The isotope shift is arithmetic

Substituting deuterium for hydrogen changes nothing about a molecule except the masses of some of its nuclei. The electrons are unmoved — the Born–Oppenheimer separation makes the electronic energy surface independent of nuclear mass, and a force constant is a curvature of that surface — so the whole of the change in the spectrum comes from arithmetic that was fixed before the substitution was made.

That makes an isotopologue the only genuine prediction a fitted force field can produce. Everything else it says about the molecule it was fitted to is a restatement of what it was fitted to; the substituted molecule is a set of numbers nobody supplied.

Where the factor of root two comes from, and where it goes #

The rule everyone learns is that deuteration divides an X–H frequency by √2. Its derivation is a diatomic:

$$\tilde\nu = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}, \qquad \mu = \frac{m_X m_H}{m_X + m_H}.$$

Hold k fixed, double the hydrogen’s mass, and the reduced mass doubles — provided the partner is infinitely heavy. For X = O the reduced mass goes from 0.948 to 1.789, a factor of 1.887 rather than 2, so the frequency ratio for a hypothetical isolated O–H oscillator is 1.374 rather than 1.414.

Water’s actual ratios are 1.3667 for the bend, 1.3868 for the symmetric stretch and 1.3648 for the antisymmetric stretch. Three different numbers, none of them √2, none of them the diatomic estimate either. The modes are not diatomics — as normal modes are not bond stretches computes, each stretch is half in each bond and the oxygen participates — so no two-body reduced mass describes any of them.

Methane provides the exception that explains the rule. Its doubly degenerate bending mode has a ratio of 1.4137, which is √2 to four figures. In that mode the carbon does not move at all — it cannot, because the mode belongs to a symmetry species that has no component the central atom could move along — so the vibration is purely a motion of four hydrogens against a fixed frame. Double every moving mass, halve every frequency squared, and the ratio is exactly √2.

The rule is right whenever the substituted atoms are the only ones moving, and wrong by a computable amount otherwise. Methane’s other three ratios, where the carbon does move, run from 1.3086 to 1.3654.

The identity that is exact #

There is one statement about isotope shifts that holds regardless of the force field, and it is the only thing in this essay that is not model-dependent.

Take the product of all the frequency ratios between two isotopologues. The force constants cancel out of it completely, and what is left is fixed by the masses and the moments of inertia:

$$\prod_{i} \left(\frac{\tilde\nu_i}{\tilde\nu_i'}\right)^{2} = \prod_{a}\left(\frac{m_a'}{m_a}\right)^{3} \left(\frac{M}{M'}\right)^{3} \frac{I_A I_B I_C}{I_A' I_B' I_C'}.$$

This is the Teller–Redlich product rule. It comes from the determinant of the mass-weighted Hessian rather than from its entries: the determinant of the force-constant matrix is common to both isotopologues, and everything else that survives the six zero modes is the atomic masses, the total mass carried by the three translations, and the three moments carried by the rotations.

For water the product of the three ratios is 2.586797397 and the identity requires 2.586797397 — agreement to two parts in ten billion, which is the accuracy of a Hessian taken by central differences and not the accuracy of the theorem. For ammonia the product is 6.464828 to six parts in a billion, and for methane 16.114122 to two parts in a billion.

The rule is checked for each isotope figure, and it is a genuinely independent check. Nothing in the fitting knows about it, and a mass attached to the wrong atom, a moment computed about the wrong centre, or a mode left out of the product would break it while every frequency still looked plausible. A typical catch is a moment of inertia clamped to a minimum of one so that linear molecules would not divide by zero — a fix that silently moves water’s smallest moment from 0.616 to 1 and its predicted product from 2.587 to 3.297.

What a field fitted to one molecule says about another #

Water’s field is fitted to H₂O and D₂O at once — six frequencies, four constants, a residual of seven thousandths of a per cent. HOD is then a molecule the fit has never seen, and its spectrum follows with nothing adjusted.

The computed frequencies are 1,445, 2,824 and 3,890 wavenumbers. The harmonic values quoted for HOD are 1,440, 2,824 and 3,890. The two stretches agree to better than a hundredth of a per cent and the bend to a third of a per cent, and nothing was fitted to any of them.

