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Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of
Physics and Atronomy, Louisiana State University, LA 70803). This
material may not be reproduced for profit, modified or published in any
form (this includes electronic redistribution) without the prior written
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\title{Chapter 3: The Classical Theory of Crystal Diffraction}
\author{Bragg}
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%\Large
In the last two chapters, we learned that solids generally form
periodic structures of different symmetries and bases. However, given a
solid material, how do we learn what its periodic structure is? Typically,
this is done by diffraction, where we project a beam (of either particles or
radiation) at a solid with a wavelength $\lambda\approx$ the characteristic
length scale of the lattice ( $a\approx $ twice the atomic or molecular radii
of the constituents).
\begin{figure}[htb]
\centerline{\includegraphics[height=3.0in,keepaspectratio,clip=true]{scatter.pdf}}
\caption[]{\em{Scattering of waves or particles with wavelength of
roughly the same size as the lattice repeat distance allows us to
learn about the lattice structure. Coherent addition of two particles
or waves requires that $2d\sin{\theta}=\lambda$ (the Bragg condition),
and yields a scattering maximum on a distant screen.}}
\label{scatter}
\end{figure}
Diffraction of waves and particles (with de Broglie wavelength $\lambda=h/p$)
of $\lambda\approx a$ allows us to learn about the
%large scale (small $k=2\pi/\lambda$ )
periodic structure of crystals.
In a diffraction experiment one identifies Bragg peaks which
originate from a coherent addition of scattering events in
multiple planes within the bulk of the solid.
However, not all particles with de Broglie wavelength $\lambda\approx a$
will work for this application. For example, most charged particles cannot
probe the bulk properties of the crystal, since they lose energy to the
scatterer very quickly. Recall, from classical electrodynamics, the
rate at which particles of charge $q$, mass $M$, and velocity $v$ lose
energy to the electrons of charge $e$ and mass $m$ in the crystal is given
roughly by
\beq
\der{E}{x}\approx -\frac{4\pi n q^2 e^2}{mv^2} \ln{\lep\frac{m\gamma^2v^3}{qe\omega_0}\rip} \sim \frac{q^2}{v^2}\,.
\label{damp}
\eeq
As an example, consider a non-relativistic electron scattering into a solid
with $a\approx 2\A$. If we require that
$a=\lambda=h/p=12.3\times10^{-8}{\rm{cm}}/\sqrt{E}$ when $E$ is measured in
electron volts, then $E\approx 50$eV. If we solve Eq.~\ref{damp} for the
distance $\delta x$ where the initial energy of the incident is lost requiring
that $\delta E =E$, when $n\approx 10^{23}$/cm$^3$ we find that
$\delta x\approx 100\A$.
Thus, if $\lambda\approx a$, the electrons do not penetrate into the
bulk of the sample (typically the first few hundred $\A$ of most
materials are oxidized, or distorted by surface reconstruction of
the dangling bonds at the surface, etc. See Fig.~\ref{cpscat})
\begin{figure}[htb]
\centerline{\includegraphics[height=3.0in,keepaspectratio,clip=true]{cpscat.pdf}}
\caption[]{\em{An electron about to scatter from a typical material.
However, at the surface of the material, oxidation and surface reconstruction
distort the lattice. If the electron scatters from this region,
we cannot learn about the structure of the bulk.}}
\label{cpscat}
\end{figure}
Thus, electrons do not
make a very good probe of the bulk properties of a crystal (instead
in a process call low-energy electron diffraction, LEED, they may be used to
study the surface of especially clean samples. I.e. to study things like
surface reconstruction of the dangling bonds, etc.).
Thus although they are obviously easier to accelerate (electrons
or ion beams), they generally do not penetrate into the bulk and
so tell us more about the surface properties of solids which are
often not representative of the bulk.
Thus the particle of the choice to determine bulk properties is
the neutron which is charge neutral and scatters only from the nuclei.
