An occupation wave is its own Fourier series

The satellites that need a second integer began with a crystal whose atoms are pushed along by a wave whose wavelength fits no whole number of cells. It found that the crystal diffracts to sharp spots, each needing two integers to index, and that the spot of order $m$ has an amplitude given by a Bessel function $J_m$ of the wave’s amplitude. It named a second kind of modulated crystal and set it aside. In an occupational modulation no atom moves. The sites stay on their lattice, and the wave decides how much of an atom sits on each: an alloy whose composition rises and falls along the crystal, a vacancy concentration that ripples, two species ordering with a period the lattice cannot hold.

This essay computes that second kind, and it turns out to be the simpler of the two by a long way. The satellite of order $m$ of an occupation wave is the $m$-th Fourier coefficient of the wave’s profile, and nothing else. From that one sentence follow a ceiling no smooth wave can exceed, a list of orders that vanish, a proof that a step is the strongest wave possible, and the fact that the Fibonacci chain is an occupationally modulated crystal whose profile is a step.

Three ways to fill a lattice with a wave #

An occupation wave has two ingredients. The first is its wavevector $q$, the fraction of a period the wave advances from one site to the next. The second is its profile $f$, the occupancy a site receives as a function of where in its period the wave stands at that site. Site $n$ sits at the point $\operatorname{frac}(nq + \varphi)$ of the period, where $\varphi$ is the phase, and is occupied to the fraction $f$ of that point. The profile must take values between nought and one, because an occupancy is a fraction of an atom.

The picture at the head of this essay shows three profiles at the same irrational wavevector. A cosine rises and falls smoothly, so every site is partly occupied and no two sites carry the same amount. A crenel, a profile that is one on an interval and nought outside it, gives sites that are either full or empty. A crenel three tenths of a period wide fills three sites in ten, in an order that never repeats. A crenel as wide as the wavevector itself fills the sites of a lattice exactly where the Fibonacci chain puts its long letter, which is the last section’s subject.

The one formula #

The amplitude the chain scatters with at a wavevector $k$, per site, is the average over sites of the occupancy times a phase:

$$\frac{F(k)}{N} \;=\; \frac1N \sum_{n=0}^{N-1} f\big(\operatorname{frac}(nq + \varphi)\big)\, e^{2\pi i k n}.$$

At a wavevector of the form $k = h + mq$ with whole numbers $h$ and $m$, the phase $e^{2\pi i k n}$ depends on $n$ only through $\operatorname{frac}(nq + \varphi)$, the same point of the period that decides the occupancy. So the sum becomes an average over the points $\operatorname{frac}(nq + \varphi)$ of a fixed function of those points. Because $q$ is irrational, those points fill the period evenly, a fact due to Weyl that three gaps, and never four explored from a different angle. The average then becomes an integral:

$$\frac{F(h + mq)}{N} \;\longrightarrow\; e^{-2\pi i m\varphi}\int_0^1 f(x)\, e^{-2\pi i m x}\,dx \;=\; e^{-2\pi i m \varphi}\,\hat f(m).$$

The amplitude of the satellite of order $m$ is $|\hat f(m)|$, the size of the profile’s $m$-th Fourier coefficient. It does not depend on $h$: every satellite of order $m$ has the same strength, at every main reflection, because the sites are points and carry no form factor. Scattering power that falls off with angle would multiply everything by a slowly varying factor and change nothing here.

The formula is checked the direct way. For each profile the chain is built site by site, twenty thousand sites, and its scattering is summed at each satellite position. The sums agree with the Fourier coefficients to about a ten-thousandth of a site, the size the finite chain’s discrepancy allows. The comparison is between two computations that share nothing but the profile. One adds up twenty thousand complex numbers, and the other evaluates one integral in closed form.

A cosine stops at first order, and at a quarter #

The simplest occupation wave is a cosine: $f(x) = c + a\cos 2\pi x$. Its Fourier series has three terms, the mean $c$ and two coefficients of size $a/2$ at orders one and minus one. So a cosine occupation wave has first-order satellites and nothing else, however strong it is. That is the first contrast with a displacive wave, whose satellites run to every order because $J_m$ of a non-zero argument is never zero. A displacive wave with a single harmonic has an infinite series of satellites, and an occupational wave with a single harmonic has one pair.

