The node a nodeless orbital has

The calculation of the first-row hydrides found that lithium hydride has no momentum-space minimum at π/R, and that where a bond of lithium’s 2s and hydrogen’s 1s would have one, it would not be the fringe but the 2s function’s own node in momentum: a flat point in the profile that a little hydrogen tips into a dip. The node was where the essay was most exposed, and it said so. Its 2s was hydrogenic at Slater’s effective charge of 1.30, whose momentum node sits at Z/2 — 0.624 π/R at lithium hydride’s length — and it added that a better function would put the node elsewhere and that a nodeless Slater-type function would have none.

The second half of that sentence is false, and finding out why is most of what this essay is about. The first half is true, and the size of the move matters more than expected. Three lithium 2s functions go through the same construction here — the hydrogenic one, a Hartree–Fock one solved from scratch, and the nodeless Slater function — and the profile’s behaviour near π/R is different with each.

A lithium atom solved from its integrals #

The Hartree–Fock function is the one a better model would use, so it has to be the real one rather than a fit typed in from a table. Lithium’s ground state is $1s^2\,2s$, and its restricted Hartree–Fock energy is $E = 2h_{11} + h_{22} + J_{11} + 2J_{12} - K_{12}$: two core electrons in one spatial orbital, one valence electron in another, the repulsion of every pair, and one exchange term between the valence electron and the core electron of the same spin. Both orbitals are expanded in thirty-four s-type Gaussians whose exponents form a geometric series from 0.002 to about 80,000 — an even-tempered set, which reaches the limit of a function space without any exponent being chosen by hand. Every integral over one centre is closed-form, and the orbitals are made stationary by diagonalising one coupled operator until the energy stops changing.

The atom comes out at −7.4327267 hartree, against the published Hartree–Fock limit of −7.432727; the ratio of potential to kinetic energy is −2 to a part in ten million; the 2s orbital energy is −0.19632 hartree, which is −5.342 eV and, through Koopmans’ theorem, within 0.05 eV of lithium’s measured ionisation energy of 5.392. The 2s mean radius is 3.874 bohr, the tabulated value.

Against that function the hydrogenic one is visibly wrong in a particular way. Its node is at 2/Z = 1.54 bohr, nearly twice as far out as the Hartree–Fock node at 0.83, and its mean radius is 4.62 bohr against 3.87. Slater’s charge is fitted to make the orbital’s outer part roughly right, and a single exponent cannot also place the inner lobe: the screening model has no term for the inner structure and so the inner lobe is wherever the outer exponent drags it. The Hartree–Fock 2s turns over where it has to in order to stay orthogonal to a 1s core whose mean radius is 0.57 bohr, and that is half the hydrogenic distance.

There is also a value at the nucleus to compare. Every exact orbital has a cusp there: Kato’s condition says its slope divided by its value equals −Z, which for lithium is −3. The hydrogenic 2s at Z = 1.30 has the ratio −1.30, because it believes the nucleus is a charge of 1.30. The Hartree–Fock function, read across the first hundredth of a bohr, has −2.69 — not −3, because a sum of Gaussians has zero slope at the origin itself and only imitates the cusp just outside it, but close, and of the right size. The difference turns out to matter, through a route nobody would guess from the position picture.

A node position does not decide #

Take each function to momentum and the three tell a story that is not the position story.

The hydrogenic node is where the closed form puts it, at Z/2 = 0.650 bohr⁻¹, which at lithium hydride’s 1.5949 Å is 0.624 π/R. The Hartree–Fock node is at 0.926 bohr⁻¹, or 0.889 π/R — 42 per cent further out, and only 11 per cent below the fringe. That is a direct consequence of the position picture: a node nearer the nucleus in position means a node further out in momentum, since the two variables trade length for inverse length.

