The zero belongs to one determinant

A bonding orbital built from two 1s functions has a momentum profile along its bond that is exactly zero at π/R. The zero comes from a cosine, the cosine comes from the two centres, and its position depends on the separation and on nothing else, so the first zero returns the bond length to twelve digits whatever exponent the energy chose. That is a striking thing for a measurement to be able to do, and it rests on an assumption that was stated and not tested: that the electrons are in that orbital and nowhere else.

Two electrons in a hydrogen molecule are not only in that orbital. The simplest correction any calculation makes puts a little of the pair into the antibonding combination, and the valence-bond picture puts a great deal of it there. So the question is what the second orbital looks like in momentum, and what happens to the zero once it holds anything.

A sign turns the cosine into a sine #

The transform of a function moved to a new centre is the transform times a phase. Adding two identical 1s functions at plus and minus half the separation gives twice a cosine of $p_z R/2$ times the atomic transform; subtracting them gives twice a sine of the same argument, times $i$. Everything else is the normalisation, which is $\sqrt{2+2S}$ for the sum and $\sqrt{2-2S}$ for the difference, with $S$ the overlap.

So the antibonding profile along the bond is the atomic directional profile times $4\sin^2(qR/2)/(2-2S)$, in closed form, and it has its zeros exactly where the bonding profile has its maxima. At zero momentum the bonding profile is at its largest, 0.936 of an electron per unit momentum; the antibonding profile is exactly zero there, rises to a maximum at 0.751 atomic units, and falls to zero again at $2\pi/R$. The figure above draws both at the separation and exponent the one-electron energy chooses: 1.9972 bohr and 1.2393.

The complementarity is exact rather than approximate. Because $\cos^2 + \sin^2 = 1$, the two profiles weighted by their normalisations add to four atomic profiles at every momentum — $(2+2S)J_g + (2-2S)J_u = 4J_{\text{atom}}$ — and that identity is checked at six momenta to thirteen digits. It is the momentum-space version of a statement already made about energies: the two combinations are not symmetric about the atom, because the denominators are different, and the asymmetry is the overlap.

A measurement along the bond would therefore separate the two orbitals by one number. The bonding electron’s profile is largest at zero momentum; the antibonding electron’s is zero there. No energy, no model of the atom and no fitted parameter enters that distinction — only whether the density along the bond peaks or vanishes at the origin.

Opposite senses were the bonding orbital’s #

The bonding orbital’s surprise was that it is stretched along the bond in position and flattened along the bond in momentum: a ratio of second moments of 2.056 in one picture and 0.600 in the other. The mechanism was the cosine cutting the momentum distribution off at π/R while nothing cut it off across the bond. It is worth checking whether that is a property of two-centre orbitals in general or of that one.

It is a property of that one. The antibonding combination is prolate in position, at 4.335 — more so than the bonding orbital, because its nodal plane between the nuclei pushes density out past each atom along the axis. And it is prolate in momentum as well, at 1.731. The sine removes density at low momentum along the bond, where the cosine kept it, so the distribution along the bond is pushed outward rather than cut off, and the second moment along the axis grows.

The numbers are tied together the same way the bonding orbital’s were. The profile across the bond is a Bessel-weighted quadrature with the interference term now subtracted, and both profiles carry 0.99999 of an electron by entirely different routes. Their second moments give $\langle p^2\rangle = 2.5161$, and twice the antibonding kinetic energy computed from the matrix elements alone is 2.5211: the same 0.2 per cent of quadrature resolution the bonding orbital showed.

So the statement “a bond is prolate in one picture and oblate in the other” names a bonding orbital, not a bond. An electron in the antibonding orbital is prolate in both, and a density built from both orbitals has an anisotropy that depends on how the electrons are shared between them. That is where the zero’s fate is decided too.

A filled pair keeps the fringe and turns it over #

The obvious guess about a filled pair — one electron in each orbital, or two in each — is that the cosine and the sine cancel and leave two atoms. The weighted identity above almost says so. It does not quite, because the weights in a real density are not $2+2S$ and $2-2S$: each orbital is occupied by whole electrons, and each carries its own normalisation.

Per electron the filled pair’s profile along the bond is

$$J(q) = J_{\text{atom}}(q)\,\frac{1 - S\cos qR}{1 - S^2}.$$

The fringe survives with its phase reversed and its contrast equal to the overlap. At zero momentum the pair sits at $1/(1+S)$ of the atom, 0.683; at π/R, where the bonding orbital alone was zero, it rises to $1/(1-S)$, 1.867. The overlap here is 0.464, so the modulation is nearly half the atomic profile. The bonding orbital alone swings from 1.37 of the atom to zero; the pair swings the other way, by less, and never reaches zero anywhere.

