A fault in the deep layers costs exactly the fine peaks

The period-doubling chain is aperiodic and diffracts in nothing but sharp peaks, and the account of where they sit rested on one construction. Number the letters from zero. Letter $n$ is $b$ exactly when the power of two dividing $n + 1$ is odd. Sorting the letters by that power splits the chain into layers, layer $j$ being every site whose $n + 1$ is $2^j$ times an odd number, and every layer is periodic, with period $2^{j+1}$, and carries a single letter. The chain is the union of infinitely many periodic chains, and its diffraction is the sum of theirs: a peak at every fraction $m/2^s$, the same height $\frac23 \cdot 2^{-s}$ for every odd $m$, and nothing in between.

A chain like that, ordered without being periodic, invites a question a crystallographer would ask of any layered structure. Real stacks are not perfect. The layers of a close-packed metal or of silicon carbide’s many polytypes are put down one at a time and now and then put down wrongly, and the streaks a faulted stack makes found what that does to an ordinary crystal: some rows of reflections stay sharp, and others smear. The period-doubling chain is a stack in a stranger sense, its layers interleaved rather than piled, the deep ones ever sparser. What does a fault in the deep layers do to it?

The answer turns out to be exact, and it depends sharply on how the fault is made. Redraw the deep layers site by site and the chain loses precisely its fine peaks, each one, and nothing else. The intensity they carried comes back as a perfectly flat floor, to the last digit of a closed form. Redraw the same layers once per layer and the chain loses nothing at all.

Deep sites, and two ways to spoil them #

Call a site deep at depth $r$ when $2^r$ divides $n + 1$. The deep sites are layers $r$, $r + 1$, $r + 2$ and so on, together, and they form one arithmetic progression, every $2^r$-th site. A fault at depth $r$ redraws some of their letters at random and leaves the shallow layers alone, which is the arithmetic version of a stack whose first few layers were laid down carefully and whose later ones were not.

There are two natural ways to draw the new letters, and the two figures below this section keep them apart. In a site fault, each deep site independently has its letter redrawn, with probability $q$, as $b$ with probability $p$ and $a$ otherwise. When $q = 1$ every deep site is redrawn. In a layer fault, each deep layer draws one letter, at random, and every site of that layer carries it. The site fault breaks the layers up into independent coin tosses. The layer fault keeps every layer periodic and only changes what each one says.

The perfect chain itself fixes the most natural value of $p$. Among the deep sites at depth $r$, half belong to layer $r$, a quarter to layer $r + 1$, and so on, and the odd layers carry $b$. If $r$ is odd the first deep layer carries $b$, and the share of $b$ among the deep sites is $\frac12 + \frac18 + \cdots = \frac23$. If $r$ is even it is $\frac14 + \frac1{16} + \cdots = \frac13$. Call this share $d$. A fault that draws $b$ with probability $d$ keeps the composition of the deep sites and of the whole chain unchanged on average. It is the faithful fault, and it is the one a physical mechanism that does not care about the chemistry would produce.

The fine peaks go, and nothing else moves #

A random chain’s diffraction separates into two parts, and the average scatters sharply and the rest does not is the general statement. Replace every letter by its expected value and diffract that: this gives the Bragg peaks. What remains, the variance of each letter about its expectation, scatters as independent noise, and for letters drawn independently from site to site it spreads evenly over every frequency, a flat floor whose total is the average variance per site.

For the site fault both parts can be written down. The shallow sites keep their letters, so their expected value is their letter. A deep site keeps its own letter with probability $1 - q$ and is redrawn with probability $q$, so its expected value is $(1 - q)$ times its own letter plus $q p$. The expected chain is therefore the perfect chain with every deep letter scaled by $1 - q$, plus a constant $qp$ on the deep sites.

The second piece is a comb, a constant on every $2^r$-th site, and a comb of period $2^r$ diffracts only at multiples of $2^{-r}$. So it touches the coarse peaks, those at $m/2^s$ with $s \le r$, and no others. The first piece is the perfect chain’s deep part scaled down, and the deep part is the only part that reaches a fine peak. The layer sum in the earlier account gave layer $j$ an amplitude at denominator $2^s$ only when $j \ge s - 1$, and a fine peak has $s > r$, so every layer that reaches it is deep. A fine peak sees nothing but the deep layers, and sees them scaled by $1 - q$. With $q = 1$ it sees nothing at all.

