Oblate in the picture nobody draws

A picture of a σ bond is a shape stretched between two nuclei, and the shape is right. Computed on the simplest two-centre orbital already in hand — two 1s functions added, at the separation and exponent the energy chooses for a one-electron diatomic — the second moment along the bond axis is 1.449 square bohr against 0.705 across it, on the density every contour here is drawn from. A ratio of 2.06: prolate, as drawn.

The same orbital, written in momentum, is flattened along the bond.

A bond is a cosine #

The transform of a function moved to a new centre is the transform times a phase. So for the symmetric combination of two identical functions at plus and minus half the separation along the bond, the molecular momentum function is twice $\cos(p_z R/2)$ times the atomic momentum function, divided by the square root of 2 + 2S with S the overlap. Nothing else happens. The atomic function is untouched, and every difference between a bond and two separate atoms in this picture is one cosine whose period is set by the bond length.

That is a stronger statement than the position-space analogue, where the two atomic densities overlap in a region and the cross term is a function of position with no simple form. Here the cross term is exactly a cosine, it is exact for any atomic function whatever, and it depends on the geometry alone.

The cosine also has zeros, and the atomic factor does not. A 1s momentum density falls as the eighth power of p and never vanishes; cos² vanishes at π/R, 3π/R and so on. So the molecular momentum density has nodes that neither atom has, at places fixed by the separation.

Only one direction keeps them #

A Compton experiment measures one component of the momentum, integrating over the other two. Whether the fringes survive depends on which component.

Measure along the bond. The cosine’s argument is $p_z$, which is the component being measured and not one of the two being integrated over, so it comes straight out of the integral: the profile along the bond is four times cos²(qR/2) times the atomic directional profile, divided by 2 + 2S, in closed form. The atomic directional profile times a cosine squared — so it has exact zeros, at exactly the momenta the density does.

Measure across the bond. Now $p_z$ is one of the variables integrated over. Writing $\cos^2$ as $(1 + \cos p_z R)/2$ and integrating the oscillating half over a circle in the perpendicular plane gives a Bessel function: the θ integral of cos(Rρ sin θ) around the circle is $2\pi J_0(R\rho)$. What is left is an atomic term plus a Bessel-weighted one, and it has no zeros at all.

One orbital, two directions, and only one of them is ever zero. That asymmetry is not a property of the orbital’s shape in the ordinary sense; it is a property of which variable the interference lives in.

The zero is the bond length #

The first zero is at $q_1 = \pi/R$ and therefore $R = \pi/q_1$, and that is exact for every atomic function and every exponent. Sweeping the separation from 1.4 to 4.0 bohr, the exponent the energy chooses contracts by about a quarter across the range and the overlap falls from 0.72 to 0.19 — and the recovered separation is right to twelve digits at every point.

So a momentum measurement can read a bond length with nothing fitted and no model of the density, on one assumption: that the orbital is a symmetric combination of two centres. The radial function does not enter, the exponent does not enter, the overlap does not enter. Everything the zero knows is the geometry.

That is worth setting beside what the position picture offers. There a bond length is read from a diffraction pattern, and the diffraction pattern is a transform of the total density, which is the same object this argument is transforming. The two routes are the same physics; what differs is that the momentum route puts the geometry into a zero and the position route puts it into a maximum, and a zero is a place a measurement can find without a model of the background.

Two pictures, two opposite shapes #

The anisotropies come out at 2.056 in position and 0.600 in momentum. One above one, one below.

The mechanism is the cut-off. Along the bond the momentum density is multiplied by a function that reaches zero at 1.573 atomic units and stays small beyond it, so nothing much of $\langle p_z^2\rangle$ is collected past there. Across the bond nothing cuts anything off and the distribution runs out to the atomic width, which at this exponent is around 1.24 — comparable, and unbounded rather than terminated.

The comparison of those two numbers is the physically interesting part and it is not a coincidence. The atomic momentum width is about ζ, the exponent; the cut-off is π/R. Both are fixed by the same variational problem — ζ and R are chosen together to minimise one energy — so their being within a quarter of each other at equilibrium is a consequence of that single minimisation rather than of anything about this argument. It also means the two effects are the same size at equilibrium, which is why the momentum anisotropy is 0.6 rather than 0.1 or 0.95.