The mode shapes are the more interesting half of that prediction. In H₂O the two stretching modes are half in each bond and cannot be otherwise, because the two bonds are related by a symmetry operation. In HOD they are 100 per cent in one bond and 100 per cent in the other. No bond changed, no force constant changed, and the modes went from completely delocalised to completely localised — because what forbade localisation was a symmetry, and one substitution removed it.

Ammonia, where a substitution changes the shape of the spectrum #

Ammonia’s six modes fall into four distinct frequencies — two of species a₁, two of species e and doubly degenerate. Deuteration moves them all down, by ratios of 1.3594, 1.4011, 1.3549 and 1.3372.

The four ratios are not close to each other and none of them is √2. The largest, 1.4011, belongs to the degenerate bend — the mode in which the nitrogen moves least — and the pattern from methane repeats: the closer a mode comes to being a motion of the hydrogens alone, the closer its ratio comes to √2.

That gives the rule a usable form. Rather than “X–H frequencies drop by √2”, the accurate statement is: the ratio measures how much of the mode is the substituted atom’s motion, and it runs between 1 and √2 as that share runs from nothing to everything. An isotope shift is a measurement of a mode’s composition, which is exactly the quantity normal modes are not bond stretches computes directly.

Reading a ratio as a composition #

The ratios computed above are not a scatter of numbers. Laid beside the mode compositions they line up, and the correspondence is close enough to be used.

A mode’s frequency ratio on substitution measures how much of the mode is the substituted atom’s motion. Methane’s degenerate bend is entirely a motion of hydrogens against a stationary carbon and its ratio is 1.4137, which is √2 to four figures. Its symmetric stretch moves the carbon not at all either — the carbon sits at the centre of four bonds stretching together, so it has nowhere to go — and yet its ratio is 1.3654 rather than √2, because the bonds are stretching and the mass that resists is the reduced mass of the C–H pair rather than the hydrogen alone.

Ammonia makes the same point with a spread. Its four ratios are 1.3549 for the symmetric stretch, 1.3372 for the degenerate stretch, 1.3594 for the umbrella and 1.4011 for the degenerate bend, and the order tracks how much the nitrogen participates: most in the two stretches, least in the bend that swings the hydrogens tangentially past a nearly stationary nitrogen.

A set of isotope ratios is therefore a coarse measurement of the mode compositions, available without any force field at all — which is why the ratios were the standard tool for assigning spectra long before anybody could compute a composition directly. The composition and the ratio are two views of the same quantity, and both can now be computed and compared.

Where the substitution is small and still decisive #

Deuterium is the extreme case — a mass doubled — and much of the practical use of isotope shifts is at the other end, where the change is a few per cent and the shift a few wavenumbers.

A shift of that size is still decisive for an assignment, for two reasons. It is large compared with the precision a band position is measured to, so the direction and magnitude are unambiguous. And it is selective: a mode that does not move the substituted atom does not shift, so a substitution that moves one band and leaves another where it was has settled which is which.

That selectivity is what makes the technique work on molecules far too large to compute. Labelling one carbon in a protein shifts the bands that carbon moves in, by amounts that no force field is needed to interpret — only the observation that a band moved.

Why this matters beyond the spectra #

Two consequences worth drawing out, both of which reach past vibrational spectroscopy.

Zero-point energy is an isotope effect. The vibrational ground state of a harmonic oscillator sits at half a quantum above the bottom of the well, and half a quantum is proportional to the frequency. Deuterating a molecule lowers every frequency and therefore lowers its zero-point energy, which changes bond dissociation energies, equilibrium constants and reaction rates without changing a single force constant. The kinetic isotope effect is that arithmetic applied to a transition state, and the numbers it needs are exactly the ones in these figures.

A ground-state structure is not an equilibrium structure. The same zero-point motion that shifts the energies also means a molecule’s average geometry is not the geometry at the bottom of its well, and an isotopologue averages differently because it moves less. That is the reason a bond length out of a spectrum gets slightly different bond lengths from different isotopic pairs, and it is the same physics seen from the rotational side.

Why the rule survives changing the model #

There is one more thing worth saying about the product rule, and it is the reason it has outlived every force field it was ever applied to.

The rule follows from a determinant. The product of the non-zero eigenvalues of the mass-weighted Hessian is the pseudo-determinant of the force-constant matrix divided by the masses, corrected by the norms of the six motions that were removed — three translations carrying the total mass, three rotations carrying the moments. The force-constant matrix appears in that expression only through its determinant, and that determinant is common to both isotopologues, so it cancels.