The second contrast is a ceiling. An occupancy must lie between nought and one, so $c - a \ge 0$ and $c + a \le 1$, and together these force $a \le \tfrac12$. No cosine occupation wave can give a first-order satellite stronger than a quarter of a site. A composition wave in an alloy can do no better than swing the occupancy from nought to one and back, and even then its satellite is a quarter of what a fully occupied site scatters at a main reflection. The ceiling is lower still away from half occupancy. A wave about a mean occupancy of one fifth can swing only between nought and two fifths, so its amplitude is at most a fifth and its satellite at most a tenth. The strongest cosine needs the average site exactly half full, which is why the most striking composition waves are found near equal proportions of the two species. A reported first-order satellite stronger than a quarter, in a crystal whose modulation is known to be occupational and sinusoidal, is a contradiction. Either the modulation is not a pure cosine, or something in the structure is being displaced as well.

A step skips the orders its width divides #

A crenel’s Fourier coefficients are just as easy to write down:

$$|\hat f(m)| \;=\; \frac{|\sin(\pi m \Delta)|}{\pi |m|},$$

for a crenel of width $\Delta$. They fall off as one over the order, so a crenel has satellites of every order, weaker and weaker. And they vanish exactly where $\sin(\pi m\Delta)$ does, at the orders $m$ for which $m\Delta$ is a whole number.

A crenel a third of a period wide therefore has no third-order satellite, no sixth and no ninth, while the first, second, fourth and fifth are all there. The missing orders are not an approximation or a weakness: the sum over sites at those positions comes out under a thousandth of a site, the noise level of a finite chain. A missing order in an occupationally modulated crystal measures the width of the occupied fraction of the period, and it does so exactly, since a whole number times the width must be a whole number. In practice a crenel width is usually tied to composition. An alloy in which one species occupies a third of the sites, arranged by a step-like wave, would lose every third satellite whatever its wavevector.

This is the same arithmetic that makes a superlattice’s extra reflections appear with intensities set by a difference of scattering powers. A commensurate crenel is an ordered superstructure. Its satellites land on rational positions and become superlattice reflections, and its missing orders become the superlattice reflections that ordering does not produce. The incommensurate case keeps the same intensities and moves them to irrational positions.

A step is the strongest wave there is #

The ceiling for a cosine raises an obvious question: what is the strongest first-order satellite that any occupation wave can make? The first Fourier coefficient is an average of $f(x)e^{-2\pi i x}$ over the period. With $f$ confined to values between nought and one, that average is largest when $f$ is one wherever $\cos 2\pi x$ is positive and nought wherever it is negative, which is a crenel of width one half. Its first coefficient is $1/\pi$, about 0.318.

An occupation wave cannot beat a step. Any profile between nought and one has a first-order satellite of at most $1/\pi$, and the half-width crenel reaches it, a quarter as strong again as the best cosine. Four hundred random profiles, each with sixty-four independent occupancies across the period, reach at best a tenth, which is what averaging many uncorrelated values does to a single Fourier coefficient. The figure’s curve, $\sin(\pi\Delta)/\pi$ against the crenel’s width, is symmetric about one half. A crenel covering most of the period is the same as one covering little of it, with the roles of full and empty sites exchanged.

The extremal property has a practical reading. An observed first-order satellite near a third of a main reflection, in an occupational modulation, says the profile is nearly a step, because nothing else can get close to that strength. The higher orders then confirm it, since a half-width step has its first, third and fifth orders in the proportions $1, \tfrac13, \tfrac15$ and none of even order. That is the same logic by which the satellites that need a second integer read a displacement amplitude off the ratio of first- to second-order Bessel functions.

The Fibonacci chain is a crenel #

The third profile in the opening figure was chosen with care. Its width is $1/\varphi$, the reciprocal of the golden ratio, and its wavevector is the same $1/\varphi$. A crenel whose width equals its wavevector is a very particular wave: site $n$ is full exactly when the point $\operatorname{frac}(n/\varphi + \rho)$ lands in the last $1/\varphi$ of the period. Written as a sequence of two letters, L for a full site and S for an empty one, that is the definition of a mechanical word, the record of a line of golden slope crossing a square lattice. And a mechanical word at the golden slope is the Fibonacci chain.

The census checks it letter by letter. The Fibonacci chain built by substitution, repeating L → LS and S → L, agrees with the crenel of width $1/\varphi$ at a phase of 0.618 over the first two thousand sites without a single disagreement. So a Fibonacci sequence of two species on a lattice is an occupationally modulated crystal, and its diffraction is a crenel’s: main reflections at the whole numbers with intensity $1/\varphi^2$ of a fully occupied site, satellites of first order at $h \pm 1/\varphi$ with intensity $\sin^2(\pi/\varphi)/\pi^2$, and every higher order in its place, each needing two integers to index.

