The one intensity symmetry does fix

Intensities are the quantity symmetry declines to supply, and every symmetry argument is careful to say what it is not computing — which bands exist is exact and how strong they are is not — and the division has held.

It has one exception, and the exception is sharp enough to be worth an essay of its own. Symmetry fixes no Raman intensity and it fixes one intensity ratio, exactly, with no model in it at all.

What a Raman experiment can measure besides a position #

A Raman band has a position, a strength, and a third quantity that an infrared band does not have. Scattered light from a plane-polarised beam comes back partly with the original polarisation and partly perpendicular to it, and the ratio of the two is the depolarisation ratio. It is dimensionless, it is measured by turning an analyser, and it is quoted in tables beside the wavenumber — the only intensity-like quantity in this subject that a spectrum reports without an instrument calibration in it.

What it is a ratio of, for a molecule tumbling freely in a gas or a liquid, is two rotational invariants of the polarisability derivative $\boldsymbol{\alpha}'$ — the mean

$$a = \tfrac{1}{3}\operatorname{tr} \boldsymbol{\alpha}'$$

and the anisotropy $\gamma^2$, which is a sum of squared differences of its components. The ratio is

$$\rho = \frac{3\gamma^2}{45a^2 + 4\gamma^2},$$

and both quantities are unchanged by any rotation of the molecule, which is what lets them describe a sample with every orientation in it. This is the same rotational average that lets a spectrum count environments rather than atoms: what survives it is what a fluid sample can report.

Three quarters, and why it cannot be anything else #

The trace of a tensor is an invariant. A quantity that is invariant under every operation of the group transforms as the totally symmetric representation, and its derivative along a coordinate of any other species therefore vanishes identically — the integrand is odd about something the molecule has. That is the one theorem every selection rule is a case of, applied to a quantity rather than to a transition.

So $a = 0$ for every non-totally-symmetric mode, and the ratio collapses:

$$\rho = \frac{3\gamma^2}{4\gamma^2} = \frac{3}{4}.$$

No force field enters that. No charge, no polarisability, no bond length, no geometry beyond the symmetry.

The computation is done here with a bond-polarisability model — the oldest and crudest account of Raman intensities, in which the molecule’s polarisability is a sum of one cylindrical contribution per bond, with a value and a slope along each — and the point of using a crude model is exactly that the result does not depend on it. The situation is the reverse of the infrared case, where a fixed-charge model gets every forbidden derivative exactly right and every allowed intensity badly wrong; here it gets one ratio exactly right for a reason that has no charges in it.

Eighteen bands, and where the exactness comes from #

It is worth seeing the claim as a list rather than as an argument. Across five molecules there are eighteen Raman-active bands, and fourteen of them are not totally symmetric: three of methane’s t₂ stretches, three more of its t₂ bends, two of its e bends, four of boron trifluoride’s e′ modes, two of ammonia’s e pairs, and one antisymmetric stretch each from water and sulfur dioxide. All fourteen come back at three quarters.

They have nothing else in common. They span three point groups, five molecules, frequencies from 480 to 3,943 wavenumbers, and relative intensities from a hundredth of the strongest band to a fifth of it. The one thing they share is that their species is not the totally symmetric one, and that is the only thing the argument uses.

The depolarisation argument divides a census of modes in exactly one place — the totally symmetric row against every other — and that division is the same in every group. It is worth noticing that no other property of the mode enters: not its frequency, not its intensity, not which atoms move in it.

The tripwire, and what it caught #

A check that merely looked for three quarters would be a poor check, because a broken calculation can produce three quarters everywhere. So the whole thing is run twice: once with the fitted bond parameters and once with a set chosen to be absurd — a fluorine ten times as polarisable across its bond as along it, slopes of the wrong sign, a nitrogen four hundred times more polarisable perpendicular than parallel.

This earned its keep immediately. The first version of the polarisability model gave each bond a fixed value along and across it, with no dependence on the bond length at all — which is a reasonable-looking simplification and is fatal. With it the trace of the whole tensor is a constant, so the mean derivative is zero for every mode including the symmetric ones, every depolarisation ratio comes out at three quarters, and the output looks entirely regular: the right modes are active, the intensities order sensibly, and the one number the essay is about is wrong in a way that agrees with itself.

What separates the two versions is not whether they produce three quarters. Both do. It is whether anything is left over.

