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The Patterson function is used to solve the
phase problem In physics, the phase problem is the problem of loss of information concerning the phase that can occur when making a physical measurement. The name comes from the field of X-ray crystallography, where the phase problem has to be solved for the det ...
in
X-ray crystallography X-ray crystallography is the experimental science determining the atomic and molecular structure of a crystal, in which the crystalline structure causes a beam of incident X-rays to diffract into many specific directions. By measuring the angles ...
. It was introduced in 1935 by Arthur Lindo Patterson while he was a visiting researcher in the laboratory of Bertram Eugene Warren at MIT. The Patterson function is defined as :P(u,v,w) = \sum\limits_ \left, F_\^2 \;e^. It is essentially the
Fourier transform A Fourier transform (FT) is a mathematical transform that decomposes functions into frequency components, which are represented by the output of the transform as a function of frequency. Most commonly functions of time or space are transformed ...
of the intensities rather than the structure factors. The Patterson function is also equivalent to the
electron density In quantum chemistry, electron density or electronic density is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point. It is a scalar quantity depending upon three spatial va ...
convolved with its inverse: :P\left(\vec\right) = \rho\left(\vec\right) * \rho\left(-\vec\right). Furthermore, a Patterson map of ''N'' points will have peaks, excluding the central (origin) peak and any overlap. The peaks' positions in the Patterson function are the interatomic distance vectors and the peak heights are proportional to the product of the number of electrons in the atoms concerned. Because for each vector between atoms ''i'' and ''j'' there is an oppositely oriented vector of the same length (between atoms ''j'' and ''i''), the Patterson function always has centrosymmetry.


One-dimensional example

Consider the series of delta functions given by :f(x) = \delta(x) + 3 \delta(x-2) + \delta(x-5) + 3 \delta(x-8) + 5 \delta(x-10); \, then the Patterson function is :\begin P(u) = &5 \delta(u+10) + 18 \delta(u+8) + 9 \delta(u+6) + 6 \delta(u+5) + 6 \delta(u+3) + 18 \delta(u+2) + 45 \delta(u) + \\ & + 18 \delta(u-2) + 6 \delta(u-3) + 6 \delta(u-5) + 9 \delta(u-6) + 18 \delta(u-8) + 5 \delta(u-10). \end


References


External Links

* {{DEFAULTSORT:Patterson Function Crystallography