Radiation is often also used. Here the choice is only a matter
of the wavelength used. X-rays are chosen since then $\lambda\approx a$
\section{Classical Theory of diffraction}
In this theory of diffraction we will be making three basic
assumptions.
\begin{enumerate}
\item That the operator which describes the coupling of the
target to the scattered "object" (in this case the
operator is the density) commutes with the Hamiltonian.
Thus, this will be a classical theory.
\item We will assume some form of Huygens principle: that every
radiated point of the target will serve as a secondary
source spherical wavelets of the same frequency as the source and
the amplitude of the diffracted wave is the sum of the
wavelengths considering their amplitudes and relative
phases. (For light, this is equivalent to assuming that it is
unpolarized, and that the diffraction pattern varies
quickly with scattering angle $\theta$ so that the
angular dependence of a unpolarized dipole,
$1+(\cos{\theta})^2$, may be neglected.)
\item We will assume that resulting spherical waves are not
scattered again. In the fully quantum theory which we
will derive later for neutron scattering, this will
correspond to approximating the scattering rate by
Fermi's golden rule (first-order Born approximation).
\end{enumerate}
The basic setup of a scattering experiment is sketched in
Fig.~\ref{setup}.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.5in,keepaspectratio,clip=true]{setup.pdf}}
\caption[]{\em{Basic setup of a scattering experiment.}}
\label{setup}
\end{figure}
Generally, we will also assume that $|R| \gg |r|$, so that
we may always approximate the amplitude of the incident waves on the target
as plane waves.
\beq
A_P=A_O e^{i\lep\k_0\cdot(\R+\r)-\omega_0 t\rip}\,.
\label{plane}
\eeq
Then, consistent with the second assumption above,
\beq
A_B(R')\propto \int d^3r A_P \rho(\r)\frac{e^{i\k\cdot(\R'-\r)}}{\left|\R'-\r\right|}\,,
\eeq
which, after substitution of Eq.\ref{plane}, becomes
\beq
A_B(R')\propto A_O e^{i\lep\k_0\cdot\R+\k\cdot\R'-\omega_0 t\rip}
\int d^3r \rho(\r)\frac{e^{-i(\k-\k_o)\cdot\r}}{\left|\R'-\r\right|}\,.
\eeq
At very large $R'$ (ie.\ in the radiation or far zone)
\beq
A_B(R')\propto \frac{A_O e^{i\lep\k_0\cdot\R+\k\cdot\R'-\omega_0 t\rip}}
{R'}
\int d^3r \rho(\r) e^{-i(\k-\k_o)\cdot\r}\,.
\eeq
Or, in terms of the scattered intensity $I_B\propto \left| A_B\right|^2$
\beq
I_B\propto \frac {\left| A_O\right|^2}{R'^2} \left|\int d^3r \rho(\r) e^{-i(\k-\k_o)\cdot\r}\right|^2
\,.\eeq
The scattering intensity is just the absolute square of the Fourier transform
of the density of scatterers. If we let $\K=\k-\k_0$ (cf.~Fig.~\ref{scatter}),
then we get
\beq
I_B(\K)\propto \frac {\left| A_O\right|^2}{R'^2} \left|\int d^3r \rho(\r) e^{-i\K\cdot\r}\right|^2=\frac {\left| A_O\right|^2}{R'^2} |\rho(\K)|^2
\,.\eeq
From the associated Fourier uncertainty principle $\Delta k\Delta x\approx \pi$,
we can see that the resolution of smaller structures requires larger
values of $\K$ (some combination of large scattering angles and short
wavelength of the incident light), consistent with the discussion
at the beginning of this chapter.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.5in,keepaspectratio,clip=true]{scatterloss.pdf}}
\caption[]{\em{Since the measured scattering intensity
$I(\K)\propto |\rho(\K)|^2$ the complex phase information is lost.