The reason the cancellation fails has already appeared in a different form. A complete set of orbitals built from a non-orthogonal basis does not sum to the identity on the basis functions; it sums to the inverse of the overlap matrix. For two functions that inverse has an off-diagonal element of $-S/(1-S^2)$, and the filled pair’s density is exactly the atomic densities scaled by $1/(1-S^2)$ plus that element times the cross term. Transformed, the cross term is the cosine. So the inverted fringe is the inverse overlap matrix, and the reason a filled angular shell has no shape while a filled two-centre pair keeps one is that spherical harmonics are orthogonal and two atomic functions on different centres are not.

What the dip at zero momentum costs #

A density with less weight at low momentum has more kinetic energy, and the filled pair has less weight at low momentum than two atoms do. The excess can be written down without any quadrature.

The kinetic energy of the pair per electron is the mean of the bonding and antibonding values, $(T_{aa} - S\,T_{ab})/(1-S^2)$, and subtracting the atom’s $\zeta^2/2$ leaves, after the exchange integral is written out,

$$\Delta T = \frac{\zeta^2 S\, w^2 e^{-w}}{3(1-S^2)}, \qquad w = \zeta R.$$

Every factor in it is positive, so the excess is positive at every separation. At the bonding orbital’s equilibrium it is 0.156 hartree an electron; at 1.4 bohr, with the overlap at 0.60, it is 0.33; by five bohr it has fallen to 0.005. The line in the figure is the closed form and the dots are the same number from the two kinetic matrix elements, which agree to the last digit printed.

That is closed-shell repulsion, and in momentum it has a picture. Two filled shells brought together cannot both keep their electrons in the low-momentum part of the distribution along the axis between them, because orthogonality to each other forbids it; the dip at zero momentum is where they would have been, and the kinetic energy is the price of moving them out. The energy-level version of this argument says the antibonding level rises further than the bonding level falls. The momentum version says where, in the distribution, the extra energy lives: on the bond axis, below π/R.

An antibonding occupation fills the zero #

Between one determinant with both electrons in the bonding orbital and a filled pair lies every two-electron density a calculation might produce. In this basis each is described by two natural occupations, $n_g$ and $n_u$, adding to two. Per electron the profile along the bond is half of $n_g J_g + n_u J_u$.

At $q = \pi/R$ the cosine is zero and the sine is one, so the bonding term vanishes identically and the antibonding term is all that is left:

$$J(\pi/R) = n_u\,\frac{J_{\text{atom}}(\pi/R)}{1 - S}.$$

The depth of the profile at π/R is the antibonding occupation, times a number the atom fixes. It is zero only when that occupation is zero, and it is linear in it everywhere — checked at five occupations from zero to a tenth, to fourteen digits.

The position of the minimum is a separate matter and a worse one. The atomic profile falls steeply through π/R — as the sixth power of momentum at large q — so a small positive floor added at π/R is lower just beyond it than at it, and the minimum slides outward. At an antibonding occupation of two hundredths it sits at 1.608 instead of 1.573, and π over that momentum reads the separation as 1.954 bohr instead of 1.997: two per cent short. At five hundredths it reads 1.878. And past a critical occupation of 0.104 the slope never turns positive before the next fringe; the minimum has become a shoulder and there is nothing to read at all.

So the two properties of the zero part company the moment the determinant is not single. Its depth becomes a measurement of the antibonding occupation, and a clean one, because it is linear and needs only the atomic profile and the overlap. Its position stops being a measurement of anything exact; it becomes a bond length biased short by an amount that grows with the same occupation. What made the zero attractive — a geometry with no model of the density — survives only in the limit where nothing is in the antibonding orbital.

Where the valence-bond bond stands #

The occupation a real hydrogen molecule has in its antibonding orbital is small. Natural-orbital analyses of accurate wavefunctions, reported since the 1960s, put it at about two hundredths at the equilibrium separation, so the minimum along the bond would survive there and the bond length it reports would be short by a couple of per cent. That figure is quoted rather than computed here: the two-electron calculations in these essays are on lattice models and not on the continuum hydrogen molecule.

What can be computed is the other standard description built from the same two functions. The Heitler–London wavefunction puts one electron on each atom and symmetrises. Rewritten in orbitals it is $(1+S)\,\sigma_g^2 - (1-S)\,\sigma_u^2$ up to normalisation, so its natural antibonding occupation is $(1-S)^2/(1+S^2)$: 0.236 at the overlap used above.

That is above the critical occupation, and it stays above it at every separation from 1.4 to 4 bohr. The critical occupation rises as the atoms separate — 0.060 at 1.4 bohr, 0.307 at four — because a smaller overlap makes the fringe shallower relative to the atomic slope. The Heitler–London occupation rises faster, from 0.117 to 0.659. So the valence-bond bond never has a minimum along the bond at all, at any separation this sweep reaches: its profile per electron is $J_{\text{atom}}(1 + S\cos qR)/(1+S^2)$, a fringe of contrast $S$ riding on a slope too steep for it to turn.

