Why does diffraction give a Fourier transform? |ki| = |ks| = 2π/λ ki ks r l1 = (λ/2π) × ki·r l2 = −(λ/2π) × ks·r Total additional pathlength is (λ/2π) × (ki − ks)·r Total phase shift is (ki − ks)·r ki ks Q = ki − ks −ks Total phase shift is Q·r sensitive to the actual wave-vectors ki and ks , but only to their difference. Because of this, we can formulate our scattering equations terms of single wave-vector pattern Q = kiis−not ks . One interesting property by this ·in equation is athat the interference ψ1 (x) +highlighted ψ2 (x) = A exp(ik (21) s x) × {1 + exp[i(ki − ks ) · r]}. This is precisely the same Q we have already in to BL1 and BL2. Rewriting equation sensitive to the actual wave-vectors ki and ks , butmet only their difference. Because of this, (21) in this light, we can say that equations the scatteredterms beamof has been modified by a Q factor we formulate our scattering single = ki is − not ks . Onecan interesting property highlighted by thisinequation isathat the wave-vector interference pattern This is precisely the same Q we have already met in to BL1 and BL2. Rewriting equation sensitive to the actual wave-vectors ki and ks , but only their difference. Because of this, (21) in this light, weour can say that the scattered beam of hasa been modified by a factor we can formulate scattering equations in exp(iQ terms wave-vector Q = ki −(22) ks . F (Q) = 1 + exp(iQ · r) = · 0) +single exp(iQ · r). This is precisely the same Q we have already met in BL1 and BL2. Rewriting equation (21) in this light, we F can say the scattered beam ·has been modified a factor It is straightforward to extend this idea of + arbitrarily point particles at (Q) = 1that + exp(iQ ·tor)a=collection exp(iQ 0) exp(iQ · many r). by (22) positions r1 , r2 , r3 , . . . to give the general “scattering factor”: It is straightforward to extend idea to collection at F (Q) = 1 this + exp(iQ · r)a = exp(iQ · of 0) arbitrarily + exp(iQ ·many r). point particles (22) positions r1 , r2 , r3 , . . . to give the general “scattering factor”: F (Q) = exp(iQ · r1 ) + exp(iQ · r2 ) + exp(iQ · r3 ) + . . . (23) It is straightforward to extend �this idea to a collection of arbitrarily many point particles at positions r1 , r2 , r3 , . . . to = give theexp(iQ general factor”: · rj“scattering ). (24) F (Q) = exp(iQ · r1 ) + exp(iQ · r2 ) + exp(iQ · r3 ) + . . . (23) j � = exp(iQ · rj ). (24) IB Mineral Sciences Indeed, Module by converting over· rall space we ·allow generalisaj to an F (Q)this = sum exp(iQ ·and rintegral ) + exp(iQ r319 ) + a. .further . (23) B: Reciprocal Space, Symmetry 1Crystallography 2 ) + exp(iQBH � tion to arbitrary particle distributions (i.e. allowing for non-point particles): exp(iQ (24) j ).colours in the Fourier images, but not we are saying is that we = measure the intensity ·ofrthe IB Mineral Sciences Indeed, the by colours converting this sum an integral over alla space allow a further generalisathemselves. This makes difficult to reconstruct real-spacewe description j toit very � Module B: Reciprocal Space, Symmetry and Crystallography BH 19 of structureparticle from what isdistributions an incomplete “colour-blind” reciprocal-space description.particles): Known tion to arbitrary (i.e. allowing for non-point lySpace, Symmetry and Crystallography as the “phase problem”, this issue forms = the subject the final ·lecture F (Q) ρ(r)ofexp(iQ r) dr.in our course, (25) arevery saying is that we measure intensity of the colours Fourier images, but One important in solving this problem isover an understanding of symmetry in not Indeed, BL9. by we converting thishelp sum to the an integral allin the space we allow a further generalisa� the colours themselves. makes very learning difficult to reconstruct a real-space description crystalline materials. We will This spend someit time about symmetry and its effect on tion to arbitrary particle distributions (i.e. allowing for non-point particles): of structure from is an incomplete reciprocal-space description. Known diffraction patterns — what the subject lectures BL5 through to BL8. · r) dr. Fof(Q) =“colour-blind” exp(iQ (25) l in Space, Symmetry and Crystallography This equation should be entirely familiar. What have shown here