
\chapter{The Reciprocal Lattice and Diffraction}
\label{C5}

\section{The Reciprocal Basis}

The basis vectors {\bf a}$_i$ were defined in Chapter~1; a  vector {\bf u} in real space can then be written as a linear combination of the basis vectors:
$${\bf u} = u_1 {\bf a}_1  + u_2{\bf a}_2 + u_3{\bf a}_3,$$
where $u_1$, $u_2$ and $u_3$ are its components, when {\bf u}
is referred to the basis A.

  The reciprocal lattice  
constitutes a special co-ordinate system, designed originally to simplify
the study of diffraction phenomena.
If we consider a lattice, represented by a basis symbol `A' and an
arbitrary set of non-coplanar basis vectors
{\bf a}$_1$, {\bf a}$_2$ and {\bf a}$_3$, then the corresponding reciprocal basis A$^*$ has
basis vectors {\bf a}$^*_1$, ${\bf a}^*_2$ and ${\bf a}^*_3$, defined by
the following equations:
 
$${\bf a}^*_1  = ({\bf a}_2\wedge {\bf a}_3) / ({\bf a}_1 .{\bf a}_2\wedge {\bf
a}_3) $$
$${\bf a}^*_2 = ({\bf a}_3\wedge {\bf a}_1 ) / ({\bf a}_1 .{\bf a}_2\wedge {\bf
a}_3)$$
$${\bf a}^*_3 = ({\bf a}_1 \wedge {\bf a}_2) / ({\bf a}_1 .{\bf a}_2\wedge {\bf
a}_3)$$ 
The term $({\bf a}_1 .{\bf a}_2\wedge {\bf a}_3)$ represents the volume of
the unit cell formed by ${\bf a}_i$, while the magnitude of the vector
$({\bf a}_2\wedge {\bf a}_3)$ represents the area of the $(1~0~0)_{\rm A}$ plane (Fig.~\ref{fig:reciprocal}). Since $({\bf a}_2\wedge {\bf a}_3)$ points along the normal to the
$(1~0~0)_{\rm A}$ plane, it follows that ${\bf a}^*_1$ also points along the normal
to  $(1~0~0)_{\rm A}$ and that its magnitude $\vert{\bf a}^*_1\vert$ is the reciprocal of the
spacing of the $(1~0~0)_{\rm A}$ planes (Fig.~\ref{fig:reciprocal}).

\begin{figure}[!ht]
  \begin{center}
    \begin{minipage}[t]{0.42\linewidth}
\raisebox{-3.4cm}{\includegraphics[width=0.98\linewidth]{Diffraction/reciprocal.eps}} 
    \end{minipage}\hfill
    \begin{minipage}[t]{0.55\linewidth}
      \caption{The relationship between {\bf a}$_1^*$ and {\bf a}$_i$. The vector  {\bf a}$_1^*$ lies normal to the basal plane wxyz and the volume of the parallelepiped formed by the basis vectors  {\bf a}$_i$ is given by  {\bf a}$_1.${\bf a}$_2\wedge$ {\bf a}$_3$. The area of the basal plane is $\vert${\bf a}$_2\wedge$ {\bf a}$_3\vert$. The height is the spacing between planes parallel to the basal plane, given by $1/\vert {\bf a}^*_1\vert$.}  \label{fig:reciprocal}
    \end{minipage}
  \end{center} \end{figure} 

The components of any vector referred to the
reciprocal basis represent the Miller indices of a plane whose normal is along
that vector, with the spacing of the plane given by the inverse of the
magnitude of that vector.
For example, the vector $({\bf u};{\rm A}^*) = (1~2~3)$ is normal to planes with Miller
indices (1~2~3) and interplanar spacing $1/\vert {\bf u}\vert$. 
  We see  that
 \begin{equation}{\bf a}_i.{\bf a}^*_j = 1 \quad\hbox{when}\quad i = j, \quad\hbox{and}\quad {\bf
a}_i.{\bf a}^*_j = 0 \quad\hbox{when}\quad  i \neq j\end{equation}


