The ordering a manifold picks

The sweep of the orderings closed on a question it could not run. The five Hermitian forms in use span half a per cent in ammonia’s splitting, which settles that the ambiguity is small; and small is not the same as decided. Its closing paragraph named the thing that would decide it: writing the full kinetic energy in curvilinear coordinates and reducing it onto the umbrella path gives a definite operator with its own extra potential, and whether that operator lands inside the cluster of orderings in use is a calculation somebody should do.

The operator is available without a force field, because the reduction that produces it has already been done. The mass function is the reduction.

A reduction is a manifold #

Take the classical kinetic energy of the reduced problem, half μ(x) times ẋ squared. That is a metric: it says how far apart two nearby values of the coordinate are in the only sense that matters dynamically, and a coordinate with a metric on it is a one-dimensional Riemannian manifold.

A Riemannian manifold has a distinguished Laplacian. It is the Laplace–Beltrami operator — in one dimension, minus a half times μ to the minus a half, times the derivative of μ to the minus a half times the derivative — and it is distinguished because it is the only second-order operator built from the metric alone that is invariant under changes of coordinate. Nothing about the choice is a taste.

It is also not a member of the von Roos family, and that is why the connection was not obvious. The family consists of the Hermitian orderings with respect to the ordinary measure dx; the Laplace–Beltrami operator is Hermitian with respect to the metric’s own measure, $\sqrt{\mu}\,dx$. They act on different spaces, so they cannot be compared until one of them is carried across.

The map that carries it is the unitary multiplication by μ to the quarter power, which is an isometry between the two spaces and therefore preserves the whole spectrum. Under it the Laplace–Beltrami operator becomes an ordinary Hermitian operator on dx with the right principal part — so it must be a von Roos member. Which one is arithmetic.

Which member, read off rather than derived #

The coefficients could be got by hand in a dozen lines of algebra. They are got here by finite differences instead, and the reason is the same reason that sweep checked its own ordering algebra that way: a derivation produces two numbers and a wrong derivation produces two wrong ones with nothing to say so.

So the transformed operator is applied to a smooth positive test function by nested central differences, BenDaniel–Duke is applied the same way, the difference is divided by the function, and the result is fitted to the two basis functions every ordering is a combination of — the squared first derivative of the mass over its cube, and the second derivative over its square.

Fitting rather than solving is what makes the answer checkable. A fit over forty-one positions returns two coefficients and a residual, and a residual above the finite differences’ own accuracy would mean the operator contains a third function the family does not have. The residual is two parts in a hundred million of the potential being fitted.

The square coefficient comes out at −0.218750 and the curvature coefficient at 0.125000. Those are −7/32 and 1/8 to six decimals.

Working back: the curvature coefficient is minus a quarter of the sum of the exponents, so the sum is −1/2; the square coefficient is half of the sum plus the product, so the product is 1/16. A pair with sum −1/2 and product 1/16 is the roots of a quadratic whose discriminant is exactly zero — a double root at −1/4.

That is Mustafa and Mazharimousavi’s ordering, and it is in that list of five.

And it is not the one to reach for #

BenDaniel–Duke sits at the origin of the plane. It is the ordering that adds no extra potential at all, it is the one every treatment of a position-dependent mass writes down first, and it is what the sweep used as its reference point precisely because both of its coefficients vanish.

It is not the covariant one. The two are a quarter of the plane apart in each exponent, and the covariant operator adds a potential that is negative where the mass is varying fastest.

So the ordering question is closed rather than bounded, and what that half a per cent now measures is something else. It was reported as the size of an ambiguity: five defensible choices, and here is the spread between them. Read with the covariant point identified, it is the size of the error the other four make — which is a different statement about the same number, and a more useful one, because an error has a direction.

There is something mildly deflating about the result and it should be said. The answer is that the choice does not matter much and now cannot be argued about; nothing in ammonia’s splitting moves by more than a fraction of a per cent, and the forty-three per cent correction the mass construction was worth is untouched. What has changed is that a paragraph of hedging is replaced by a point.

What the extra potential actually looks like #

The covariant correction is a function of position and it is worth reading, because its shape says where the choice of ordering can matter and where it cannot.

