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X-ray Crystallography and the Fourier Transform

When a monochomatic X-ray diffracts off a crystal it performs part of a mathematical operation: the Fourier transform...

Tony Phillips
Stony Brook University
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Introduction

X-ray crystallography has been essential, since the beginning of the 20th century, to our understanding of matter; recently, as knowledge of the chemical composition of proteins has progressed, the determination of their 3-dimensional structure has become indispensable for the correct interpretation of their functions. Our main access to this information is X-ray crystallography. (X-rays are used because their wavelengths are on the order of inter-atomic distances in molecules, in the range 1-100 Å ; one Å is $10^{-10}$m). Mathematics enters into the process at two stages. The study of space groups tells what periodic configurations are possible in three-space; in fact these are often called "crystallographic groups" by mathematicians. The other connection, the subject of this column, is the surprising and pleasing fact that when a monochomatic X-ray diffracts off a crystal it performs part of a mathematical operation: the Fourier transform (developed in the 19th century in completely different contexts); when the incidence angle is varied, the complete transform is produced. The flaw in this lovely picture is that we cannot measure all the details of the diffracted wave; otherwise the entire molecular structure could be calculated by inverting the Fourier transform.
   There are many online resources devoted to X-ray crystallography; I have especially profited from Randy Read's Protein Crystallography Course and from Kevin Cowtan's Interactive Structure Factor Tutorial. Thanks also to my Stony Brook colleagues Peter Stephens and Miguel Garcia-Diaz for helpful correspondence and remarks.

Fourier series: temperature distribution in a wire

Fourier series and the Fourier transform were invented as a method of data analysis. For example, let us follow Jean-Baptiste Joseph Fourier (1768-1830) in studying the time evolution of the temperature distribution in a circular loop of circumference $a$, given an initial distribution of temperature $f(x), 0\leq x\leq a$; (we require $f(a)= f(0)$). We start by calculating what are now known as the Fourier coefficients of $f$: $$a_0 = \frac{1}{a}\int_0^af(x)~dx,$$ $$a_h = \frac{2}{a}\int_0^af(x)\cos~ \frac{2\pi h}{a}x~dx,$$ $$b_h = \frac{2}{a}\int_0^af(x)\sin~ \frac{2\pi h}{a}x~dx.$$ It can be proved, if $f$ is sufficiently well-behaved, that the linear combination $$a_0 +\sum_{h=1}^{\infty}a_h\cos~ \frac{2\pi h}{a}x + \sum_{h=1}^{\infty}b_h\sin~ \frac{2\pi h}{a}x$$ (the Fourier series of $f$; when $a=2\pi$ these formulas simplify pleasantly) converges everywhere to $f(x)$.

(Taken as an initial distribution separately, each of the $\cos~ \frac{2\pi h}{a}x$ and $\sin~ \frac{2\pi h}{a}x$ functions determines a simple solution, as does a constant function; the linearity of the heat equation allows these separate solutions to be combined, with coefficients $a_h$ and $b_h$, to give the complete solution to the problem.)

It is useful to simplify the formulas by using Euler's identities $\cos~ t = \frac{1}{2}(e^{it} + e^{-it})$, $\sin~ t = \frac{1}{2i}(e^{it} - e^{-it})$ and grouping terms to yield $$f(x) = \sum_{h=-\infty}^{\infty}c_h e^{i\frac{2\pi h}{a}x}$$ and $$c_h = \frac{1}{a}\int_0^a f(x)~e^{-i\frac{2\pi h}{a}x}~dx;$$ in general, the $c_h$ are complex numbers.