Thus, a scattering experiment does not provide enough information to
invert the transform
$\rho(\r)=\int \frac{d^3r}{(2\pi)^3} \rho(\K) e^{+i\K\cdot\r}$.}}
\label{scatterloss}
\end{figure}
In experiments the intensity $I$ as a function of the scattering angle
$\K$ is generally measured. In principle this is under-complete
information. In order to invert the Fourier transform (which is
a unitary transformation) we would need to know both the real
and imaginary parts of
\beq
\rho(\K)=\int d^3r \rho(\r) e^{-i\K\cdot\r}\,.
\eeq
Of course, if the experiment just measures $I\propto |\rho(\K)|^2$, then
we lose the relative phase information (i.e. $\rho(\K) =\rho_Ke^{i\theta_K}$
so that $I\propto |\rho_K|^2$, and the phase information $\theta_K$
is lost).
So, from a complete experiment, measuring $I(\K)$ for all scattering angles,
we do not have enough information to get a unique $\rho(\r)$ by inverting
the Fourier transform.
Instead experimentalists analyze their data
by proposing a feasible model structure (i.e. a $\rho(\r)$ corresponding
to some guess of which of one the 14 the Bravais lattice and the basis),
Fourier transform this, and compare it to the experimental data. The
parameters of the model are then adjusted to obtain a best fit.
\section{Scattering from Periodic Structures}
Given this procedure, it is important to study the scattering
pattern that would arise for various periodic structures. The density in
a periodic crystal must have the same periodicity of the crystal
\beq
\rho(\r+\r_n)=\rho(\r)\;\;\;{\rm{where}}\;\; \r_n=n_1\aone+n_2\atwo+n_3\athree
\eeq
for integer $n_1,n_2,n_2$.
This also implies that the Fourier coefficients of $\rho$ will be
chosen from a discrete set. For example, consider a 1-d periodic structure
\begin{figure}[htb]
\centerline{\includegraphics[width=6.0in,keepaspectratio,clip=true]{oneD_periodic.eps}}
\caption[]{\em{ }}
\label{}
\end{figure}
\beq
\rho(x+na)=\rho(x)\,.
\eeq
Then we must choose the $G_n$
\beq
\rho(x)=\sum_n\rho_ne^{iG_n x}\,,
\eeq
so that
\beqa
\rho(x+ma)&=&\sum_n\rho_ne^{iG_n (x+ma)}=\sum_n\rho_n e^{iG_n(x)}
e^{iG_n ma}\nonumber \\
&=&\sum_n\rho_ne^{iG_n(x)}=\rho(x)\,,
\eeqa
I.e. $e^{iG_n ma}=1$, or $G_n= 2 n\pi/a$ where $n$ is an integer.
This may be easily generalized to three dimensions, for which
\beq
\rho(\r)=\sum_\G \rho_\G e^{i\G\cdot\r}
\eeq
where the condition of periodicity $\rho(\r+\r_n)=\rho(\r) $ means that
\beq
\G\cdot\r_n=2\pi m\;\;\;m\in{\cal{Z}}
\eeq
where ${\cal{Z}}$ is the group of integers (under addition).
Now, lets consider $\G$ in some three-dimensional space and decompose it
in terms of three independent basis vectors for which any two are not
parallel and the set is not coplanar
\beq
\G=h\gone+k\gtwo+l\gthree\,.
\eeq
The condition of periodicity then requires that
\beq
\lep h\gone+k\gtwo+l\gthree\rip\cdot n_1\aone=2\pi m\;\;\;m\in{\cal{Z}}
\eeq
with similar conditions of the other principle lattice vectors $\atwo$ and
$\athree$. Since $\gone$, $\gtwo$ and $\gthree$ are not parallel or coplanar, the only
way to satisfy this constraint for arbitrary $n_1$ is for
\beq
\gone\cdot\aone=2\pi \;\;\;\gtwo\cdot\aone=\gthree\cdot\aone=0
\label{gcon}
\eeq
or some other permutation of 1 2 and 3, which would just amount to
a renaming of $\gone$, $\gtwo$, and $\gthree$. The set ($\gone$, $\gtwo$, $\gthree$)
are called the basis set for the reciprocal lattice. They may be constructed
from
\beq
\gone=2\pi\frac{\atwo\times\athree}{\aone\cdot(\atwo\times\athree)}
\;\;\;
\mbox{plus cyclic permutations}\,.