That identification settles, in one dimension, something the extra dimension that makes it periodic stated as a matter of vocabulary: an incommensurately modulated crystal has continuous atomic surfaces in superspace, a quasicrystal has intervals. Here the interval is visible. The crenel is the interval, the profile of the occupation wave is the atomic surface, and the Fibonacci chain is the modulated crystal whose surface is as discontinuous as a surface can be. What makes it a quasicrystal and not a modulated crystal is where the scattering goes. In a cosine occupation wave the main reflections carry nearly everything and the satellites are a perturbation. In the golden crenel the first-order satellites carry nearly a quarter as much as the main reflections, and by Parseval’s identity the whole set of satellites together carries $\Delta - \Delta^2 = 0.236$ of a site against the main reflections’ $0.382$, about three fifths as much. The distinction is a matter of degree along one family of profiles, not a difference in kind.

The chain here has equal site spacing, which the geometric Fibonacci chain does not. That chain has long and short tiles of lengths in the golden ratio, so its atoms are displaced as well as distinguished. That is a displacive and an occupational modulation at once, and its satellites combine a crenel’s coefficients with Bessel-like factors from the displacement. The pure occupational case is the simplest member of the family, and it already has the whole two-integer spectrum.

Reading the profile back, and what cannot be read #

If the satellites are the profile’s Fourier coefficients, the profile ought to be recoverable from them by a Fourier synthesis in one variable, the position within the period. Up to a point it is. A diffraction experiment measures the size of each satellite and not its phase, and a profile is fixed by its coefficients only with their phases. The set of satellite amplitudes is the Patterson function of the profile, not the profile: it records the autocorrelation of the occupancy along the period, and every profile with the same autocorrelation gives the same satellites.

Some of the profiles it cannot tell apart are obvious. A profile and its mirror image, $f(x)$ and $f(-x)$, have Fourier coefficients of the same size, because reflecting a function conjugates its coefficients. So an occupation wave that rises slowly and falls sharply cannot be told from one that rises sharply and falls slowly, from the satellite amplitudes alone. That is the law that hides handedness in a new setting: the same conjugation that makes a diffraction pattern centrosymmetric makes the profile’s direction invisible. A shift of the profile along the period changes every phase and no amplitude, which is only the statement that the phase of the wave is arbitrary. And there are less obvious pairs, profiles that are genuinely different and share every amplitude, of the kind two structures, one Patterson constructs for arrangements of atoms.

The crenel escapes all of this by being so constrained. A profile known to take only the values nought and one on a single interval is fixed by its width, and the width is readable from the amplitudes in two independent ways: from the ratio of first-order to main reflections, which is $\sin(\pi\Delta)/(\pi\Delta)$, and from the orders that vanish. Two readings that agree are a strong test that the profile really is a step, and two that disagree say that it is not. A crystallographer fitting a modulated structure faces this choice in practice, between a smooth profile with few parameters and a step with one, and the satellites decide between them as long as enough orders are measured to see whether they fall off as one over the order or stop.

What the census has to refuse #

The last refusal builds the cosine that would give a first-order satellite of 0.3 and finds that it needs occupancies from minus a tenth to one and a tenth, which no site can have. It is the ceiling stated as a test, and a structure model that fitted such a satellite with a single cosine would fail it in exactly this way. The fourth test is the step’s extremality, checked against the analytic curve and against random profiles. A random profile beating $1/\pi$ would mean the bound had been derived wrongly.

Still open: waves that move atoms and fill sites at once #

Real modulated crystals rarely keep their modulations pure. A composition wave strains the lattice, because atoms of different sizes want different spacings, so an occupational wave drags a displacive wave along with it. The satellites of the combination are neither the Fourier coefficients of the profile nor Bessel functions of the displacement but a mixture. The two waves interfere order by order, so that a satellite can be strengthened on one side of a main reflection and weakened on the other. That asymmetry between satellites at $h + mq$ and $h - mq$ is a standard signature of combined modulation in measured patterns, and the arithmetic here, which gives the two sides equal strength, cannot produce it. Computing it needs both kinds of wave in one chain, with a stated relation between them, and it would say how large a strain wave must be before the satellites stop being a portrait of the occupation profile.