as the “phase problem”, this issue subject we of the final lecture in our course, swe the Fourier but measure theimages, intensity ofnot the colours informs thetheρ(r) Fourier images, but notis quite remarkable: the time being, will concern ourselves a nothing little more with of of diffraction BL9. One verywe important help in solvingisthis problem is but anthe understanding symmetry in namely, For that the scattering process ageometry Fourier transformation itself. In a � experiments, and the relationship between the time various quantities The first struct a real-space description es. This makes it very difficult to We reconstruct a real-space description crystalline materials. will spend some learning aboutinvolved. symmetry and itsthing effect on diffraction experiment, the scattered wave characterised by kswetells aboutremarkable: the Fourier we will doshould is topatterns relate scattering vector Q to BL5 the various diffraction concepts first us This equation beour familiar. What we shown here is quite F (Q) ρ(r) exp(iQ · r) dr. (25) diffraction —entirely the subject of lectures through to have BL8. we the intensity of colours ink=the Fourier images, butlaw. not waves for different ocal-space description. Known t is measure an incomplete “colour-blind” reciprocal-space description. Known encountered in IA:the namely, the Bragg d-spacing and the the Bragg component atthe the wave-vector Q =diffraction − nothing kangle . little By2θ,more measuring scattered iis sa namely, that scattering process butwith atheFourier transformation itself. In a For the time being, we will concern ourselves geometry diffraction Reminding ourselves of the scattering geometry illustrated in the diagram on pageof BH 16, es. final This makes it very difficult to reconstruct a real-space description experiments, and the relationship between the various quantities involved. The first thing he in our course, em”, thislecture issue forms the subject of the final lecture in our course, wave-vectors k (and, effectively, for different Q), we can actually piece together a picture of s the relationship diffraction experiment, the scattered wave characterised by ks tells us the Fourier we begin with This equation should familiar. we have shown here is about quite remarkable: we will do is to be relateentirely our scattering vector QWhat to the various diffraction concepts we first reciprocal space. Diffraction isan anBragg experimental technique that allows us to view reciprocal t is an incomplete “colour-blind” reciprocal-space description. Known at the the wave-vector Q = ki diffraction − knothing measuring the scattered different understanding of symmetry rtant help incomponent solving this problem isprocess understanding of symmetry s . By encountered inin IA: namely, the angle 2θ, d-spacing and the Braggin law. waves for namely, that scattering is but a Fourier transformation itself. In a Reminding ourselves of the scattering geometry illustrated incan the diagram on page BH 16, space directly. wave-vectors ks (and, effectively, piece together picture of 2 the 2 Q), em”, this spend issue forms the subject ofabout final in actually our course, Q2 =scattered |Q| =for|kdifferent ks |2 = kilecture + ks2 weand (26) out and itsexperiment, on Wesymmetry will some time learning symmetry its onus diffraction the wave characterised byeffect k about a the Fourier i− s tells weeffect begin with the relationship 2 space. Diffraction an experimental technique that allows to view reciprocal == ki2k+ k− (27) us waves s cos 2θ s − component atThere the wave-vector Q k2k.ikBy the Namely, scattered different inreciprocal solving this problem isisan understanding of symmetry in Well, almost. isthrough (as always) one fundamental problem. that whenforwe detect �measuring �i � BL8. —rtant the help subject of lectures BL5 to BL8. 