The two textbook descriptions of the same bond therefore make qualitatively different predictions about one measurable curve. They are known to differ about ionic terms and about dissociation, which are energies. Here they differ about whether a feature exists. The truth lies between them and much nearer the orbital end, which is the same thing the exact two-site calculation found about double occupancy at short bond lengths.

What the method was #

Every profile along the bond is a closed form: the atomic 1s directional profile $8\zeta^5/3\pi(\zeta^2+q^2)^3$ times a cosine or sine squared, over the orbital’s normalisation. The antibonding orbital’s profile across the bond is a quadrature over the perpendicular plane with a Bessel weight — $J_0$ computed from its own integral representation — and the position-space moments are a three-dimensional Gauss rule over the normalised density.

The exponent at each separation is the one that minimises the bonding orbital’s one-electron energy, and the antibonding orbital uses the same exponent on purpose: the question is what a sign does to one pair of functions, and a re-optimised exponent would change two things at once. The kinetic matrix elements are $T_{aa} = \zeta^2/2$ and $T_{ab} = -\zeta^2 S/2 + \zeta k$, with $k$ the exchange attraction integral.

A minimum is located as a change of sign in the slope, scanned outward from 0.2 atomic units in steps of a quarter of a thousandth and refined by golden section inside the bracketing step. It is deliberately not found by minimising near π/R, because above the critical occupation the profile is monotone there and a minimiser would return the edge of its window and call it a minimum. The critical occupation is a bisection on whether that scan finds one.

The checks, all run wherever these figures are drawn: both antibonding profiles carry one electron; their second moments add to twice the antibonding kinetic energy; the antibonding profile along the bond is zero at the origin where the bonding one peaks; the weighted sum of the two profiles is four atoms at six momenta; the antibonding orbital is prolate in both pictures and the bonding one is not; the depth at π/R matches its closed form at five occupations; the minimum reads the separation exactly at zero occupation and short at two hundredths; the critical occupation separates profiles that have a minimum from profiles that do not; and the Heitler–London occupation exceeds it at six separations. The refusal is the filled pair: its kinetic excess must equal the closed form, its dip and rise must be $1/(1\pm S)$, and one electron in each orbital must reproduce it exactly. A filled pair with a zero in it would be reporting the formula rather than the electrons.

The limits of two functions #

One exponent, two centres, 1s functions only. The fringe positions and the linear depth law survive any atomic function, because they come from the phases and the normalisations. The critical occupation does not: it is a competition between the fringe and the atomic slope, and a basis with polarisation functions changes the slope. The critical values above are this basis’s, and a better basis could move them either way.

The natural occupations are treated as free. A two-electron wavefunction in a two-function basis has exactly two natural orbitals here only because symmetry makes them the bonding and antibonding combinations; that is true for a homonuclear pair and would not be for a heteronuclear one, where the occupations and the orbitals move together.

And the profile is one electron pair’s. Anything larger has other occupied orbitals, whose profiles add a background at π/R that is not zero. Against that background the depth law still holds for the σ pair’s own contribution, but separating it from the rest is exactly the problem a total Compton profile poses and this calculation does not solve.

What a zero is evidence of #

The pattern in this result is general and easy to miss. A quantity that is exactly zero in a model is usually zero because of a single term, and exactness is the model’s most fragile prediction. Here it took one determinant. Any admixture — two hundredths of an electron — replaces the zero with a floor proportional to the admixture, and a floor under a steep slope moves the minimum. The position of a feature and the depth of a feature are then different measurements with different sensitivities, and only one of them was the geometry.

The generous reading is that the zero was never the useful object. The depth is, because it is linear in something a single determinant sets to zero and a correlated wavefunction does not — an occupation number, read off a momentum profile at a point the geometry names. That turns a failed bond-length reading into a correlation measurement, and it was available only because the orbital picture had put the zero somewhere exact first.

Still open: a polarised bond, and a second pair #

The obvious open question is what polarisation does to the critical occupation. Everything above uses one 1s function per atom, and a real bonding orbital mixes in a p function along the bond whose momentum density is zero at the origin and peaks further out. That steepens nothing at π/R directly, but it changes the atomic slope the fringe is competing with — and since the critical occupation is exactly that competition, the question of whether the hydrogen molecule’s two hundredths leave a readable minimum, or sit on the edge of losing it, depends on it. It is a two-function calculation at each separation, and it would also settle whether 0.600 is the bonding orbital’s momentum anisotropy or only an upper bound on it.

The nearer question is a second pair. A molecule with a σ pair and a π pair has two profiles along the bond, and the π pair’s is zero at the origin for its own reason — the p function’s momentum density — rather than because of any fringe. At π/R the σ pair contributes its antibonding occupation and the π pair contributes whatever its atomic profile is there, which is not small. Whether the depth law is still readable once a second pair’s background sits under it is the difference between a statement about hydrogen and a statement about bonds.