A dipole is not what an infrared spectrum sees

An infrared band exists when light can drive a vibration, and light drives a vibration through the molecule’s dipole. That much is in every account, and it is where the account usually stops — leaving the impression that a polar molecule has an infrared spectrum and a non-polar one does not.

The quantity that matters is not the dipole. It is the derivative of the dipole along the mode: how fast the charge distribution shifts as the atoms move. A molecule with no dipole at all can have a large one, and a molecule with a large dipole can have a mode along which it does not change.

Carbon dioxide, which has none and shows three #

The clean case is the linear triatomic.

In the symmetric stretch both oxygens move outward together. The molecule stays symmetric at every instant, so its dipole is zero at every instant, so the derivative is zero — the band does not exist, and this is the mode a Raman spectrum sees instead.

In the antisymmetric stretch one bond lengthens as the other shortens. The molecule is momentarily unsymmetrical and has a momentary dipole, which oscillates. The band is strong.

The bend does the same thing in a perpendicular direction, twice over, which is why the 673 cm⁻¹ band is a degenerate pair.

None of that has anything to do with the dipole moment of the equilibrium structure, which is zero and stays zero.

The two routes, agreeing on every mode #

The interesting part of the calculation is not the derivatives but the check on them.

Which modes can be seen is decided by symmetry: a mode is infrared active exactly when its species carries one of the coordinate functions x, y or z, and that is read off a character table with no geometry in it at all. The derivative is a geometric calculation with no group in it: charges on atoms, displaced along an eigenvector of a mass-weighted Hessian.

The agreement is across twenty-eight modes of five molecules, and it is exact rather than approximate: the forbidden derivatives come out at machine noise. That is the same kind of statement as an overlap that symmetry forbids coming out at 10⁻¹⁷ — not a small number, a cancellation.

One molecule’s zeroes are larger than the rest and the reason is worth recording, because the same limit has turned up twice elsewhere in this collection. Boron trifluoride’s silent symmetric stretch comes out at 1.4 × 10⁻¹⁰ rather than 10⁻¹⁶, because its coordinates are stored to four decimal places and its threefold symmetry is therefore only exact to about that. The zero is a property of the symmetry, and the symmetry is only as good as the file it came out of.

What water’s dipole does not decide #

Water is the molecule with a dipole, and it is worth checking what its dipole buys it.

Every mode of water is active, and that is a consequence of its symmetry being low: in C2v the three vibrations are A1, A1 and B2, and all three of those species carry a coordinate function. A molecule with only three atoms and no symmetry to speak of has nothing that can be forbidden.

So the correlation people notice — polar molecules have infrared spectra — is real and is a correlation between two consequences of the same cause. Low symmetry allows a dipole and allows most modes to be active. High symmetry forbids the dipole and forbids some modes. What it does not do is forbid all of them, and the number it forbids has nothing to do with the dipole.

The dipole moments themselves are worth stating rather than drawing, because the point they make is a list: three of carbon dioxide, methane and benzene have no dipole at any origin, and all three have infrared spectra. A quantity that is exactly zero in three of the four cases cannot be what the spectrum is reporting.

The silent modes, counted #

It is worth putting the silent modes in one place, because between them they make the general rule visible.

Carbon dioxide’s symmetric stretch, 1354 cm⁻¹, silent because the molecule stays centrosymmetric throughout it.

Methane’s two E deformations, 1580 cm⁻¹, and its totally symmetric stretch, 3027 cm⁻¹ — three silent modes out of nine.

Boron trifluoride’s symmetric stretch, 888 cm⁻¹, silent for the same reason as carbon dioxide’s: the molecule keeps its full symmetry while it happens.

Every one of these is a mode along which the molecule’s symmetry is preserved. That is the general rule and it is a stronger statement than the character-table version, because it can be seen: a distortion that keeps every symmetry element the molecule has cannot create a dipole, since the same symmetry forbade one in the first place.

The totally symmetric stretch of any molecule with a centre or with enough symmetry to forbid a dipole is therefore always silent. The molecules where that is not true are exactly the molecules with a dipole already — where the totally symmetric mode changes a dipole that is not zero, and is usually one of the strongest bands.

So the same fact, stated twice: the modes an infrared spectrum cannot see are the modes that preserve whatever symmetry forbids the dipole. Where there is no such symmetry, nothing is forbidden and everything is visible.

