The top that reports all three

The symmetric top ends in a negative result. A symmetric top has two rotational constants; its rigid-rotor spectrum consists of lines at $2B(J+1)$ for every J and every K; and the second constant, A, cancels out of every one of them exactly. Ammonia’s A of 6.3406 cm⁻¹ appears in none of its microwave lines at any J and any K.

The obvious next question is what happens when the symmetry that produced the cancellation is removed. The answer is that everything changes at once: the formula goes, the degeneracies go, and the missing constant becomes visible in almost every level.

There is no formula #

For a symmetric top the rigid-rotor Hamiltonian is diagonal in the basis labelled by J and K, so the energies can be written down. For an asymmetric one the same Hamiltonian, written in the same basis, has off-diagonal elements connecting K to K ± 2:

$$\langle J,K|H|J,K\rangle = A K^2 + \tfrac{1}{2}(B+C)\left[J(J+1) - K^2\right]$$

$$\langle J,K{\pm}2|H|J,K\rangle = \tfrac{1}{4}(B-C)\sqrt{[J(J{+}1)-K(K{\pm}1)][J(J{+}1)-(K{\pm}1)(K{\pm}2)]}$$

The off-diagonal element is proportional to $B - C$, which is zero exactly when the top is symmetric. So the whole difference between having a formula and not having one is one subtraction.

K survives only as a label for what a level becomes in the two limiting cases, which is why an asymmetric top’s levels are written $J_{K_aK_c}$ — carrying the K it would have if the molecule were prolate and the K it would have if it were oblate. Two labels, because there is no single one.

What the levels do when a constant is changed #

The symmetric top’s question, asked of a molecule that is not symmetric.

The two survivors are worth naming because they are not an accident of the arithmetic. The ground state is zero for every rotor whatever its constants. The level at $B + C$ is the one with a single quantum of rotation about the a axis unavailable to it — the $1_{01}$ level — and its energy is $B + C$ exactly, with no A in it, for any asymmetry whatever.

Everything above them carries A. So a spectrum of water determines all three constants, and a spectrum of ammonia determines two of its three (its two equal moments count once).

That is the difference between the two kinds of top in one sentence: a symmetric top hides its unique constant in one level of every J, and an asymmetric top hides it in two levels altogether.

That last figure is the sharp one and it deserves its own sentence. The transition is not gradual. A top with $B - C = 10^{-9}$ has the same count of blind levels as water does — two — even though its spectrum would be indistinguishable from a symmetric top’s at any achievable resolution. The count is a statement about exact degeneracy and the spectrum is a statement about resolvable splittings, and the two part company as soon as the question is asked precisely.

How large the splittings are #

The abstract statement — every degeneracy is split — is worth turning into numbers, because the numbers say whether an experiment can see it.

At J = 2, water’s five levels are 71.08, 80.03, 95.24, 133.61 and 134.85 cm⁻¹. The prolate symmetric-top formula built from the same molecule’s A and the mean of its B and C would put them at 72.31, 87.64 twice, and 133.61 twice. So one of water’s pairs is split by 15.2 cm⁻¹ — the two levels the formula puts on top of each other are nowhere near each other — and another by 1.23.

Formaldehyde at the same J has 7.32, 15.26, 15.73, 40.020 and 40.023. Its first pair is split by 0.47 and its second by three thousandths of a wavenumber. Sulfur dioxide’s second pair at J = 2 is split by one thousandth.

That is the practical meaning of κ. A near-prolate top’s low-K pairs split visibly and its high-K pairs do not, so a spectrum shows a symmetric top’s pattern with a few lines doubled — and the doubling is precisely where the third constant is measured from. For a molecule as asymmetric as water there is no symmetric-top pattern left to perturb.

The splittings also grow with J, because the off-diagonal element carries a factor that grows with $J(J{+}1)$: formaldehyde’s second pair splits by 0.003 at J = 2 and by 0.011 at J = 3. So a near-symmetric molecule’s third constant is measured at high J, where the asymmetry has had the most opportunity to act.

The label that stops working #

A symmetric top’s levels are labelled by K, the projection of the angular momentum on the unique axis, and K is a good quantum number because the Hamiltonian does not mix different values of it.

For an asymmetric top it mixes K with K ± 2, so no level has a K. The convention that replaces it is to label each level with two K values — the one it would have if the molecule were made prolate and the one it would have if it were made oblate, keeping the order of the levels fixed as the molecule is deformed to each limit. Water’s ground state is $0_{00}$, its next level is $1_{01}$, and so on.

