Past exam of the condensed matter physics major option course of the University of Oxford / 2016 / 1 / d

With $A=S_{x_{\mathrm{Si}}}(h,k,l)$ and $B=S_{x_{\mathrm{Mn}}}(h,k,l)$,For $l=0$, the transformation in part c maps $(h,k,0)$ onto its Friedel pair with only a phase factor. Cyclic permutations give the same result when either other index vanishes. Friedel's law therefore survives for every such reflection, including any systematic absences.

For the $(111)$ reflection set $t=2\pi x$. ThenSince the purely positional sums obey $S_x(-\mathbf{h})=S_x(\mathbf{h}{)}^{\ast }$, the scalar resonant Friedel-pair intensity difference isup to the same positive experimental scale factor. In particular,Different positional phases can therefore produce a nonzero resonant intensity difference at $(111)$, allowing the opposite structural handednesses to be distinguished. The claimed violation is generic, not an unconditional identity: the source supplies no numerical internal coordinates. For a concrete noncentrosymmetric counterexample, take $x_{\mathrm{Si}}=1/8$ and $x_{\mathrm{Mn}}=5/8$. Their orbit sums at $(111)$ are $\sqrt{2}(1-i)$ and $-\sqrt{2}(1-i)$, so the difference vanishes even if $f^{\prime \prime }\ne 0$; each generic chemical orbit is noncentrosymmetric, so this cancellation does not require inversion symmetry. Accidental cancellation, or $f^{\prime \prime }=0$, must be excluded before asserting a violation for a particular specimen.