The pair that cancels only at zero overlap

A σ combination of two atomic functions has a directional Compton profile along the bond that carries a factor of 2 ± 2 cos(qR), and the sign is decided by whether the orbital’s parity matches its atomic function’s. Cosine combinations are exactly zero at q = π/R; sine combinations are not. So the depth of the minimum there counts something: how many of a molecule’s occupied combinations carry a sine, weighted by how many electrons are in each.

That is a count read off a measurement, which is the most attractive kind of result these essays produce, and it has one problem. A directional Compton profile needs an oriented molecule. What a gas-phase experiment measures is the profile averaged over every orientation, and the closing question of the essay before it was whether any of the count survives that average.

The quantity is worth restating in the form the average will act on. A momentum profile is the momentum density integrated over a plane — every momentum whose component along the scattering direction equals q — and the momentum density of a bond is the atomic one times an interference factor that depends on the momentum’s component along the bond. When the scattering direction is the bond those are the same component, which is the whole reason a directional profile carries the factor intact.

None of it does, and the reason is not the one that was expected.

Three factors where there was one #

The average of cos(p·R) over directions of R at fixed momentum is $j_0(pR)$, the spherical Bessel function, whose first zero is at $pR = \pi$ — exactly where the directional zero was. That is the whole of what the earlier essay anticipated, and it is right for an s combination.

It is wrong for the others, because the cross term is not alone in the average. The momentum density of a two-centre combination is the atomic radial function squared, times the orbital’s angular density, times 2 ± 2 cos(p·R) — and the angular density depends on the same angle the interference does, measured from the bond. So the average couples them, and the average of a product is not the product of the averages.

Using $\langle P_2(\cos\theta)\,\cos(x\cos\theta)\rangle = -j_2(x)$, with $\cos^2\theta = (1 + 2P_2)/3$ for a pσ orbital and $\sin^2\theta = 2(1 - P_2)/3$ for a pπ one:

• an s combination averages to $2 \pm 2\,j_0(pR)$;
• a pσ combination to $2 \pm 2\,[\,j_0(pR) - 2\,j_2(pR)\,]$;
• a pπ combination to $2 \pm 2\,[\,j_0(pR) + j_2(pR)\,]$.

The first zeros are at pR = 3.1416, 2.0816 and 4.4934 — that is π, 0.663π and 1.430π. The directional profile gives every orbital in the molecule the same zero at π/R; the average gives each angular kind its own, and they are nowhere near each other. That alone would make a count harder to read, since the contributions no longer vanish together.

And the exact zero is gone entirely #

It is worse than that, and the reason is structural rather than about the factors.

A Compton profile is an integral of the momentum density over a whole plane — every momentum whose component along the scattering direction is q. In the directional case the factor cos(qR) depends only on that component, so it is constant over the plane of integration and can be taken outside: the profile inherits the factor and inherits its zero.

In the averaged case the factor is a function of the momentum’s magnitude, not of its component. It is inside the integral, the integral runs over every magnitude from q upward, and passing through a zero of the integrand does nothing to the integral. So the averaged profile has no feature at π/R at all.

What is left to look at is the difference between the molecule’s averaged profile and the profile of the same electrons on two free atoms — which is the same profile with every interference factor set to zero. Everything that does not involve the bond cancels out of it exactly.

The signature is large, not small #

The first surprise is the size. The interference part at zero momentum is 34.2 per cent of Li₂’s whole profile, 31.0 of $\mathrm{B_2}$'s, 21.7 of $\mathrm{C_2}$'s, 13.8 of $\mathrm{N_2}$'s, 6.2 of $\mathrm{O_2}$'s and 2.3 of $\mathrm{F_2}$'s.

A Compton profile is measured to about a per cent of its peak, so every one of those is far above the noise. The parity information is not marginal in the averaged profile — it is a leading term at the profile’s peak, and what has been lost is only the exact zero, which was a place rather than a magnitude.

The sign is worth reading too, because it says what bonding does to a momentum distribution. Every one of those numbers is positive: the molecule’s profile at zero momentum is higher than two free atoms’. A bonding combination spreads the electron out in position and therefore concentrates it at low momentum, so the profile’s peak rises — which is the momentum-space statement of delocalisation, and it is the one quantity in this essay that a reader could have predicted.

The fall along the series has an obvious reading and it is nearly right. Lithium has one bonding combination occupied and nothing else; fluorine has every valence combination occupied, bonding and antibonding alike, so the interference terms of each pair cancel and almost nothing is left. The interference part is therefore a measure of how unpaired the occupation is — which is what a parity mismatch count means.