The coarse peaks need one more step. At a frequency $m/2^s$ with $s \le r$, every deep site sits at the same phase, because the deep sites are $2^r$ apart and $2^r$ times the frequency is a whole number. The deep sites therefore contribute to a coarse peak only through their total, the number of $b$ among them, and not through which ones carry it. Scaling the deep letters by $1 - q$ and adding $qp$ changes that total by $q(p - d)$ per deep site, so every coarse amplitude moves by

$$\Delta A = q\,(p - d)\,2^{-r},$$

and the move is zero exactly when $p = d$. A faithful fault leaves every coarse peak where and as high as it was, and removes every fine one. On 65,536 letters faulted below depth 4, the five coarse peaks out to $\frac1{16}$ agree with the perfect chain’s to within $10^{-3}$, about twice the scatter one random chain of that length carries, while the peak at $\frac1{32}$, which held $4.3 \times 10^{-4}$ of the intensity, is reduced to a bin of the floor.

The books balance at every depth #

What the fine peaks carried is known exactly from the perfect chain. At denominator $2^s$ there are $2^{s-1}$ peaks of intensity $\frac49 \cdot 4^{-s}$, which is $\frac29 \cdot 2^{-s}$ together, so everything finer than $2^r$ carries $\frac29 \cdot 2^{-r}$. And what the floor holds is known exactly from the variance. A deep site that carried $b$ is $b$ afterwards with probability $1 - q + qp$; one that carried $a$ is $b$ with probability $qp$. Averaging the variance of a coin over the deep sites and multiplying by their density $2^{-r}$ gives the floor, and at $p = d$ the algebra collapses to

$$\text{floor} = q\,(2 - q)\; d\,(1 - d)\; 2^{-r}.$$

For a complete fault, $q = 1$, that is $d(1 - d) \cdot 2^{-r}$. The parity of $r$ changes $d$ from a third to two thirds, and $d(1 - d)$ does not notice: it is $\frac29$ either way. So the floor is exactly the fine peaks’ intensity, $\frac29 \cdot 2^{-r}$, at every depth.

The balance had to come out, and that is what makes it a check rather than a discovery. Parseval’s identity says the total intensity of the $b$ sites, per site, is their density, a third, and a faithful fault keeps the density. The coarse peaks were shown unchanged by the phase argument, so whatever the fine peaks lose has nowhere to go but the floor. But the two sides of the ledger were computed by routes that share nothing. The fine peaks’ share came from the layer sums of the perfect chain, which never mention a variance. The floor came from the variance of a coin, which never mentions a layer sum. That $d(1 - d)$ lands on $\frac29$ for both parities is the two routes agreeing, and an error in either, a wrong height for the perfect chain’s peaks or a wrong variance for the redrawn letters, would have broken it. An unfaithful fault breaks it legitimately: the coarse peaks move, Parseval then balances by moving intensity between the coarse peaks and the floor, and the floor no longer matches the fine peaks’ share.

The measurement agrees. One random chain of 32,768 letters per depth gives ratios of measured floor to lost intensity of 1.002, 1.000, 0.983, 1.026, 1.007, 1.010 and 0.986 for depths one to seven. At depth eight the deep sites number only 128 and the ratio is 0.925, which is one random chain’s scatter rather than a disagreement. The closed form has no free parameter to absorb a discrepancy, so a single slip in the variance, such as forgetting that a site can be redrawn to the letter it already had, would have shown at every depth at once.

Only one density spares the coarse peaks #

The faithful density is not a detail. Draw the deep letters at any other density and the coarse peaks move by $q(p - d) \cdot 2^{-r}$, all of them in the same direction along the deep comb’s phase. At depth 4 the zero peak, the average density, moves from 0.3333 to 0.3438 when $b$ is drawn with probability one half, and the measured chain gives 0.3431. The peak at $\frac1{16}$ moves further in proportion, because it is smaller.

The two lines in the figure cross the line of the perfect chain at different places, and the reason is the same parity that sets $d$. Below an even depth the first deep layer carries $a$, the deep sites are a third $b$, and drawing more $b$ than that weakens the peak at $\frac1{16}$, because the shallow layer that subtracts from it is a $b$ layer. Below an odd depth everything reverses. A coin with the right bias for one depth is the wrong coin for the next. An unbiased coin, the obvious choice for a model of random faulting, is faithful at no depth at all, and at every depth it shifts the coarse peaks by a sixth of $2^{-r}$.

