The fourth line wants an isotope, not a shape

Four numbers carry every claim made about ammonia’s inversion. NH₃’s ground doublet is 0.7935 wavenumbers and its first excited doublet 35.81; ND₃’s are 0.0531 and 3.35. Everything built on them so far has used them two at a time.

The essay that drew both lines against the square root of the mass ended with a division of the model’s error into two parts by their isotope signature. A barrier lower for ND₃ than for NH₃ — the transverse zero-point term — closed the gap between the ground lines and improved the untouched excited isotope ratio threefold. And a quartic fitted to each molecule’s own ground line left both excited lines too large, by nine per cent for NH₃ and fifteen for ND₃: same sign on both molecules, so that part was read as the shape of the well. The question it left was the obvious one. Give the well a third shape parameter, fit it to three of the four lines, and see whether it predicts the fourth. That is the only way to test a shape rather than accommodate it, since any family with enough parameters will pass through whatever it is fitted to.

The test has been run, and the answer is that a third parameter spent on the shape does not predict the fourth line. The same parameter spent on the isotope does, and what neither can reach is a single number the four lines define between them.

The family the question named has an edge where the lines push #

The lead named a particular family: the Gaussian hump on a harmonic well, which is Manning’s form and the one the textbook barrier of 2020 wavenumbers comes out of, with its force constant released so that the minimum is no longer pinned to the measured geometry. Its three parameters are the hump’s height $B$, its width parameter $C$, and the position of the minimum $x_0$. Pinned at the measured geometry and fitted to NH₃’s two lines it gives a barrier of 2098 wavenumbers with $C x_0^2 = 0.172$; that is the starting point for every fit below.

Left out, NH₃’s ground line is predicted at 0.869 against 0.7935 — 9.6 per cent high — by a fit that converges. The other three fits do not converge at all. Each walks $C x_0^2$ down by two to three orders of magnitude, to between $1.5 \times 10^{-4}$ and $2.2 \times 10^{-3}$, and stops with one of its three target lines still missed by 5 to 13 per cent.

That is a family running out, not a fitter failing. As $C \to 0$ with $B C^2$ held finite, the hump’s Taylor series stops at its quartic term and the Gaussian well becomes $A(x^2 - x_0^2)^2$ — the plain quartic, which is the family’s boundary rather than one of its members. The three stalled fits are asking for a barrier more quartic than the quartic, and the Gaussian hump cannot supply one. So the question as posed cannot be answered in the family it named, and the honest move is to re-ask it in a family where the quartic sits in the middle instead of at an edge.

A family with the quartic inside it #

The replacement is the simplest one with the properties that matter:

$$V(x) = H \left(1 - y^2\right)^2 e^{s y^2}, \qquad y = x / x_0 .$$

It has its minima at $\pm x_0$ for every value of $s$, because $(1 - y^2)^2$ has a double root there and the exponential never vanishes. Its barrier is $H$. It is positive everywhere, which a polynomial correction is not — a version with $(1 + s y^2)$ in place of the exponential dips below the bottom of the wells inside the region the molecule can reach, and under the planar mass that loses two of the four fits to a spurious outer state. And at $s = 0$ it is exactly the quartic, which is checked point by point. Measured as a fraction of its own height, a negative $s$ narrows the barrier at every level and softens the outer walls; a positive $s$ does the reverse.

In this family every one of the four leave-one-out fits converges, under both defensible reduced masses — the bond-conserving one, which has the ligands sliding outward as the apex descends, and the conventional planar one. And every prediction misses. Under the bond-conserving mass the left-out lines come back at +9.6, −13.3, −11.4 and +43.7 per cent; under the planar mass at +9.2, −11.0, −10.3 and +18.6.

So the over-shoot on the excited lines is not a shape the well can supply. A well with three shape parameters and no isotope term, fitted to any three of the four lines, predicts the fourth no better than the two-parameter wells did, and in the worst case — ND₃’s excited line left out — five times worse.

Every fit that spends it on the shape moves the nitrogen #

The misses are not the only thing wrong with those fits, and the second thing explains the first.

The measured structure — a bond of 1.0124 Å at an angle of 106.67° — puts the nitrogen 0.382 Å above the plane of the hydrogens. Under the bond-conserving mass the four shape-alone fits put it at 0.435, 0.544, 0.462 and 0.655 Å, and lower the barrier to 1429, 777, 1254 and 452 wavenumbers to compensate. Under the planar mass they move it less, to 0.396–0.510 Å, and they still move it.

The reason is an identity the essay on the splitting’s sensitivities proved and checked to six decimals: for a well of fixed shape, substituting $x = \alpha y$ shows that the splitting depends on the mass and the geometry only through the combination $\mu x_0^2$. A free minimum is a free mass. Moving the nitrogen outward by a factor α makes both molecules behave as though they were $\alpha^2$ times heavier, which changes their isotope ratio at a fixed barrier — and the isotope ratio is precisely the thing a single isotope-independent well cannot get right. So the shape-alone fits are not testing a shape. They are using the geometry to smuggle in a mass correction, and paying for it in a structure the microwave spectrum rules out.

