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3. Special Reflections

In a space group with n equivalent positions (R_ix + t_i), i = 1, 2, $\dots$, n the structure factor can be calculated through summing up the contributions to it in the following way:

$$F(\textbf{h}) = \sum^{N/n}_{j=1} f_j \cdot \exp (2{\pi}i \cd... ...}_i \cdot \textbf{h} \cdot \textbf{x}_2) + \cdots \textbf{x}_n)$$ (4)

i.e. first all the atoms of the first equivalent position are added, then those atoms related to the first ones through R₂ + t₂ are added, and so on until all atoms within the unit cell are added.

According to (1) every equivalent position can be written as Rx + t. If this expression is inserted into (4), and we for the sake of simplicity look at a space group with 2 equivalent positions, we obtain:

$$F(\textbf{h}) = \sum^{N/2}_{j=1}f_j \exp (2{\pi}i\textbf{hx}... ...\sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{h}[\textbf{Rx} + t]).$$ (5)

The second of these sums can be rewritten, since

$$\exp (2{\pi}i\textbf{h}[\textbf{Rx} + \textbf{t}]) = \exp (2{\pi}i[h\textbf{Rx} + \textbf{ht}])$$

$$\phantom{J\exp (2{\pi}i\textbf{h}[\textbf{Rx} + \textbf{t}])} = \exp (2{\pi}i\textbf{hRx} + 2{\pi}i\textbf{ht})$$ (6)

$$\phantom{M=I\exp (2{\pi}i\textbf{h}[\textbf{Rx} + \textbf{t}])} = \exp (2{\pi}i\textbf{hRx})\cdot \exp (2{\pi}i\textbf{ht}).$$

If (5) and (6) are combined we get:

$$F(\textbf{h}) = \sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{hx... ...tbf{ht}) \cdot \sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{hRx}).$$ (7)

It is obvious that in the general case the contributions from the two parts of the structure differ, both in amplitude and phase. If, however, the two contributions are equally large, i.e. have identical amplitudes, there will be several interesting situations. The amplitudes of two (or more) parts of the structure are equal if and only if $\textbf{hR}_i = \textbf{h}$ or $\textbf{hR}_i = -\textbf{h}$ for at least one $\textbf{R}_i$, $i \neq 1$.

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