And the direction of the position anisotropy is what makes the pair worth reporting. Nothing about a picture of a bond suggests that its momentum distribution is squashed in the direction the density is stretched. The reason the intuition fails is the ordinary one: a broad function has a narrow transform, and broad along the bond becomes narrow along the bond. Written out that way it is obvious; it is not obvious from a drawing, because a drawing has no second variable in it.

The checks, and the one that is a refusal #

The two profiles are computed by completely different routes — one a closed form, one a Bessel-weighted quadrature over a mapped radial grid — so their agreeing about anything is evidence. Both integrate to 0.99999 electrons, which is the only available check that the perpendicular integral is the same orbital as the parallel one.

The second moments are tied to the energy. For a cylindrically symmetric orbital $\langle p^2\rangle = \langle p_z^2\rangle + 2\langle p_x^2\rangle$, and the virial theorem makes $\langle p^2\rangle = 2\langle T\rangle$ at the variational minimum. The two profiles give 1.1733 and the energy calculation, which knows nothing about either profile, gives 1.1753 — 0.17 per cent apart, which is the quadrature’s own resolution.

The refusal is the separated limit. Pull the two centres apart at a fixed atomic exponent and the orbital becomes two atoms, so the momentum anisotropy has to return to one. It does: 0.600 at two bohr, 1.0000 at forty, with the overlap down to 2 × 10⁻¹⁵.

The arithmetic of that limit is worth a sentence because it is not the obvious one. The fringes do not disappear as R grows — their period shrinks, so there are more of them, not fewer. What happens is that they crowd together much faster than the atomic density varies, so their average over any interval of the profile becomes a constant one half; the half is the same in both directions and cancels out of the ratio. A molecular signature vanishing by becoming too fine to see is a different mechanism from one vanishing by becoming small, and only the first is what happens here.

Had the anisotropy stayed at 0.6 at forty bohr, everything above would be a statement about the algebra of a cosine rather than about a bond.

What was computed, and how #

The orbital is the σ combination of two 1s Slater functions, with the exponent at each separation chosen by minimising the one-electron energy — the same variational calculation that found the orbital contracting by a quarter when the bond forms. The overlap is the closed form $e^{-w}(1 + w + w^2/3)$ with $w = \zeta R$.

The profile along the bond is the closed form above. The profile across it is a one-dimensional quadrature with a Bessel weight, and the Bessel function is computed from its own integral representation rather than from a series — $\tfrac{1}{\pi}\int_0^{\pi}\cos(x\sin\theta)\,d\theta$, by the same Gauss rule everything else here uses, which cannot be wrong in the way a truncated series can.

The position moments are a three-dimensional quadrature over the normalised density on a grid mapped to the exponent’s own scale.

Eleven things are checked: that each profile carries one electron; that the two second moments add to twice the kinetic energy; that the profile along the bond has several exact zeros in range and is exactly zero at each; that the profile across the bond is positive at each of them and, at the first, is a substantial fraction of its own peak, so the contrast is one an experiment could see; that $\pi/q_1$ recovers the separation; that the position anisotropy is above one and the momentum anisotropy below it; and the separated limit, in two parts — that the overlap has gone at forty bohr and that the anisotropy has returned towards one.

What the fringes do to the density itself #

The profiles above are integrals, and it is worth looking at the momentum density they integrate before leaving it.

Along the bond the density is the atomic one multiplied by $\cos^2(p_z R/2)$. At the equilibrium separation the first zero of that factor sits at 1.573 atomic units, and the atomic density there has fallen to about a fifth of its value at the origin — so the first fringe removes a real part of the distribution rather than a tail. The second zero, at 4.719, sits where the atomic density is down by three and a half orders of magnitude and removes nothing worth counting.

So a molecule of this size has exactly one interference feature that matters, and it is the first one. That is a statement about a coincidence of scales: the fringe spacing is 2π/R and the atomic width is about ζ, and the two are within a factor of two of each other only because ζ and R come out of one minimisation. A much longer bond would put the first fringe deep inside the atomic distribution and produce several visible features; a much shorter one would put it out in the tail and produce none.

