are incorportated. Thi i called the Laue method, and it i frequently ued to find the orientation of ingle crytal pecimen. Method 3: powder diffraction. Thi method ue monochromatic x-ray, but intead of tudying a large ingle crytal it tudie a powder a collection of many many microcopic crytal oriented in random direction. From the tandpoint of olid tate phyic, a microcopic crytal i effectively infinite becaue

$For a given incidence angle on thi powder, θ, there might be a crytallite that ha the correct orientation of crytal plane to produce a diffraction ignal.$

4 it ha o many unit cell that the boundarie do not change the propertie much. For a given incidence angle on thi powder, θ, there might be a crytallite that ha the correct orientation of crytal plane to produce a diffraction ignal. A θ i varied, one pick up all of the poible crytal-plane pacing (d) in that crytal. The erie of peak from a powder diffraction experiment give the fingerprint of a material. Laue pattern of NaCl (table alt) ingle crytal along 4-fold ymmetric axi Powder x-ray diffraction pattern of NaCl

One way to do thi (thi i a HW quetion) i the Wigner-Seitz cell, which i defined by the following procedure: 1.

5 Brillouin zone For a direct primitive lattice, there i often more than one way to define a primitive cell (the parallelogram defined by vector which can be ued to acce every point in the lattice). One way to do thi (thi i a HW quetion) i the Wigner-Seitz cell, which i defined by the following procedure: 1. Pick one lattice point and draw line to connect thi to all nearby lattice point 2. Draw perpendicular biector through all of thee line 3. The hape formed from the interection of the perpendicular biector i the Wigner-Seitz cell

6 When thi ame procedure i done for the reciprocal lattice the name of the cell i the firt Brillouin zone. Thi concept will become important in later chapter when we add electron into the picture. The Brillouin zone contain all of the wavevector which can be Bragg reflected by the crytal. Structure factor The final piece for uing what we have learned about the reciprocal lattice and diffraction to learn about the repeating periodic tructure of crytalline olid i including the bai. Thu far, we have only been conidering the lattice, but a we learned in chapter 1, a unit cell in a crytal conit of a lattice and a bai. Conider a crytal of identical unit cell, each with electron denity inide them given by n(r), which i a function of poition inide the cell (later we will ubtitute atomic poition in there). We are hining an x-ray plane wave at thi crytal with wavefunction e ik r and meauring a diffracted beam with wavefunction e ik r, where k = k. Alo, we are getting a diffraction ignal, o k k = G where G i a vector of the reciprocal lattice. The diffraction amplitude for N unit cell may be given by N multiplied by the diffraction amplitude for a ingle cell: F G = N cell dv n(r)e ig r = NS G What thi integral phyically doe i it conider every poition r inide the unit cell and introduce a phae delay from that infiniteimal volume element. We are conidering a pecific G, o the um over all poible G (lat lecture) i omitted. S G i the tructure factor. n(r) can be decompoed into contribution from every atom inide the unit cell. For convenience, we alway put one of the atom at the origin. n(r) = n j (r r j ), where there are atom in the bai Plug thi into the expreion for the tructure factor above Define a new variable ρ = r r j S G = dvn j (r r j )e ig r = dvn j (r r j )e ig (r r j) e ig r j S G = e ig r j dvn j (r r j )e ig (r r j)

7 e ig r j dvn j (ρ)e ig ρ Define the atomic form factor f j = dvn j (ρ)e ig ρ. Thi number repreent the cattering power of the j-th atom in the cell. It will be equal for two atom of the ame type (e.g. two odium atom at two different poition in a unit cell). It will alo depend on what exactly i being cattered x-ray, neutron, or electron. S G = f j e ig r j The cattering intenity i proportional to S G 2 o it i ok if S G i not real. In term of primitive lattice vector, the poition of each of the atom in the bai i given by: r j = x j a 1 + y j a 2 + z j a 3 where x j, y j, z j are fractional G r j = (ν 1 b 1 + ν 2 b 2 + ν 3 b 3 ) (x j a 1 + y j a 2 + z j a 3 ) = 2π(ν 1 x j + ν 2 y j + ν 3 z j ) S G (ν 1 ν 2 ν 3 ) = f j e 2πi(ν 1x j +ν 2 y j +ν 3 z j ) Practice Structure factor of BCC lattice. For thi exercie, we will ue the conventional cubic unit cell of the BCC lattice, which a 2 atom per bai. Atom 1: r 1 = 0 Atom 2: r 2 = 1 2 ax ay az S G (ν 1 ν 2 ν 3 ) = f j e 2πi(ν 1x j +ν 2 y j +ν 3 z j ) = fe 0 + fe iπ(ν 1+ν 2 +ν 3 ) The ubcript ha been dropped on the f becaue two identical atom are ued in thi exercie, meaning they have the ame form factor. The econd term can be +f or f, depending if the integer in the exponent add up to a poitive or negative number 0 if ν S G,BCC (ν 1 ν 2 ν 3 ) = { 1 + ν 2 + ν 3 i odd 2f if 0 if ν 1 + ν 2 + ν 3 i even