The model, and what it cannot do #

The charges are a quoted correlation turned into numbers — each atom’s Pauling electronegativity against the molecule’s mean, scaled, with the residue shared out so the molecule is neutral. Nothing here computes a density, and the numbers are the same ones the multipole table uses, from the same rule in the same file.

That model has one virtue and one serious limitation.

The zeroes are exact. Whether a derivative vanishes is a symmetry question, and moving fixed charges along a mode answers it correctly however wrong the charges are: the contributions cancel term by term for a forbidden mode, whatever numbers are attached to the atoms. So the pattern of activities is right.

The magnitudes are not. A real molecule’s charges move as it vibrates — density flows from one atom to another as a bond stretches — and that charge flux contributes to the derivative alongside the fixed-charge term. For many bands the two are comparable, and for some the flux term is larger and has the opposite sign. So the relative intensities above are indicative and no more.

That distinction is worth holding on to because it is the same one this field keeps meeting: a crude model can be exactly right about what vanishes and badly wrong about what does not, and the two halves of its output deserve different amounts of trust.

Ammonia, where everything is visible and nothing is equal #

Ammonia is the case between the two extremes and is worth a section because it shows what the derivative does when nothing is forbidden.

All six modes are active because in C3v the vibrations are 2A1 ⊕ 2E and both species carry coordinate functions. So symmetry says nothing at all about which bands appear, and everything about the spectrum’s appearance is in the derivatives.

That is the ordinary case, and it is the reason intensities matter. For a molecule with high symmetry, the interesting question is which bands exist and the answer is exact. For a molecule with low symmetry every band exists and the interesting question is how strong — which is the question a fixed-charge model answers badly and a real calculation answers with difficulty.

The umbrella mode at 1020 cm⁻¹ is the strong one in a real ammonia spectrum, and this model puts it at 0.76 of the largest. That is the right ordering for the wrong reason: the real intensity of that band owes a great deal to the lone pair’s density following the nitrogen, which is charge flux and is not in this model at all.

What a band’s strength actually is #

For completeness, the quantity a spectrum measures is the integrated absorption of a band, and it is proportional to

$$A \propto \left| \frac{\partial \boldsymbol{\mu}}{\partial Q} \right|^2$$

evaluated at the equilibrium geometry, with Q the normal coordinate. The dipole moment itself does not appear, and neither does the frequency except through the density of states.

Two consequences follow that are worth stating separately from the arithmetic.

A weak band is not a forbidden one. A derivative can be small without being zero, and the difference matters: a forbidden band is absent at every temperature, in solid, liquid and gas alike, at every concentration, while a weak one appears when the sample is thick enough. Confusing the two is how a symmetry argument gets falsified by an experiment that was actually confirming it.

An intensity is a derivative and derivatives are hard. Computing band positions well is routine; computing intensities well is not, because a derivative of a property is more sensitive to the wavefunction than the property is — which is the same second-order argument that makes the energy the easiest thing to get right and everything else harder.

The count of bands a spectrum shows is smaller than the count of modes for two separate reasons, and it is worth keeping them apart: degeneracy, which merges modes that have the same frequency, and silence, which removes modes that have no derivative. This essay is entirely about the second, and nothing in it changes a frequency.

The classical derivative and the quantum matrix element #

There are two ways to say what makes a band, and they give the same answer for a reason worth stating once.

The quantum statement is that a transition between two vibrational states has an intensity proportional to the square of the transition moment ⟨v′|μ|v″⟩ — the dipole operator taken between the two states. That is the same object selection rules are one theorem is about, and its vanishing is a symmetry question exactly as the derivative’s is.

The classical statement is the one used above: expand the dipole in the normal coordinate,

$$\boldsymbol{\mu}(Q) = \boldsymbol{\mu}_0 + \frac{\partial \boldsymbol{\mu}}{\partial Q}Q + \dots$$

and the transition moment between adjacent vibrational levels picks out the linear term, because ⟨v+1|Q|v⟩ is non-zero and ⟨v+1|1|v⟩ is not. So the constant term — the dipole moment itself — contributes nothing to any transition, and the first derivative is what survives.

That is the arithmetic behind the whole essay in one line: μ₀ multiplies an overlap between orthogonal states and drops out. A molecule’s dipole moment cannot appear in any vibrational intensity, however large it is.

The quadratic term does not vanish either, and it is what gives overtones their intensity — ⟨v+2|Q²|v⟩ is non-zero. Overtones are weak because the second derivative is small compared with the first, not because they are forbidden, and a band whose fundamental is forbidden can have an allowed overtone. Nothing in this collection computes one.

