Photoelectron — the series · Molecular Geometry & Bonding

methane: 2 valence bands. The measured valence photoelectron bands of methane, each labelled with the symmetry species of the orbital it comes from, and beside them the species the valence basis spans — the central atom's s and p functions and one s on each ligand, reduced in the molecule's own group. A band carrying a species the reduction does not produce would stop this figure being drawn.

What a photoelectron spectrum measures

The bands of a photoelectron spectrum are routinely read off as orbital energies. They are ionisation energies, which is a different quantity — and the identification rests on two errors of about an electronvolt each that happen to have opposite signs.

part 1 · wrong

water: 4 valence bands. The measured valence photoelectron bands of water, each labelled with the symmetry species of the orbital it comes from, and beside them the species the valence basis spans — the central atom's s and p functions and one s on each ligand, reduced in the molecule's own group. A band carrying a species the reduction does not produce would stop this figure being drawn.

Water's lone pairs are not a pair

Every course draws two equivalent lone pairs on water, pointing away from the hydrogens like a pair of ears. Its photoelectron spectrum shows the two bands they would produce at 12.6 and 14.7 electronvolts, two point one apart, in different symmetry species.

part 2 · bonding

ammonia: 3 valence bands. The measured valence photoelectron bands of ammonia, each labelled with the symmetry species of the orbital it comes from, and beside them the species the valence basis spans — the central atom's s and p functions and one s on each ligand, reduced in the molecule's own group. A band carrying a species the reduction does not produce would stop this figure being drawn.

Fewer bands than electrons

Methane has eight valence electrons and two photoelectron bands. Ammonia has eight and three; water has eight and four. The count is not of electrons, not of bonds and not of orbitals — it is of the symmetry species the occupied orbitals fall into, and it falls as the symmetry rises.

part 3 · spectra

Three answers to one question. The energy to remove an electron from a half-filled four-site system, computed three ways against the repulsion: exactly, by solving a self-consistent field twice — once for the molecule and once for the ion — and by reading the highest occupied orbital energy straight off the molecule, which is Koopmans' theorem. All three agree exactly at zero repulsion. The theorem always sits above the two-calculation answer, because letting the ion relax can only lower it; the exact answer sits above both, because the molecule is more correlated than its ion. The two errors have opposite signs and do not cancel: the residue grows to 6.03.

Koopmans' theorem is exact for nothing

Reading an ionisation energy off an orbital energy neglects two things that pull in opposite directions, and the cancellation between them is quoted as the reason it works. Compute all three energies in a model where the exact answer is available and the cancellation is real, partial, and gone by the time the repulsion is twice the hopping.

part 4 · wrong

Three bands of one spectrum, and the bond length behind each. Nitrogen's three photoelectron bands, drawn as the vibrational intensity distributions computed from the measured bond lengths and vibrational constants of the three states of the ion. Each band's lines add to one. The middle band is spread over five lines because the electron removed came out of a strongly bonding orbital and the bond lengthened by 77.22 thousandths of an ångström; the outer two keep 92 and 88 per cent of their strength in a single line.

The width of a band is a bond length

Nitrogen's three photoelectron bands are one sharp line, a progression of five, and a line with a shoulder. Computed from the measured bond lengths of the three states of the ion, the intensities come out at 0.917, 0.263 and 0.880 in the first line of each — because removing a weakly bonding electron lengthens the bond by 18.7 thousandths of an ångström, a strongly bonding one by 77.2, and an antibonding one shortens it by 23.7.

part 5 · spectra

A photoelectron band is a filter, and the group chooses the filter. The Huang–Rhys factor of every vibration of five molecules under a change of geometry that lengthens every bond alike — which is what removing an electron from a non-degenerate orbital does. On a logarithmic scale spanning sixteen decades, eight modes carry the whole of it and the rest sit on the floor at arithmetic noise. Which ones is decided by the point group: only a totally symmetric vibration can appear, whatever the size of the change.

A band is a filter on the modes

A photoelectron band's vibrational structure reports the frequencies of a few of the ion's vibrations and is silent about the rest, and which few is decided by the point group before any geometry is known. Under a change of shape that lengthens every bond alike, methane's totally symmetric stretch gets a Huang–Rhys factor of 0.905 and its other eight modes get between 10⁻²⁸ and 10⁻³⁵.

part 6 · spectra

Three lines, then a hundred. The exact removal spectrum of a 6-site Hubbard ring at half filling: every final state of the ion, at the energy it costs to reach and with the intensity the matrix element gives it. With no repulsion there are 3 lines and they are the occupied orbital energies. At U = 8 there are 100, on a molecule with 6 orbitals — so the spectrum cannot be read as a list of orbital energies, because there are more bands in it than there are orbitals to name.

