Three constants, two potentials

A Morse curve built from a diatomic’s measured $\omega_e$ and $\omega_e x_e$ falls short of its measured $\alpha_e$, by four to fifteen per cent across four molecules, and the shortfall was read as a cubic coefficient that is too small. Building a quartic from the two coefficients the measurements imply then overshot, because the relations between measured constants and potential coefficients are leading terms of series whose later terms a quartic does not have. The way past that was named: a potential that carries the higher terms by construction and has one more parameter to spend.

The simplest such potential is a Morse curve whose range varies with the separation,

$$V(r) = D\left(1 - e^{-a(1 + \lambda\xi)(r - r_e)}\right)^2,\qquad \xi = \frac{r - r_e}{r_e}.$$

At λ = 0 it is the Morse curve. The question it was proposed to answer is whether, fitted to $\omega_e$, $\omega_e x_e$ and $\alpha_e$ together, it lands on the cubic the leading-order relation reads off $\alpha_e$ — which would make the shortfall is the cubic a statement about the potential rather than about a truncation.

For one molecule of four the answer is yes. For the other three the family cannot be fitted at all.

Holding two constants, scanning for the third #

The fit is made in two stages, because the family has three parameters and the targets are not equally easy to reach. At each value of λ, D and a are solved so that the potential’s $\omega_e$ and $\omega_e x_e$ are exactly the measured ones — to a part in ten billion, by Newton’s method on the two relative residuals. What that leaves free is $\alpha_e$, and it is read from the same four solved levels as every $\alpha_e$ in this argument, on the same grid. Scanning λ from +0.1 to −0.5 then traces the family’s $\alpha_e$ as a single curve per molecule, and the measured $\alpha_e$ either lies on it or it does not.

At λ = 0 the solve must return the Morse curve itself — its D and a to the grid’s resolution, and the $\alpha_e$ already computed for it. It does, for all four molecules, which is the check that the family extends the curve it claims to extend.

Three molecules the family cannot reach #

For hydrogen chloride $\alpha_e$ rises from the Morse value of 0.2775 cm⁻¹ as λ goes negative, reaches 0.2940 at λ = −0.3, and falls again. The measurement is 0.3072. The best the family can do is 4.3 per cent short, and it closes 56 per cent of the Morse curve’s shortfall. Deuterium chloride’s maximum is 0.1085 against 0.1133 — also 4.3 per cent short, also 56 per cent closed. Hydrogen fluoride’s is 0.7250 at λ = −0.25, against 0.798: 9.1 per cent short, 37 per cent closed.

So no member of this family reproduces any hydride’s three constants. The earlier essay’s proposal — fit the family and read the cubic it lands on — has no landing point for them. That is not a failure of the fitting: the maximum is interior to the scanned range, the $\omega_e$ and $\omega_e x_e$ are held exactly all the way along it, and the curve turns back down on the far side of it.

Member by member, the rotation constant is the cubic #

The earlier essay’s worry was that the leading-order relation between $\alpha_e$ and the cubic coefficient is only the first term of a series, and that a quartic built on it overshot because its later terms were wrong. The family is the test of that worry, because every member carries a full set of later terms, all of them smooth.

Read each member’s own cubic off its curve and put it into $\alpha_e = -(6B_e^2/\omega_e)(1 + a_1)$, and the result is that member’s solved $\alpha_e$ — to a tenth of a per cent for carbon monoxide, to 0.3 per cent for the chlorides, and to 0.7 per cent for hydrogen fluoride, across the whole scan from λ = +0.1 to −0.5. Along this family $\alpha_e$ is its cubic. The later terms that wrecked the quartic are present in every member and cost next to nothing here, because in a smooth potential they come in the proportions a smooth potential has, not the ones a truncation leaves.

That settles the earlier question in the sense it was asked. The shortfall is the cubic is a well-posed statement: for a potential whose higher terms are those of a smooth curve, the measured $\alpha_e$ and the cubic determine each other to within a fraction of a per cent, and the quartic’s overshoot was the truncation’s.

It also says exactly why the hydrides cannot be reached. With $\omega_e$ and $\omega_e x_e$ held, the family’s cubic has a limit. Hydrogen chloride’s gets no more negative than −2.306, at λ = −0.3, and its measured $\alpha_e$ asks for −2.365; deuterium chloride’s limit is the same −2.306; hydrogen fluoride’s is −2.139 against −2.253. The ceiling on $\alpha_e$ is a ceiling on the cubic, and the cubic is capped because holding $\omega_e x_e$ ties the quartic to the cubic, and this family can only move the two together along one curve in their plane. Carbon monoxide’s needed cubic, −2.697, lies inside its family’s range, whose limit is −2.707.

In the plane of the two coefficients the picture is simple. Hydrogen chloride’s measured $\omega_e x_e$ and $\alpha_e$ ask, through the leading-order relations, for a cubic of −2.365 and a quartic of 3.665. The family, held to the measured $\omega_e$ and $\omega_e x_e$, traces one curve through that plane as λ moves, and its most negative cubic, −2.306, comes with a quartic of 3.323. The point the measurements name is off the family’s curve, and no value of its one free parameter reaches it.