\begin{figure}[htp]
\centering
\subfigure(a){
\includegraphics[width=0.27\textwidth]{Diffraction/cubic_P_001.eps}}\hfill
\subfigure(b){
\includegraphics[width=0.23\textwidth]{Diffraction/cubic_P_011.eps}}\hfill
\subfigure(c){
\includegraphics[width=0.27\textwidth]{Diffraction/cubic_P_111.eps}}
      \caption{Sections of the reciprocal lattice of a cubic-P lattice. (a) Section normal to [001]. (b) Section normal to $[01\overline 1]$. (c) Section normal to $[\overline 11\overline1]$. Notice how the distance varies inversely with the spacing of the planes. The origin of the reciprocal lattice is not labelled, but if necessary, can be designated 000. }  \label{fig:cubicp}
\end{figure}


\section{Crystallography of diffraction}

Consider waves of length $\lambda$ incident on planes of atoms. The beams reflected from  {\em different} planes in the parallel set illustrated in  Fig.~\ref{fig:Bragg} must also be in phase to avoid destructive interference. Therefore, the path difference between beams $a$ and $b$, i.e. the distance {\em xyz}, must be an integral number of wavelengths. Since $xyz=2d\sin\theta$, the diffraction condition is
\begin{equation}n\lambda=2d\sin\theta \label{eqn:Bragg}\end{equation}
where $n$ is an integer and this equation is the Bragg law, with $\theta$ the Bragg angle.
\begin{figure}[!ht]
  \begin{center}
    \begin{minipage}[t]{0.62\linewidth}
\raisebox{-4cm}{\includegraphics[width=0.98\linewidth]{Diffraction/Bragg.eps}} 
    \end{minipage}\hfill
    \begin{minipage}[t]{0.35\linewidth}
      \caption{Electromagnetic waves incident on a set of parallel crystal planes with an interplanar spacing $d$. The angle of emergence of the scattered waves is the same as that of incidence.}  \label{fig:Bragg}
    \end{minipage}
  \end{center} \end{figure} 

The Bragg law can be  expressed conveniently in terms of vectors in reciprocal space. In Fig.~\ref{fig:Ewald}, the Bragg law is satisfied if \begin{equation}{\mathbf k'} - {\mathbf k}= {\mathbf g}\end{equation} Here {\bf g} is a reciprocal lattice vector beginning at the origin (0,0,0) and representing the crystal which is being illuminated by the radiation {\bf k}. From the geometry of the triangle, it is evident that $\sin\theta = \frac{1}{2}g/k$ so that $\sin\theta = (0.5/d)/(1/\lambda)$ which on rearrangement gives the Bragg law. 


\begin{figure}[!ht]
\centering
\subfigure(a){
\includegraphics[width=0.35\textwidth]{Diffraction/Ewald.eps}}\hfill
\subfigure(b){
\includegraphics[width=0.45\textwidth]{Diffraction/Ewald2.eps}}
      \caption{(a)  The Ewald sphere construction in reciprocal space.  {\bf k} represents the direction of the incident beam, ${\mathbf k'}$ that of the diffracted beam and {\bf g} the normal to the diffracting planes. Both  {\bf k} and ${\mathbf k'}$ have a magnitude $\lambda^{-1}$ and {\bf g} has a magnitude $d^{-1}$. The beam is incident on a crystal which is represented by its reciprocal lattice with origin at 0,0,0. (b) Another illustration of the Ewald sphere superimposed on the reciprocal lattice. In this case only the 101 and $10\overline 1$ planes are in Bragg condition. The reciprocal lattice vectors which do not touch the sphere are not in Bragg orientation.} \label{fig:Ewald}
\end{figure}


\begin{figure}[!ht]
\centering
\subfigure(a){
\includegraphics[width=0.27\textwidth]{Diffraction/cubic_F_001.eps}}\hfill
\subfigure(b){
\includegraphics[width=0.23\textwidth]{Diffraction/cubic_F_011.eps}}\hfill
\subfigure(c){
\includegraphics[width=0.27\textwidth]{Diffraction/cubic_F_111.eps}}
      \caption{Electron diffraction patterns from austenite (cubic-F).  (a) Zone axis parallel to [001]. (b) Zone axis parallel to $[0\overline11]$. (c) Zone axis parallel to [111]. Notice that in any pattern, one only needs to index two reciprocal lattice vectors in order for all the others to be determined. In the cases illustrated, the symmetry of the patterns is consistent with that of the zone axes. Contrast these patterns with the reciprocal lattice sections presented in Fig.~\ref{fig:cubicp}.}  \label{fig:cubicf}
\end{figure}