2 s � 2 2π 2π space directly.k (and, effectively,=for2 different − 2are cos (28) wave wave-vectors Q),measuring we2θcan actually together picture of s a scattered such as FQ(Q), what|kiwe is not piece the itself,a but some 2 about We will spend some wave time learning = |Q|2 = λsymmetry − ks |2 =λki2 + ks2 and its effect on (26) 2 2 reciprocal space. Diffraction is an experimental technique that allows us view reciprocal 16π 2 2 2square with the geometry of diffraction we will concern ourselves a little more with the geometry diffraction intensity I(Q) that is (as proportional to the ofproblem. itsof modulus: I(Q) ∝when |F (Q)| .detect This = k + k − 2k (27) Well, almost. There always) one fundamental Namely, thatto we i ks cos 2θ i = BL8. sin θ,s (29) — the subjectspace of lectures BL5 through to � �2 λ 2 � �2 directly. 2π 2π are is afirst real positive value, whereas the amplitude Fisfirst (Q) a wave complex number. By aintensity scattered wave such as F (Q), what not is the itself, but some involved. The entities relationship between thething various quantities thing − 2 measuring cosThe 2θ (28) = 2 weinvolved. λ 2 λ 2 where we that have used the identity costo 2θ = 2 2sinsquare θ. magnitude Q of information measuring intensity rather than1 −with amplitude weConsequently, have surrendered allI(Q) about the intensity I(Q) isis proportional the ofproblem. itsthe modulus: ∝when |F (Q)| . detect This we will concern ourselves a little more the geometry of diffraction 16π us diffraction concepts we first Well, almost. There (as always) one fundamental Namely, that we e our scattering vector Q to the various diffraction concepts we first 2 scattering vector is related to the Bragg θ byθ,the equation = angle sin (29) 2 phase ofthe the amplitude. In terms of our Fourier transform illustrations of BL1, what λcat/duck intensity is awave real positive whereas the amplitude Fis(Q) a complex number. By a scattered as value, F quantities (Q), what weinvolved. are and measuring notisthing the wave itself, but some eamely, relationship between the such various TheBragg first d-spacing thediffraction Bragg law. the and Bragg angle 2θ,amplitude d-spacing the law. 4π measuring intensity rather than we the 2 2 have surrendered all information about intensity I(Q) isthe proportional of its modulus: I(Q) =costhe sin=θ.square (30) wherethat we have used the identity 1Q−to 2θ 2 sin θ. Consequently, the magnitude Q of ∝ |F (Q)| . This enofour scattering vector Q to various diffraction concepts we first λ the diagram on page BH 16, the scattering inour the diagram on page BH 16, scatteringillustrated vector related toof Bragg angle θ by Fourier the equation phase ofgeometry the Inisterms cat/duck transform BL1, what intensity is the aamplitude. real positive value, the whereas the amplitude F (Q) isillustrations a complex of number. By BH 19 BH 19 BH 19 Note that because the experimental limit of θ is 90 (i.e.and 2θ = 180 )Bragg then there is a amely, the Bragg diffraction angle 2θ,amplitude d-spacing the law. ationship measuring intensity rather we surrendered all information about the 4π maximum measurable value than of Q: namely, Q = 4π/λ. kθ.s haveon Q= sin (30) of the scattering illustrated inseeour the diagram page BH 16, λ phase ofgeometry the amplitude. of cat/duck Fourier transform illustrations of BL1, what Bragg’s law told us thatIn we expect to a diffraction peak corresponding to the Bragg kterms i angleNote θ whenever there were a set of atomiclimit planes the distance λ/2there sin θ. is a that because the experimental of θseparated is 90 (i.e.by 2θ = 180 d) = then ationship 2θ ◦ ◦ max max ◦ Recasting this in terms of Q,value we have maximum measurable of Q: namely, Qmax = 4π/λ. max ◦ 2told us that 2 to see a diffraction peak corresponding to the Bragg Q2 = |Q|2 = |kBragg’s = ki2 +wekexpect (26) i − klaw s |(26) 2π sQ atomic = .planes separated by the distance d = (31) angle θ whenever there were a set ofQ λ/2 sin θ. d 2Recasting 2 this in terms of Q, we have θ (27) = ki + ks − 2ki ks cos 2θ (27) 2 complete 2 sense in terms of our picture of the reciprocal lattice. this � Q2 = |Q|2 = Hopefully |k� k�should |22 =make k� (26) 2 i −know, ithat+wekexpect 2π We now components in reciprocal space when2π s only toQsee 2π s from BL2, = . (31) ever2Q is a reciprocal lattice vector Q = ha +kb +lc . The corresponding magnitude of Q 2 (28) cos 2θ − 2 costhe2θlatticed (and (28) = 2 = iskquite k − 2k k cos 2θ (27) i s easy to calculate in the case where reciprocal lattice) is orthogonal: i + s λ λ Hopefully this should make complete sense in terms of our picture of the