\eeq
It is easy to see that this construction satisfies Eq.~\ref{gcon}, and that there is a one to one correspondence between the lattice and its reciprocal
lattice. So, the reciprocal lattice belongs to the same point group as
the real-space lattice\footnote{One should note that this does not mean
that the reciprocal lattice must have the same Bravais lattice structure
as the real lattice. For example, the reciprocal of a fcc lattic is bcc
and vice versa. This is consistent with the the statement that the
reciprocal lattice belongs to the same point group as the real-space
lattice since fcc and bcc share the $O_h$ point group}.
\subsection{The Scattering Intensity for a Crystal}
Lets now apply this form for the density
\beq
\rho(\r)=\sum_\G \rho_\G e^{i\G\cdot\r}
\eeq
to our formula for the scattering intensity
\beq
I_B(\K)\propto \frac {\left| A_O\right|^2}{R'^2} \left|\int d^3r
\sum_\G \rho_\G e^{-i(\K-\G)\cdot\r}\right|^2
\eeq
The integral above is simply
\beq
V\delta_{\G,\K}=\left\{\begin{array}{l} \mbox{$V$ if $\G=\K$} \\
\mbox{$0$ if $\G\neq \K$}
\end{array}\right.\,,
\eeq
where $V$ is the lattice volume, so
\beq
I_B(\K)\propto \frac {\left| A_O\right|^2}{R'^2}
\left|\rho_\G\right|^2V^2\delta_{\G,\K}
\eeq
This is called the Laue condition for scattering.
The fact that this is proportional to $V^2$ rather than $V$ just
indicates that the diffractions spots, in this approximation, are
infinitely bright (for a sample in the thermodynamic limit). Of
course, this is because the spots are infinitely narrow or fine. When
real broadening is taken into account, $I_B(\K)\propto V$ as expected.
Then as $G = h\gone+k\gtwo+l\gthree $, we can label the spots with the
three integers ($h,k,l$ ), or
\beq
I_{hkl}\propto \left| \rho_{hkl}\right|^2\,.
\eeq
Traditionally, negative integers are labled with an overbar, so
$-h\to\overline{h}$. Then as $\rho(\r)$ is real, $\rho_\G=\rho_{-\G}$,
or
\beq
I_{\overline{h}\overline{k}\overline{l}}=I_{hkl}
\;\;\;\mbox{Friedel's rule}
\eeq
Most scattering experiments are done with either a rotating crystal,
or a powder made up of many crystalites. For these experiments, Friedel's
rule has two main consequences
\begin{itemize}
\item For every spot at $\k-\k_0=\G$, there will be one at $\k'-\k_0=-\G$.
Thus, for example if we scatter from a crystal with a 3-fold symmetry axis,
we will get a six-fold scattering pattern. Clearly this can only happen,
satisfy the Laue condition, and have $|\k|=|\k_0|$, if the crystal is rotated
by $\pi$ in some axis perpendicular to the
three-fold axis. In fact, single-crystal experiments are usually done
either by mounting the crystal on a precession stage (essentially like
an automotive universal joint, with the drive shaft held fixed, and the
joint rotated over all angles), or by holding the crystal fixed and
moving the source and diffraction screen around the crystal.
\item The scattering pattern always has an inversion center, $\G\to-\G$
even when none is present in the target!
\end{itemize}
\subsection{Bragg and Laue Conditions (Miller Indices)}
Above, we derived the Laue condition for scattering; however, we began
this chapter by reviewing the Bragg condition for scattering from
adjacent planes. In this subsection we will show that, as expected,
these are the some condition.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{miller.pdf}}
\caption[]{\em{
Miller indices identification of planes in a lattice.
