Polarization is out of the plane

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1 Physics 055 Lecture 2 Recall from last time the various crystal classes and the dierent translational and rotational symmetry operations which made this enumeration possible. You should also be familiar with the distinction between the CONVENTIONL and the PRIMITIVE unit cells. (Refer to the 2-D Centered Rectangular Bravais Lattice.) Example: Consider the 2-D close-packed structure formed by a uniform collection of marbles. a 2 a + a 2 a For this system a \hexagonal" lattice (lattice vectors ~a and ~a 2 ) is formed with a single \marble" per unit cell. Notice that the Wigner-Seitz cells form hexagons. If identical 2-D planes of marbles are placed directly on top of one another, a simple (primitive) hexagonal 3-D lattice is formed. a 3 unit cell a 2 marble center a Simple Hexagonal (one marble basis) In this case j~a j = j~a 2 j = j~a 3 j =60,==90 Note: this system is no longer close-packed. Let us dene this stacking as //. Looking back at our 2-D hexagonal planes, notice that marbles could be packed more tightly if the marbles are displaced (~a 3 + ~a 2 ) (denoted by B's) or 2(~a 3 + ~a 2 ) (denoted by C's) from one layer to the next so that the marble in the next layer reside in the hollows forms by three neighboring marble in the layer. 7

2 B B C C B B C C B B C C Notice that the B's and C's themselves form a hexagonal 2-D lattice. Now assume that the rst two layers stack to B. For the third layer, either or C translations (assume close-packing) are now possible. If the stacking is BB..., the hexagonal close packed structure (HCP) has been formed. It has a hexagonal 3-D lattice and a two marble basis j ~a j=j ~a 2 j, j ~a 3 j= q 8=3 j ~a j, =60,==90 with marble at (0,0,0) and marble 2 at ( 3 ; 3 ; 2 ) in terms of (~a ;~a 2 ;~a 3 ). If instead the stacking is BCBC..., we have a primitive rhombohedral lattice with a single atom basis j~a j = j~a 2 j = j~a 3 j and = = =60. lternately, we could form a hexagonal unit cell with a 3 marble basis. In addition, this packing is also the same as the conventional face-centered cubic structure (FCC) which has a cubic unit cell and a four atom basis! Question: How are directions in the \real" space lattice designated? For this, a common notation of [u v w] in brackets is used to select a real space vector ~R = u~a + v~a 2 + w~a 3. P Example: Cubic lattice using [3 2 ] ( 3 R ~P is perpendicular to the plane formed by the [3 2 ] lattice vector ~R =3~a + 2 ~a 2 +~a 3 Notice that we form a plane which intercepts the axes at said points. How do we describe the plane? ( u v ) Lowest Common Denominator w )x6=(236) 2 R P For cubic symmetry []? (). In general, this is NOT true. 8

3 Experimental Methods Now that we have dened a system of crystals, it is important to develop a series of experimental procedures in order to study \real" crystals. For this there are a wide variety of techniques to study surfaces and the bulk. \Real" space images! Transmission electron microscopy (TEM) Scanning tunneling microscopy (STM) Diraction Images! -ray diraction, neutron scattering We will discuss diraction techniques: Sources of x-rays - conventional, synchrotron Sources of neutrons - reactors, spellation Conventional x-ray source Filament - e s h ν Low energy electron diraction (LEED) Metal target Fe, Co, Cu, g, Mo, Cr Two types of radiation:. Bremsstrahlung - \de"acceleration of incident e's 2. Characteristic line spectra - photons from atomic transition n=2 to K n=3 to K However, the incident e's must excite s electrons to the rst allowable transition KeV K α Synchrotron Light Source Bremsstrahlung light cone E H Photon Intensity K β a electrons/positrons move in a ring Bremsstahlung λ increasing cceleration Potential limit Photon Intensity (log scale) Photon Energy (kev) 9

4 Neutrons: Reactor - Thermal neutrons from ssion events E neutron =0.025 ev = 25 mev :5 Spellation - High energy protons impinge on a uranium pile. Reactor (steady state beam) High Flux Neutron neutrons sample Reactor collimators Spellation source (pulsed beam) Proton accelerator high energy neutrons Moderator sample Uranium target low energy neutrons Scattering of x-rays: This can be solved classically by using Maxwell's equations, or quantum mechanically. In either case the calculation is involved, and will not be reproduced here. (See e.g., -Ray Diraction, zaro, Kaplow, Kato, Weiss, Wilson, Young.) The J. J. Thomson model of scattering due to acceleration of the electrons follows. The incident photon E ~ = E ~ 0 expfi(!t kz)g, can be described as a travelling wave and the electron is considered to be xed by an attached spring with spring constant K. y-z plane y E H z x-z plane -x electron k is the spring constant 0