reciprocal lattice. � � � � by Pythagoras we 2 have 2 We2π that we expect only to see components in reciprocal space when2now know, from BL2,2π 16π Q is a 2 reciprocal Q = ha of Q − 2 lattice vectorcos 2θ+kb +lc . The corresponding magnitude (28) = (29) = 2 ever sin θ, (29) is2quite easy to calculate in the case where the lattice (and reciprocal lattice) is orthogonal: λ λ λby Pythagoras we have 16π 2 = sin2 θ, 2 (29) sequently, the1 − magnitude of θ. Consequently, the magnitude Q of the identity cos λ2θ2 = 2Qsin quation is related to the Bragg angle θ by the equation the identity 1 − cos 2θ = 2 sin2 θ. Consequently, the magnitude Q of is related to the Bragg 4π angle θ by the equation (30) Q= sin θ. (30) λ 4π Q= 2θmax = 180◦ ) then there is θ. a he experimental limit ofλ θsin 90◦ (i.e. 2θmax = 180◦ ) then there (30) is a e value of Q: namely, Qmax = 4π/λ. he experimental limit of θ is 90◦ (i.e. 2θmax = 180◦ ) then there is a ak corresponding the hat we expect to to see a Bragg diffraction peak corresponding to the Bragg e value of Q: namely, Q max = 4π/λ. d by theadistance d = λ/2 sin θ.separated by the distance d = λ/2 sin θ. re were set of atomic planes hatofwe to see a diffraction peak corresponding to the Bragg ms Q, expect we have re were a set of atomic planes separated by the distance d = λ/2 sin θ. ms of Q, we have 2π Q = (31) . (31) d 2π Q =inlattice. . (31) picture of the reciprocal make complete sense dterms of our picture of the reciprocal lattice. entsthat in reciprocal whenBL2, we expectspace only to see components in reciprocal space when∗ our picture of the reciprocal lattice. make sense in terms corresponding of∗ +lc Q of latticecomplete vector Qmagnitude = ha∗ +kb . The corresponding magnitude of Q BL2, that expect onlythe to lattice see components in reciprocal wheneciprocal lattice) is orthogonal: late in thewe case where (and reciprocal lattice) isspace orthogonal: lattice QSciences = ha∗ +kb∗ +lc∗ . The corresponding magnitude of Q ve IBvector Mineral late inModule the case the lattice reciprocal is orthogonal: B: where Reciprocal Space,(and Symmetry and lattice) Crystallography ve ∗ ∗ ∗ ∗ ∗ ∗ � ∗ 2 �1/2 (ha ) + (kb∗ )2 + (lc∗ )2 � 2 2 �1/2 4π h 4π 2 k 2 4π 2 l2 = + + , a2 b2 c2 Q = BH 20 (32) (33) which, when substituted into equation (31) gives us the (hopefully) familiar expression for d-spacing in an orthogonal crystal: 1 h2 k 2 l2 = 2 + 2 + 2. d2 a b c (34) Qmax = 4π / λrad 10-1 10 Radio 10-3 10-5 Microwaves People Eye 10-7 Infra-red Needle 10-9 Ultraviolet Cell 10-11 10-13 X-Rays Molecule Atom Microscope Synchrotron Nucleus Particle Colliders x-rays (λ ~ 1 Å) neutrons (λ ~ 1 Å) electrons (λ ~ 0.02 Å) Kα Kβ Lα K L M Kα Kβ λmin lab x-rays 10-15 Gamma Rays Intensity 103 Wavelength (metre) Δrmin = 2π / Qmax = λ / 2 Wavelength produce electromagnetic radiation. This mechanism gives rise to a broad distribution of x-ray energies commensurate with the energies of the incident electrons. The second (and typically more useful) mechanism of x-ray production occurs when an electron strikes one of the metal atoms and causes an inner-shell electron to be ejected. An outer shell electron then relaxes into the now-partially-vacant inner shell, and emits an x-ray photon of a characteristic energy. For us, the important case is that of electrons falling from the L to the K shells (i.e. from the second to the first shell), and this transition is labelled Kα. This transition will have different energies (and hence produce x-rays of different wavelengths) for different metal targets: ˚ Transition Wavelength (A) TiKα1 2.47851 FeKα1 1.93597 CuKα 1.54184 AgKα 0.561603 WKα2 0.20901 ˚ Transition Wavelength (A) CrKα1 2.28962 CoKα1 1.78896 MoKα 0.711445 TaKα1 0.215947 AuKα1 0.180195 The traditional method of recording and measuring x-ray diffraction patterns is to use photographic film that has been optimised for x-rays. This is ideal for obtaining a wide sweep of the diffraction pattern in one measurement. The “greyness” of each reflection can be measured by optical methods and converted into a relative intensity. Several diffraction photographs may be required in order to record the range from very strong to very weak, since the strong reflections can easily