The localisation transformation, demonstrated

The claim this site makes most often is that a basis is not a thing. It is a claim about a transformation, and until this essay it was made by describing the transformation rather than by performing it.

The two descriptions #

Methane has eight bonding electrons in four occupied orbitals, and there are two standard ways to write them.

The localised set. Four equivalent bond orbitals, each an sp³ hybrid on the carbon pointing at one hydrogen, mixed with that hydrogen’s 1s. This is what a chemist draws, it matches the structural formula, and every member of the set looks like every other.

The canonical set. One orbital of a₁ symmetry — carbon’s 2s combined with the in-phase sum of all four hydrogens — and three degenerate orbitals of t₂ symmetry, each carbon 2p combined with a hydrogen combination that follows the same Cartesian direction. Two energies, in a one-to-three ratio, which is what the photoelectron spectrum shows.

They disagree about how many kinds of orbital there are, about how many energies there are, and about whether an orbital sits on one bond or on four. They are related by an orthogonal matrix.

The route taken here, which is backwards #

The textbook route starts from the canonical orbitals and localises them. This one starts from the bonds and finds the canonical set, and the reason is that in that direction nothing has to be assumed.

Build four bond orbitals. An sp³ hybrid pointing along each C–H direction, plus that hydrogen’s 1s in some proportion. The proportion is a free parameter and stays free.

Make them orthonormal. Real bond orbitals are not orthogonal to each other, and the four here overlap substantially. Löwdin’s symmetric orthogonalisation fixes that, and it is the right choice rather than a convenient one: Gram–Schmidt would privilege whichever orbital is handled first, so four bonds the molecule cannot tell apart would come out as four a reader could. Löwdin’s is provably the orthonormal set closest to the one it started from, and it treats all four alike.

Apply the tetrahedral matrix. Row zero is the totally symmetric combination; rows one to three follow x, y and z. It is orthogonal, and the orthogonality is checked row by row rather than taken from the algebra.

Look at what comes out. The first canonical orbital has the same coefficient on all four hydrogens and no carbon 2p at all. Each of the other three carries exactly one 2p and no 2s, with hydrogen coefficients proportional to that direction’s components of the bond vectors.

Nobody imposed either property. They are what a₁ and t₂ consist of, and they arrived from a matrix multiplication.

That means two independent calculations, sharing nothing, reach the same answer by different routes. The character-table treatment generated methane’s twenty-four operations and reduced a character. This one built four bonds and multiplied by a four-by-four matrix. Both say a₁ ⊕ t₂.

The matrix that produces the four hybrid directions has four rows orthonormal to better than one part in 10¹⁵, and the tetrahedral transformation used later in this essay is a matrix of the same kind acting on a different space. Both are rotations, and neither has an energy attached.

Why the density cannot change #

The reason is one line of algebra and it is worth having, because the numerical demonstration is only convincing to somebody who knows what it is demonstrating.

The first-order density matrix in the atomic basis, for an orthonormal set of occupied orbitals with coefficients $C$, is

$$P = n\, C C^{\mathsf T}.$$

Replace $C$ by $CV$ for an orthogonal $V$. The density becomes $n\,CVV^{\mathsf T}C^{\mathsf T}$, and $VV^{\mathsf T}$ is the identity, so it is $n\,CC^{\mathsf T}$ again.

Not similar. Identical. Every observable that is a functional of the density — and that is a great many of them — is therefore unchanged by any orthogonal mixing of the occupied orbitals, whatever it does to the individual orbitals.

It is worth noticing what that argument does not require. It does not require the orbitals to be solutions of anything, or the molecule to be methane, or the transformation to be the tetrahedral one. It requires only that the occupied set be orthonormal and the mixing be orthogonal, which is why the conclusion is so much broader than the demonstration below and why it was safe to state long before anybody performed it.

The demonstration, and the number it prints #

The bonding density along the line from carbon to a hydrogen, computed from the localised set and from the canonical set, is one curve rather than two: the largest disagreement across every point sampled is around 10⁻¹⁶, which is the arithmetic rather than the physics.

The largest disagreement is around $10^{-16}$ of the peak density. That is the last bit or two of a double-precision number, which is the arithmetic and not the physics.

The matrix version prints a similar figure: the two density matrices differ by about $10^{-16}$ in their largest element, and both integrate to eight electrons.

Both are run at four different values of the free mixing parameter, and the agreement is required at every one. That matters, because a demonstration that worked only at one carefully chosen value would be demonstrating something about the value.

The check that lets it fail #

An invariance test that always passes proves nothing, and this one would always pass if the arithmetic were merely reporting that $CC^{\mathsf T}$ equals itself.