That gives the coarse peaks a use. If the fine peaks of a faulted limit-periodic structure are gone, they say nothing about the fault except its depth. The coarse peaks carry the fault’s composition. A measured zero peak or first coarse peak that departs from the perfect structure’s value by $\Delta A$ fixes $q(p - d)$, which is how far the faulting process prefers one letter over the composition it replaced.

A partial fault is a step, not a slope #

A real stacking fault does not redraw every deep site. Take $q$ less than one, with the faithful density. Every fine peak’s amplitude is scaled by $1 - q$, so its intensity is scaled by $(1 - q)^2$, and every coarse peak is untouched. The floor rises as $q(2 - q)$ to meet exactly what the fine peaks lose, $1 - (1 - q)^2$ of their share.

The shape of that loss is worth setting beside the one every crystallographer knows. Thermal vibration also takes intensity out of Bragg peaks and spreads it between them, and it does so by the Debye–Waller factor, which for a displacement of mean square $\langle u^2 \rangle$ is $\exp(-4\pi^2 k^2 \langle u^2 \rangle)$ in one dimension. It grows with the frequency, smoothly, so high-angle reflections fade first and a single temperature factor fitted across all of them describes the loss. A warm plane crystal is the extreme case, where the factor goes to zero for every reflection.

A deep fault does something no smooth function of $k$ can mimic. Below depth 3, the peak at $\frac1{16}$ and the peak at $\frac7{16}$, much further out, lose the same fraction, while the peak at $\frac14$, between them in frequency, loses nothing. The damage is indexed by the denominator, not by the distance from the origin. A refinement that fitted a temperature factor to such a pattern would find the fit impossible: intensities would need to drop at $\frac1{16}$, recover at $\frac18$ and $\frac14$, and drop again at $\frac5{16}$. That impossibility is itself the diagnosis, the signature of disorder organised by layer rather than by position.

The same partial-fault arithmetic gives a measure of how faulted a sample is from a single ratio. Every fine peak divided by its perfect value is $(1 - q)^2$, at every fine denominator. Four averaged chains reproduce it to within a few hundredths at four denominators and five values of $q$, the worst point being 0.276 against 0.25 at denominator 128, where the peak is weakest and a single bin’s scatter weighs most.

Once per layer, every peak survives #

The question the earlier account left open was whether randomness could produce scattering that is neither peak nor smooth floor, and specifically whether a random choice made once per layer might. It cannot, and the reason is one line. A layer that carries one letter at all its sites is still periodic. A chain built from periodic layers, whatever letter each one carries, is still a Toeplitz sequence, and the argument that gave the perfect chain pure point diffraction never used which letters the layers carried. Every realisation of the layer fault is a limit-periodic chain in its own right.

The heights change, and they can be predicted from the letters drawn. Write $c_j$ for the letter of layer $j$, one for $b$ and zero for $a$. The amplitude at $m/2^s$ is, up to a phase,

$$\Bigl|\,-c_{s-1}\,2^{-s} + \sum_{j \ge s} c_j\,2^{-j-1}\Bigr|,$$

the same layer sum as before with the perfect chain’s alternating letters replaced by whatever was drawn. It is still the same for every odd $m$ at one denominator, so the peaks still come in equal families, and the families’ heights are now a lottery.

The figure shows one regularity that the formula explains. In the first draw the layers from three to six all carry $b$, and the peaks at $\frac1{16}$, $\frac1{32}$, $\frac1{64}$ and $\frac1{128}$ come out with one height. Each step from $s$ to $s + 1$ drops the subtracting term of layer $s - 1$ and turns layer $s$’s positive term into a subtracting one, so the amplitude changes by $2^{-s}$ times the difference of the two layers’ letters. It does not change at all when the two layers carry the same letter, $b$ or $a$. A run of identical letters makes a plateau of equal peaks, and the perfect chain, whose letters alternate, is the one arrangement in which no two generations are equal.

A peak can vanish under a layer fault, but only in one way: if every layer from $s - 1$ onward draws $a$, the sum is zero and the peaks at denominator $2^s$ and beyond are all gone. That realisation is not a random chain at all but a periodic one, the approximant of period $2^{s-1}$, and it has probability zero. With probability one, a layer fault keeps every peak the perfect chain had and puts nothing anywhere else. The answer to the earlier question is no. A random choice per layer gives peaks and only peaks, and a random choice per site gives peaks and a floor. Neither gives anything in between.