saturate the film. The alternative method of measuring x-ray intensities is to use a detector (or “counter”). This has the advantage over films of giving the intensity directly and more accurately. The disadvantage is that a single detector will only record a single point in reciprocal space at any one time. This problem can be overcome by using area detectors that contain a large number of pixels, each of which is capable of recording its own signal. These are now the most widely used form of detector. One fundamental limitation on the use of x-ray tubes is that the maximum intensity is limited by the need to prevent overheating of the target. X-ray production is not very efficient, and most of the energy of the incident electron beam is lost as heat. This can be overcome in part by using rotating targets, so that the electron beam does not strike a single area of the target. However, there is still a mechanical limitation on the maximum intensity produced. Moreover, the wavelength of x-rays produced by a given x-ray tube is determined by the nature of its metal target, and so is not variable. nucleus electron cloud ∗ ∗ anes tice planes can becan evaluated be evaluated via thevia dotthe product dot product d∗h1 k1 l1 d · d∗h1∗hk21kl21l2· .dAlso, = 1/|d d= | 1/|d∗ | h2 k2 l2 . dAlso, ween ng between (hkl) planes, (hkl) planes, and is and given is explicitly given explicitly by theby formula the formula 2 ∗ 2 2 2∗ 2∗ 2 2 ∗ 2 ∗ ∗ ∗ ∗ ∗ h+(ak∗2)(b +) k+(bl (c ) + ) l+(c2klb ) +c 2klb cos ∗αc∗∗ + cos2lhc α∗ + a 2lhc cos∗βa∗∗ + cos2hka β ∗ +∗ b2hka cos ∗γb∗∗, cos γ ∗ , 2 2 ∗ ∗ re , γthe are internal the internal anglesangles of the of reciprocal the reciprocal lattice lattice unit cell. unit This cell. equation This equation re hefamiliar more familiar expression expression Z f (Q) AH 25 “form factor” or 2 scattering 2 2 factor”2 “atomic 1 IB MINERAL h1 kSCIENCES h l k2 l2 = + =Solids:++ Module A: Degrees of Order in The Role+ of Diffraction d2 ad22 ba2 2 c2b2 c2 0 AH 25 (1) (1) Q AH 25 IB MINERAL SCIENCES AH 25 between lattice planes can be evaluated via the dot product d∗h1 k1 l1 · d∗h2 k2 l2 . Also, d = 1/|d∗ | Module A: Degrees of Order in Solids: The Role of Diffraction nal tems. is thesystems. spacing between (hkl) planes, and is given explicitly by the formula ∗ 2 2 ∗be ∗ ∗ dot product ∗ ∗∗ ∗ ∗ d = ∗1/|d∗ | between evaluated · d∗∗ 2 k22hka (d∗ )2lattice = h2 (aplanes ) + kcan (b )2 + l2 (c∗ )2 + via 2klbthe c cos α∗ + 2lhcd ah1 kcos b cos γ , l2 . Also, 1 l1 β h+ is the spacing between (hkl) planes, and is given explicitly by the formula where α∗ , β ∗ , γ ∗ are the internal angles of the reciprocal lattice unit cell. This equation ∗ 2 (d∗ )2to=the h2 (a ) +familiar k 2 (b∗ )2 expression + l2 (c∗ )2 + 2klb∗ c∗ cos α∗ + 2lhc∗ a∗ cos β ∗ + 2hka∗ b∗ cos γ ∗ , reduces more nsform of the of unit thecell unit cell ∗ 2 where , β ∗ , γ ∗aare internal angles lattice unit cell. equation 1 of h2 the kreciprocal l2atom lins contains aαgroup group oftheatoms, of atoms, where where each each isatom defined is This defined by a position by a position and has and a has a = + + (1) 2 2 2 reduces to the more familiar expression d a b c2 AH 25 AHIf 25 the position IB MINERAL MINERAL SCIENCES SCIENCES AH 25by AH 25 ctron of electron density. density. If= the position ofIB2 each ofatom eachisatom represented is represented a Dirac by a delta Dirac delta 2 of Order 2Solids: for orthogonal systems. Module Module A: Degrees Degrees inl⊗ in Solids: The Role TheofRole Diffraction of Diffraction 1 A: h k of Order IB Mineral Sciences = 2+ + the (1) as being ne think can of think the of electron theSpace, electron density density of whole the BH whole group group as being described described by by 2 d2 and a Crystallography bof c2 Module Reciprocal Sciences Symmetry 37 IBB:Mineral between between lattice lattice planes planes can becan evaluated be evaluated via thevia dotthe product dot product d d ·d · .dAlso,. dAlso, = 1/|d d =| 1/|d | Fourier transform of the unit cell Module B:between Reciprocal Space, Symmetry and Crystallography 37 orthogonal ffor tion the of delta the functions delta functions with the with functions the functions describing the BH electron density density issystems. the