Highlighted by the solid lines are the parallel planes formed by
($1,2,2$) translations along the principle lattice vectors
($\aone,\atwo,\athree$), respectively. Typically these integers are labeled
by ($m,n,o$), however, the plane is not typically labeled as the
($m,n,o$) plane. Rather it is labeled with the inverses
$h'=1/m\;\;\;k'=1/n\;\;\;l'=1/o$. Since these typically are not integers,
they are multiplied by $p$, the smallest integer such that
$p(h',k',l')=(h,k,l)\in{\cal{Z}}$.
In this case, $p=2$, and the plane is labeled as the ($2,1,1$) plane.
Note that the plane formed by ($2,4,4$) translations along the principle
lattice vectors is parallel to the ($2,1,1$) plane.
}}
\label{miller}
\end{figure}
Consider the real-space lattice shown in Fig.~\ref{miller}.
Highlighted by the solid lines are the parallel planes formed by
($1,2,2$) translations along the principle lattice vectors
($\aone,\atwo,\athree$), respectively. Typically these integers are labeled
by ($m,n,o$), however, the plane is not typically labeled as the
($m,n,o$) plane. Rather it is labeled with the inverses
\beq
h'=1/m\;\;\;k'=1/n\;\;\;l'=1/o\,.
\eeq
Since these typically are not integers, they are multiplied by $p$,
the smallest integer such that
\beq
p(h',k',l')=(h,k,l)\in{\cal{Z}}\,.
\eeq
In this case, $p=2$, and the plane is labeled as the ($2,1,1$) plane.
On may show that the reciprocal lattice vector $\G_{hkl}$ lies
perpendicular to the ($h,k,l$) plane, and that the length between adjacent parallel planes $d_{hkl}=2\pi/G_{hkl}$. To show this, note that the
plane may be defined by two non-parallel vectors $\v_1$ and $\v_2$ within
the plane. Let
\beq
\v_1=m\aone-n\atwo=\aone/h'-\atwo/k'\;\;\;\v_2=o\athree-n\atwo=
\athree/l'-\atwo/k'\,.
\eeq
Clearly the cross product, $\v_1\times\v_2$ is perpendicular to the
($h,k,l$) plane
\beq
\v_1\times\v_2= -\frac{\athree\times\aone}{h'l'}
-\frac{\aone\times\atwo}{h'k'}
-\frac{\atwo\times\athree}{k'l'}\,.
\eeq
If we multiply this by $-2\pi h'k'l'/\aone\cdot(\atwo\times\athree)$, we
get
\beqa
&& \frac{-2\pi h'k'l'\v_1\times\v_2}{\aone\cdot(\atwo\times\athree)}=\nonumber\\
&&\frac{2\pi p}{p}\leb
k'\frac{\athree\times\aone}{\aone\cdot(\atwo\times\athree)}+
l'\frac{\aone\times\atwo}{\aone\cdot(\atwo\times\athree)}+
h'\frac{\atwo\times\athree}{\aone\cdot(\atwo\times\athree)}
\rib=\nonumber\\
&&\G_{hkl}/p
\eeqa
Thus, $\G_{hkl}\perp$ to the ($h,k,l$) plane. Now, if $\gamma$ is the angle
between $\aone$ and $\G_{hkl}$, then the distance $d_{h'k'l'}$ from the origin
to the ($h',k',l'$) plane is given by
\beq
d_{h'k'l'} = m|\aone|\cos{\gamma}=
\frac{|\aone|\aone\cdot\G_{hkl}}{h'|\aone||\G_{hkl}|}=
\frac{2\pi h}{h'G_{hkl}}=
\frac{2\pi p}{G_{hkl}}
\eeq
Then, as there are $p$ planes in this distance (cf.~Fig.~\ref{miller}),
the distance to the nearest one is
\beq
d_{hkl}=d_{h',k',l'}/p=2\pi/G_{hkl}
\eeq
With this information, we can reexamine the Laue scattering condition
$\K=\k-\k_o=\G_{hkl}$, and show that it is equivalent to the more intuitive
Bragg condition. Part of the Laue is condition is that
$|\K|=K=|\k-\k_0|=G_{hkl}$, now
\beq
K=2 k_0 \sth = \frac{4\pi}{\lambda}\sth
\;\;\;\mbox{and}\;\;\;
G_{hkl}=2\pi/d_{hkl}
\eeq
thus, the Laue condition implies that
\beq
1/d_{hkl}=2\sth/\lambda
\;\;\;\mbox{or}\;\;\;
\lambda=2d_{hkl}\sth
\eeq
which is the Bragg condition.
