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Next: Suggestions for Further Reading Up: Rotation Matrices and Translation Vectors in Previous: 6. Systematic Enhancement, Epsilon

7. Phase Shifts

In analogy with equivalent positions, and as an effect of these, there are equivalent reflections. Two equivalent reflections, h and h$^{\prime}$, always have the same amplitudes, i.e. |F(h)| = |F(h$^{\prime}$)|, but their phases may differ. The phases are, however, related to each other in an easily deduced way. The difference in phase between two equivalent reflections is called phase shift. How the phase shift arises and how great it is will be shown now.

Two reflections h and h$^{\prime}$ are equivalent if there exists an Ri such that $\textbf{h}^{\prime} = \pm\textbf{hR}_i$.Due to Friedel's law h is always equivalent to -h. The phases of two equivalent reflections are related as:

\begin{displaymath}
F(\textbf{h})^{\prime} = F(\textbf{hR}) =
\exp (2\pi i\textbf{ht}) \cdot F(\textbf{h}).\end{displaymath}

Proof :

The proof is carried out for a 2-fold symmetry for the sake of simplicity. A similar strategy can be used for higher symmetries. If x and Rx + t are equivalent positions, then R-1 (x - t) is also an equivalent position, since just as well as x gives rise to x$^{\prime}$, x$^{\prime}$ gives rise to x, by the same symmetry operation:

\begin{displaymath}
\textbf{R}^{-1}([\textbf{Rx} + \textbf{t}]-\textbf{t})=\text...
 ...tbf{Rx})=(\textbf{R}^{-1}\textbf{R})\cdot\textbf{x}=\textbf{x}.\end{displaymath}

R-1 is the inverse matrix of R, i.e. R- 1R=I. We shall make use of this when we calculate F(h$^{\prime}$) = F(hR):

\begin{displaymath}
F(\textbf{h}^{\prime}) = \sum^N_{j=1} f_j \exp (2{\pi}i\textbf{hRx}).\end{displaymath} (8)
In analogy with (5) we now sum over half the unit cell:

\begin{displaymath}
F(\textbf{h}^{\prime}) = \sum^{N/2}_{j=1} f_j \exp (2{\pi}i\...
 ...2{\pi}i\textbf{hR}[\textbf{R}^{-1}\{\textbf{x}-
\textbf{t}\}]).\end{displaymath} (9)
The second sum in (9) is easily simplified to (10) making use of RR- 1 = I.

\begin{displaymath}
\sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{h}[\textbf{x}-\tex...
 ...extbf{ht})\cdot \sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{hx}).\end{displaymath} (10)
The term exp ($-2{\pi}i$ht) can be brought outside the summation since it is a constant. Note the minus sign of the exponent! We now get

\begin{displaymath}
F(\textbf{h}^{\prime}) = \exp (- 2{\pi}i\textbf{ht})\cdot \s...
 ...\textbf{hx}) + \sum^{N/2}_{j=1} f_j \exp (2{\pi}i\textbf{hRx}).\end{displaymath} (11)
(11) is compared to (7):

\begin{displaymath}
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}).\end{displaymath} (7)
The two expressions (11) and (7) are identical except for the phase term exp ($-2{\pi}i$hr) which is applied on both sums of (7) in order to get (11). Thus F(h) and F(h$^{\prime}$) have equal amplitudes but differ in phase by exp ($2{\pi}i$ht).

The phase shift is called $S_\textbf{h}$ and is equal to exp ($-2{\pi}i$ht). Note the minus sign!

\begin{displaymath}
F(\textbf{h}^{\prime}) = \exp (-2{\pi}i\textbf{ht})\cdot F(\textbf{h}).\end{displaymath} (12)
This is often written in other forms, denoting the phase of $F(\textbf{h})
\varphi (\textbf{h})$:

\begin{displaymath}
\varphi(\textbf{h}^{\prime}) = \varphi(\textbf{h}) + S_\text...
 ...\prime}) = \varphi(\textbf{h}) - 360^{\circ} \cdot
\textbf{ht}.\end{displaymath}

Since the phase shift depends on the translation vector, all phases of equivalent reflections derived through symmetry operations with translation vector = 0, are equal. An example of equivalent reflections with different phases will be given. Derive the phases of all reflections equivalent to (h 0 1) in space group P3121. The equivalent reflections were derived in section 4.

\begin{displaymath}
(3\,0\,1), (0\,-3\,1), (-3\,3\,1), (0\,3\,-1), (-3\,0\,-1)\,\mbox{ and }\,(3\,-3\,-1).\end{displaymath}

If the first reflection, (3 0 1), is +60$^{\circ}$, then the second becomes

\begin{displaymath}
60^{\circ} - 360^{\circ} \cdot (3 \cdot 0 + 0 \cdot 0 + 1 \c...
 ...rc} \cdot \frac{1}{3} = 60^{\circ} -
120^{\circ} = -60^{\circ}.\end{displaymath}

In a similar way the phases of the other reflections are 60$^{\circ} -
240^{\circ} = 180^{\circ}, 60^{\circ} - 0^{\circ} = 60^{\circ}, 60^{\circ} -
120^{\circ} = -60^{\circ}$ and $60^{\circ} - 240^{\circ} = 180^{\circ}$.

Note that the fifth reflection ($-3\, 0 -1$) also is the Friedel pair of (3 0 1). Due to Friedel's law the phase of any reflection must be minus that of its Friedel pair. In all cases where a symmetry operation generates an equivalent reflection which is also its Friedel mate, we have two indications of the phase value. If the phase of (hkl) is $\varphi$ then the phase of (-h -k -l) is $-\varphi$ due to Friedel's law, and the phase is $\varphi - S_h$ due to the phase shift. We now have a system of equations:

\begin{displaymath}
(- h - k - l) = - \varphi \phantom{+S_h}\end{displaymath}

\begin{displaymath}
(-h -k -l) = \varphi + S_h\end{displaymath}

with the solution $-\varphi = \varphi + S_\textbf{h}$,i.e. $\varphi = S_\textbf{h}/2$ (modulo 180$^{\circ}$ since $S_\textbf{h}$ of course is modulo 360$^{\circ}$). This alternative way of deriving phase restrictions is clearly more relevant than that of section 4.

In a similar way the systematically absent reflections can be shown to be exactly those reflections which have two contradictory phase indications. In P21 (0 k 0)-reflections with k odd are extinct. The equivalent positions of P21 are (x, y, z) and ($-x, \frac{1}{2} + y, -z$). The reflections (hkl) and (-h, k, -l) are equivalent and the phase shift is k/2. A reflection like (0 3 0) is thus equivalent to itself, but the equivalent reflection generated has a phase differing from the original one by 180$^{\circ}$. The phase of (0 3 0) is at the same time $\varphi$ and $\varphi
+ 180^{\circ}$, which of course is only possible if the amplitude of the reflection is 0!


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Next: Suggestions for Further Reading Up: Rotation Matrices and Translation Vectors in Previous: 6. Systematic Enhancement, Epsilon

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