A warm plane crystal has no Bragg peaks

The average scatters sharply and the rest does not split the scattering of a disordered crystal into two parts. The average structure gives Bragg reflections. The variance about it spreads over the whole of reciprocal space as diffuse scattering. The total is exactly one unit per site whatever the disorder does. That essay computed the split for occupational disorder, two kinds of atom on a perfect lattice, and named the other great source of disorder in a single sentence. Thermal motion is an average over time rather than over sites: the atoms at their mean positions give Bragg reflections attenuated by a Debye–Waller factor, and the variance is thermal diffuse scattering.

This essay computes that sentence, and in doing so finds that it is only half true. In a crystal in space it holds. In a crystal confined to a plane there is no Debye–Waller factor at all, because an atom’s thermal wandering grows without limit as the crystal grows. The Bragg reflections then are not sharp peaks with a thermal background underneath. They are power-law cusps whose heights fall with the crystal’s size, at a rate set by the temperature and by how far each reflection is from the origin.

A harmonic crystal, computed exactly #

The model is the simplest crystal with a thermal motion worth the name: a square lattice of atoms joined by springs, stiffness $k_1$ to the four nearest neighbours and $k_2 = k_1/2$ to the four next-nearest, the second set of springs being what keeps a square lattice from shearing freely. Its vibrations are phonons. A phonon of wavevector $q$ has frequencies set by the eigenvalues of a two-by-two dynamical matrix $D(q)$ built from the springs, and in classical equilibrium at temperature $T$ every mode carries a mean-square amplitude $kT\,D(q)^{-1}$.

On a crystal of $N \times N$ atoms with periodic edges, the allowed wavevectors are a grid of $N^2$ points in the zone, and every displacement correlation is a sum over that grid, exact and complete. Because each displacement is a sum of independent Gaussian modes, the scattering of the harmonic crystal is exact too:

$$I(Q) = \sum_R e^{iQ\cdot R} \exp\!\Big(-\tfrac12\big\langle (Q\cdot(u_R - u_0))^2 \big\rangle\Big)$$

per atom, where the average in the exponent is the thermal correlation between the displacements of two atoms a lattice vector $R$ apart. Temperature enters only through the ratio $kT/k_1$. Nothing is sampled and no atom is moved: every number below comes from sums over the phonons.

Where the diffuse scattering gathers #

The simplest part of the scattering to compute is the part that comes from one phonon at a time. Expanding the exponential to first order gives the one-phonon diffuse intensity, $kT\,Q^{\mathsf T} D(q)^{-1} Q$ with $q$ the distance from $Q$ to the nearest reciprocal lattice point.

The picture at the head of this essay is that intensity, and it has three features, each read off the formula. It gathers under every Bragg reflection, rising as one over the square of the distance to it, because the acoustic phonons’ frequencies fall to nought at the zone centre and a soft mode carries a large amplitude. It is weak near the origin and strong far out, because the factor $Q^{\mathsf T} \dots Q$ grows as the square of the scattering vector: displacements matter more to waves of shorter wavelength. And the lobes around each reflection are not round. A small step from the (1,0) reflection along $k$ gives 2.7 times the intensity of the same step along $h$, because the step along $k$ probes transverse phonons, which are softer. That anisotropy is how an experiment reads a crystal’s elastic constants from its diffuse scattering. Its sensitivity to the springs is the same sensitivity the order a diffuse pattern measures found to occupational correlations, in a different variable.

The wandering that does not stop #

The Debye–Waller factor is $e^{-\langle (G\cdot u)^2\rangle}$, the damping of each Bragg reflection by the mean-square displacement of one atom along its scattering vector. For it to exist, that mean-square displacement has to be finite.

In the plane it is not. The square lattice’s mean-square displacement along one axis is 0.68 in units of $kT/k_1$ at sixteen atoms on a side, 0.83 at thirty-two, 0.98 at sixty-four, 1.12 at a hundred and twenty-eight and 1.27 at two hundred and fifty-six. It gains 0.147 at every doubling, to three figures, which is a straight line against the logarithm of the size. An atom in a plane crystal wanders further the larger the crystal is, without limit. The simple cubic lattice with the same springs gives 0.31 at four atoms on a side, 0.34 at eight, 0.36 at sixteen and 0.367 at twenty-four, and it is levelling off.

