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

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 $F(0)$.

The dominant contributions are $2F(10)\cos (2\pi x)+2F(01)\cos (2\pi y)$. Their extrema occur at $x=0$ or $\frac{1}{2}$ and independently $y=0$ or $\frac{1}{2}$, modulo unit translations. The four candidate positions areWhich is a maximum rather than a minimum or saddle depends on the two signs. Higher Fourier terms sharpen these broad maxima into atomic peaks.

The sign choices that make each dominant cosine positive at the selected maximum are:

Equivalently, at a selected maximum $(x_0,y_0)$, the signs are $(-1{)}^{2x_0}$ and $(-1{)}^{2y_0}$.

The third term is $2F(11)\cos [2\pi (x+y)]$. To reinforce a maximum at $(x_0,y_0)$ from the preceding table, its sign should beHence the direct-methods triplet relationship is $S(10)S(01)S(11)\simeq +1$ for the strong reflections under consideration. This maximizes constructive density at shared atomic positions. For weak or competing reflections the relation is probabilistic rather than an unconditional algebraic theorem.

Moving the origin by $\mathbf{t}$ changes a structure-factor phase by $-2\pi \mathbf{h}\cdot \mathbf{t}$ in the corresponding coordinate convention. The triplet phaseis unchanged because the three origin shifts cancel. In a centrosymmetric origin convention, the associated sign product $S(\mathbf{h})S(\mathbf{k})S(\mathbf{h}+\mathbf{k})$ is therefore invariant.

For example shifting $x$ by half a period flips both $S(10)$ and $S(11)$, leaving $S(01)$ unchanged; shifting $y$ flips the other corresponding pair. A general origin may make the factors complex, but the phase-invariant statement remains valid.

Set $\mathbf{k}=\mathbf{h}$ in the direct-methods triplet relationship. It becomes $S(\mathbf{h}{)}^{2}S(2\mathbf{h})\simeq +1$, and since the square of a nonzero sign is one, $S(2\mathbf{h})\simeq +1$. This recovers the direct-methods pair relationship, with the same strong-amplitude qualifications.