It is nearly right, and the way it fails is the point of this essay.

The order is already slightly wrong, and it is the tell. $\mathrm{F_2}$ has every valence combination filled, so a count says its interference is exactly zero, and it is 2.3 per cent. $\mathrm{O_2}$ has one antibonding pair half filled, so a count says a small positive number, and it is 6.2 per cent — larger than $\mathrm{F_2}$'s by nearly three times, where the count says the ratio should be infinite. Neither discrepancy is large enough to notice on its own; both are the same effect.

The cancellation is exact only at zero overlap #

Resolve nitrogen’s interference part by atomic function. The 1s pair is filled — two electrons in the bonding combination, two in the antibonding — so its imbalance is zero, and it contributes −0.00002 per cent of the profile. That is the cancellation working: bonding plus antibonding of one atomic function is twice the atom, exactly.

The 2s pair is also filled. Its imbalance is also zero. It contributes −24.7 per cent of the profile, which is larger in magnitude than everything else in the molecule put together and has the opposite sign.

The two combinations are not normalised alike. A bonding combination of two functions overlapping by S is divided by $\sqrt{2 + 2S}$ and the antibonding one by $\sqrt{2 - 2S}$, so two electrons in each give

$$\frac{2(2 + 2\cos)}{2 + 2S} + \frac{2(2 - 2\cos)}{2 - 2S},$$

whose interference terms cancel only when S = 0. The identity that does hold is $(2 + 2S)J_+ + (2 - 2S)J_- = 4J_{\text{atom}}$ — the cancellation is exact when the two combinations are weighted by their norms, and an occupation is not a norm.

The 1s pair cancels because its overlap is 7 × 10⁻⁵: two deeply contracted core functions on atoms two bohr apart barely see each other. The 2s pair does not, because its overlap is 0.496.

Plotted against the overlap, every balanced pair in the six molecules falls on the same story: residue below 10⁻⁷ at overlaps of order 10⁻⁵, residue of tens of per cent at overlaps near a half, and the bound the arithmetic gives is a fixed fraction of S times the profile.

So it is not a count, and the miscount is the majority #

Split each molecule’s interference part into the contributions of functions with an imbalance — what a count adds up — and the contributions of functions without one, which a count omits.

For Li₂ the count is the whole of it: 34.2 per cent counted, nothing omitted, because the only pair with an overlap is the one carrying the imbalance. For every other molecule the omitted part is larger. $\mathrm{B_2}$: +6.3 per cent counted against −37.3 omitted. $\mathrm{N_2}$: +10.9 against −24.7. $\mathrm{F_2}$: +2.2 against −3.1.

Lithium is the exception for a reason worth naming, since it is the one case where the reading works. Li₂ has three occupied combinations — the 1s pair and the 2s bonding — so its only overlapping pair is the one with the imbalance, and there is nothing left over to spoil the count. The reading is exact for the molecule with the fewest electrons and fails for every molecule that has a filled valence pair, which is every other molecule in the series and nearly every molecule there is. The pattern is the one an accounting of what a bond order counts keeps producing: a quantity that is a count in the simplest case and a weighted sum in every other.

The two have opposite signs, so the count does not merely miss part of the signal; it gets the sign of the total wrong for five of the six molecules. A measurement of the interference part at zero momentum, read as a parity count, would report a negative mismatch for every molecule from boron to fluorine.

That is a firm no to the question the essay before it asked, and it is a no with a mechanism rather than an estimate: the obstruction is the difference between the two combinations’ normalisations, it is proportional to the overlap, and it is not a feature of the average at all. The directional profile has the same defect and hides it, because there the interference factor multiplies the whole profile and the reading is taken at a point where the factor is zero — which is where the normalisations cannot matter.

What the average costs and what it buys #

There is one thing the averaged profile does better than the directional one, and it is worth setting against everything above.

A directional measurement needs an oriented sample, which for a gas means an alignment technique and for a solid means a single crystal with the bonds in a known direction — and in a solid the bonds are not the only thing in the momentum density. The averaged profile needs neither: it is what a gas-phase X-ray scattering experiment produces without any preparation, and it has been measured for every molecule discussed here.

And the quantity it gives is a magnitude rather than a position. A zero at π/R is a beautiful thing and reading it requires knowing R to place the measurement, and knowing that nothing else contributes there. The interference part at zero momentum requires nothing except a profile of two free atoms to subtract, and free-atom profiles are the best-known quantities in the subject. The zero read a bond length with nothing fitted; the average reads a number with one subtraction.