Electron density distribution in a crystal

   The dual operations of integration: $f(x)~\rightarrow~ \{c_h\}$ and summation: $\{c_h\}~\rightarrow~f(x)$ can be realized in other contexts. In this column we will consider the function $\rho(x,y,z)$ that gives the electron density distribution in the crystalline state of some compound. Suppose, for simplicity, that the unit building block, corresponding to a molecule of the compound, is a rectangular solid; say with edge-lengths $a, b$ and $c$; these $a$ by $b$ by $c$ solids are stacked in three-space so as to give a structure repeating every $a$ units in the $x$-direction, every $b$ units in the $y$ and every $c$ units in the $z$. Then the function $\rho(x,y,z)$ will be triply periodic, with periods $a, b$ and $c$, and consequently can be represented as a triple Fourier series $$\rho(x,y,z) = \sum_{h=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} \sum_{l=-\infty}^{\infty} c_{hkl}e^{i(\frac{2\pi h}{a}x + \frac{2\pi k}{b}y +\frac{2\pi l}{c}z)},$$ where $$c_{hkl}=\frac{1}{abc}\int_0^a\int_0^b\int_0^c \rho(x,y,z)e^{-i(\frac{2\pi h}{a}x + \frac{2\pi k}{b}y +\frac{2\pi l}{c}z)} ~dx~dy~dz,$$ directly generalizing our formulas for the circle. If the coefficients $c_{hkl}$ are known, the electron density distribution can be calculated, and then the structure of the molecule can be determined. It is therefore remarkable that the diffraction patterns formed when the crystal is bombarded with X-rays contain precious information about the $c_{hkl}$. Roughly speaking, we can imagine the complex numbers $c_{hkl}$ placed at the vertices $(h,k,l)$ of a 3-dimensional lattice; each X-ray diffraction pattern projects this lattice onto the plane of the image plate. If we label by $p_{\Theta}$ the projection produced by a beam meeting the crystal at a generalized angle $\Theta$, a vertex $(h,k,l)$ which is good position with respect to $\Theta$ (this condition also depends on the wavelength $\lambda$ of the radiation) will appear on the plate as a spot at location $p_{\Theta}(h,k,l)$ and of intensity proportional to the square of the absolute value $|c_{hkl}|$. Varying $\Theta$ will bring a new set of vertices into good position; eventually the lattice can be reconstructed, along with the absolute value of the coefficients at the vertices.

  • The phase problem. The absolute values $|c_{hkl}|$ alone, although they contain a great deal of information about the molecule in question, do not allow $\rho(x,y,z)$ to be completely reconstructed. A simple example comes from temperature distributions like those studied by Fourier. Consider (with circumference $a = 2\pi$) the temperature distributions $f(x) = \cos ~x + \cos~3x = \frac{1}{2}(e^{ix}+e^{-ix}) + \frac{1}{2}(e^{3ix} + e^{-3ix})$ and $g(x) = \cos ~x - \cos~3x = \frac{1}{2}(e^{ix}+e^{-ix}) - \frac{1}{2}(e^{3ix} - e^{-3ix})$. So the coefficents for $f$ are all $\frac{1}{2}$, while those for $g$ are $c_{-3} = c_3 =-\frac{1}{2}, c_{-1} =c_1 = \frac{1}{2}$. The absolute values are the same, but the distributions are different, as shown below. X-ray crystallographers have devised many ways to get around this limitation, called "the phase problem;" they are beyond the scope of this column.

     

    2 heat distributions
    The temperature distributions $f(x) = \cos ~x + \cos~3x$ (red) and $g(x) = \cos ~x - \cos~3x$ (blue), on a circular loop of circumference $2\pi$ (plotted radially) are quite different but have Fourier coefficients with the same absolute values.

The reciprocal lattice

X-rays interact with a crystal through interaction with parallel families of planes. Suppose as before that the unit cell in the crystal is an $a\times b\times c$ rectangular parallelipiped (when $a,b$ and $c$ are all different, this structure is called orthorhombic). Every triple $(h,k,l)$ of integers defines a family of planes through the crystal, defined by the equation $$\frac{xh}{a}+ \frac{yk}{b}+\frac{zl}{c} = n ~~~~ (n \mbox{ an integer}).$$ Let us change coordinates to $X = \frac{x}{a}$, $Y=\frac{y}{b}$, $Z=\frac{z}{c}$. (We are now in the reciprocal lattice; for a non-rectangular crystal the change of coordinates is only slightly more complicated). Then the equation defining the planes becomes $$Xh+Yk+Zl=n,$$ for $n$ an integer. For each $(h,k,l)$ this family of parallel planes fills up the crystal, in the sense that each unit cell vertex lies in (exactly) one of them, as is easy to check. We call these the lattice planes. Graphically, it is easier to represent the analogous concept in two dimensions, so we suppress $z$ and $l$ for the moment. Our crystal is then an array of $a\times b$ rectangles; in the $(X,Y)$ coordinates these measure $1\times 1$. The pair $(0,1)$ gives the lines $Y = n$, for $n$ an integer. These lines are parallel to the $X$-axis, and slice through the base of each unit cell. On the other hand, $(1,1)$ gives the lines $X + Y = n$, or $Y = n -X$, $(-2,1)$ gives $Y = n + 2X$, and $(2,-3)$ gives $Y = -n/3 + (2/3)X$. (See image).

lattice lines
Lattice lines: the lines corresponding to $(1,1)$ are shown in blue, those for $(-2,1)$ in orange, and those for $(2,-3)$ in purple. Not all lines are shown.