\section{Intensities}

The electrons from an individual atom can coherently scatter an X-ray beam; the atomic scattering factor $f$ is the ratio of the amplitude scattered by an atom to that scattered by one electron. We are interested in coherent scattering by all the atoms in a unit cell, in which case the resultant for reflections from  $hkl$ planes is given by
a summation known as the {\em structure factor}:
\begin{eqnarray}
F_{hkl}&=&\sum_1^{n~atoms} f_n \exp\{2\pi i(h{u_n} + k{v_n} + l{w_n})\}\\\nonumber
&\equiv& \sum_1^{n~atoms} f_n [\cos\,2\pi (h{u_n} + k{v_n} + l{w_n}) +i\sin\, 2\pi (h_{u_n} + k{v_n} + l{w_n})]
\end{eqnarray}
where $f_n$ is the scattering factor of atom $n$ and the sum is over all atoms in the unit cell.\symbolfootnote[2]{$e^{\pi i} = e^{3\pi i}=-1$,~~~ $e^{2\pi i}=e^{4\pi i}=+1$,~~~ $e^{\pi i/2}=i$,~~~ $e^{3\pi i/2}=-i$,~~~ $(1+i)(1-i)=2$} The magnitude $\vert F\vert$ is now the ratio of the amplitude scattered by a unit cell to that scattered by one electron. $\vert F\vert^2$ is proportional to the scattered intensity. Notice that for a centrosymmetric system, the sine term can be neglected since that function is not symmetric about zero. As an example, for the structure of Cu (cubic close-packed, with four atoms in the cell at 000, $\frac{1}{2}$$\frac{1}{2}$0, $\frac{1}{2}$0$\frac{1}{2}$, and 0$\frac{1}{2}$$\frac{1}{2}$),
\begin{equation}
F=f[1+e^{\pi i(h+k)} + e^{\pi i(h+l)} + e^{\pi i(k+l)}] \label{eqn:structure-factor-fcc}\end{equation}
so that the $\{100\}$ and $\{110\}$ reflections would have zero structure factor.


Techniques in which the atomic scattering factor scales with the atomic number include electron and X-ray diffraction. Reflections such as $\{100\}_\delta$ that are dependent on the differences in the atomic scattering factors of unlike species, will therefore be weak. In neutron diffraction, the scattering factors can be quite different for atoms with similar atomic numbers so superlattice reflections can be more readily detected.

\section{Diffraction from Thin Crystals}

When an incident beam deviates by $\Delta\theta$ from the ideal Bragg angle $\theta$ consistent with equation~\ref{eqn:Bragg}, it is possible to find another beam from within the depth of the crystal which has a path difference of $\frac{1}{2}\lambda$ and which will destructively interfere with it. This depth, needed to find the wave which is out of phase, increases as $\Delta\theta$ becomes smaller; hence for thin crystals, it may not be possible to destroy deviant rays. For thin crystals it becomes possible to obtain diffracted intensity even when the incident beam is not at the exact Bragg orientation.

