Two integers made one exponent

When boron trifluoride’s three B–F bonds are stretched unequally, its degenerate Raman bands split and each pair’s depolarisation ratio moves off three quarters by an amount that depends on which way the distortion points. The sum of those departures over every band barely depends on direction at all — flat to four parts in ten thousand where the largest single band varies by ninety-eight per cent — and the reason is a theorem: a quadratic form invariant under a three-fold rotation is isotropic on its two-dimensional representation.

Four parts in ten thousand is not zero, and the attempt to identify the leftover found it scaling as the amplitude to the 1.248. That is neither the first power a cubic term would give nor the second a quartic would, and fitting both together left residuals that marched — high, low, high — in a pattern a correct model does not leave. Reversing a distortion changed the sum by about as much as the residual was, which established an odd term, and the question left was what else was there.

Two routes were proposed. One separates the terms by symmetry rather than by fitting; the other asks whether the reading is analytic at all. Both can be settled, and a third question — whether the flat quantity is one anybody measures — turns out to matter more than either.

Reversal is an exact filter #

A distortion of amplitude $a$ in direction $\theta$, reversed, is the same distortion in direction $\theta + 180°$. So the average of the two readings contains only the even powers of $a$ in the sum’s expansion, and half their difference contains only the odd powers:

$$E(\theta) = \tfrac12[S(\theta) + S(\theta + 180°)], \qquad O(\theta) = \tfrac12[S(\theta) - S(\theta + 180°)].$$

This is not a fit. It is an identity about any function with a power series, and $E + O$ reproduces the forward reading at every angle and every amplitude to rounding. What it buys is that the cubic and quartic terms, which a single sweep mixes, are now in different numbers.

The halves also separate by shape. At an amplitude of 0.04 the even half follows $\cos 6\theta$ and the odd half follows $\sin 3\theta$, each to better than half a per cent of its own size. That is what symmetry requires of them. An even function of the distortion on a representation of a three-fold group can only vary with a six-fold period, because reversing is itself a rotation by 180 degrees on the stretch plane; an odd function picks up a sign under that rotation, and its lowest allowed harmonic is three-fold. The odd half is zero along the direction that stretches one bond against another — where reversing only swaps two equivalent bonds — and largest halfway between, exactly where the reversal test found its largest asymmetry.

Why the halves had to have these shapes #

The shapes are a second, independent check that the split is the right one, and they connect to a result about this probe that looked unrelated.

The probe is exactly blind to the totally symmetric stretch, which is why the sweep runs over the plane of stretches that sum to zero. That plane carries the two-dimensional representation of the molecule’s three-fold symmetry, and on it a rotation by 120 degrees of the molecule is a rotation by 120 degrees of the distortion. Every reading must therefore repeat with a period of 120 degrees in $\theta$. Reversal adds a rotation by 180. An even function is unchanged by both, so its period is their greatest common divisor, 60 degrees: $\cos 6\theta$ is the lowest harmonic it can have. An odd function changes sign under the 180-degree rotation and not under the 120-degree one, so its lowest harmonic is $\sin 3\theta$ or $\cos 3\theta$, and the mirror plane that makes zero and sixty degrees equivalent picks the sine.

So the harmonic content was fixed before anything was computed, and the computed halves have it to half a per cent. What symmetry does not fix is the size of each, and that is what the amplitudes measure. It is the same division of labour the intensity symmetry fixes makes for a single band: symmetry decides that a number is three quarters or that a shape is six-fold, and a calculation decides everything else.

Each half has an integer #

Swept over nine amplitudes from 0.01 to 0.16, the even half’s anisotropy scales as the amplitude to 2.005, and a single power describes all nine points to 0.6 per cent. That is the quartic term, cleanly, with nothing else in it.

The odd half scales as the amplitude to 0.981 below 0.04, where four points are described by one power to half a per cent. That is the cubic term. Above 0.04 the odd half stops following a power: it rises more slowly, peaks at 0.113 and falls.

The combined residual’s 1.248 is then no mystery. It is two straight lines of slopes one and two, added, and fitted as though they were one. The two halves are equal at an amplitude of 0.084, inside the swept range; below it the cubic dominates and the local slope is near one, above it the quartic does and the slope climbs towards two. A single exponent fitted across a crossover reports where the sweep happened to sit, not a property of either term. Fitted over the five amplitudes up to 0.04 alone, the combined residual gives 1.16; over the three from 0.08, 1.35. Neither is an exponent of anything.