So the same calculation is handed a mixing that is not orthogonal — one bond orbital with a third of another added to it — and required to report a changed density.

It does, by about a fifth. The figure prints that number alongside the invariance, and the two together are the claim: the density is invariant under this transformation, and it is not invariant under transformations in general, so the invariance is a property of orthogonality rather than of the arithmetic.

Three further checks guard the construction. The four raw bond orbitals must all overlap each other equally, because if they did not they would not be four equivalent bonds and the whole exercise would be describing something else. The metric must be positive definite, so a singular set is refused. And the canonical orbitals’ symmetry properties — the equal hydrogen coefficients in a₁, the single 2p in each t₂ — are checked rather than admired.

What does differ, and by how much #

The two sets are not the same set, and it is worth measuring how different they are, because the number is large.

The participation ratio comes out at 1.07 for a localised orbital and exactly 4 for a canonical one — a factor of nearly four in how many hydrogens each orbital occupies, and nothing at all in what they produce together. That is the whole of what a change of basis can and cannot do.

A factor of nearly four in how spread out the orbitals are, and nothing at all in the density they produce. That pair of statements is the essay in its shortest form.

It also settles what kind of experiment could tell the two apart. Not one that measures the density — X-ray diffraction, for instance, sees the same thing either way. What distinguishes them is an experiment that asks which one-electron function an electron came out of, and ionisation is exactly such an experiment, which is why methane’s photoelectron spectrum shows two bands and four equivalent bonds would predict one.

What was computed, and how #

The basis is eight functions: carbon 2s and three 2p, and a 1s on each of four hydrogens. Building its overlap matrix takes three numbers and a good deal of geometry.

Two of the three are numerical integrals — carbon 2s with a hydrogen 1s at the bond distance, and carbon 2p pointing along the bond with the same 1s — evaluated on the same Gauss–Legendre grid the overlap calculation uses. The third, hydrogen with hydrogen, has a closed form.

Everything else follows from a symmetry argument used as a labour-saving device. A p orbital’s overlap with a spherical function on another centre is the sigma overlap times the cosine of the angle between the p axis and the line of centres, exactly — the perpendicular component is odd about the plane containing that line and integrates to nothing. That is the exactly-zero argument doing work rather than making a point.

The numbers that go in are stated. The functions are hydrogen-like; carbon’s effective nuclear charge is 3.25, from Slater’s rules; hydrogen’s is 1; and the C–H distance is the measured 1.087 ångström. None of the four is fitted to anything here.

One detail surfaced that a purely algebraic treatment would have missed. A hydrogenic 2s has a radial node, and with carbon’s effective charge that node sits at about 0.6 bohr while the hydrogen is at 2.05 — so across the whole bonding region carbon’s 2s is negative, and a hybrid written with the conventional positive s coefficient overlaps the hydrogen with the wrong sign. The two contributions subtract, and what was meant to be a bond orbital is built from an antibonding interaction. The repair is to take whichever sign makes them add, decided by looking at the two integrals rather than by adopting a convention.

The surprise: the freedom is larger than the demonstration #

The transformation used here is orthogonal because both descriptions are orthonormal and orthogonality is what preserves that. But the density’s invariance is broader than orthogonal mixing.

Any non-singular linear mixing of the occupied orbitals leaves the density alone, once the metric is handled properly — the density is a property of the occupied subspace, and a subspace does not know which basis has been chosen inside it.

That is a stronger statement and a less visual one, and it sharpens the essay’s claim considerably. The question “which orbitals are the real ones” is not merely underdetermined between two candidates. It is underdetermined between infinitely many, because the object that is real is a four-dimensional subspace of function space and every basis for it produces the same density.

Which makes the canonical set’s status precise. It is not more real. It is the basis that diagonalises a particular operator, and that operator is the one ionisation couples to — a fact about the experiment rather than about the molecule.

What it costs #

Three overlap integrals, an eight-by-eight matrix, a four-by-four eigendecomposition for the Löwdin orthogonalisation, and a matrix product. The whole thing runs in under a fifth of a second, most of which is the two numerical overlaps.

The honest cost is elsewhere, and it is worth stating flatly.

There is no Hamiltonian anywhere in this. No energy is evaluated, nothing is minimised, and nothing is self-consistent. What has been demonstrated is a statement about bases, which is exactly as strong as it sounds and no stronger.

The mixing parameter is free and stays free. How much hydrogen enters each bond orbital cannot be determined without a real calculation. Every result is invariant to it, which is why the demonstration works and also why it says nothing about methane’s energy.