How to tell a peak from a floor #

On a finite chain every frequency bin holds some intensity, and a peak at one size and a strong bin of floor can look alike. The test that separates them is what the length does. A Bragg peak’s amplitude is an average over the chain and converges to a fixed value. A floor’s amplitude in one bin is a sum of independent random terms divided by the length, so it shrinks as one over the square root of the length.

At $\frac3{64}$ the perfect chain holds $1.0 \times 10^{-2}$ from 1,024 letters to 65,536, the layer-faulted chain holds $1.3 \times 10^{-2}$, and the site-faulted chain falls from $4 \times 10^{-3}$ to $3 \times 10^{-4}$, following the square-root line with the irregularity that a single random sum carries. The same test was used in neither a peak nor a bump to decide that the Thue–Morse chain has no peaks away from the integers, and it is the only test a finite computation can offer. What it shows is a trend, and the proofs that the floor is flat and the peaks are exactly where the formula says are the algebra above.

What the account has to refuse #

The refused claim is the natural misreading of the first figure: that a fault from depth 4 merely weakens the peak at $\frac1{32}$. Its amplitude after a complete site fault is $4.7 \times 10^{-4}$ on 65,536 letters, against $2.1 \times 10^{-2}$ in the perfect chain, and it falls further with length. The other tests guard the arithmetic. The coarse peaks are compared at the faithful density and at a density of one half, where they must move by the predicted amount and not otherwise. The floor is compared with the closed form at six depths, and the layer fault’s heights with the letters actually drawn.

What the figures cannot show #

Every figure is one random chain, or four, of a few tens of thousands of letters, and the statements are about averages over infinitely long chains. The decomposition into peaks from the average and a floor from the variance holds almost surely, for almost every realisation, in the infinite limit; a finite chain shows it with a scatter that the figures show honestly rather than average away. Nothing here measures how fast a given realisation converges, and the per-layer heights in particular depend on infinitely many letters, of which a chain of $2^{15}$ letters sees fifteen.

The conventions matter. The scatterer is a unit on every $b$ site and nothing on $a$; with scattering lengths $f_a$ and $f_b$ every intensity above scales by $(f_b - f_a)^2$ apart from the zero peak, and the floor scales the same way. The faults are independent from site to site. A fault that is correlated along a layer, the same wrong letter for a stretch and then a different one, is neither of the two cases here, and the figures say nothing about it. And the chain is one-dimensional. The streaks a faulted stack makes is the three-dimensional analogue for close packing, where faults are correlated between neighbouring layers and some rows of reflections are spared by an integer condition. Here the condition is the denominator, and the sparing is total.

Who worked it out #

Stacking faults in layered crystals were first analysed by Hendricks and Teller and by Wilson in 1942. The split of a random structure’s diffraction into the diffraction of its average and a diffuse part from its variance is standard. Baake, Birkner and Moody worked it through rigorously for random point sets in 2010, and Baake and Grimm’s Aperiodic Order of 2013 sets out the period-doubling chain’s own spectrum. Toeplitz sequences in general, with any letters on the layers, were studied by Jacobs and Keane from 1969, and the pure point diffraction of a layer-faulted chain follows from the same construction as period doubling’s. The faithful density, the itemised ledger, the step loss and the plateaus of equal peaks are computed here.

Still open: a fault that runs along a layer #

The two faults here are the extremes of one family. A site fault draws a fresh letter at every deep site; a layer fault draws one letter for a whole layer and keeps it. In between is a fault that keeps a letter for a stretch of a layer and then redraws it, the way a real fault persists over a domain before the stacking recovers. Such a layer is a periodic comb multiplied by a slowly varying random sequence, and its diffraction should be the comb’s peaks broadened into humps of width set by the stretch, centred on the fine positions rather than spread flat. That would be scattering that is neither a peak nor a floor, at exactly the frequencies how much of the hat is a crystal would call the aperiodic part, and the transition from humps to peaks as the stretches lengthen is a computation on the same chains. It has not been made. The question also has a version for the Fibonacci chain, whose structure is not a stack of periodic layers at all, where it is not even clear what a deep fault would mean.