So the trade is a clean one and it is not the trade the earlier essay expected. What the average destroys is the count — the arithmetic that turned a depth into an integer — and what it keeps is the effect of bonding on the momentum distribution, at full strength, in a quantity an ordinary experiment measures.

What the model is, and where it breaks #

Minimal basis, Slater exponents, no s–p mixing. One function per shell per atom, with exponents from Slater’s rules and occupations from the standard filling order. Real $\mathrm{N_2}$ mixes 2s and 2pσ substantially, which redistributes the imbalance between two functions with different averaged factors and different overlaps; the 2s pair’s residue would then be shared with the pσ pair rather than sitting entirely in one place. The existence of the residue does not depend on that, because it depends only on the two combinations having different norms.

One determinant. Every occupation here is an integer, and the zero the directional profile has belongs to a single determinant — correlation puts a small occupation into the antibonding combination and fills it in. In the averaged picture there is no zero to fill in, so what correlation does instead is move the imbalances off integers, which moves the counted part and not the residue.

The average is over orientations and not over anything else. A real gas measurement also averages over rotational and vibrational states, and vibration changes R, which moves every interference factor’s argument. The effect is to smear the oscillations at large q and to leave the value at $q = 0$ nearly untouched, since $j_0(0) = 1$ whatever R is — so the numbers this essay reads, which are all at zero momentum, are the ones least sensitive to it.

And the two profiles being compared are the same quadrature. The averaged profile and the two-free-atom profile differ only in whether the interference factor is included, so every source of error in the radial functions, the quadrature and the normalisation appears in both and cancels out of the difference. That is why the difference can be quoted to a tenth of a per cent while neither profile is trustworthy to better than one.

And the sum rule is the check. Twice the integral of the averaged profile must be one per electron, and it is the same quadrature that normalised the density, so a profile that failed it would be reporting its own integration error. It comes out between 0.9997 and 1.0001 across the six molecules.

A normalisation is not a weight #

The habit: when a cancellation is expected between two states, check whether it is a cancellation between occupations or between amplitudes.

Bonding and antibonding of one atomic function cancel, and that is true in a specific currency: weighted by their norms. An occupation is a different currency, and the two agree only when the norms do — which is when the overlap vanishes. Every textbook statement that a filled bonding–antibonding pair “is the same as two atoms” carries that qualification, and it is usually invisible because the quantity being discussed is an energy, where the same overlap appears in the other factor and the statement fails in the other direction.

The corollary is about which pair to look at. The core pair is the one that looks dangerous — it holds a third of nitrogen’s electrons and its profile is broad and featureless — and it is the one that cancels perfectly, precisely because it is contracted. The valence pair is the one that looks harmless and it is the one that ruins the reading. Depth of contraction and size of overlap run opposite ways, so the intuition that a core is the thing to worry about is exactly inverted here.

Who measured what, and when #

Compton profiles of the first-row diatomics have been measured since the 1970s, by X-ray scattering on gases, and are quoted to about a per cent of the peak. The relation between a directional profile and the momentum density is Compton’s own; the spherical-average identities $\langle P_l(\cos\theta)\, e^{ix\cos\theta}\rangle = i^l j_l(x)$ are standard. The parity rule for the directional factor is the earlier essay’s.

What is computed here is the orientational average of each combination’s interference factor, the averaged profiles of the six diatomics against two free atoms, and the decomposition of the interference part by atomic function with the overlap residue identified and measured.

The numbers worth carrying are −24.7 and +10.9 — one molecule’s interference part, split into the part a count omits and the part it adds up.

Still open: a vibrational average, and the mixing #

The obvious open question is what a vibrational average does to the oscillations beyond the first swing. Every reading above is at zero momentum, where the interference factor is one for any bond length and the average over R is therefore harmless. The first zero crossing sits between 0.28π and 0.77π in qR across the six molecules, and there the factor’s argument matters: averaging over the ground-state spread in R, which an earlier essay computes for several molecules from their own force fields, would smear the crossing by an amount that is a fraction of a per cent for nitrogen and larger for lithium, whose bond is soft. Whether the crossing survives as a readable feature is a quadrature over a known distribution and needs nothing new.

The nearer question is the mixing. A minimal basis with no s–p mixing puts nitrogen’s whole 2s imbalance at zero and its whole 2pσ imbalance at two, and real $\mathrm{N_2}$ has neither: its σ orbitals are mixtures, so both functions carry a fractional imbalance and both carry a residue. The apparatus for a mixed σg already exists here, parameterised by the p weight, and extending it to the averaged profile would say whether the residue that ruins the count gets larger or smaller when the imbalance is shared between two functions whose overlaps differ by a factor of two. It could go either way, which is why it is worth computing rather than arguing.