Both of those are available to check and neither is a molecule. A bond of eight bohr between two 1s functions is not bound, and a bond of half a bohr is two nuclei on top of each other. The one regime where the interference is visible is the regime where bonds exist, and the reason is that both quantities are set by the same balance between kinetic and potential energy — which is a satisfying thing to find and is not a prediction, since the coincidence was noticed after the numbers were computed rather than before.

Where this stops #

One electron and two centres. This is $\mathrm{H}_2^+$ with a single Slater function per atom, which is the simplest object that has a bond in it. Nothing here needs more — the interference factor is exact for any atomic function, so the zeros and their positions survive a better basis untouched — but the anisotropy numbers do not: a real bonding orbital has polarisation functions in it, and a p contribution changes both second moments.

The zeros need a symmetric combination. For two different atoms the two phases do not combine into a cosine and the profile along the bond has minima rather than zeros. So the bond-length reading is a homonuclear statement, and how deep the minima are in a heteronuclear case is a calculation this has not done.

And a real measurement sees every electron. A directional Compton profile of a molecule is a sum over all its occupied orbitals, and the σ bonding orbital’s zeros sit on top of contributions from everything else, none of which vanishes there. Whether a zero survives as a visible minimum in a total profile depends on how much density the other orbitals put at that momentum in that direction — which for $\mathrm{H}_2$ is nothing, since there is no other orbital, and for anything larger is the question.

The generalisation #

The habit is to compute a quantity’s anisotropy in both variables before describing its shape.

A shape is a statement about a function and a function has two pictures. Saying that a bonding orbital is extended along the bond names the position picture and sounds like a property of the orbital; the momentum picture disagrees, and both are correct. The general rule is not subtle — a broad function has a narrow transform — but it is almost never applied to a direction, because a drawing shows one variable and the reflex is to read the drawing as the object.

The corollary is about where to look for a geometry in a measurement. Position-space methods put a structural parameter into the position of a maximum, which has to be found against a background — and a maximum in a density is already a quantity three different questions give three answers to. A transform puts the same parameter into the period of an interference, which locates a zero — and a zero is easier to find precisely, because it does not need the background to be modelled. That is the whole of why diffraction works, and it is worth noticing that a momentum measurement is a second instance of it rather than a curiosity.

Who found it, and when #

The interference factor in a molecular momentum density is as old as the LCAO idea; Coulson worked out directional Compton profiles for $\mathrm{H}_2$ in 1941. The Bessel-weighted form of the perpendicular profile follows from the standard integral of a cosine around a circle. What is computed here is the closed form for the variationally chosen orbital, both directional profiles at several separations, the two anisotropies side by side, and the separated limit as a check on the mechanism.

The number worth carrying is not 0.600. It is the pair 2.056 and 0.600: one orbital, two pictures, one ratio above one and one below, and only the first of them ever drawn.

Still open: the second orbital, and a heteronuclear bond #

The obvious open question is the antibonding combination, which costs nothing and answers a question this essay raises without noticing. Subtracting the two centres rather than adding them replaces the cosine with a sine, whose zeros are at 2π/R and whose value at the origin is zero — so the antibonding orbital’s profile along the bond vanishes at zero momentum, which the bonding one’s maximises. A measurement along a bond axis would therefore distinguish a bonding electron from an antibonding one by whether the profile peaks or vanishes at q = 0, with no reference to any energy. Whether that survives in a real two-electron system where both orbitals are occupied is the follow-up, and the first half is one sign change in the formula above.

The nearer question is the polarisation. The whole calculation uses one 1s per atom, and the exponent it optimises contracts by a quarter — which is the basis set’s only way of describing what the bond does to the atoms. A real bonding orbital mixes in a p function pointing along the bond, and a p function’s momentum density is zero at the origin and peaks away from it. So adding polarisation moves density out of the low-momentum region along the bond, in the same direction the interference factor already does, and the momentum anisotropy should fall below 0.6. By how much is a two-function calculation at each separation, and it would say whether 0.600 is the answer or an upper bound.