5 ~F e = e[ E ~ + (_xh)] kx c m! x= e ~ E k~x! 0 = q k=m! x= e m ~ E! 2 0 ~x Solution: ~x = x 0 e i!t^x x 0 =- ee 0 m a =j ~ x j=!2 ee 0 m(! 2 0! 2 ) (! 2 0! 2 ) Since m nucleus >> m electron No contribution by the nucleus. The electron is accelerated in the x-direction and radiates energy. For exact details refer to any text on classical electrodynamics (e.g., Jackson, Reitz & Milford). Time and usefulness do not allow us to delve further. There are however two components to the incident radiation: Polarization is out of the plane Es φ.. x φ E s Ei y θ z Ei y x θ z Polarization is in the plane Es Ei x θ φ z (ssume that your eye /detector is in the x-z plane) Note: is the angle between the scattered beam and the direction of the accelerated electron is the angle from the ^z-axis E s = a e sin Rc 2 E s E 0 = e 2! 2 sin Rc 2 m(! 2 0! 2 ) I s I 0 = E2 s E 2 0 e 4! 4 sin 2 =+ R 2 m 2 c 4 (! 0 2! 2 ) 2

6 [Note: If!! 0 (i.e., at an absorption edge) Thompson theory breaks down.] In general,!>>! 0 and the x-ray intensity / (a \light bulb") R 2 [Incoherent (i.e., Compton scattering) is also present.] I s = e 4 R 2 m 2 c 4 sin2 I 0 s was just shown by viewing only in the x-z plane: For x-rays polarized in the x-dir + = =2 For x-rays polarized in the y-dir = /2 Thus for an unpolarized beam I = I 2 0(x z)+ I 2 0(y z) I s =! 2 e2 hsin i 2 =2+sin 2 (=2 ) mc 2 R 2 2 I s = R 2 e 2 mc 2! 2 +cos 2 2 Polarization factor The polarization factor is just one of many necessary \corrections." 2

7 Physics 055 Lecture 3 Title: Scattering Due to toms Last time we calculated the scattering of photons (using unpolarized incident radiation) from a single classically bound electron. I s =! I o R e 2 2 +cos 2 2 mc 2 2 Thus the total scattering cross section is given by: T = ZZ Is I o R 2 sindd = e mc 2! 2 =0: cm 2 : In reality atoms are a collection of electrons. These electrons are spatially separated and to rst order non-interacting. Thus, their scattering will not superimpose completely coherently. For example, assume two electrons such that one is at the origin whilst the other is located at ~r i. Scattered radiation k Incident radiation s k 0 r i p 2 p electron at origin (reference point) The incoming wave is described by ~E i = E ~ 0 exp ni(! o 0 t ~ k0 ~r) 3

8 and the outgoing wave is given by ~E f = E ~ s exp ni(! o s t ~ ks ~r) Recall " photon =h! =hjkjc; = 2 jkj ; = c Since each electron independently scatters the incoming photon eld, there is necessarily a path dierence p +p 2 between the scattered waves. This path dierence implies a phase dierence. p = ^k 0 ~r i p 2 =+^k s ~r i pp +p 2 = ^ks ^k0 ~ri phase dierence ssume elastic scattering 0 = s so path 2 = = ~ ks ~ k0 ~ri =~s~r i and ~ k s ~ k0 is the change in the \momentum" of the scattered waves (i.e., diraction vector). Thus we have a new and important quantity called the scattering vector or diraction vector. ~s ~ k d ~ k s ~ k0 To get the total scattered amplitude for a single atom, it is clearly necessary to take the superposition of scattering from all the electrons. The amplitude for the i th electron is i = f e exp i(~s ~r i ) where f e is the amplitude for an electron at the origin. Thus, the total amplitude is = where Z is the atomic number. Z i= I intensity = =f 2 e i = f e exp i(~s ~r i ) i j I=f 2 e e i~s~r j i;j i e i~s~r i e i~s(~r i ~r j ) This summation is for the superposition of point-like e's. This distribution clearly changes with time. In actuality, the time averaged distribution is required. Replacing the point-like e's with a charge distribution gives Z = f e V (r)e i~s~r dv 4