Two isotopologues, one ceiling #

The earlier comparison found that how much the terms past leading order matter is ordered by one number, $(\beta r_e)^2 B_e/\omega_e$, and it is tempting to read the ceilings the same way: carbon monoxide has the smallest measure and is reached, hydrogen fluoride the largest and is furthest from reach.

The two chlorides say that reading is not the right one. Hydrogen chloride’s measure is two fifths larger than deuterium chloride’s — deuterium’s mass halves $B_e$ and changes $\omega_e$ by a factor of $\sqrt{2}$ — and they close the same 56 per cent of their shortfalls to a tenth of a per cent. They share a potential, since an isotope changes the masses and not the electronic energy, and the family’s ceiling is a property of how its shapes compare with that potential. The measure mixes the potential with the masses; the ceiling does not.

The Morse curve’s own shortfall says the same thing a step earlier. Built from each isotopologue’s measured $\omega_e$ and $\omega_e x_e$, it falls short of hydrogen chloride’s $\alpha_e$ by 9.67 per cent and of deuterium chloride’s by 9.63 — the same shortfall to four hundredths of a point, from spectroscopic constants that differ by factors of $\sqrt{2}$ and two. The shortfall belongs to the difference between a Morse shape and the chloride potential, and the family inherits it: its ceiling is where its own shapes run out against that potential, and a change of mass moves every number in the problem except that one.

One molecule it reaches twice #

Carbon monoxide’s $\alpha_e$ rises from the Morse value of 0.01672 through the measured 0.01750, peaks near λ = −0.35 at 0.01761, and falls back through 0.01750. So two members of the family fit all three of its constants, at λ = −0.2214 and λ = −0.4534.

Both land on the leading-order cubic. The relation $\alpha_e = -(6B_e^2/\omega_e)(1 + a_1)$ reads −2.69672 off the measured $\alpha_e$; the two fitted curves carry −2.69671 and −2.69687. Their quartics are 4.5094 and 4.5095 against the leading-order reading of 4.5027, which differs by a sixth of a per cent. For this molecule, then, the earlier question has the answer the proposal hoped for: a potential that carries every higher term and is fitted to all three constants has exactly the cubic the truncated relation assigned it. The shortfall is the cubic was a well-posed statement about carbon monoxide.

And the two fits are the same curve where the constants look. The lowest four levels of carbon monoxide — the ones every constant here is read from — lie between the classical turning points 0.90 and 1.13 bond lengths of the fourth of them, and across that interval the two potentials differ by at most 1.9 cm⁻¹, a quarter of a thousandth of the energy there. The matching coefficients are the same statement in another form: to the order the three constants reach, the two potentials are one potential.

That is a small interval. It is a quarter of a bond length wide, it contains the equilibrium separation and a narrow skin around it, and the three spectroscopic constants are, in effect, a description of that skin. What a vibrational constant measures is the shape of the well where the molecule vibrates; where it does not vibrate, which is most of the curve, the constants are silent.

Where the two potentials part #

Beyond the region the lowest levels sample they are nothing alike. The first is still rising at three bond lengths. The second reaches a maximum at 2.10 bond lengths and falls back to zero at 3.2: it does not dissociate at all. The fitted D parameters are 1.35 and 2.15 times the Morse curve’s, which says nothing about depth once the curve can turn over, and the two predict $\omega_e y_e$ of 0.018 and 0.011 cm⁻¹ — a fourth measured constant would choose between them, and the three used here cannot.

Three measured constants fix the bottom of a well and choose between these two potentials not at all. That is the surprising connection in this essay, and it is the same fact the earlier quartic ran into from the other side. The quartic had the right coefficients and the wrong higher terms; these two have the same right coefficients and entirely different higher terms, and the constants cannot see the difference because the constants are the coefficients.

Every member that helps turns over #

The second carbon monoxide fit turning over is not an accident of that molecule. It is what the family does whenever λ is negative, and negative λ is the only direction that raises $\alpha_e$.

The range $a(1 + \lambda\xi)$ shrinks with the separation when λ is negative, so the exponent $a(1 + \lambda\xi)(r - r_e)$ rises to a maximum at $\xi = 1/2|\lambda|$ and falls back to zero at $\xi = 1/|\lambda|$. The potential follows it: a maximum, a return to the bottom energy, and then a climb without limit. Hydrogen chloride’s member at λ = −0.3, where its $\alpha_e$ peaks, turns over at 2.67 bond lengths; the member at −0.5 turns over at two.