Both basis functions of the family are built from the mass’s derivatives, and the bond-conserving mass is flat near the pyramid and steep near the flat geometry — the ligands barely move outward when the apex is near its minimum and move fast when it is near the plane. So both basis functions are small at the minima and large at the top of the barrier, and the covariant potential is concentrated exactly where the tunnelling wavefunction is thinnest.

That is the reason the effect is small and also the reason it is not zero. A correction concentrated under a barrier does not shift a level much, because the amplitude there is exponentially small; it does change the action, because the action is an integral under the barrier and is what the splitting depends on exponentially. The two effects pull in opposite directions and the surviving fraction of a per cent is what is left.

And it is why the sign matters more than the size. The covariant potential is negative where the mass is varying fastest, so it lowers the effective barrier, so it raises the splitting — and every one of the four other orderings in the family gets that direction wrong or right by accident. A spread was reported and no direction could be, because a spread has none.

What the reduction still leaves out #

Closing the ordering question exposes what it was standing in front of, and the thing behind it is larger.

The whole von Roos family is about quantising a given one-dimensional problem. It assumes the reduction has already happened — that the molecule really does move on a one-dimensional manifold with metric μ(x). It does not. The bonds are not rigid; they are stiff, and a constraint imposed as the limit of a stiff potential is not the same as a constraint imposed by hand.

What the difference leaves behind is the zero-point energy of everything frozen. Five of ammonia’s six vibrations are transverse to the umbrella path, their frequencies change along it, and half their sum is a term added to the potential. It is not in the von Roos family, and the reason is structural rather than numerical: the family’s two basis functions are built from the mass profile, which the constraint fixes, while the zero-point term depends on the stiffnesses the constraint threw away. No choice of α and γ can produce it.

It is also isotope-dependent, where an electronic potential is not — which makes it the natural candidate for the one defect four essays of this argument have not shifted.

How large it would have to be #

Computing the term needs frequencies at every point along the path, which is a force field nobody here has. Its scale does not: the transverse modes are every vibration but the umbrella, and both molecules’ fundamentals are measured. Half the sum of the transverse frequencies is 6740 wavenumbers for NH₃ and 4957 for ND₃, a difference of 1783.

So the question is asked the other way round, which is the habit these essays keep with a quantity they cannot compute. Not what the correction is, but how large it would have to be.

A barrier was fitted to each molecule separately and the two answers differ: 2116.9 against 2019.9 wavenumbers under the bond-conserving mass, 2329.9 against 2220.5 under the rigid-triangle one. One electronic potential has to serve both, so the gap — 97 and 109 wavenumbers — is the amount an isotope-dependent term would have to supply.

Against 1783, that is 5.4 and 6.1 per cent.

A few per cent is an entirely ordinary amount for five vibrational frequencies to change between a pyramidal geometry and a planar one. Ammonia’s stretches stiffen slightly and its bends soften as the molecule flattens, the changes are of that order, and they are of that order differently in the two isotopologues because the frequencies are.

So the mechanism is the right size, which is the strongest statement available without the force field. It is not the same as being right: a term of the right magnitude and the wrong sign would be no help, and the sign is exactly what the measured fundamentals cannot supply, since they are all at the pyramid. What this establishes is that the residual is not too large for the one obvious missing term to explain, and four essays of sweeping parameters had not established even that.

What was computed, and how #

The mass function is the bond-conserving construction of the essays before it, with its closed-form first and second derivatives — a finite difference of a function that diverges at its domain’s edge being the worst possible place to take one. The test function is a Gaussian times a low-order polynomial, positive throughout the sampling range so that dividing by it introduces nothing. The sampling runs from a seventh of the pyramid height to about one and a half times it, which is the region the molecule occupies.

Ten things are checked. That the fitted square coefficient is −7/32 and the curvature coefficient 1/8, each to three parts in a thousand. That the fit’s residual is below two parts in a thousand of the potential, so the operator is in the family rather than near it. That the exponents are a double root, and that the root is at −1/4. That the matching ordering is the one named, and that BenDaniel–Duke is a different point by more than a tenth in both coefficients. That a constant mass has vanishing first and second derivatives, so every ordering coincides and the covariant one adds nothing — the refusal, and the case a routine reporting its own finite differences would fail. And that every construction wants a lower barrier for the deuterated molecule, with the required correction between one and fifteen per cent of the transverse zero-point difference.