9 Thus a new quantity, f atom f e tomic scattering form factor, is realized. Notice that as ~s! 0, (towards the forward scattering direction) the phase dierences go to zero. Thus, intuitively one expects a smoothly decreasing intensity with increasing angle. Now for the scattering from an arbitrary collection of atoms all atoms T = atom = atoms k where f k = f e R k (r)e i(~s~r) d~r for each atom f k e i~s~r k ~r k position of the k th atom f k atomic form factor for the k th atom The total observed intensity is I T = T T= k;` f k f ` ei~s(~r k ~r`) Notice that at no point has the assumed periodicity of a crystalline lattice centered into this discussion! This will be done next time. However, we will go a bit further. SSUME a collection of identical atoms: I T = k=` + k6=` fk f ` ei~s(~r k ~r`) I T = k=` f k f k ei~s(~r k ~r ) k + f k f ` ei~s(~r k k6=` ~r`) ssume there are N identical atoms I T = Nf 2 + k6=` f k f ` ei~s(~r k ~r`) () The rst term has no phase or angular dependence. It represents a coherent background and is called the self-scattering term. nother venture: ssume we replace the discrete sum of T function: with an atomic distribution Z T (~s) =f (~r)e i~s~r dv V where f is the individual atom form factor and (~r) is the atomic number density. Then Z I (~s) =(~s) (~s)=jfj 2 (~r)(~r 0 ) e i~s(~r ~r0) d~rd~r 0 5

10 R r r Origin Then the second term in Eqn. (on the previous page) becomes I 0 (s) =jfj 2 Z Z (~r)(~r+ ~ R)e i~s~r d~rd(~r+ ~ R) We can dene a new quantity P ( ~ R) Z (~r) Let ~r 0 = ~r+ ~ R ~r + R ~ d~r pair distribution function Now I 0 (~s) =jfj 2 Z ( ~ R)e i~s~r d ~ R Clearly, I 0 (~s) is just the Fourier transform of the pair distribution function. IMPORTNT - This quantity is what a diraction experiment extracts From this information and insight the actual atomic positions can often be ascertained. 6

11 Physics 055 Lecture 4 Title: Coherent Scattering from a crystal. Scattering from a periodic arrangement of atoms and the construction of the reciprocal lattice. Recall the total atomic scattering amplitude Now invoke lattice periodicity (~s) = all atoms k f k e i~s~r k R = n ~a + n 2 ~a 2 + n 3 ~a 3 nd also assume (at present) there is just one atom per unit cell Thus ~r k = n ~a + n 2 ~a 2 + n 3 ~a 3 f k = f (~s) =f all atoms k e i~s(n ~a +n 2 ~a 2 +n 3 ~a 3 ) This sum is somewhat awkward to follow ifn ;n 2 ;n 3!. Let us examine the sum in the range 0 n ;n 2 ;n 3 N-. Hence there are only N 3 atoms to deal with. (~s) =f N n =0 N e i(~sa )n n 2 =0 N e i(~s~a 2 )n 2 n 3 =0 e i(~s ~a 3)n 3 Now each sum can be simplied if the following substitution is made: 0 finite crystal N- N n=0 e i(~s~a)n = 7 = n=0 e i(~s~a)n n=n e i(~s~a)n e i(~s~a)n e i(~s~a)n n=0 = e in (~s~a) n=0 n=0 e i(~s~a)n e i(~s~a)n = e in (~s~a) e i(~s~a) e i(~s~a) e i(~s~a) e in (~s~a) e i(~s~a) = 2( cos ~s ~a)

