Patterson space origin

A Patterson coordinate system is defined based on unit vectors u, v, and w, which are parallel with the axes of the real unit cell of the crystal. Each point (u, v, and w) in Patterson space defines the end point of a vector u having a unique direction and length from the origin of Patterson space to that point, that is, u = (u,v, w). Every point, or vector, in Patterson space (u, v, w) will have associated with it a value P(u). 

There are other circumstances in which some of the atomic coordinates in a crystal can be discovered by evaluation of the Patterson function over a particular plane or along a particular line. For instance, it may be known, from a consideration of the space-group and the equivalent positions in the unit cell, that there is one particular atom at the origin of the cell and others somewhere on the plane y = L The... 

The analysis of the Patterson function requires extensive use of symmetry. Consider all possible interatomic vectors (calculated as u,y = [x, -Xy]) originating from an atom in the general site position of the space group P2i/m, which are listed in Table 2.18. Only three of the vectors (shaded in the first row of the table) are unique, and the relationships between them are established by the combination of symmetry elements in the unit cell. 

The Patterson map, or Patterson unit cell within the bounds of 0 to 1 along u, v, and w, must contain not only vectors from any atom x, y, z to x, y, z but also the vectors in the opposite directions from atoms x, y, z to jt, y, z. Hence every atomic relationship in real space is represented by a pair of centrosymmetrically related vectors u,v,w(x — x,y-y,z-z) and - u, —v, —w (x - x, y — y,z — z). This tells us that Patterson maps, like reciprocal space and the diffraction pattern, always contain a center of symmetry at their origin even when the real crystallographic unit cell does not. 

Apart from the high peak at the origin (the sum of the vectors from each atom to itself), there should be one high peak in the Patterson map at v = A, and the position of this peak will give values forx and z of the heavy atom in the unit cell (with the screw axes at x = 0, z = 0). They coordinate of one atom is arbitrary in this particular space group. This is evident in the general positions of the space group (listed above) since if there is an atom aty there is also one at Vi + y, but no infor-... 

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