The discriminant is not tested for its sign, and that is deliberate. It is exactly zero at the answer, so its sign is decided by the fit’s last digits; a test on the sign would be a test of the finite differences. It is tested against a threshold relative to the squared sum of the exponents, which is the scale it would have if the two exponents were genuinely different.

Where this stops #

The covariant operator is the right operator for the manifold and the manifold is a modelling choice. Which metric the reduction has is decided by which mass construction is used, and that choice is worth forty-three per cent where this one is worth a fraction of a per cent. Identifying the Laplacian of a metric does not make the metric right.

And the Laplace–Beltrami operator is the natural quantisation of a manifold, not the demonstrated limit of a stiff potential. Those are different claims and the literature on constrained quantisation is about the difference: a thin-tube limit produces the covariant operator plus terms from the extrinsic geometry and from the transverse zero point, and only in special cases does the extra vanish. What is established here is which von Roos member the covariant operator is; what is not established is that the covariant operator is what a molecule has.

The zero-point estimate uses fundamentals as though they were harmonic frequencies. Ammonia’s stretches are anharmonic by a few per cent, so the 1783 is itself good to a few per cent — which is the same order as the fraction being computed from it. The comparison survives because it is an order-of-magnitude comparison and is stated as one; a claim at the per-cent level from these numbers would not.

The generalisation #

The habit is to ask whether an ambiguity is an ambiguity of the problem or of the quantisation, because those have different resolutions.

The ordering question looks like an ambiguity of quantisation — a classical expression with several quantum forms — and is presented that way in every treatment of it. For a mass that came from somewhere it is not. A reduction hands over more than a function: it hands over a metric, and a metric determines the Laplacian. Five candidate orderings and no way to choose is what the problem looks like when the metric has been forgotten in the step that produced it.

The corollary is about what a resolved ambiguity uncovers. Closing the ordering question here does not make the model better — the numbers barely move — and it does make visible the term that was hiding behind it, which is larger and is not in the family at all. A parameter swept is a parameter that cannot be blamed, and the useful thing about closing a small question is that it shortens the list of places the residual could be.

Who found it, and when #

Von Roos’s family is 1983 and collects the orderings proposed before it; BenDaniel and Duke’s is 1966 and Mustafa and Mazharimousavi’s is 2006. The Laplace–Beltrami operator as the quantisation of a curved configuration space is DeWitt, 1957. The thin-tube limit and the extra potentials it leaves are Jensen and Koppe, 1971, and da Costa, 1981. The frequencies are the measured fundamentals of both isotopologues. What is computed here is the identification of the covariant operator with a specific von Roos member for the bond-conserving mass function, and the size the missing transverse term would have to have.

The number worth carrying is −1/4, twice: the two von Roos exponents a manifold picks, in a family where the point everybody writes down first is the origin.

Still open: the sign, and the other pyramid #

The obvious open question is the sign of the correction, which the measured fundamentals cannot give because they are all at the pyramid. The gap wants ND₃’s effective barrier lower than NH₃’s, so the transverse zero-point difference has to fall going from the pyramid to the plane by more in NH₃ than in ND₃ — and since the difference is essentially the hydrogen stretches’ isotope shift, that amounts to asking whether the stretches soften as the molecule flattens. Frequencies at the planar geometry are a force field’s output and nothing else’s, so the question is genuinely blocked; what is not blocked is a bound. The stretches cannot soften by more than they do at dissociation, which is a limit, and the limit would say whether the required 5.4 per cent is comfortably inside the available room or at the edge of it.

The nearer question is phosphine, which has been in this argument since its third essay and has no measured splitting at all. Its transverse frequencies are measured and its barrier is six times ammonia’s, so the transverse zero-point term is computable on the same footing and the correction’s relative size can be compared. If the required fraction comes out similar for a molecule whose barrier is six times larger, the term is behaving like a property of the reduction; if it comes out wildly different, the 5.4 per cent here is a coincidence of two numbers rather than a mechanism. That is one more molecule’s worth of quoted frequencies and nothing new to build.