And the nodeless Slater function has a node in momentum too. Its radial part is $r\,e^{-\zeta r}$, and the transform can be done by hand:

$$\phi(p) \;\propto\; \frac{3\zeta^2 - p^2}{(\zeta^2 + p^2)^3},$$

which vanishes at $p = \sqrt{3}\,\zeta$. With ζ = 0.65 — the same Slater charge of 1.30 divided by the principal quantum number — the node is at 1.126 bohr⁻¹, or 1.080 π/R, eight per cent above the fringe. A function that is positive at every radius changes sign in momentum. The essay that counted nodes in both variables found that a hydrogenic orbital has the same number of radial nodes in momentum as in position; that is a fact about the hydrogenic family, two Laguerre and Gegenbauer polynomials of the same degree, and the Slater function shows it is not a law about orbitals.

The sum and the slope at the nucleus #

So what decides whether an s function has a node in momentum? Two values, one at each end of the transform.

At zero momentum the transform is the function’s integral, $\sqrt{2/\pi}\int R\,r^2\,dr$, so its sign is the sign of the function in sum. At large momentum the transform is controlled by what happens at the nucleus. Integrating $\int R(r)\,r\sin(pr)\,dr$ by parts twice leaves a leading term

$$\phi(p) \;\to\; -\,2\sqrt{2/\pi}\;\frac{R'(0)}{p^4},$$

so the far tail carries the sign opposite to the slope at the nucleus. A continuous function that starts with one sign and ends with the other crosses zero. An s function has an odd number of momentum nodes exactly when its integral and its slope at the nucleus have the same sign.

The hydrogenic and Hartree–Fock 2s functions are positive at the nucleus and falling there, as every function with a cusp is, and their outer lobes are larger than their inner ones, so they are negative in sum. Sum negative, slope negative: a node. The Slater function starts at zero and rises, and it is positive throughout, so it is positive in sum. Sum positive, slope positive: a node. The Hartree–Fock 1s is positive, falling and positive in sum — the signs disagree, the transform keeps one sign, and it has no node. Each of these is computed rather than argued, by scanning the momentum function and counting the sign changes.

The rule has a reading that connects two things usually kept apart. For any exact orbital Kato’s cusp makes the slope at the nucleus equal to −Z times the value there, so the far tail always carries the sign of the value at the nucleus, and a 2s function has a node in momentum because it is positive at the nucleus and negative in sum — because its inner lobe is smaller than its outer one. The radial node in position is how a function gets that way, but it is not what the momentum node counts. A nodeless function with no cusp at all, which is what $r\,e^{-\zeta r}$ is, gets a momentum node by the other route: it starts at zero with a positive slope.

Lithium hydride’s bond does not notice #

The obvious first test is the molecule. Put each 2s on lithium, keep lithium’s 2pσ and hydrogen’s 1s and Hoffmann’s parameters exactly as before, and solve the three-function problem again.

The overlap with hydrogen’s 1s changes — from −0.376 with the hydrogenic function to −0.468 with the Hartree–Fock one, whose outer lobe is more compact and sits nearer hydrogen, and +0.476 with the Slater one, whose sign convention is opposite because it is positive in sum. The orbital barely moves. It is 91.3 per cent hydrogen with the hydrogenic 2s, 89.3 with the Hartree–Fock one and 89.1 with the Slater one, and its profile falls monotonically in every case, reading 10.2, 8.7 and 7.9 per cent of its peak at π/R. The earlier verdict survives the function: lithium hydride’s bond is too polar to have a minimum, and no choice of 2s brings it within reach. The band of shares in which a minimum exists has to be read for that, and it does not come near nine-tenths hydrogen with any of them.

The band halves, and then collapses #

The band is where the function matters. Hold the functions and the length fixed, take the bonding combination of lithium 2s and hydrogen 1s at every mixing angle from one degree to eighty-nine, and record where the profile has a minimum below 1.9 π/R.

With the hydrogenic 2s a minimum exists from 0.7 per cent hydrogen to 45 per cent, and it sits at 0.527 to 0.595 π/R. With the Hartree–Fock 2s the band runs from 0.8 per cent to 22 — half as wide — and the minimum sits at 0.731 to 0.834. With the Slater 2s it exists only from 2.6 per cent to 6.5, and sits at 0.898 to 0.955.