That reading makes a prediction: replace the free geometry with an honest isotope term, and the fits should stop needing to move anything.

The same parameter, spent on the isotope #

The isotope term is the transverse zero-point correction in its simplest form — ND₃’s whole potential lowered by a fraction δ, NH₃’s left as it is — and the minimum goes back to the measured 0.382 Å. Three parameters again: height, shape and δ. One of them is now isotope-dependent and none of them is the geometry.

Every left-out line comes back within five per cent under the bond-conserving mass: +4.6, −3.3, −4.5 and +4.0 per cent. Under the planar mass within eight: +7.9, −5.6, −7.3 and +6.7. Under both masses and for every line the miss is smaller than the shape-alone miss for the same line, by a factor of two to eleven.

The fitted terms say what the shape has taken over from the isotope. With the quartic, the barrier shift needed to close the two ground lines was 97 wavenumbers. With the shape free it is 57 to 86, depending on which line is left out: a softer well absorbs part of what the isotope term used to carry, and the shapes the fits settle on are all on the soft side, $s$ between −0.058 and −0.080. That is the answer to the half of the question the transverse decomposition left open — the fraction of the frozen modes’ zero-point difference the correction has to be is smaller once the well is allowed a shape, by up to two fifths. The five per cent that was carried across to phosphine was sized on the quartic’s 97, so it was an upper end rather than a value.

And it is the combination that predicts, not either part. Shape alone misses by tens of per cent while dragging the molecule out of its measured structure. Isotope term alone, on the quartic, leaves the excited lines nine and fifteen per cent high. Both together, at the measured geometry, predict whichever line is held back within five.

What is left is one number #

Five per cent is not nothing, and the four errors have a pattern that says what it is. Leaving out NH₃’s ground line gives a prediction that is high; NH₃’s excited line, low; ND₃’s ground line, low; ND₃’s excited line, high. Plus, minus, minus, plus — under both masses.

Four measured lines and three parameters leave exactly one combination of the lines that no setting of the parameters reaches, and every leave-one-out error is that one combination seen from a different side. Here it has a name. With the height and the isotope term fitted to the two ground lines, the shape sets both excited lines at once, and the quantity left over is the double ratio: ND₃’s excited-over-ground doublet ratio divided by NH₃’s.

$$D = \frac{\Delta_1(\mathrm{ND_3}) / \Delta_0(\mathrm{ND_3})}{\Delta_1(\mathrm{NH_3}) / \Delta_0(\mathrm{NH_3})} = \frac{63.09}{45.13} = 1.398 \ \text{measured.}$$

It is a well-chosen quantity for this question because so much cancels out of it. Within one molecule the ratio of two doublets cancels most of the barrier’s height; between the two molecules a uniform isotope lowering cancels to first order as well. What survives is how the barrier’s width changes between the energy of a ground state and the energy of an excited one, compared across two masses — which is a statement about shape and mass together, and about nothing else in the model.

Along the curve it rises monotonically with $s$. At the quartic, $s = 0$, it is 1.475. At the shape that also matches NH₃’s excited line, $s = -0.058$, it is 1.454: four per cent above the measurement. To bring it down to 1.398 needs $s = -0.257$ — but there NH₃’s own excited line is 25.7 wavenumbers against 35.81, twenty-eight per cent low. The double ratio wants a shape four and a half times further from the quartic than NH₃’s own excited doublet will allow. Under the planar mass the curve sits higher throughout, and on the whole range from −0.4 to 0.2 it never comes down to the measurement at all.

The shape the ratio is asking for #

On a full plot of the potential the three wells are hard to tell apart: all three rise to about two thousand wavenumbers from the same two minima. Drawn as fractions of their own barrier heights, with each well’s own levels marked between their turning points, they separate in exactly the place that matters.

At the ground state’s energy the half-width of the barrier — the distance from the planar geometry to where the potential falls to the state’s energy — is 0.701, 0.700 and 0.698 of the apex height for the quartic, the shape that matches NH₃’s excited line, and the shape that would match the double ratio. The same to a third of a per cent. The ground doublets see one barrier in all three wells, which is why the height and the isotope term can be fitted to them at every shape without the shape mattering.

At the excited state’s energy the three part company: 0.378, 0.384 and 0.405. The softer walls let the excited doublet settle lower in the well relative to its barrier — from 73 per cent of the barrier’s height in the quartic to 72 and then 67 — and lower in the well the barrier it tunnels through is wider. ND₃’s excited doublet sits lower again, at 59, 57 and 52 per cent, because the heavier molecule’s levels are packed closer to the bottom.

That is the residual, mechanically. The double ratio compares how NH₃’s excited state, at about three quarters of the barrier, and ND₃’s, at a little over half, each see the barrier’s width, relative to two ground states that see the same width in every well. A one-parameter family moves both excited states by one rule: softening enough to bring ND₃’s excited line into line with NH₃’s widens NH₃’s own barrier too far, and its doublet falls from 35.8 to 25.7. What the four lines appear to ask for is a barrier whose width at three quarters of its height and at a half can be set separately — a second shape parameter.