spacing is the spacing between (hkl) planes, (hkl) planes, and is and given is explicitly given explicitly by theby formula thedescribing formula the electron � � � unit cell contains a=group of atoms, where atom is has a γ , about (dThe (dhF)Fourier (a=)transform h+(a k=)(b+) k+ (bl − (c ) (ha + ) each lof + 2klb )the c 2klb cosof αcFdefined cos +the 2lhc α(Q)+aby 2lhc cosadensity βaposition + cos2hka β +and b2hka cosabout γb ,cos atom, ρThe The ρjof(r). Fourier transform electron electron density a single a single atom atom × (48) (Q) δ[Q +(ckb ++lc )] j (r). transform Fourier of) the unit cell distribution electron density. If the position of each atom is represented by a Dirac delta � � F , β ,(Q = hkl) =are F the (hkl) (49) cell. where where α α γ , β are , γ the internal internal angles angles of the of reciprocal the reciprocal lattice lattice unit unit This cell. equation This equation ering cfunction, scattering factor factor freduces (Q) fhkl) (Q) which weexpression derived derived earlier. one can think of thewhich electron density of we the� whole group asearlier. being described by jFto �= the = 0 familiar The unit cell contains a group ofjfamiliar atoms, where each atom is defined by a(50) position and has a reduces the(Qto more more expression ∗ h1 k1 l 1 ∗ 2 ∗2 2 ∗ 2 2 ∗2 2 ∗ 2 2 crystal ∗ ∗ crystal ∗∗ ∗ 2∗ 2 ∗ 2 2 ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗ unitcell ∗ ∗ ∗ ∗∗ h1hk21kl21l2 ∗ ∗∗ ∗ h2 k2 l 2 ∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ hkl ∗ unitcell × Funitcell (Q) Fcrystal (Q)the=functionsδ[Q − (ha∗ +the kb∗electron + lc∗ )] density the convolution of the delta functions with describing distribution of electron density. If the position of each atom is by a Dirac delta 2 hkl 22 2 2 represented 2 � � h1 electron kh l k density l about each atom, ρj (r).� The Fourier� transform1 of the about a single atom � � (1) (1) � = 2lc2 + ++ 2 2group + ∗=whole function, onej can thinkjFcrystal of the electron density of as being described 2×2 F(hkl) (hkl) = (51) δ[Q −hkl) (ha∗ d+2 kb )] ba unitcell j by j dFunitcell cb (Q) c2 Ffcrystal = =∗ +athe is the atomic scattering factor which we derived earlier. j (Q) (Q hkl the� convolution the delta functions with the functions describing the electron density � forof =systems. Funitcell (52) orthogonal for orthogonal systems. Fcrystal (Q (hkl) �= hkl) = 0 � jatom j position about each ρj (r). The transform of delta the electron density a single atom An at atom, the r�j isFourier represented by the function δ(r − rabout has Fourier j ), and � � is the atomic scattering factor f (Q) which we derived earlier. transform exp(iQ · rFourier ). The electron density associated with the atom at position r can be Fourier transform transform of the of unit the cell unit cell j j j j j j j � 2 2 represented by the convolution ρ (r) ⊗ δ(r − r ), and the resultant Fourier transform by the j j � �+ lzj )] exp(−Bj sin θ/λ ) (hkl) = by the f (Qdelta + ky � position �unit hkl ) exp[2πi(hx jδ(r An atom at the rcontains is Frepresented function −isjrdefined has Fourier unit Thecell contains a group a group of atoms, of atoms, where where each atom eachisatom defined byand a position by a position and has and a has a j� cell j ), product fj (Q)j The × exp(iQ · r ). j j j � transform exp(iQ · rdistribution electron density associated the atom atisposition cana delta be delta distribution of electron of electron density. density. If the position If the position ofwith eachof atom each isatom represented represented by ar�jDirac by Dirac j ). The � 2 described 2by function, can one can of think the of electron the electron density density of the of whole the whole group group as being as being described by Fthink (hkl) = b exp[2πi(hx + ky + lz )] exp(−B sin θ/λ ) represented byfunction, the ρof (r) ⊗ δ(r − r ), and the resultant Fourier transform by the The actual unit cellconvolution willone consist several atoms, the contributions from which are simply j j j j j jj � of the of the convolution the convolution delta the functions delta functions with the with functions the functions describing describing the electron the electron density density j product fj (Q) × The exp(iQ · rresult ). added together. end for the Fourier transform of the unit cell is the “structure j IB Mineral Sciences about about each atom, each atom, ρj (r). The ρj (r).Fourier The Fourier transform transform of the of electron the electron density density about about a single a single atom atom factor” Module B: Reciprocal Space, Symmetry and Crystallography � 2 earlier. 