Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 58 / a

Reality of the electron density implies $F(-h)=F(h{)}^{\ast }$. Writing $F(h)=|F(h)|e^{i{\varphi }_h}$ and pairing positive and negative indices givesThe sign in the cosine follows from the supplied negative-sign Fourier-synthesis convention. $F(0)$ is the mean electron density in the fractional-coordinate normalization.

With an origin at an inversion center, the structure factors are real, so their phases are either $0$ or $\pi$. Defining $S(h)=\mathrm{sgn}F(h)$ givesCentrosymmetry converts the phase problem into a sign problem; it does not imply that every structure factor is positive.

For a strong $h$ term, maxima lie where $S(h)\cos (2\pi hx)=+1$. At either possible set of such maxima, $\cos (4\pi hx)=+1$: doubling a phase of either $0$ or $\pi$ gives zero modulo $2\pi$. A strong $2h$ term reinforces those positive maxima when $S(2h)=+1$. The direct-methods pair relationship is thereforeA negative strong second harmonic would oppose the maxima favored by the first and tend to create implausible negative or flattened density elsewhere. The relationship is a strong-reflection inference, not a sign identity for arbitrary weak coefficients.

The inference is most useful when the contributing normalized amplitudes are large. Positivity gives a precise illustration: let $f_h=F(h)/F(0)$ and regard $\rho (x)/F(0)$ as a probability density over a period. From $⟨{\cos }^2(2\pi hx)⟩\ge ⟨\cos (2\pi hx){⟩}^2$,Thus $|f_h|>1/\sqrt{2}$ forces $F(2h)>0$. If $F(h)$ is weak, this bound allows either sign; a small $F(2h)$ also contributes too little to the density sketch to impose a persuasive sign constraint. Large unnormalized intensity alone should not be confused with a large amplitude relative to $F(0)$.