Those subscripts are therefore not quantum numbers at all. They are the results of two thought experiments, recorded as a name — which is a convention doing exactly what a mode composition’s percentage does in the neighbouring field, and with the same virtue: it is unambiguous once its construction is stated, and meaningless without it.

The one thing that does survive as a genuine label is the parity of K under the two limits, which is what makes the four Wang blocks and is why each J’s matrix factors into pieces. That factoring is what makes the diagonalisation cheap and is the only symmetry left in the problem.

Where the molecules actually sit #

Ray’s parameter is $\kappa = (2B - A - C)/(A - C)$, which is −1 for a prolate symmetric top and +1 for an oblate one. It can be computed for any molecule with a known structure, because the moments come out of the coordinates and the masses with no bonding argument anywhere in them.

Water at −0.4322 is about as asymmetric as a molecule gets — the value is not far from zero, which is the most asymmetric a top can be. Formaldehyde at −0.9618 and sulfur dioxide at −0.9414 are near-prolate: their B and C differ by about ten per cent and their level pattern is a symmetric top’s with each degenerate pair split into two close lines.

Ammonia comes out at 0.9999905 rather than exactly 1, and the shortfall is worth a sentence because it is not physics. Its stored coordinates are quoted to four decimal places, so its two equal moments agree to about a part in a million rather than exactly — the same rounding that makes its threefold axis only threefold to 10⁻⁴. A test on κ tight enough to notice would call a symmetric top asymmetric for a reason that is about the file the structure came out of.

The classification the constants produce is worth naming once, because the boundary in it is not a boundary in anything measurable. A molecule counts as a near-prolate or a near-oblate top according to which pair of its three moments is closer together, and nothing changes discontinuously as that comparison flips — it is a change of which approximate labels are convenient, not a change in any spectrum.

Two checks the matrix has to pass #

The energies above are eigenvalues, and an eigenvalue is exactly the sort of thing that comes out plausible and wrong. Two independent checks run while the figures are drawn.

At B = C the levels must be the prolate formula, exactly. The off-diagonal elements vanish, the matrix is diagonal, and every level must equal $A K^2 + B[J(J{+}1) - K^2]$ to the last bit a double holds. That is checked for every J up to 4, and it is the check that the matrix elements above have the right coefficients — a wrong factor of two in the off-diagonal term would leave this passing and everything else wrong, but a wrong diagonal would not.

At A = B the levels must be the oblate formula, which is a different expression built from the same matrix. Passing both is a stronger statement than passing either.

And the trace is a sum that needs no diagonalisation at all. The sum of the eigenvalues is the sum of the diagonal elements, which does not depend on the off-diagonal ones — so the check holds at every asymmetry and is completely independent of the two limits above. It is the cheapest check available and it is the one that would catch a solver returning a permuted or rescaled spectrum.

The input to everything above is three moments of inertia per structure, and nothing about bonding appears anywhere in them: a rotational spectrum measures mass and distance, and what it returns is a set of distances if — and only if — enough independent constants are available to solve for them.

What was computed, and from what #

Nothing in this essay is quoted. The three constants come from the structure: the moments of inertia are computed from the atomic positions and the isotopic masses, sorted, and converted by one constant — exactly as in the rigid-rotor case, where the point was that no bonding argument appears anywhere in a rotational spectrum.

The levels come from those constants and nothing else. Each J is a matrix built from the three numbers, and the eigenvalues are taken by the Jacobi method, the same route that serves Hückel matrices and force fields — a method checked against closed forms where they exist, which is why it can be trusted here where there is no closed form to check against.

The sensitivity is computed the crude way, on purpose: A is raised by one per cent, every level is recomputed, and the differences are divided by the change. Nothing is differentiated analytically, so a level that does not move is a level whose energy came out bit-identical from two separate diagonalisations of two different matrices — which is a stronger statement than a derivative being zero.

The comparison against measurement is available and is deliberately not made here. Water’s constants from a microwave spectrum are 27.881, 14.522 and 9.278 cm⁻¹; the structure in this collection gives 27.377, 14.587 and 9.516, which is within two per cent on each. The difference is the vibrational averaging this model does not have and the rounding in the stored coordinates, in that order of size — and quoting the computed values against the measured ones as though the gap were an error would misdescribe both.

What the rigid rotor leaves out #

Centrifugal distortion. A real molecule stretches as it spins, and the correction is not a fitting parameter — it follows from the vibrational frequencies and agrees with measurement. For an asymmetric top the distortion has five independent constants rather than one, and fitting them is most of what a modern microwave analysis does.