In transmission electron microscopy we are constrained to work with thin foils. It follows that diffraction becomes possible even when the incident electron beam is not at the exact Bragg orientation. We express this on the Ewald sphere construction by extending the reciprocal lattice points in the direction normal to the thin foil (Fig.~\ref{fig:thin}, \ref{fig:thin2}, \ref{fig:xray}). The fact that electrons have a very small wavelength compared with X-rays also helps achieve a greater chance of diffraction because the Ewald sphere is then relatively flat.

\begin{figure}[!ht]
\centering
\subfigure(a){
\includegraphics[width=0.45\textwidth]{Diffraction/two_zones.eps}}
\subfigure(b){
\includegraphics[width=0.45\textwidth]{Diffraction/ReO3_111.eps}}
\caption{(a) Ewald sphere construction for diffraction from a thin foil. The origin of the reciprocal lattice is at `O' and the layer designated `P' is the next layer normal to the incident beam, a distance $1/t^*$ from the first layer.  (b) $\langle 111\rangle$ axis ${\rm ReO_3}$ showing the higher order Laue zone (the second layer of the type sketched in (a).) - pattern courtesy of P. Midgley.}
\label{fig:thin}
\end{figure}



\begin{figure}[!ht]
\centering
\subfigure(a){
\includegraphics[width=0.25\textwidth]{Diffraction/thin_1.eps}}\hfill
\subfigure(b){
\includegraphics[width=0.25\textwidth]{Diffraction/thin_2.eps}}\hfill
\subfigure(c){
\includegraphics[width=0.25\textwidth]{Diffraction/thin_3.eps}}
      \caption{Fe with cubic-F structure. (a) Electron diffraction from a foil which is 200~nm thick, zone axis [111]. (b) Diffraction from the same foil after thinning it to just 50~nm thickness, zone axis [111]. Because of this minute thickness, it becomes possible to pick up reflections from planes of the form $\{224\}$. (c) as (a) but zone axis [1~1~0.9]. The Ewald sphere is now tilted away from some of the reciprocal lattice points on the left hand side.}  \label{fig:thin2}\end{figure}

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{Diffraction/thin_xray.eps}
      \caption{The effect of crystal size (in all directions) on an X-ray diffraction pattern. The material concerned is halite. The peaks broaden as the crystal gets smaller.}  \label{fig:xray}
\end{center}
\end{figure} 

\section{Neutron Diffraction}

Neutrons are scattered by the atom nucleus and hence the scattering factor is not sensitive to scattering angle; unlike X-rays, the scattering power does not diminish with the Bragg angle $\theta$, so good results are obtained even at high $\theta$. Also unlike X-rays, there is little correlation between the scattering factor and atomic number. Because the interactions are with the nucleus, it  becomes possible to distinguish between isotopes because they now have different neutron scattering factors (for example, hydrogen and deuterium).

X-rays typically penetrate a few micrometers of the surface of a sample, whereas neutrons can penetrate several centimetres. Residual stress is that which is locked in a sample that is otherwise at equilibrium, with no externally applied stress. Fig~\ref{fig:weld} shows residual stress contours that are determined by measuring lattice spacings, comparing them with a stress-free sample to calculate strains, and then converting strains into stresses using the elastic properties of the material. 

\begin{figure}[!ht]
  \begin{center}
    \begin{minipage}[t]{0.62\linewidth}
\raisebox{-4cm}{\includegraphics[width=0.98\linewidth]{Diffraction/weld.eps}} 
    \end{minipage}\hfill
    \begin{minipage}[t]{0.35\linewidth}
\caption{Stress distribution in a steel weld determined using neutron diffraction. This would not be possible using X-rays. The numbers on the contours represent MPa.}
\label{fig:weld}
    \end{minipage}
  \end{center} \end{figure} 

\clearpage\section{Summary}

It takes a leap of imagination to understand why it is necessary to deal with a space in which dimensions are inverted. But the fact is that in crystallography, it can be regarded simply as a lattice in which all reciprocal lattice vectors represent normals to planes in real space, with the magnitudes of these vectors representing the inverse of the spacing of those planes. Whereas a plane in real space is defined by a pair of vectors that lie in that plane, in reciprocal space the plane is represented by a vector. A plane in reciprocal space represents all real space planes that share a common direction (a zone axis).

Imagine now that we wish to find a direction that is parallel to a plane normal. This is trivial for a cubic system where plane normals and directions with the same indices are parallel. So a [123] direction is parallel to the normal of the (123) plane. This is not generally true for non-cubic systems, but there exists a metric tensor which makes it easy to find vectors in reciprocal space that are parallel to those in real space (and vice versa), a subject reserved for advanced study. 