12 or (~s) = f h e in (~s~a ) ih ih e in (~s~a 2 ) e in (~s~a 3 )i e i~s~a 2 e i~s~a 2 e ~s~a 3 2( cos ~s ~a )2( cos ~s ~a 2 )2( cos ~s ~a 3 ) The intensity I = * gives I = jfj 2 ( cos N ~s ~a )( cos ~s ~a ) :::: ( cos N ~s ~a 3 )( cos ~s ~a 3 ) ( cos ~s ~a ) 2 :::: ( cos ~s ~a 3 ) 2 I = jfj 2 ( cos N ~s ~a )( cosn (~s ~a 2 )) ( cosn (~s ~a 3 )) ( cos ~s ~a )( cos ~s ~a 2 )( cos ~s ~a 3 ) or I = jf j 2 sin 2 (N (~s~a )=2)sin 2 (N~s~a 2 =2)sin 2 (N~s~a 3 ) sin 2 (~s~a =2)sin 2 (~s~a 2 =2) sin 2 (~s~a 3 =2) The behavior of the intensity is relatively straightforward. ~s ~a 2 =2n and ~s ~a 3 =2m. Thus (for one set of terms) Let N=0 and let both sin 2 N~s ~a 2 =2 sin 2 ~s ~a 2 =2 = sin 2 Nn sin 2 n = N2 (using L'Hopital's rule) or since sin(nn)! 0 expand top and bottom in Taylor's series sin x x if x! 0 (choose n =0) Thus sin 2 Nn sin 2 n (Nn)2 (n) 2 = N 2 I = jfj 2 sin2 N~sa sin 2 ~s ~a N 4 Plot: Numerator = sin 2 5(~s ~a ) and Denominator = sin 2 (~s ~a )(0~s~a 2) ~s ~a ( radians) ~s ~a ( radians) 8

13 Now for sin 2 5~s ~a 00 sin 2 ~s ~a So ~s ~a ( radians) I(0) = jfj = jfj 2 N 6 Now compare this intensity with the self scattering whichwas proportional to the number of atoms or N 3 (in this case) when N atoms! In addition, as N becomes large the intensity due to nite size eects (so called \F" peaks) become unobservable. However, these fringes can be seen in very small crystals using electron diraction. They can also be seen in surface scattering from crystals with steps. s N!, the curves become -functions with FWHM (Full widths at half maximum) of 2/N and height N 6 at the positions ~s ~a = 2h ~s ~a 2 = 2k h; k; ` are integers ~s ~a 3 = 2` 9 >= This is the Laue condition! >; This set of expressions are called the Laue diraction conditions. When ~s~a ;~s~aa 2 ;~s~a 3 are all multiples of 2, the amplitudes interfere constructively. This condition is equivalent to Bragg's law n = 2d sin. (Constructive interference a scattered waves) To review: One atom per unit with a crystal length of N- on a side 9

14 N 3 sites (~s) =f i e i~s(n ~a +n 2 ~a 2 + n 3 ~a 3 ) gives I = jfj 2 sin2 (N~s ~a )=2:::: sin 2 (~s ~a )=2::::: If we restrict our sum to ~s's such that the Laue conditions are met then ~s ~a i =2` and all of the scattering is coherent. (i =,2,3) (~s) =fn 3 If we dene a normalizing f = f 0 =N 3, then (~s) =f 0 (N independent) What if there are two atoms per cell tom is located at n ~a + n 2 ~a 2 + n 3 ~a 3 While tom B is at n ~a + n 2 ~a 2 n 3 ~a 3 + ~ where ~ is not a lattice vector. N3 sites The total (~s) =f j N 3 e i~s(n ~a +n 2 ~a 2 +n 3 ~a 3) sites + f B j e i~s(n ~a +n 2 ~a 2 +n 3 ~a 3 + ~ ) Restrict sum to those that satisfy the Laue condition. The set of ~s's which satisfy the Laue condition are designated ~ G and are called the reciprocal lattice vectors. ( G)=f ~ N 3 +f B N 3 e i~ G ~ f 0 =f=n 3 ( G)=f ~ +f 0 0 B ei~ G ~!there are two terms here, one for each atom within the unit cell ( the basis). Generalizing gives unit cell atoms ~G = j f j e i ~ G~r j and I ~G = structure factor ~r j is the relative position of atoms within the unit cell. 20

Solid State Physics Lecture 3 Diffraction and the Reciprocal Lattice (Kittel Ch. 2)

$Solid State Physics Lecture 3 Diffraction and the Reciprocal Lattice (Kittel Ch. 2)$ Solid State Physics 460 - Lecture 3 Diffraction and the Reciprocal Lattice (Kittel Ch. 2) Diffraction (Bragg Scattering) from a powder of crystallites - real example of image at right from http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow.html