\begin{figure}[htb]
\centerline{\includegraphics[height=3.0in,keepaspectratio,clip=true]{braggsquare.pdf}}
\caption[]{\em{
Comparison of the Bragg $\lambda=2d_{hkl}\sth$ and Laue $\G_{hkl}=\K_{hkl}$
conditions for scattering.
}}
\label{braggsquare}
\end{figure}
Note that the Laue condition is more
restrictive than the Bragg condition; it requires that {\em{both}} the
magnitude and the direction of $\G$ and $\K$ be the same. However,
there is no inconsistency here, since whenever we apply the Bragg
condition, we assume that the plane defined by $\k$ and $\k_0$ is
perpendicular to the scattering planes (cf.\ Fig.~\ref{braggsquare}).
\subsection{The Structure Factor}
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{molbasis.pdf}}
\caption[]{\em{
Examples of lattices with non-trivial bases. The CuO$_2$ lattice
(left) is characteristic of the cuprate high-temperature
superconductors. It has a basis composed of one Cu and two O atoms imposed
on a simple cubic lattice. The BCC lattice(right) can be
considered as a cubic lattice with a basis including an atom at the corner
and one at the center of the cube.
}}
\label{molbasis}
\end{figure}
Thus far, we have concentrated on the diffraction pattern for a periodic
lattice ignoring the fine structure of the molecular of the basis. Examples
of non-trivial molecular bases are shown in Fig.~\ref{molbasis}. Clearly
the basis structure will effect the scattering (Fig.~\ref{formfac}). For
example, there will be interference from the scattering off of the Cu and
two O in each cell. In
fact, even in the simplest case of a single-element basis composed of a
spherical atom of finite extent, scattering from one side of the atom
will interfere with that from the other. In each case, the structure
of the basis will change the scattering pattern due to interference of the
waves scattering from different elements of the basis. The structure
factors account for these interference effects. The information
about this interference, and the basis structure is contained in the
atomic scattering factor $f$ and the structure factor $S$.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{formfac.pdf}}
\caption[]{\em{
Rays scattered from different elements of the basis, and from different
places on the atom, interfere giving the scattered intensity additional
structure described by the form factor $S$ and the atomic form factor
$f$, respectively.
}}
\label{formfac}
\end{figure}
To show this reconsider the scattering formula
\beq
I_{hkl}\propto\left|\rho_{hkl}\right|^2
\eeq
The Fourier transform of the density may be decomposed into an integral
over the basis cell, and a sum over all such cells
\beqa
\rho_{hkl}&=&\frac{1}{V}\int d^3r\rho(\r)e^{-i\G_{hkl}\cdot\r}
=\frac{1}{V}\sum_{\rm{cells}}\int_{\rm{cell}} d^3r\rho(\r)e^{-i\G_{hkl}\cdot\r}
\nonumber\\
&=&\frac{1}{V}\sum_{n_1,n_2,n_3}
\int_{\rm{cell}} d^3r\rho(\r)e^{-i\G_{hkl}\cdot(\r+\r_n)}
\eeqa
where the location of each cell is given by $\r_n=n_1\aone+n_2\atwo+a_3\athree$.