2BH 37 is the atomic is the atomic scattering factor factor fj (Q) which fj (Q) which wecontributions derived we derived earlier. The actual unit cell will consistscattering of several atoms, the which are simply B � j = 8π �ufrom j F (Q) = fj (Q) exp(iQ · rj ). added together. end result transform of unit cell−isrδ(r � � the “structure AnThe atom An at atom the at position thefor position rthe is Fourier r�j is represented by the� by delta thethe function delta function δ(r and r�j ),has and Fourier has Fourier jrepresented j ), − �j � � � factor” � transform transform exp(iQexp(iQ · rj ). The · rj ).electron The electron density associated associated with the with atom the at atom position at position r�j can rbe � j can be ∗ ∗density ∗ × replicated Funitcell (Q) (48) Fcrystal (Q)of=atoms δ[Q (ha +unit kb +cell lc )] is In forming a crystal, the group in− one atFourier eachFourier lattice point, represented represented by theF by convolution the convolution ⊗ρδ(r (r)−⊗rδ(r −rjr). the and resultant the resultant transform transform by theby the (Q) =hkl jρfj j(r) (Q) jexp(iQ j ), ·and j ), j j j � and mathematically this operation corresponds to the convolution of the(49) unit cell with the Fcrystal (Q× hkl) ×=exp(iQ product product fj (Q) f=j (Q) exp(iQ ·Frunitcell · r�j ). j ). j(hkl) lattice. Thus the Fourier transform is the product of the (50) Fourier transform Fcrystal (Q �= hkl) = j0of the crystal j In forming a crystal, theactual group ofcell atoms in cellatoms, is at from eachwhich lattice point, The actual The unit cell unit will consist will consist of one several ofunit several atoms, thereplicated contributions the contributions from which are simply are simply of the unit cell (derived here)�and that of the lattice (i.e. the reciprocal lattice). Since the addedadded together. together. The end The result end result for thefor Fourier the Fourier transform transform of the of unit the cell unit is cell the “structure is the “structure and mathematically this operation corresponds to the convolution of the unit cell with the F (hkl) = f for (Qhkl )all exp[2πi(hx + lzj )] exp(−B θ/λ2 ) (51) j + kyj when reciprocal lattice is zero-valued Q except Q isj sin a 2reciprocal lattice vector, we factor” factor”transform j lattice. Thus the Fourier of the crystal is the product of the Fourier transform � � ∗ ∗ ∗ need only consider the values�Q = 2π(ha += kb +f=jlc The be expressed in terms F (Q) F lz (Q) (Q)).jexp(iQ 2 2r·jrjcan F (hkl)here) = bj exp[2πi(hx kyj + θ/λexp(iQ ) ). · rj ). (52) j + the j )] exp(−B of the unit cell (derived and that of lattice (i.e.fsinj (Q) the reciprocal lattice). Since the j jlattice vectors to give of their components xj , yj , zj parallel to the real-space j reciprocal lattice is zero-valued for all Q except when Q is a reciprocal lattice vector, we In forming In forming a crystal, a crystal, the group the group of of atoms in one unit onecell unit is cell replicated is replicated at eachatlattice each lattice point, point, ∗ ∗atoms ∗ in ∗ need only consider the values = operation 2π(ha +corresponds kblc∗ ∗+ rjthe be expressed in terms Q ·mathematically rj = Q 2π(ha + kb + ) ·lc (xj).a + b +can zconvolution j c) and mathematically and this this operation corresponds toThe theyjto convolution of the of unit thecell unit with cellthe with the of their components xThus zj=parallel to+the real-space lattice vectors toofgive j , yj ,the lattice.lattice. Thus Fourier the Fourier transform transform crystal the crystal is the is product the product the of Fourier the Fourier transform transform 2π(hx ky +oflzthe ).of crystal (48) osition the position r is represented r is represented by theby delta the function delta function δ(r − rδ(r ), − and r ),has and Fourier (49)has Fourier � (50) � xp(iQ r ). The · r ).electron The electron density density associated associated with the with atom theat