So the identity holds for any force field whatever: a good one, a bad one, a deliberately wrong one. That makes it useless as a test of a model and ideal as a test of arithmetic and assignment, which is what it was always used for. A spectroscopist with a set of assigned frequencies for two isotopologues could check the assignment against the rule before knowing anything about the potential, and a failure meant a band had been mis-assigned or missed.

It functions the same way here. Every isotope figure is checked against it, and the clamped moment of inertia described above is exactly the kind of fault it exposes — a product that moves from 2.587 to 3.297 while every frequency stays exactly right.

What the model cannot do here #

The harmonic approximation is doing more work in an isotope calculation than elsewhere. Anharmonicity is a property of the potential, so it is common to both isotopologues — but the two sample different parts of it, because the heavier one has a smaller zero-point amplitude. Comparing harmonic frequencies, as this essay does throughout, keeps that out of the arithmetic. Comparing observed band centres would not: H₂O’s antisymmetric stretch is observed 187 wavenumbers below its harmonic value and D₂O’s is observed about 100 below, so an isotope ratio of band centres carries an anharmonic correction of about two per cent that has nothing to do with mass.

No intensity is computed. Deuteration changes how strongly a band absorbs as well as where it sits, because the dipole derivative along a mode changes when the mode changes shape. HOD’s two stretches are localised where H₂O’s were not, and that alters their intensities substantially. None of that is in this calculation.

Tritium is available and unused. The mass table here carries it, and the arithmetic would work; there is no measured spectrum in this collection to check against, and a prediction with nothing to test it against is not this site’s business.

Drawn as a spectrum, the deuterated molecule’s bands sit below the ordinary one’s by factors near the square root of two, and the product of the ratios is the number the rule fixes. The individual factors depend on the whole force field; their product does not, which is why the rule tests an assignment rather than a model.

What an exact identity can and cannot test #

An identity that holds whatever the force constants are is a strong statement, and it is worth being precise about what it can be used for, because the obvious use is not available.

It cannot test a force field. Any field whatever — the right one, a wrong one, a family of them — satisfies the product rule exactly, because the rule follows from the masses and the geometry and never touches the constants. A fitted field reproducing it is reproducing something it could not have failed.

What it does test is the assignment.

A vibrational spectrum arrives as a list of bands, and deciding which band is which mode is a judgement made by the spectroscopist. It is not always easy: bands overlap, weak ones are missed, and two modes of similar frequency can be swapped. The product rule takes the frequencies of two isotopologues, multiplies them together within each symmetry species, and compares the ratio against a number computed from the masses alone. Swap two bands between species and the ratio fails.

So the rule is a check on the bookkeeping rather than on the physics, and it is used that way: a published assignment is expected to satisfy it, and a failure sends the spectroscopist back to the spectrum rather than to the model.

It has a second use of the same kind. If one frequency of an isotopologue is unobserved — too weak, or hidden under another band — the product rule determines it from the others, exactly, without any force field at all. That is the rare case of a spectroscopic quantity obtained by arithmetic rather than by measurement, and it is available precisely because the identity has no adjustable content.

Which is the general shape of an exact relation’s usefulness. It cannot discriminate between models, because every model satisfies it. It can discriminate between readings of a measurement, because a misread spectrum is one that violates a relation the molecule cannot violate.

Who found the rule #

Edward Teller and Otto Redlich arrived at the product rule independently in 1935, Redlich publishing it and Teller having communicated it privately somewhat earlier — the joint name is the usual compromise. It arrived at a moment when isotopic substitution was becoming the main experimental tool for vibrational assignment, deuterium having been isolated only three years before, and its value was precisely that it required no force field: a spectroscopist could check an assignment against it before knowing anything about the potential.

That is still the best way to use it. It is one of very few results in this subject that constrain a measurement without any model at all, and the other example of the kind is the vanishing-integral theorem — selection rules are one theorem — which likewise says what must be true whatever the energies turn out to be.

Where to read on #

The composition argument goes next to the place it is used most and questioned least: the frequency is not the bond strength separates the two quantities in a frequency, and finds a molecule whose two identical bonds give frequencies a thousand wavenumbers apart. After it, group frequencies, and where they stop asks when a frequency may be attributed to a bond at all.