X-ray diffraction: how a monochromatic plane wave performs Fourier analysis on the electron density distribution.

diffraction diagram
This diagram labels the vectors involved in the calculation of diffraction. An incoming plane wave ${\bf k_0}$ diffracts off an electron at $O$ as a wave ${\bf k}$ with the opposite incidence angle $\theta$, and also off another electron at relative position ${\bf r}$. The wave vectors ${\bf k_0}$ and ${\bf k}$ have the same magnitude, which we set for convenience to be $\frac{1}{\lambda}$, the reciprocal of the wavelength. The diffracted rays travel different lengths: the path through $O'$ is longer by $AO' + O'B$ than the path through $O$, as shown. $AO'$ is the projection of ${\bf r}$ along ${\bf k_0}$ and has length ${\bf r}\cdot{\bf k_0}/|{\bf k_0}| = \lambda {\bf r}\cdot{\bf k_0}$. $O'B$ is minus the projection of ${\bf r}$ along ${\bf k}$, and so has length $-\lambda {\bf r}\cdot{\bf k}$. This gives, for the difference in path lengths, $\lambda {\bf r}\cdot({\bf k_0}-{\bf k})$ or $\lambda {\bf r}\cdot{\bf s}$. Since $\lambda$ is the wavelength, the ray traveling through ${\bf r}$ will have its phase retarded by $2\pi{\bf r}\cdot{\bf s}$; corresponding to a factor $e^{-2\pi i {\bf r}\cdot{\bf s}}$.
    The length of the vector ${\bf s}$ is $2|{\bf k}| \sin\theta = 2\sin\theta/\lambda$ ; so ${\bf r}\cdot{\bf s}$, which is equal to the component of ${\bf r}$ in the direction of ${\bf s}$ times the length of ${\bf s}$, is equal to $2d\sin\theta/\lambda$; constructive interference, which corresponds to ${\bf r}\cdot{\bf s} = n$, an integer, thus also corresponds to $2d\sin\theta = \lambda n$. This is Bragg's law; its discovery in 1912 started the field of X-ray crystallography.
The diffraction corresponding to a diffraction vector ${\bf s}$ and a single electron at position ${\bf r}$ multiplies the amplitude of the scattered wave by a phase factor $e^{-2\pi i {\bf r}\cdot{\bf s}}$. If $\rho({\bf r})$ is the electron density function in the crystal, the effect on ${\bf s}$ will sum to $$ F({\bf s}) = \int_{\mbox{crystal}}\rho({\bf r})e^{-2\pi i {\bf r}\cdot{\bf s}}~ d{\bf r}.$$ So the structure factor $F({\bf s})$ appears as the Fourier transform of the electron density function $\rho({\bf r})$. (Straightforward diffraction experiments only measure the (square of) the absolute value of $F({\bf s})$, which shows up as the intensity of the spot corresponding to ${\bf s}$. If the phases were also known, the Fourier transform could be inverted to give an exact picture of the electron density $\rho({\bf r})$).
    It is possible to rewrite this integral in terms of the $(h,k,l)$ vectors in reciprocal space. For each such vector ${\bf H}$ we write, using the reciprocal coordinates ${\bf X} = (X,Y,Z)$ $$ F(h,k,l) = \int_{\mbox{unit cell}}\rho({\bf X})e^{-2\pi i {\bf X}\cdot(h,k,l)}~ d{\bf x}.$$ In this case the inverse Fourier transform $$\rho({\bf X})=\int F(h,k,l)e^{2\pi i {\bf X}\cdot(h,k,l)}~ dV$$ where $dV$ is volume in reciprocal space, can be approximated by a Fourier series $$\rho({\bf X})=\sum_{h=-\infty}^\infty~\sum_{k=-\infty}^\infty~\sum_{l=-\infty}^\infty F(h,k,l)e^{2\pi i {\bf X}\cdot(h,k,l)}$$ which can be compared with the Fourier series for $\rho(x,y,z)$ given at the start of this column.
    Here is a nice example, from Kevin Cowtan's Interactive Structure Factor Tutorial. The example is 2-dimensional, and shows how rapidly the structure factors, with their phases, converge to the target structure. The target electron density function looks like this:

target

Cowtan's simulation leads to the approximate Fourier synthesis of the target from just the seven largest structure factors: those corresponding to $(h,k) = (0,1), (1,0), (-1,2), (-2,1), (1,2), (3,-2), (3,1)$. Here is how the synthesis proceeds, step by step, each time adding in the next structure factor. These images are from his tutorial, and are used with permission. The unit cell (not orthorhombic!) is outlined in dots.

step 1   step 2  

step 3   step 4

step 5   step 6

step 7

 

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Tony Phillips
Stony Brook University
Email Tony Phillips

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