The reason is a count of long waves. The mean-square displacement is a sum over phonons of one over the square of each frequency, and an acoustic phonon’s frequency is proportional to its wavevector. In the plane the number of phonons with wavevector near $q$ grows as $q$, so the sum near the zone centre behaves like $\int dq/q$, which grows as the logarithm of the longest wavelength the crystal can hold. In space the number grows as $q^2$, the integrand is finite, and the sum converges. Long, soft, cheap waves are too numerous in two dimensions for their amplitudes to add up to anything finite. The argument was made by Peierls in 1934 and later extended by Mermin and Wagner to a general theorem. Here it is a measured straight line.

Every Bragg peak fades #

If the Debye–Waller factor goes to nought as the crystal grows, the Bragg reflections go with it, and the exact intensity shows how.

A Bragg reflection from a crystal of $N^2$ atoms normally has an intensity proportional to $N^2$ times the square of the structure factor, so its intensity divided by $N^2$ is a constant, the peak’s strength per atom. For the warm square lattice at $kT/k_1 = 0.02$ that ratio is 0.585 at the (1,0) reflection for sixteen atoms on a side and 0.464 for sixty-four. It keeps falling, as a power of $N$: the exponent is 0.168. The (1,1) reflection falls with an exponent of 0.336, the (2,0) with 0.675 and the (2,1) with 0.846. The reflection’s total intensity grows as $N^{2-\eta}$, not as $N^2$, and for a large enough crystal the peak per atom is as small as anyone likes.

That places the warm plane crystal exactly in the gap that neither a peak nor a bump found for a chain of a quite different kind, whose strongest reflection grew as the length to the power one and a half. A Bragg peak grows as the square of the number of scatterers and a diffuse bump as the first power. Anything between is a third kind of scattering, and a two-dimensional crystal at any temperature above nought makes it at every reflection.

The exponents are not arbitrary. At a fixed temperature they are proportional to $|G|^2$: the ratios for the four reflections are one, two, four and five to within half a per cent, which are the ratios of their squared distances from the origin. At a fixed reflection they are proportional to the temperature: the lines at $kT/k_1$ of 0.01, 0.02 and 0.04 have slopes in the ratio one, two and four. So $\eta = c\,T|G|^2$, with a constant $c$ set by the lattice’s elastic stiffness. The reflections near the origin are nearly sharp and the ones far out are hardly peaks at all. At the temperature used here the (2,1) reflection’s exponent is close to one, which would already make it hard to tell from the diffuse background around it in a crystal of a few thousand atoms on a side.

Why the exponent is the wandering, read at one reflection #

The proportionality has a one-line reason, and it ties the two figures before this one together. The exponent in the scattering formula is the variance of $G\cdot(u_R - u_0)$, the difference in phase that two atoms a distance $R$ apart impart to a wave scattered at the reflection $G$. Because the mean-square displacement grows as the logarithm of the crystal’s size, the variance of the relative displacement of two atoms grows as the logarithm of their separation, at the same rate. So the factor $\exp(-\tfrac12\langle\dots\rangle)$ falls as a power of $R$, and summing a power of $R$ over a crystal of $N^2$ atoms gives a power of $N$.

The numbers agree to three figures. The mean-square displacement along one axis grows at 0.2123 $kT/k_1$ per unit of $\ln N$. The fading exponent of the (1,0) reflection at $kT/k_1 = 0.02$, divided by $|G|^2 = 4\pi^2$ and by the temperature, is 0.2125. The exponent of a reflection’s fading is the rate at which an atom’s wandering grows, multiplied by the square of the reflection’s scattering vector and by the temperature. That is why a reflection twice as far from the origin fades four times as fast, and why doubling the temperature doubles every exponent. The small difference between the two numbers is the finite crystal: the fading exponent is fitted over sizes from sixteen to sixty-four, where the logarithm has not quite settled.

What the fading means at practical sizes #

A power of the size is a slow way to vanish, and it is worth putting numbers to it. At $kT/k_1 = 0.02$ the (1,0) reflection keeps 0.585 of the intensity a perfect crystal would give it at sixteen atoms on a side. Continued at its measured exponent of 0.168, it would keep about 0.29 at a thousand atoms and 0.20 at ten thousand. That is diminished, but it is plainly still a peak standing well above its wings. The (2,1) reflection starts at 0.075 at sixteen atoms and, with an exponent of 0.846, would keep about 0.002 at a thousand and 0.0003 at ten thousand. That is a reflection lost in the diffuse wings around it long before the crystal reaches the size of a real grain.