So the family raises $\alpha_e$ by changing the curve out where no low-lying level reaches, and the change that raises it is one that stops the curve being a bond. The lowest four levels of hydrogen chloride reach no further than 1.29 bond lengths, the outer turning point of the fourth, which is why $\alpha_e$ can be read from them at all; but a potential that is correct at the bottom and has a barrier and a second well further out is not a better model of hydrogen chloride than the Morse curve, whatever its $\alpha_e$.

That changes what the earlier proposal can mean. A Morse curve with a linearly varying range carries the higher terms by construction, as it was meant to, and the ones it carries are the ones that make it turn over. It is a device for testing what the constants determine, and on that it has been decisive. It is not a potential for the molecule.

What was held, and what was read #

The four molecules are the earlier comparison’s: ¹H³⁵Cl, ²H³⁵Cl, ¹H¹⁹F and ¹²C¹⁶O, with the same measured $\omega_e$, $\omega_e x_e$, $B_e$ and $\alpha_e$ and the same finite-difference solver — four thousand points from 0.6 to 2.6 bond lengths, the lowest four levels, $\omega_e$, $\omega_e x_e$ and $\omega_e y_e$ from their three spacings, $\alpha_e$ from the rotational constants of the lowest three. The scan runs from λ = +0.1 to −0.5, which keeps the range factor positive over the whole grid. Members below about −0.31 already turn over inside it, above the levels used; past −0.625 the curve’s second zero — where it falls back to the bottom energy — would lie inside it too, and the lowest levels would no longer be the only ones the solve could find.

The checks, made wherever these figures are drawn: every member holds the measured $\omega_e$ and $\omega_e x_e$ to 10⁻⁸; carbon monoxide has exactly two fitting members, both with the leading-order cubic to a part in a thousand and the quartic to a per cent, differing in the fitted D by more than half and in $\omega_e y_e$ by more than a third; each hydride’s $\alpha_e$ has an interior maximum between four and ten per cent below the measurement and closes less than two thirds of the Morse shortfall; the two chlorides close the same share to a per cent though one’s measure is two fifths larger than the other’s. The refusal is λ = 0, which must return the Morse curve’s own D, a and $\alpha_e$.

What one family can and cannot show #

One family. A Morse curve with a linearly bent range is one way to add a shape parameter, and its failure on the hydrides is a statement about it. A range that bends quadratically, or a curve with a different long-range form, has a different ceiling, and nothing here says that no three-parameter potential can reach hydrogen chloride’s constants — only that this one, the one proposed, cannot.

The measured constants as given. Each is quoted to the precision the earlier essays used. Hydrogen fluoride’s larger gap is well outside any uncertainty in its $\alpha_e$; the chlorides’ 4.3 per cent is too.

Four levels. Every constant here is read from the lowest four levels, and $\omega_e y_e$ from their spacings. A higher level would sample further out, and for the negative-λ members it would eventually meet the barrier. The constants are a statement about the bottom by construction.

A shortfall that has an answer for one molecule #

The earlier essays put a question: is the shortfall is the cubic a statement about the potential, or about a truncation of the relation between the potential and the measurement? For carbon monoxide it is about the potential — a potential with every higher term, fitted to the constants, carries exactly the truncated relation’s cubic, and does so twice. For the three hydrides this family cannot say, because it cannot be fitted to them; the relation might still be exact and the family simply the wrong shape.

What both halves share is the finding that three constants see the bottom of the well and only the bottom. The mean displacement an earlier essay found a harmonic field could not produce is a statement about that bottom and survives everything here. A dissociation energy, a barrier, a second minimum — anything more than one and a half bond lengths out — is outside what these measurements can decide.

Who fitted what #

Bent-range and variable-range Morse potentials have a long history as flexible forms for fitting spectroscopic data, and their tendency to misbehave at large separation is well known to anyone who has extrapolated one. Dunham’s expansion of 1932 is the relation between the constants and the coefficients, and Pekeris’s closed form for the Morse curve’s $\alpha_e$ is the one the earlier essays compared against. The measured constants are the standard values for the four molecules.

What is computed here is the one-parameter family held exactly to two measured constants, its $\alpha_e$ along the whole range of λ for four molecules, the two members that fit carbon monoxide and the coefficients they carry, the ceilings the hydrides reach, and where every helpful member turns over.

Still open: a fourth constant, and a family that dissociates #

The obvious open question is $\omega_e y_e$. Carbon monoxide’s two fitting members predict 0.018 and 0.011 cm⁻¹, far enough apart that a measured value would choose between them — and if it chose the second, it would be choosing a potential that does not dissociate, which is the sharpest possible statement that a fourth constant is still a statement about the bottom of the well. Setting both fits against carbon monoxide’s measured higher constants is a comparison with numbers already published.

The nearer question is a family that cannot turn over. A range that varies as $a(1 + \lambda\xi)/(1 + \mu\xi)$, or an exponent in $1 - e^{-a(r - r_e)}$ raised to a power, keeps dissociation while adding a shape parameter, and it would say whether the hydrides’ ceilings belong to the requirement of reaching $\alpha_e$ or only to this family’s way of reaching it.