The vibrational state. Every constant here belongs to the equilibrium structure. What a spectrum measures belongs to a vibrational state, and the two differ by an amount that is a real effect rather than an error — the subject of an essay in the shape field.

Selection rules. Which of these levels a transition can connect depends on which component of the dipole moment is non-zero, which depends on the point group. Water has a dipole along one principal axis only, so its spectrum is a b-type spectrum and shows a subset of what the levels would permit. Nothing above uses a selection rule, because the claim being made is about the levels rather than about which of them are joined.

No formula is the ordinary case #

There is no formula for any of them is stated here as a property of water, and it is worth saying that this is the general situation and the symmetric top is the exception — because the difference shaped how the whole field developed.

A symmetric top’s energies are a closed expression in two quantum numbers, so a spectrum can be assigned by hand: compute the expected positions, match them against the lines, read off the constants. That is how rotational spectroscopy began and it is why linear molecules and symmetric tops were done first.

An asymmetric top has no such expression. Its energies are the eigenvalues of a matrix whose size grows with the rotational quantum number, and beyond the smallest values there is no closed form at all — a general polynomial of degree five or more has no solution in radicals, and the eigenvalues are exactly that.

So an asymmetric top’s spectrum can only be analysed by fitting: guess the three constants, build the matrix, diagonalise it, compute the transitions, compare against the lines, adjust and repeat.

That is a completely different kind of work from reading a formula, and it needs an instrument the early spectroscopists did not have. Asymmetric tops — which is nearly every molecule with more than three atoms — were not routinely analysed until computing was cheap enough to diagonalise a matrix for every trial set of constants.

Two consequences follow that are visible in the literature.

The early structural results are almost all linear molecules and symmetric tops, and the impression that microwave spectroscopy is a technique for small symmetric species is a historical residue rather than a limitation.

And an asymmetric top’s constants come with a fit rather than with an assignment. There is a residual, there is a covariance between the fitted constants, and the three numbers are correlated in a way a formula-derived pair never was — which is the same difficulty met from the other side in a susceptibility curve whose four parameters are not independent.

So the next difficulty is already visible in this finding. Twenty-three levels of twenty-five moving when one constant is changed is exactly the condition under which a fit is well determined, and it is the good case; the awkward cases are the levels that barely move, because those are where the constants trade against one another.

The two levels that do not move at all are worth one further remark, because they are not a defect of the fit and cannot be improved by measuring them better. The ground state and the level at $B + C$ are independent of $A$ exactly, for the same kind of reason the symmetric top’s spectrum was blind to one constant: a term cancels rather than becoming small. So a spectrum consisting only of transitions between those two would determine $A$ no better than the symmetric top’s did, however precisely it was measured.

That is the useful form of the contrast between the two kinds of top. Breaking the symmetry does not make every line sensitive to every constant; it makes most of them sensitive, and the determination comes from the twenty-three rather than from the two.

Who did this, and when #

The matrix above is Wang’s, from 1929, and the factoring into four blocks is his too — which is why the transformation that produces them carries his name. What made the asymmetric rotor usable was not the theory but the arithmetic: diagonalising a matrix of side 2J+1 by hand for every J up to twenty is a serious undertaking, and it was done.

King, Hainer and Cross published tables of the eigenvalues in 1943, parameterised by κ, so that a spectroscopist could interpolate rather than diagonalise. Those tables were the standard tool for thirty years and are the reason Ray’s parameter is quoted at all: it is the argument the tables were indexed by.

That is a piece of history worth keeping in view when a figure here diagonalises five matrices in a few milliseconds. The classification of tops, the two-subscript labels, the near-prolate approximations — all of it is apparatus built to avoid an eigenvalue problem that is now free. What has not changed is which quantities are determined by which measurements, and that is the part of the subject this essay is about.

Still open: the inverse problem #

The natural open question is the inverse problem, which is what a microwave spectroscopist actually does: given a set of measured line positions, recover A, B and C, and from them the structure. The same inversion is simple for a linear triatomic, where one constant and one isotopic substitution give two equations in two unknowns.

For an asymmetric top the inversion is harder in an interesting way. Three constants determine three moments of inertia, and a molecule with more than three structural parameters is underdetermined by a single isotopologue — so the standard method substitutes one atom at a time and uses the change in the moments to locate that atom. That is Kraitchman’s method, it is exact within the rigid-rotor model, and it needs the computed levels of an asymmetric top rather than the symmetric top’s formula.