Then since $\G_{hkl}\cdot\r_n=2\pi m$, $m\in{\cal{Z}}$,
\beq
\rho_{hkl}=\frac{N}{V}\int_{\rm{cell}} d^3r\rho(\r)e^{-i\G_{hkl}\cdot\r}
\eeq
where $N$ is the number of cells and $\frac{N}{V}=1/V_c$, $V_c$ the volume of
a cell. This integral may
be further subdivided into an integral over the atomic density of each
atom in the unit cell. If $\alpha$ labeles the different elements of the
basis, each with density $\rho_\alpha(\r')$
\beq
\rho_{hkl}=\frac{1}{V_c}\sum_\alpha e^{-i\G_{hkl}\cdot\r_\alpha}
\int d^3r'\rho_\alpha(\r')e^{-i\G_{hkl}\cdot\r'}
\eeq
The atomic scattering factor f and the structure factor may then
be defined as parts of this integral
\beq
f_\alpha=\int d^3r'\rho_\alpha(\r')e^{-i\G_{hkl}\cdot\r'}
\eeq
so
\beq
\rho_{hkl}=\frac{1}{V_c}\sum_\alpha e^{-i\G_{hkl}\cdot\r_\alpha}
f_\alpha=\frac{S_{hkl}}{V_c}
\eeq
$f_\alpha$ describes the interference of spherical waves emanating from
different points within the atom, and $S_{hkl}$ is called the structure
factor. Note that for lattices with an elemental basis $S=f$.
If we imagine the crystal to be made up of isolated atoms like that shown
on the right in Fig.~\ref{formfac} (which is perhaps accurate for an ionic
crystal) then, since the atomic charge density is spherically symmetric about
the atom
\beqa
f_\alpha&=&\int d^3r'\rho_\alpha(\r')e^{-i\G_{hkl}\cdot\r'}
=-\int r'^2dr'd(\cth)d\phi\, \rho_\alpha(r')e^{-G_{hkl}r'\cth}\nonumber\\
&=&
4\pi\int r'^2dr' \rho(r')\frac{\sin{G_{hkl}r'}}{G_{hkl}r'}\,.
\eeqa
As an example, consider a spherical atom of charge $Ze^-$, radius $R$,
and charge density
\beq
\rho_\alpha(r')=\frac{3Z}{4\pi R^3}\theta(R-r')
\eeq
then
\beqa
f_\alpha&=&
\frac{3Z}{R^3}\int_0^R r'^2dr' \frac{\sin{G_{hkl}r'}}{G_{hkl}r'}\nonumber \\
&=&
-\frac{3Z}{(G_{hkl}R)^3}
\lep \sin{(G_{hkl}R)} - (G_{hkl}R)\cos{(G_{hkl}R)}\rip
\eeqa
This has zeroes whenever $\tan{(G_{hkl}R)}=G_{hkl}R$ and a maximum when
$G_{hkl}=0$, or in terms of the scattering angle $2k_0\sth=G_{hkl}$,
when $\theta=0,\pi$. In fact, we have that $f_\alpha(\theta=0)=Z$.
This is true in general, since
\beq
f_\alpha(\theta=0)=f_\alpha(\G_{hkl}=0)=\int d^3r'\rho_\alpha(\r')=Z\,.
\eeq
Thus, for x-ray scattering $I\propto Z^2$. For this reason, it is often
difficult to detect small-$Z$ atoms with x-ray scattering.
\subsubsection{The Structure Factor of Centered Lattices}
Now let's look at the structure factor. An especially interesting
situation occurs for centered lattices. We can consider a BCC lattice as a
cubic unit cell $|\aone|=|\atwo|=|\athree|$, $\aone\perp\atwo\perp\athree$
and a two-atom basis $\r_\alpha=0.5\alpha(\aone+\atwo+\athree)$, $\alpha=0,1$.