Two of the three behave as the hydride calculation described. As the hydrogen share falls toward zero, the minimum runs up toward the node from below: the hydrogenic minimum is at 0.588 π/R at 0.7 per cent hydrogen against a node at 0.624, the Hartree–Fock minimum at 0.834 at 0.8 per cent against a node at 0.889. Those are the node’s minima, born out of the flat point. Their depth tells the same story — the hydrogenic minimum dips to 3 to 9 per cent of the peak, the Hartree–Fock one to under 2 — because the further the node sits into the momentum tail, the less density there is to dip.

The third does not. The Slater function’s node is above the fringe, and the old account says the minimum should then settle on π/R. It does neither thing. At 0.8 and 1.7 per cent hydrogen there is no minimum at all; at 2.6 per cent one appears at 0.955 π/R, 12 per cent below the node and 5 per cent below the fringe; by 6.5 per cent it has moved to 0.898 and at 7.5 it is gone. Its depth is between a thousandth and a seven-hundredth of the peak. No 2s–1s combination with the Slater function has a minimum at the fringe or at the node.

Rising or falling at the node #

The reason is the one line the hydride calculation used for the flat point, followed one step further. A single s function’s profile has slope $-2\pi q\,|\phi(q)|^2$, so it is flat at the node and falling everywhere else. Add a little hydrogen and two things are added to that curve: hydrogen’s own profile, which is smaller and falls everywhere, and the interference between the two centres, which carries the cosine of $qR$ and rises and falls across the fringe. Near the flat point only slopes matter, and the hydrogen’s own term is quadratic in its small coefficient while the interference is linear in it. So for a small admixture, whether the flat point becomes a minimum is decided by the sign of the interference’s slope at the node.

If the interference rises through the node, it lifts the far side of the flat point and leaves a dip on the near side, and a minimum is born there at the smallest admixture — moving away from the node as the square root of the coefficient, because the flat point is a double zero of the slope. If it falls, it adds to the descent and the flat point becomes an ordinary falling stretch. The interference’s crest is not at π/R. It is set by both functions, and it sits at 0.811 π/R with the hydrogenic 2s, 0.938 with the Hartree–Fock 2s and 1.054 with the Slater one. The hydrogenic node is far below its crest, the Hartree–Fock node just below its own, and the Slater node 2.5 per cent past its crest — so the interference’s slope at the node is strongly positive, weakly positive and slightly negative in turn. The band’s width follows the same order. Why each band closes where it does is not computed here; the natural reading is that hydrogen’s own falling profile, growing as the square of its share, overtakes the interference’s rise.

That explains the Slater function’s odd band too. Its interference still rises below 1.054 π/R, and below the node the pure 2s profile, though not flat, falls very slowly — its slope at 0.95 π/R is under a hundredth of its slope at 0.6. With a few per cent of hydrogen the interference’s rise can just outpace that slow descent, and a shallow minimum appears on the shoulder. It is not the node’s; it is what happens to a nearly level stretch of curve, and the stretch is nearly level because the node is near.

So the competition the hydride calculation described was the wrong pair. It was set up as the node against the fringe at π/R, the feature the polarity reading was built on. It is the node against the interference’s crest, which moves with the functions, and the fringe is not where anything happens in any of the three.

Every number, and how each is held #

The atom is checked against four published numbers — the energy to two millionths of a hartree, the 2s orbital energy to a hundred-thousandth, the 2s mean radius to a thousandth of a bohr, and the virial ratio to a hundred-thousandth — and its node against a second basis: twenty-four Gaussians with a coarser ratio put the Hartree–Fock 2s node within a thousandth of a bohr⁻¹ of the thirty-four-function value, so the node is the function’s and not the expansion’s. The refusal is a basis that cannot hold the core: the same series stopped at fourteen functions, with nothing tighter than an exponent of 2, gives −6.6056 hartree, 0.83 hartree above the limit, and is rejected.