It is worth setting that beside the exponent this reading is built on. The departure of a single band from three quarters goes as the square of the distortion, because a ratio of quadratic invariants squares what it measures, and that two held along every direction while the coefficient varied thirty-fold. A clean integer with a direction-dependent prefactor was the right shape there; the residual looked like a broken version of the same law and was really two laws at once.

The pattern the two-term fit left #

The earlier two-term fit took the relative residual as $Aa + Ba^2$, which is exactly the model the halves now confirm — a first-order odd term and a second-order even term. It halved the misfit and left a marching pattern, and the halves say why.

Divided by the amplitude, the odd half is constant at 0.0094 through 0.04 and then falls — to 0.0083 at 0.08, 0.0072 at 0.113 and 0.0048 at 0.16, about half its small-amplitude value. That is a fifth-order odd term of the opposite sign, and it is the one thing the two-term model does not contain. The even half, divided by the square of the amplitude, stays within two per cent of 0.0975 to the end of the range — rising only to 0.0990 at 0.16 — so almost all of the misfit is in the odd half.

The halves’ anisotropies are also very nearly additive — at the largest amplitude the combined residual is $3.306 \times 10^{-3}$ and the two halves’ add to $3.303 \times 10^{-3}$ — so the two-term model’s shape was right and its odd coefficient was wrong at large amplitude. The pattern was the quintic, and it could not have been found by adding a third term to the fit, because a cubic in $a$ and a fifth-order odd term have different enough shapes over this range to trade off against the quartic rather than show up cleanly.

The absolute value never acts #

The other proposed source of non-integer behaviour was the reading itself. Each band contributes $|0.75 - \rho|$, and an absolute value is not analytic where its argument changes sign. If some band’s ratio crossed three quarters during the sweep, the reading would have a kink, and a kink in a power series can produce any apparent exponent.

It cannot happen, and the reason is not a property of boron trifluoride. A depolarisation ratio measured with linearly polarised light is $3\gamma^2/(45\bar\alpha^2 + 4\gamma^2)$, where $\bar\alpha$ is the mean of the polarisability derivative and $\gamma^2$ its anisotropy. Both invariants are non-negative, so the ratio is largest when the mean vanishes, and then it is exactly three quarters. No tensor gives more. A band’s departure $0.75 - \rho$ is therefore never negative, and the absolute value is the identity on everything it is ever applied to.

The sweep confirms it rather than merely being consistent with it: across every amplitude and every direction, forward and reversed, the least departure of any band is $-1.1 \times 10^{-16}$, which is a rounding error on a band sitting exactly at three quarters. Those bands exist — along a direction that keeps a mirror plane, one member of each split pair keeps a purely anisotropic tensor — and they touch the bound without crossing it.

So of the two algebraic suspects the earlier calculation named, the absolute value is eliminated by a one-line bound, and the ratio’s denominator is left. The denominator’s own expansion is what produces the higher odd and even terms in the first place; the halves show it produces them as integers.

The flat quantity is not the observed one #

The sum of each band’s departure is a convenient quantity — it is what the quadratic-form argument makes flat — but it is not what a spectrum reports when the split is too small to resolve. An unresolved pair of bands is measured as one feature, and its depolarisation ratio is the ratio of the pair’s total intensities in the two polarisations. Intensities add; ratios do not.

Computed from the same Raman derivatives, the unresolved pair’s departure is between 14 and 28 times less isotropic than the sum, at every amplitude swept. At 0.01 its anisotropy is $2.9 \times 10^{-3}$ against the sum’s $1.05 \times 10^{-4}$. It scales as the first power of the amplitude, to 1.003, and it is almost entirely odd: the reversal difference accounts for all of it but a few per cent.

That changes what the flatness means. The quadratic-form argument makes a sum of quadratic forms isotropic, and the sum of departures is such a sum at leading order. An intensity-weighted ratio is not: its numerator and denominator are each quadratic forms, but their ratio is not one, and the cubic term that the sum suppresses survives in the ratio at full size. The flatness is a property of the stand-in, not of the observable — which is the question the earlier calculation flagged as the one that had something resting on it, and the answer is that the thing resting on it does not hold for a measurement.