atom position at position rj can rbe j can be e by convolution the convolution ρ (r) ⊗ρδ(r (r)−⊗rδ(r ), and − r ), the and resultant the resultant Fourier Fourier transform transform by the by the (51) Q) xp(iQ × exp(iQ · r ). · r ). (52) unit l will cell consist will consist of several of several atoms,atoms, the contributions the contributions from which from which are simply are simply he her.end Theresult end result for thefor Fourier the Fourier transform transform of the of unit thecell unit is cell the “structure is the (53) “structure � � F (Q) = F (Q) f=(Q) exp(iQ f (Q) exp(iQ · r ). · r ). aal,crystal, the group the group of atoms of atoms in onein unit onecell unit is cell replicated is replicated at eachatlattice each lattice point, point, matically ly this operation this operation corresponds corresponds to the convolution to the convolution of the unit of thecell unit with cellthe with the Fourier s the Fourier transform transform of the of crystal the crystal is the product is the product of the of Fourier the Fourier transform transform cell erived (derived here) and here)that andofjthat thej of lattice (i.e. the (i.e. reciprocal the reciprocal lattice).lattice). Since the Since the j the lattice of the of unit thecell unit (derived cell (derived here) and here) that andof that the of lattice the lattice (i.e. the (i.e. reciprocal the reciprocal lattice). lattice). Since the Since the ∗ ∗ ∗ Q for ·lattice rj = 2π(ha + kb + )except ·factor, (x +we ywhen jb reciprocal reciprocal lattice is zero-valued is zero-valued forQ alllc for Q all Qj a except when when Q+is isQ ageneral reciprocal lattice vector,vector, we we lattice attice sSubstituting zero-valued is zero-valued all for Q except all except when Q iszajQc)reciprocal athe reciprocal islattice aform reciprocal lattice vector,vector, we we this expression back into the structure obtain need only needconsider only=consider the values the values Q = 2π(ha Q = 2π(ha + kb + kb lc ). + The lc ).r The expressed be expressed in terms in terms 2π(hx � j ∗+ kyj + lz∗j ). ∗∗ ∗ can rbecan onsider r the values the values Q =components 2π(ha kb lcreal-space ).lattice +The lclattice ).rjvectors The be expressed be expressed in terms in terms of their of components their x ,= yf , (Q zx 2π(ha ,parallel y+ , z kb parallel to the+ to real-space the vectors tocan givetor give j can F (hkl) =Q j hkl ) exp[2πi(hxj + kyj + lzy )]. Substituting this expression backj into the structure factor, we obtain the general form Q · the r Q=· rreal-space 2π(ha = the 2π(ha + kbreal-space + kb lc ) lattice + · (x lc a) + · (x y ba lattice + zy c) b + z c)vectors ponents ts xj , yj , zxjj ,parallel yj , zj parallel to to vectors to giveto give � ∗ j j j jj j j ∗∗ ∗∗ ∗ j j j ∗ ∗ ∗∗ ∗ j jj jj j = are 2π(hx =spherical, + kyjj + + lz ky + lz+ j2π(hx jky j ). j (Q) Here, we have assumed that=the atoms that F (hkl) fj (Q +j ). hkl ) exp[2πi(hx j so j flz y )]. depends only on the j magnitude of Q, and not on its direction. ∗ expression ∗ into ∗structure ∗the structure ∗ factor, Substituting Substituting this expression this back backthe into factor, we obtain we obtain the general the general form form Q · rj Q=· rj2π(ha = 2π(ha + kb + kb lc ) + · (xlcj a) + · (x yjjba + zyjjc) b + zj c) � � Here, we have assumed that the Fatoms are spherical, so thatj +fjky (Q) depends (hkl)F=(hkl) f=j (Qhklf)j (Q exp[2πi(hx + lz kyy j)].+ lzy )].only on the hkl ) exp[2πi(hx jj + magnitude of Q, and not on itsjdirection.jj j j jj j = 2π(hx = 2π(hx + ky + + lz ky).+ lz ). Here, Here, we have we assumed have assumed that the that atoms the atoms are spherical, are spherical, so thatsofjthat (Q) fdepends only on only theon the j (Q) depends magnitude magnitude of Q, and of Q, not and onnot its direction. on its direction. gxpression this expression back into back the into structure the structure factor, factor, we obtain we obtain the general the general form form � � F (hkl)F=(hkl) f=j (Qhklf)j exp[2πi(hx (Qhkl ) exp[2πi(hx + lz kyy )]. j + kyjj + j + lzy )]. j j ave sumed assumed that the that atoms the atoms are spherical, are spherical, so thatsofjthat (Q) fdepends only on only theon the j (Q) depends
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