More bands than there are orbitals

A photoelectron spectrum is read as a list of orbital energies, one band per occupied orbital. Computed exactly for a six-orbital ring, it has three bands with no repulsion and a hundred with eight — and by then fifty-three per cent of the intensity is in lines that no orbital corresponds to. The total intensity is three at every repulsion, exactly, because that is a sum rule and not a fit.

part 7 · spectra

The test that works until it does not. How many times stronger the weakest fundamental is than the strongest satellite, against the repulsion, on a half-filled ring of six. It starts at 23.8 and falls to 1.15 — a spectrum whose tallest satellite is as tall as its shortest band. The marked repulsion is where the other test fails as well: satellites start appearing inside the range the fundamentals span, so neither height nor position sorts the spectrum.

A hundred lines and no way to sort them

A spectrum with a hundred lines has six fundamentals in it somewhere. Sorting by height works until the tallest satellite is as tall as the shortest band, and sorting by position works until satellites start arriving between the bands — and on a ring of six both stop working at the same repulsion.

part 8 · spectra

The boundary belongs to the gap

A satellite stops being tellable from a fundamental somewhere, and it can be located on one ring at one filling. Four systems put it at repulsions of 2, 4 and 8 — ordering exactly with each one's own one-electron gap and not with its band width — and the fourth, whose gap is zero, has no boundary at all: its satellites are indistinguishable at every repulsion including none.

part 9 · spectra

How much stronger a fundamental is than a satellite, against the repulsion. The weakest fundamental divided by the strongest satellite, for a six-site ring and chain at every filling from a third to a half, against the on-site repulsion. Below the line at two the two kinds of line cannot be told apart by their height. The half-filled systems cross it and the third-filled ones do not — not at any repulsion up to sixty-four times the hopping, where the third-filled ring is still at 8.3.

A satellite that never loses its place

The repulsion at which a satellite stops being tellable from a fundamental orders exactly with the one-electron gap across four systems. Changing the gap by the filling instead is the sharper test, and the ordering does not survive it: a six-site chain has a larger gap at half filling and a smaller boundary. Below half filling there is no boundary at all, at any repulsion up to sixty-four times the hopping.

part 10 · wrong

The contrast, out to a repulsion of sixteen thousand. The ratio of the weakest fundamental to the strongest satellite, against the on-site repulsion, for eight systems with two electrons each. Both axes logarithmic. The dashed line at two is the factor the intensity test needs. Every curve flattens above it and none of them crosses, at any repulsion — including a repulsion sixteen thousand times the hopping.

A contrast with a closed form

Below half filling the satellite test flattens instead of failing, and the value it flattens at could be above or below the factor of two the test needs. It is — on all eight systems, by between 1.25 and 3.7 times. And on a ring the limit is (1 + 2cos(π/n))², to six figures, on every ring tried.

part 11 · spectra

The contrast at three fillings, and the floor two of them reach. The intensity contrast on a ring of 6 against the on-site repulsion, at three fillings. At two electrons it settles on a number well above the factor of two the test needs. At half filling it falls through two and lands on exactly one from U = 64 upward — and every point where it reads exactly one is a point where the cut between fundamental and satellite falls between two lines of identical weight. Those are drawn hollow.

A ratio of exactly one is a tie

Does the intensity contrast fall below two at half filling? It does — it falls to exactly one. But one is the floor of a ratio between two ranked quantities, and it is reached here because the cut between fundamental and satellite lands between two lines of identical weight. The guard installed to catch that case tests the wrong degeneracy, and the guard installed to license the extrapolation cannot tell an exact answer from a divergent one.

part 12 · spectra

The same three fillings, on a ring and on a chain. The intensity contrast against the on-site repulsion at two, four and six electrons, for both geometries. Two of the ring's three curves flatten onto exactly one and stay there — the hollow marks, where the rank cut falls between two degenerate lines. The chain's corresponding curve approaches the same value from above without reaching it, because a chain of six has no exactly degenerate removal lines at any repulsion at all.

The number the tie got right

On a ring of six the contrast at half filling comes out exactly one, and the one is an artefact — the rank cut had landed between two lines of identical weight, so the ratio was a quantity divided by itself. The chain of six has no such pair anywhere, at any repulsion, at any filling. Its contrast at half filling converges to one anyway.

part 13 · spectra

Six removal lines, each smooth, and the contrast is whichever two sit at the cut. The weight of each of the six strongest removal lines of the half-filled chain of six, followed from U = 16 upward by continuity in energy and labelled by the energy it tends to. Every line is monotone from U = 19. The contrast is the third strongest over the fourth, so it changes whenever two lines exchange those ranks: at U = 28.92 the line tending to +0.45 overtakes the one tending to −1.80 and the two weights are equal, and at U = 157 the fourth and fifth exchange. Lines at ±E, drawn in one colour, converge to one weight.

The limit of one is a parity

A half-filled chain of six has no degenerate removal lines, and its intensity contrast still goes to one — after dipping near U = 32 and rising again to U = 128. Followed line by line, every removal line is smooth and monotone; the dip is an exact tie between two lines at U = 28.916, and the rise ends where two satellites change places. At large repulsion the lines pair up at ±E with equal weights, so the contrast goes to one exactly when the cut falls inside a pair. On a chain of four it does not, and the limit is 1.3125.

part 14 · wrong