A Hartree–Fock treatment of methane in a small Gaussian basis would remove both caveats and is deliberately not done here. The two-electron integrals over Gaussians have closed forms and the iteration is standard; what is not standard is validating it, and a wrong self-consistent field that produces plausible numbers is exactly the failure this site exists to prevent. It is held back until there is room to check it against published results calculation by calculation.

Where the model stops #

Three limits.

Hydrogenic functions. A real carbon 2s in a minimal basis has no radial node in the bonding region, and the sign correction above is a symptom of using one that does. Nothing in the density argument depends on it; every picture does.

One molecule. Methane is the cleanest case because its four bonds are equivalent by symmetry, which is what makes the transformation a single matrix with no freedom in it. Localising water’s orbitals, or ethene’s, requires a criterion — Boys, Edmiston–Ruedenberg, Pipek–Mezey — and different criteria give different localised orbitals. That they all give the same density is the same theorem; that they disagree with each other is a reminder that “the localised orbitals” is not a well-defined phrase.

Nothing here says which description to use. The essay establishes that neither is more real. Which is more useful depends entirely on the question, and the answer is genuinely different for a reaction mechanism and for a spectrum.

The two descriptions are equivalent for one molecule and not for a series #

Everything above is about a single molecule, where the two sets are related by a transformation and the choice between them is a choice of axes. Put a homologous series side by side and the symmetry between the two descriptions breaks, in favour of the localised one — not because it is more correct, but because it is the only one of the two that is transferable.

A localised C–H bonding orbital in methane and a localised C–H bonding orbital in ethane are nearly the same function: the same hybrid on carbon, the same 1s on hydrogen, the same coefficients to a few per cent, with the difference confined to a small tail reaching onto the rest of the molecule. The canonical orbitals of the two molecules have nothing in common at all — methane has an $a_1$ and a $t_2$ set, ethane has orbitals delocalised over two carbons and six hydrogens, and no orbital of one is recognisable in the other.

That difference has a measurable consequence, and it is one of the oldest quantitative regularities in chemistry. The standard enthalpy of formation of the straight-chain alkanes falls by very nearly the same amount for every CH₂ group added: about 20.6 kilojoules a mole, repeated down the series, with the increments constant to about one kilojoule from propane onwards.

An additive increment per group is exactly what a transferable localised description predicts and exactly what a canonical description has no way to produce. Nothing about a set of orbitals delocalised over a whole molecule suggests that its energy should decompose into contributions per bond, and yet it does, to a part in twenty.

So the equivalence established in this essay is a statement about one system at a time, and it is worth keeping the scope attached. For a single molecule the two descriptions carry identical information. Across a family of molecules they do not carry identical convenience: one of them has parts that recur and the other has parts that do not, and the recurrence is why a chemist can predict a heat of formation from a structural formula without doing any calculation at all.

Which is the honest reason the localised picture survived the photoelectron spectra. It was never the better account of one molecule’s ionisation energies, and it is the only account that explains why molecules resemble each other.

Who found it, and when #

Lennard-Jones and Pople worked out the equivalence of localised and canonical descriptions in the late 1940s and early 1950s, and Lennard-Jones in particular pressed the point that the two are related by a transformation that changes nothing observable.

Löwdin’s symmetric orthogonalisation dates from 1950, and its defining property — that it is the orthonormal set closest to the original in a least-squares sense, and therefore the one that preserves symmetry equivalences — is exactly why it is the right tool here.

Edmiston and Ruedenberg gave a localisation criterion in 1963 and Boys another; that the criteria disagree while the density does not is the practical form of this essay’s argument, and it took another decade for that to be generally understood rather than argued about.

The historical irony is worth noting. Pauling’s hybrid picture was criticised from the 1960s onward on the grounds that photoelectron spectra show canonical orbitals — and the criticism, taken literally, is a category error of the same kind as the one it corrects. Neither set is what a spectrum shows. A spectrum shows ionisation energies, and the canonical set is the basis in which those energies are diagonal.

Two further readings are worth having, because between them they say what the transformation does and does not touch.

And the other direction: what happens when the same search is asked for its answer rather than given one.

Where to read on #

The construction being transformed is hybrids are a basis.

The measurement that distinguishes the two descriptions is hybridisation does not explain.

The independent symmetry route to the same answer is character tables and reduction.

And the general framework this sits inside is molecular orbital and valence bond theory.

What the pictures here cannot show. The coefficient matrices on this page are numbers in a basis, and no picture of an orbital appears anywhere in the argument — deliberately, because drawing the two sets would invite exactly the comparison the essay is refuting. The density plot is the only figure here showing a function of position, and the reason there is one curve on it rather than two is the whole of what is being claimed.