The table carries every number. The even column quadruples each time the amplitude doubles; the odd column doubles until 0.04 and then stops keeping up; the last column is larger than everything else throughout and doubles with the amplitude from start to finish.

How the sweep was run #

The molecule is boron trifluoride with its fitted valence force field held fixed. A distortion is a combination of the three bond stretches that sums to zero, parameterised by an amplitude and a direction in that plane; at each of six directions from 0 to 50 degrees and their reverses, at nine amplitudes from 0.01 to 0.16, the normal modes are recomputed at the distorted geometry and the Raman derivatives evaluated for each mode. The four modes that are depolarised in the undistorted molecule — the two members of each E′ pair — are followed by index.

A band’s departure is $0.75 - \rho$; the sum is over the four. The even and odd halves are the average and half-difference of each direction’s reading with its reverse. Anisotropies are the largest over the directions divided by the smallest, less one, for the sum and for the even half; the odd half is quoted as its largest magnitude over the mean of the sum. Exponents are least-squares slopes in logarithms, with the worst factor by which any point misses the fitted line reported beside each. The unresolved-pair observable adds $\bar\alpha^2$ and $\gamma^2$ over the two members of a pair, forms the ratio, and sums the two pairs’ departures.

The checks, run wherever these figures are drawn: the halves add to the forward reading at every angle and amplitude; the even half’s exponent is within two hundredths of two with a single power fitting to one per cent, and its shape is $\cos 6\theta$; the odd half’s exponent below 0.04 is within five hundredths of one, with a shape of $\sin 3\theta$; over the whole sweep the odd half peaks inside the range; the combined exponent lies between the halves’, which cross inside the sweep; no band’s departure is negative; and the unresolved-pair observable is more than ten times less isotropic than the sum at every amplitude. The refusal is the undistorted molecule, where every depolarised band must read three quarters exactly, so that nothing measured can be an offset in the reference.

What a fixed force field leaves out #

The force field does not change with the geometry. A real distortion changes the force constants as well as the positions, and a field that moved with the bonds would add terms of its own at every order. The integer exponents are a statement about how the Raman tensor and the normal modes respond to a geometric distortion in one field.

The polarisability derivatives are this model’s. The intensities come from a bond-polarisability description, so the unresolved-pair observable’s size — fourteen to twenty-eight times the sum’s anisotropy — is a model number; that it is larger, and odd, follows from its being a ratio of quadratic forms and would survive a better description.

And a pair is either resolved or not. Real bands have widths, and a partially resolved pair is neither of the two readings compared here. What a measurement at a given resolution reports lies between them.

A fitted exponent is a question about a window #

The general point is about exponents. A power law fitted to a quantity with two mechanisms reports a number that depends on where the data sit relative to the point where the mechanisms trade places, and it will pass a goodness-of-fit test anyway — the combined residual’s single power fits all nine points to seven per cent, which looks like a law. Before fitting an exponent, look for a symmetry that separates the mechanisms, because an exact filter needs no model and a fit needs the right one.

The second point is about what a symmetry argument protects. The theorem made a sum flat, and the sum was flat. It said nothing about the ratio a measurement forms, and the ratio is not flat. A sum over bands is not what an absence proves about a spectrum in the same way: the statement is exactly as strong as the quantity it is made about.

Still open: the resolution between, and a second molecule #

The obvious open question is what a real spectrometer at finite resolution would report. A pair split by a few wavenumbers and measured with a slit of comparable width is neither resolved nor unresolved, and the reported ratio interpolates between the two readings above in a way that depends on the band shape. Computing it for a Lorentzian of stated width, as a function of the splitting — which the normal-mode calculation already gives — would turn twenty times less flat into a curve against resolution, and say at what resolution the theorem’s flatness becomes visible in data.

The nearer question is whether the integers are boron trifluoride’s or the group’s. Any molecule whose degenerate stretches transform like boron trifluoride’s under a three-fold axis should give the same cos 6θ and sin 3θ shapes, but the ratio of the cubic and quartic coefficients, and so the crossover amplitude, is a property of the molecule. Ammonia’s stretches are the natural second case, since its pyramid keeps the three-fold axis while losing the mirror plane, and that loss is exactly what should allow a term the planar molecule forbids.