Now if both sites in the unit cell ($\alpha=0,1$) contain the same atom
with the same scattering factor $f$, then
\beq
\r_\alpha\cdot\G_{hkl}=0.5\alpha(\a_1+\a_2+\a_3)\cdot(h\g_1+k\g_2+l\g_3)
=\pi\alpha(h+k+l)
\eeq
so that
\beqa
S_{hkl}&=&\sum_{\alpha=0,1} fe^{-i\pi\alpha(h+k+l)}\nonumber \\
&=&
f(1+e^{-i\pi(h+k+l)})=\left\{\begin{array}{ll}
0 & \mbox{ if $h+k+l$ is odd}\nonumber\\
2f & \mbox{ if $h+k+l$ is even}
\end{array}\right.\\
\eeqa
This lattice gives rise to extinctions (lines, which appear in a cubic
lattice, but which are missing here)! If both atoms of the basis are identical (like bcc iron), then the bcc structure leads to extinctions; however,
consider CsCl. It does not have these extinctions since $f_{{\rm{Cs}}^+}
\neq f_{{\rm{Cl}}^-}$. In fact, to a good approximation
$f_{{\rm{Cs}}^+} \approx f_{{\rm{Xe}}}$ and
$f_{{\rm{Cl}}^-} \approx f_{{\rm{Ar}}}$.
However CsI (also a bcc structure) comes pretty close to having complete
extinctions since both the Cs and I ions take on the Xe electronic shell.
Thus, in the scattering pattern of CsI, the odd $h+k+l$ peaks are {\em{much}}
smaller than the even ones. Other centered lattices also lead to extinctions.
In fact, this leads us to a rather general conclusion. The shape and
dimensions of the unit cell determines the location of the Bragg peaks;
however, the content of the unit cell helps determine the relative intensities
of the peaks.
\begin{figure}
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{feco.pdf}}
\caption[]{\em{
The unit cell of body-centered cubic ordered FeCo.
}}
\label{feco}
\end{figure}
\paragraph{Extinctions in Binary Alloys}
Another, and significantly more interesting, example of extinctions
in scattering experiments happens in binary alloys such as FeCo on a centered
BCC lattice. Since Fe and Co are adjacent to each other on the periodic
chart, and the x-ray form factor is proportional to $Z$ ($Z_{\rm{Co}}=27$,
and $Z_{\rm{Fe}}=26$)
\beq
f_{\rm{Fe}}^{\rm{x-ray}}\approx f_{\rm{Co}}^{\rm{x-ray}}\,.
\eeq
However, since one has a closed nuclear shell and the other doesn't, their
neutron scattering factors will be quite different
\beq
f_{\rm{Fe}}^{\rm{neutron}}\neq f_{\rm{Co}}^{\rm{neutron}}
\eeq
Thus, neutron scattering from the ordered FeCo structure shown in
Fig.~\ref{feco} will not have extinctions; whereas scattering from
a disordered structure (where the distribution of Fe and Co is random,
so each site has a 50\% chance of having Fe or Co, independent of the
occupation of the neighboring sites) will have extinctions!
\subsubsection{Powdered x-ray Diffraction}
\begin{figure}
\centerline{\includegraphics[height=4.5in,keepaspectratio,clip=true]{ewald.pdf}}
\caption[]{\em{
The Ewald Construction to determine if the conditions are correct for
obtaining a Bragg peak: Select a point in k-space as the origin. Draw
the incident
wave vector $\k_0$ to the origin. From the base of $\k_0$, spin $\k$
(remember, that for elastic scattering $|\k|=|\k_0|$) in all possible
directions to form a sphere. At each point where this sphere intersects
a lattice point in k-space, there will be a Bragg peak with $\G=\k-\k_0$.
In the example above we find 8 Bragg peaks. If however, we change $\k_0$ by
a small amount, then we have none!
}}
\label{ewald}
\end{figure}
If you expose a columnated beam of x-rays to a crystal with a single
crystalline domain, you usually
will not achieve a diffraction spot. The reason why can be seen from the
Ewald construction, shown in Fig.~\ref{ewald}
For any given $\k_0$, the chances of matching up so as to achieve $\G=\k-\k_0$
are remote. For this reason most people use powdered x-ray diffraction to
characterize their samples. This is done by making a powder with randomly distributed crystallites. Exposing the powdered sample to x-rays and recording
the pattern. The powdered sample corresponds to averaging over all
orientations of the reciprocal lattice. Thus one will observe all peaks
that lie within a radius of $2|\k_0|$ of the origin of the reciprocal
lattice.
\edo