All four at once, and a mass the lines can now choose #

With the shape and the isotope term both free and the geometry held, there is also a single best well for all four lines, the least-squares fit, and it is worth reading because it is the model’s best account of the whole data set rather than any one held-out view of it.

Under the bond-conserving mass it leaves the four lines at +0.89, −1.18, −0.88 and +1.02 per cent, with a barrier of 2088 wavenumbers, a shape of −0.070 and ND₃ lowered by 3.6 per cent, 75 wavenumbers. Under the planar mass the residuals are +1.48, −1.95, −1.48 and +1.72, at a barrier of 2241, a much thinner top at −0.206, and a lowering of 52. Every column in the table carries the same alternation — the eight left-out errors and the eight residuals are one combination of the four lines, read through different fits.

That gives an answer the comparison of mass constructions never reached. Deuterium could not tell the masses apart when each molecule contributed one line: every well and every mass missed the isotope ratio by eighteen to twenty-nine per cent and the two masses differed by less than either missed. With two lines per molecule and the isotope term in the model, they separate. The bond-conserving mass leaves every line within 1.2 per cent where the planar one needs 2.0; it predicts held-out lines within five per cent where the planar one needs eight; and only the bond-conserving mass has any shape at all that reaches the measured double ratio. The construction that moved the answer thirty per cent toward the measurement on one line is the one the four lines prefer, and they prefer it by a factor of about 1.6 on every measure taken here.

The arithmetic behind each fit #

Each well is solved on a grid of 2400 points across the region where the coordinate exists — out to 0.995 of the bond length, since past it the ligands have folded flat — with the lowest four levels found by Sturm-sequence bisection and the variable mass in the BenDaniel–Duke ordering every earlier essay in this argument used — a choice the sweep of orderings showed is worth half a per cent, a tenth of the residual here. The prediction of the held-out ND₃ excited line moves by less than one part in ten thousand when the grid is doubled.

The fits are Newton iterations on the logarithms of the target lines, with each step capped at a tenth in every parameter; the four-line fit is the same iteration on the normal equations. A fit is called converged only when every target line is reproduced to a part in a million. The checks that bound what the numbers can mean: the shaped family equals the quartic at $s = 0$ to $10^{-12}$ of the barrier; at least three of the four Gaussian fits fail to converge and each ends with $C x_0^2$ below 0.12; every shape-alone fit converges, misses by more than nine per cent and moves the minimum outward; every isotope-term fit converges, predicts within five per cent (eight for the planar mass) and beats shape alone line by line; the sign pattern is the same under both masses and matches the four-line residuals; the double ratio rises monotonically with $s$ and sits three to four per cent above the measurement where NH₃’s excited line is matched. And the refusal: three parameters asked for all four lines must not converge, and return a least-squares well with every line off by about a per cent — a converged four-line fit would mean the lines were never independent.

What three parameters cannot say #

One coordinate. Every well here is a function of the apex height alone, with the bond lengths and every other mode frozen. The isotope term is where the frozen modes come back in, and it does so as a single fraction of the whole potential; a real zero-point correction varies along the path, so the δ fitted here is an average of something with a shape of its own.

Two families. The Gaussian hump and the exponential-shaped quartic are the families tried. A different one-parameter shape family might put the double ratio somewhere else, and nothing above proves that no one-parameter family can reach it at the shape NH₃’s excited line wants. What is established is narrower: these two cannot, and the reason one of them cannot is that the lines push it off its own edge.

The isotope term is a fraction, not a mode-by-mode correction. The transverse decomposition split the frozen zero-point difference into three bands with different shares, and nothing here uses that split. A correction built band by band would have a shape along the coordinate, which is a second isotope parameter and, with four lines, the end of having a test.

Measured lines, quoted to the precision they are quoted. The excited ND₃ interval is given as 3.35, three figures. A change of one in the last place moves the double ratio by 0.3 per cent — a tenth of the residual, not the residual.

Still open: a second shape parameter, and the mixed molecules’ lines #

The obvious open question is the one the double ratio poses directly: a shape whose width can be set separately at three quarters of the barrier, where NH₃’s excited doublet tunnels, and at a half, where ND₃’s does. That needs a second shape parameter, which with four lines leaves no test — unless more lines are brought in. The two partly deuterated molecules each have a measured ground doublet, and their place on the square-root line is already computed from wells fitted to the pure ones; a four-parameter well fitted to the four pure lines would predict two mixed lines at once, which is a test again. The care needed is the one that essay already took: the mixed molecules carry a term in their mass that neither pure one has, and a prediction that misses could be missing that instead.

The nearer question is the size of the isotope term itself. A uniform lowering of 3.6 per cent is 75 wavenumbers at the barrier and a few at the minimum; the band-by-band zero-point difference is a quantity with its own profile between the pyramid and the plane, and deriving δ from it instead of fitting it would turn one fitted parameter into a computed one — and give the four lines a three-parameter test in which only the height and the shape are free.