Each node is located twice, by scanning and bisecting the transform and against a closed form where one exists — Z/2 for the hydrogenic function to a millionth, $\sqrt{3}\,\zeta$ for the Slater one to $10^{-9}$ — and each function is normalised in both variables. The sign rule is checked as a rule: for each of the three 2s functions and for the 1s, the sign at zero momentum is compared with the integral, the integral with the slope at the nucleus, the far tail with the slope, and the count of nodes with the verdict.

The construction is the old one. With the hydrogenic 2s put through the new route, lithium hydride’s overlap, bond share and node come out exactly as the hydride calculation has them, and the overlap is now computed in momentum space — an integral of the two momentum functions against the spherical Bessel function of qR — and agrees with the position-space quadrature to a hundred-thousandth. Finally, the interference criterion is checked against the band: a minimum at the smallest share appears exactly for the two functions whose node lies below their interference’s crest.

What three functions cannot settle #

The Hartree–Fock 2s is the atom’s, not the molecule’s. In lithium hydride the valence orbital polarises and contracts, as an orbital in a bond is free to do, and what the molecule does to the inner lobe is not computed. The node moved by 42 per cent between two defensible atomic functions, so a few per cent more from the molecule would not surprise anyone, and with the Hartree–Fock node only 11 per cent below the fringe, a few per cent is the size that matters.

The Hückel parameters do not change with the function. Lithium’s valence-state energy is held at −5.4 eV in all three calculations; the Hartree–Fock atom’s own −5.34 eV is close to it, and the bond’s hydrogen share moves by two points. The 2pσ stays hydrogenic, since Hartree–Fock gives lithium no occupied p orbital to take.

The core is left out of the profile. The Hartree–Fock 1s is computed and has no momentum node, so it adds a broad, smooth, sign-definite background to any measured profile and cannot create a minimum; leaving it out favours the reading. And the whole construction belongs to one determinant and to one orientation, which is not what a gas measurement averages over — averaging alone removes a homonuclear zero.

A Gaussian expansion has no true cusp. The far tail of the Hartree–Fock transform is set, beyond a few bohr⁻¹, by the tightest Gaussian rather than by the nucleus. The node at 0.926 bohr⁻¹ sits well inside the range where the expansion and the exact function agree, which is what the second basis shows; the sign rule’s “far tail” is read at 6 bohr⁻¹, where the imitation still holds.

Kato, Clementi and Roetti, and the tail #

Kato showed in 1957 that an exact eigenfunction’s slope at a nucleus is fixed by the nuclear charge, and the consequence that a momentum density falls as $p^{-8}$ with a coefficient set by the density at the nuclei has long been standard in the study of momentum distributions. Clementi and Roetti’s 1974 tables of Hartree–Fock functions in Slater-type bases are the source most chemists know lithium’s 2s from, and later numerical work fixed the restricted limit at −7.432727 hartree. Even-tempered Gaussian sets were developed by Ruedenberg and co-workers in the 1970s precisely so that a basis could be driven to completeness without hand-tuning. The closed form of the transform of $r\,e^{-\zeta r}$ is elementary.

What is computed here is the three-way comparison: the Hartree–Fock 2s node of lithium at 0.889 π/R in lithium hydride, the sign rule applied to these functions, the band of shares for each, and the finding that what decides whether a minimum is born at the node is the interference’s slope there, with a crest that moves with the functions rather than sitting at π/R.

Still open: the molecule’s own 2s, and the crystal #

The nearest question is now the molecule’s function rather than the atom’s. A Hartree–Fock calculation of lithium hydride itself, with the same even-tempered s set on lithium plus p functions on both centres, gives a valence orbital whose lithium part can be projected out and transformed; whether its inner lobe moves the node above the interference’s crest would decide whether a better model of the molecule has the node’s minimum at all, since the Hartree–Fock atom leaves only five per cent of margin.

The crystal stays where the hydride calculation left it. Lithium hydride’s Compton profile was measured on solids, where a hydride ion has six lithium neighbours and the interference appears at reciprocal-lattice vectors rather than at π/R. The criterion found here — a node against an interference crest — carries straight over to a lattice, with a different crest for every set of planes.