The second exact number, and the group it belongs to #

There is a case where symmetry fixes the ratio for a totally symmetric mode as well, and it is worth its own paragraph because it is the most completely polarised line in vibrational spectroscopy.

In a cubic group — tetrahedral, octahedral, icosahedral — the polarisability of the molecule is isotropic: it has to be, because the group has no preferred direction and a symmetric second-rank tensor with cubic symmetry is a multiple of the identity. A totally symmetric vibration keeps the group, so it keeps the isotropy, so its polarisability derivative is a multiple of the identity too and its anisotropy is zero. The ratio is $3 \times 0 / (45a^2 + 0) = 0$.

Methane’s symmetric stretch comes out at $\rho = 0.000000000$, and it stays there under absurd parameters, because a group cannot be made anisotropic by a choice of numbers.

Boron trifluoride’s 0.031 moves to 0.285 under the absurd parameters, a factor of nine. That is what a number the model owns looks like.

What the polarised ones are worth #

The four ratios that are not fixed are still useful, and their usefulness is of a different kind. They are all well below three quarters — 0.0015, 0.031, 0.034, 0.040 for the strong symmetric stretches, and 0.144 and 0.700 for the two bending modes — so the qualitative statement that a polarised line is a totally symmetric mode is reliable even though the number itself is not.

That is a useful asymmetry and it is worth stating plainly. A measured ratio of 0.75 identifies a mode as non-symmetric and the identification is exact. A measured ratio well below 0.75 identifies a mode as symmetric and the identification is also reliable, because the alternative is forbidden. What is not reliable is reading the size of a small ratio as a measurement of anything: it is a ratio of two model-dependent quantities and it moves by a factor of nine when the model does.

Water’s bend, at 0.718, is the interesting case: a totally symmetric mode whose ratio is close enough to three quarters that a measurement with realistic error bars would not separate it. That is not a failure of the rule — the rule is one-directional — but it is a reminder that the exactness lives on one side of the line only.

Where this sits beside mutual exclusion #

Mutual exclusion is a rule that runs one way, and it is worth asking whether it runs backwards: mutual exclusion is a theorem for centrosymmetric molecules and is not a test for them, and D5h is the counterexample. The relation between that and this one is close.

Both are consequences of the same fact — that the products of coordinates transform in a definite way, and that whether a coordinate and a product of coordinates can land in the same species is decided by the group. Mutual exclusion is that fact applied to which species can carry both; the depolarisation ratio is it applied to which species can carry a trace, which is the one totally symmetric combination that every set of quadratic functions contains.

There is no D5h here. The mutual exclusion rule fails backwards because a degenerate representation can be large enough to hold both a coordinate and a product of coordinates at once. The trace argument has no such loophole: a trace is one function, not a set, and it is invariant in every group there is.

What the numbers say about the coordinates they were computed from #

One result was not looked for. Four of the five molecules give three quarters to $10^{-12}$ or better; ammonia gives $0.749999969$, out in the eighth decimal place.

That is not the integrator. It is the molecule’s stored coordinates: ammonia’s three N–H bond lengths differ from each other by $4.5 \times 10^{-5}$ ångström, because they were entered rounded to six decimal places and a threefold axis at that precision is not quite a threefold axis. The other four molecules have coordinates that are exact by construction or rounded to a place where it does not show.

So the depolarisation ratio turns out to be a free symmetry test on the input geometry, sensitive at a level almost nothing else reaches — about eight orders of magnitude finer than the continuous symmetry measure designed for the purpose, and finer than the tolerance the point-group search uses to decide what a molecule’s group even is. The tolerance in the check is therefore set from the coordinates’ own spread rather than from the arithmetic’s, which is the honest way round: a geometry rounded in the fifth decimal cannot support a claim in the twelfth.

There is one more thing the residual says, and it is a warning rather than a finding. Because the ratio is so sensitive to the geometry, a calculation that produced 0.7499 rather than 0.7500 would be reporting a broken structure and would look like a reporting a small physical effect. Nothing distinguishes the two from the number alone. What distinguishes them is that a symmetry-breaking geometry moves every forbidden ratio in a correlated way and by an amount that tracks the size of the distortion, while a real effect of the kind resonance produces moves particular species and can push the ratio past three quarters, which a distortion cannot.

Three quarters is exact for one experiment #

The value is exact and it is not universal, and the distinction matters because the older literature quotes a different number for the same molecules and both are right.