So the loss of Bragg order in a plane is not one event but a gradient in reciprocal space. Near the origin the reflections look nearly normal at any practical size. Far out they disappear first, and the boundary between the two moves inwards as the temperature rises. The exponent of the (1,0) reflection would reach two, the value at which a reflection’s total intensity stops growing with the size of the crystal at all, near $kT/k_1 = 0.24$. That is far outside the range in which harmonic springs describe any real solid, so in this model the innermost reflections never lose their peaks before the crystal would have melted.

What one phonon at a time misses #

The map at the head of this essay is the one-phonon approximation, and it fails exactly where the plane is interesting. It predicts a diffuse intensity that rises as one over the squared distance to each reflection, and in two dimensions that rise, summed over the area around the reflection, is itself logarithmically large, which is the same divergence as the wandering, seen from reciprocal space. The one-phonon picture cannot say what happens at the reflection itself. It adds a divergent background to a peak it assumes is sharp, and in the plane the peak is not sharp.

The exact scattering replaces both with a single object: a cusp whose height grows with the crystal as $N^{2-\eta}$ and whose wings fall as a power of the distance from its centre. In space the same exact formula separates cleanly into a sharp peak with a finite Debye–Waller factor and a one-phonon background that is integrable, which is why the textbook picture works there and why it is usually stated without the dimension being mentioned.

Nothing lost, only moved #

The first essay on this subject found that disorder never changes the total scattering, only where it goes. The thermal case obeys the same rule, and on the torus it can be checked exactly.

Summed over the $N^2$ allowed points of the zone around the (1,0) reflection, the intensity is $N^2$ at every temperature, to the last digit: one unit per atom, as the sum rule requires. What changes is how that total is shared. At $kT/k_1 = 0.005$ the central point holds 86 per cent of it, at 0.02 it holds 55 per cent, and at 0.08 it holds 9 per cent, with the rest in wings that fall off as a power of the distance from the reflection. In a crystal in space the central share would settle to a fixed Debye–Waller fraction as the crystal grew. In the plane it keeps shrinking, and the wings are the same power law as the fading.

What this says about the average #

The sentence this essay started from described thermal scattering as an average structure plus a variance. The computation shows exactly what goes wrong with it in a plane. The average structure is the atoms at their mean positions, and in a plane the mean position of an atom relative to a distant one is not well defined, because the variance of their separation grows as the logarithm of the distance. The crystal has order in its bonds and in the orientation of its lattice, but its positional order decays slowly with distance instead of persisting, a condition sometimes called quasi-long-range order. There is no average structure in the sense the split requires, and so there are no Bragg peaks for it to give.

That is a statement about harmonic springs in an idealised plane. Real two-dimensional crystals, such as a layer on a substrate, a free-standing sheet or a film of colloids, are held by a substrate, bend out of the plane, or melt by the unbinding of defects in ways the harmonic model does not contain. Graphene’s out-of-plane ripples are the famous example. What the harmonic model settles is that no such crystal can be both strictly two-dimensional and warm and still give sharp reflections; whatever sharpness is observed comes from something the model leaves out. It is also why the layer groups describe the symmetry of a sheet but not the sharpness of its diffraction. The symmetry survives the thermal motion, and the long-range positional order does not.

What the computation has to refuse #

The last test is the cold limit. At a temperature of nought every atom sits at its lattice site, the correlation in the exponent vanishes for every pair, and every unit of intensity is in the Bragg reflection with nothing in the wings. A model that produced diffuse scattering from a lattice at rest would be counting something other than thermal motion. The first two tests are the two dimensions against each other, and the third is the claim the whole essay rests on, that the exponent is proportional to $|G|^2$, tested by a factor of four between two reflections.

Still open: from the plane to a real layer #

The harmonic square lattice is the cleanest place to see the loss of Bragg peaks and the least realistic place for it to happen. The natural next step is a sheet that can move out of its plane as well, where the out-of-plane modes are softer still, with frequencies proportional to the square of the wavevector for a membrane without tension, and the mean-square height diverges much faster than the logarithm. For a sheet under tension or stiffened by coupling between bending and stretching, the divergence is tamed, which is part of how graphene keeps a crystal’s diffraction. Adding the out-of-plane modes to the dynamical matrix and repeating the exact sums would say how much of each reflection a real layer keeps and at what size it loses it. It is the same computation with a three-by-three matrix in place of a two-by-two one.