8 According to thi calculation, the Bragg peak from ome plane will be miing. For example, the (100) and (111) peak will be miing (odd um of 3 miller indice), while (110) and (200) will be preent (even um). We can try to undertand thi by conidering the primitive lattice of the BCC tructure. a 1 = a ( x + y + z ) 2 a 2 = a (x y + z ) 2 a 3 = a (x + y z ) 2 Conider the (100) plane, referenced to the conventional unit cell. In term of the primitive vector, the miller indice would be: Interection coordinate along a 1, a 2, a 3 =(-2, 2, 2) Invere: ( 1 2, 1 2, 1 2 ) Miller indice of (100) cubic in term of primitive tranlation vector: (-111) But thi i equivalent to the (-111) plane which i parallel to the one we were originally conidering, but located a/2 along the x direction toward the origin. Thu, when we go to the primitive lattice vector of BCC, we can ee that there are no miing Bragg peak, we jut double counted ome of them initially by uing a cell with 2 atom inide it when 1 would have ufficed. Example 2: Structure factor of the face centered cubic lattice Again, we conider the conventional unit cell.

$Putting the origin at the back left corner and having z point up and y to the right, we have atom at the following fractional coordinate (factor of a i omitted): (0,0,0); ( 1 2, 0, 1 2 ) ; (0, 1 2, 1$ For example, we counted all of the back corner atom (0,0,0) and left the other 7 to other unit cell becaue there are 8*1/8=1 corner atom; ditto for the face-centered atom which are each hared with 2

For example, we counted all of the back corner atom (0,0,0) and left the other 7 to other unit cell becaue there are 8*1/8=1 corner atom; ditto for the face-centered atom which are each hared with 2

9 Putting the origin at the back left corner and having z point up and y to the right, we have atom at the following fractional coordinate (factor of a i omitted): (0,0,0); ( 1 2, 0, 1 2 ) ; (0, 1 2, 1 2 ) ; (1 2, 1 2, 0) (When atom are hared with adjacent cell we count only enough of thoe atom to get the total number contained in one cell. For example, we counted all of the back corner atom (0,0,0) and left the other 7 to other unit cell becaue there are 8*1/8=1 corner atom; ditto for the face-centered atom which are each hared with 2 other cell, only count 3 of them) The tructure factor for the conventional FCC cell i given by: S G (ν 1 ν 2 ν 3 ) = f j e 2πi(ν 1x j +ν 2 y j +ν 3 z j ) = f[1 + e iπ(ν 1+ν 3 ) + e iπ(ν 2+ν 3 ) + e iπ(ν 1+ν 2 ) 0 if one index i odd and other two even or if two are odd and one i even = { 4f if all indice are odd or all are even Again, the atomic form factor i the ame for all term in our ummation becaue all atom are identical. Similar to what we aw in the BCC lattice, only ¼ of the miller indice in FCC lattice will yield Bragg peak (there are 2 way to elect all odd or all even indice, but 6 way to elect a mixture of odd and even) becaue the conventional unit cell contain 4 atom. Example 3: NaCl tructure The NaCl tructure can be thought of a two interweaving FCC lattice (one for Na, one for Cl)

$Set the origin at the Cl atom in the back left corner (000) The fractional coordinate of the Na atom next to it can be expreed a: 1 2 (a 1 + a 3 a 2 ) Thu, G r j = (ν 1 b 1 + ν 2 b 2 + ν 3 b 3 ) (x j$

10 To avoid having to write a um over 8 different term, let expre the coordinate in term of the primitive lattice vector. Set the origin at the Cl atom in the back left corner (000) The fractional coordinate of the Na atom next to it can be expreed a: 1 2 (a 1 + a 3 a 2 ) Thu, G r j = (ν 1 b 1 + ν 2 b 2 + ν 3 b 3 ) (x j a 1 + y j a 2 + z j a 3 ) = 2π(ν 1 x j + ν 2 y j + ν 3 z j ) S G,NaCl = f Cl + f Na e iπ(ν 1 ν 2 +ν 3 ) We cannot make the ame implification a before becaue the atomic form factor are different, but we can make general tatement:

11 If indice are all even or two odd and one even, the 2 nd term will give a poitive contribution to the um If indice are all odd or two even and one off, the 2 nd term will give a negative contribution to the um, likely making S G 2 (which we meaure) maller. Comparing to the powder x-ray diffraction image we can ee that the 111 peak i very mall compared to the nearby 200 and ditto for the 311 v 222. In general, we expect peak to get maller a we move to higher angle becaue higher-index plane contain fewer atom, but the analyi we did above allow u to gain further inight into why ome Bragg peak are larger than other nearby. If we are uing x-ray to do our diffraction experiment, the form factor i proportional to Z (the atomic number). Thi i derived ketchily in your textbook, but we will take it a a given for thi lecture. To be more pecific, it i electron which do the cattering o the number of electron on each pecie of atom i what matter. Thi might not be the ame a the atomic number for ionic olid like NaCl, where the crytal tructure i made up of charged pecie. To ummarize: Z Na = 11 Z Cl = 17 Electron on Na + = 10 Electron on Cl = 18

Keble College - Hilary 2012 Section VI: Condensed matter physics Tutorial 2 - Lattices and scattering

Tomi Johnson Keble College - Hilary 2012 Section VI: Condensed matter physics Tutorial 2 - Lattices and scattering Please leave your work in the Clarendon laboratory s J pigeon hole by 5pm on Monday of