Everything above assumes one arrangement: the incident light is linearly polarised, the scattered light is collected at ninety degrees, and the analyser is set parallel and then perpendicular to the incident polarisation. That is the standard modern experiment, and in it the two invariants combine as $3\gamma^2$ over $45a^2 + 4\gamma^2$, which is three quarters when the mean derivative vanishes.

Change the incident light to natural — unpolarised, which is what a mercury arc supplies and what every measurement before about 1965 used — and the average is taken over two incident polarisations instead of one. The denominator picks up an extra $\gamma^2$:

$$\rho_n = \frac{6\gamma^2}{45a^2 + 7\gamma^2}$$

For a non-totally-symmetric mode that is $6/7 = 0.857$, not $0.750$. The limiting value has moved, and nothing about the molecule has.

The two are related exactly, with no model and no parameters, by

$$\rho_n = \frac{2\rho}{1 + \rho}$$

which maps $0$ to $0$ and $3/4$ to $6/7$, and every polarised band in between. So a table of old ratios converts to a table of new ones by arithmetic, and a ratio quoted without its geometry is a number missing the half of its definition that decides what it means.

Two things follow.

The exactness belongs to the invariants, not to the number. What symmetry fixes is that $a = 0$ for every non-totally-symmetric mode. Every exact limiting value in this subject — three quarters, six sevenths, and the values that come out of backscattering and circular-polarisation arrangements — is that one statement pushed through a different rotational average. The arithmetic changes with the experiment; the vanishing does not.

And zero stays zero. The polarised bands are the exception in a useful way: $\rho_n = 2\rho/(1+\rho)$ sends zero to zero, so methane’s symmetric stretch is exactly unpolarised in every one of these arrangements. A vanishing anisotropy vanishes under any average, which is why the second exact number of this essay is the more robust of the two even though it belongs to fewer molecules.

The practical rule is the one a reader should carry away from a quoted ratio. A band described as “depolarised” is a claim about the invariants and travels between experiments; a band described as “0.75” is a claim about one experiment and does not. The first is what the symmetry argument supports, and it is the reason the qualitative word survives in use while the number has to be read with its geometry attached.

The conversion also explains a feature of the older tables that looks like sloppiness and is not. Ratios measured with natural light cluster near 0.86 rather than near 0.75, and a modern reader comparing them against the modern limit finds every depolarised band apparently overshooting a value it cannot exceed. It cannot exceed 6/7 either, and the bands sit against that ceiling exactly as they sit against three quarters today. Nothing was measured badly; the denominators differ by one term.

What this cannot say #

The intensities are a model’s. The bond-polarisability model has four fitted parameters per bond type, taken from the Raman intensity literature and not computed here. Every absolute intensity above inherits that, and so does every polarised ratio.

Nothing here is a cross-section. What a spectrometer records depends on the excitation wavelength, the scattering geometry, the concentration and the instrument, none of which is in this calculation. The ratio survives all of them, which is most of why it is the quantity that gets tabulated.

Resonance is excluded. Near an electronic absorption the polarisability derivative is no longer a real symmetric tensor, the argument above no longer holds, and ratios above three quarters — up to infinity for an antisymmetric tensor — are observed. That regime is real and is not this one, and nothing in this collection computes an electronic state well enough to enter it — the standing gap in this whole field.

The frequencies are a fitted force field’s. Every wavenumber quoted beside a ratio comes from the same fitted valence force field the rest of this collection uses, so the positions in these figures inherit its residual. The ratios do not depend on them at all — the depolarisation ratio of a mode is a property of its eigenvector and of the group, and moving a frequency does not move it.

And a solid is not a gas. The rotational average is what turns a tensor into two invariants. In an oriented crystal there is no average, the individual components are measurable, and the depolarisation ratio is not the quantity being measured at all.

Still open: electronic polarisation, and a ratio as a structural measurement #

The natural open question is the other exact ratio in the same family: the polarisation of an electronic transition, where the same rotational average acts on a transition moment rather than on a polarisability derivative, and where the possible values are again fixed by the species rather than by the size of anything.

The nearer question is what happens as the exactness is approached from outside. A slightly distorted molecule has slightly non-zero mean derivatives for modes that were forbidden, so its depolarisation ratios come off three quarters by an amount that measures the distortion. Ammonia’s stored coordinates already demonstrate the mechanism accidentally; doing it deliberately, with a distortion of known size, would turn the ratio into a structural measurement with no model in it — which is a rare thing in this subject and is exactly what the first sentence of this essay said symmetry does not supply.