What was checked #

The computed derivative agrees with the character table on every mode, for the four molecules whose point groups are finite — twenty-four modes, each checked separately. A linear molecule’s group is infinite and the reduction formula divides by the order, so those are held to the derivative alone.

Carbon dioxide’s dipole is exactly zero and it has both active and silent modes, which is the essay’s claim stated as three checks on one molecule.

Water has a substantial dipole and all of its modes are active, which is the other end of the comparison and is required so that the argument is not simply symmetric molecules are interesting.

The one charge-like quantity that is measured #

Everything taken apart so far — bond dipoles, partial charges, lone-pair contributions — has failed the same test: it is a construction rather than an observable, and different constructions disagree. The dipole derivative is the exception, and it is worth saying so, because it changes what kind of argument is available.

An infrared band’s integrated absorption is measurable in absolute units. It is not a peak height on an arbitrary scale; it is a cross-section, and it is proportional to the square of the dipole derivative along the mode. So a measured spectrum returns $|\partial\mu/\partial Q|$ for every active mode of a molecule, in debye per ångström times a root mass, with no model between the instrument and the number.

That makes it possible to run this essay’s calculation backwards. Instead of assigning charges and computing derivatives, take the measured derivatives of every mode and ask what set of atomic charges would reproduce them. The answer exists, it is over-determined for most molecules, and the charges it gives are measured in a sense that no electronegativity-derived charge is.

They do not agree with the electronegativity-derived ones, and the disagreements are not small. That is the strongest available form of the complaint against charges: it is not that two conventions disagree with each other, but that both disagree with a quantity an experiment returns.

Two things stop that being a clean victory for the measurement, and both are worth naming.

The derivative is not a charge, it is a tensor. Displacing an atom along $x$ can change the dipole along $y$, so what the intensities determine for each atom is a three-by-three array rather than a number. Reducing it to a scalar charge means taking some average of its elements, and which average is a convention — so the measured route ends in a choice too, one step later than the electronegativity route does.

And the derivative contains two physically different things. Moving a nucleus moves its own charge, and it also makes the electrons flow: $\partial\mu/\partial x = q + x\,\partial q/\partial x$. A fixed-charge model has the first term and not the second, which is precisely why its intensities are the weakest part of this essay. Separating them needs the electron density’s response, which is the calculation named as the next step.

So the ordering this essay ends with survives with an addition. Existence of a band: exact, from a group. Position: good, from a force field. Intensity: measurable to high accuracy and modelled badly — which is an unusual combination, and the reason infrared intensities were for decades a standard benchmark that computational methods failed.

What a symmetry argument does and does not settle #

It is worth separating the two questions this essay has answered, because only one of them needed a calculation.

Which bands exist is settled by the point group alone, exactly, and needs no charges, no force field and no geometry beyond the symmetry. The computed derivatives above agree with it on all twenty-eight modes, which is a check on the calculation rather than a discovery: a fixed-charge model gets every forbidden derivative right because the cancellation is a consequence of the symmetry rather than of the numbers attached to the atoms.

How strong each band is is settled by nothing symmetry knows. Two modes of the same species can differ by a factor of six in intensity, as ammonia’s do, and no amount of group theory distinguishes them. That is where a model is needed, and it is where this one is worst.

So an infrared spectrum divides into a part that is exactly predictable and a part that is hard, and the division is not where a reader might expect it. The positions of the bands need a force field and are reproduced well; the existence of each band needs only a group and is exact; and the intensities, which look like the easiest thing to estimate, need the electron density’s response to a nuclear displacement and are the part a calculation gets least right.

That ordering — exact, good, poor — is worth carrying into any spectrum. The features that can be argued about from a picture of the molecule are the ones that follow from its symmetry, and every quantitative statement about a peak height is a statement about a derivative that somebody computed.

Still open: how much intensity is charge flowing #

The natural open question is the charge flux the fixed-charge model omits, and it needs something not computed here: the derivative of the electron density with respect to a nuclear displacement. That is a well-defined object and a standard output of a real calculation, and putting it beside the fixed-charge term would say for each band how much of its intensity is charges moving and how much is charge flowing.

The nearer question is the one these numbers raise. Boron trifluoride’s strongest band by this model is the out-of-plane bend, which is also the mode with the smallest displacement of the fluorines — the boron moves and the heavy atoms barely do. So the largest dipole derivative belongs to the mode where the least mass moves, which is a reminder that intensity and amplitude are different quantities and that neither of them is the frequency.