Two events where there was one

The field at which a broken symmetry returns marks where a broken symmetry stops mattering. Screening splits the s from the p of a shell that a bare Coulomb potential leaves degenerate; a field couples them; and below a field of gap-over-twice-the-dipole the shift is quadratic while above it the shift is linear, which is the behaviour the degeneracy would have given. The case where that arithmetic runs out is the next shell up.

At n = 2 the shell has one s and one p and the two-state arithmetic is nearly exact. The n = 3 shell has a d as well, three different defects, and dipole elements that run over four whole numbers rather than two — so the crossover is not one field but several, and whether they are close together or spread out decides whether the linear effect returns is a single event at all.

There are two, they are a factor of 3.66 apart, and the answer is no.

Three defects #

Screening shifts an orbital by how much of it is inside the screened region, so an s is shifted most and a d least. The three quantum defects of the n = 3 shell at a screening of 0.01 are

$$\delta_s = 0.020417, \qquad \delta_p = 0.006682, \qquad \delta_d = 0.004003,$$

which fall by a factor of three and then a factor of 1.7 — a sequence rather than a pair, and one whose spacing is itself uneven. The same calculation for the s and p of the shell below finds the same two numbers, since the defect depends on the angular momentum and the screening rather than on which shell it is in.

which are three different numbers and therefore two gaps. That is the whole of what the two-state expression cannot carry.

Two coupled pairs among three levels #

A field connects orbitals whose angular momentum differs by one, so the shell’s z matrix has s–p and p–d entries and exactly nothing between s and d.

So there are three levels and two coupled pairs, and each pair has its own gap and its own dipole:

The p–d pair goes first, at a field less than a third of the s–p pair’s, because its gap is five times smaller and its dipole only a third smaller.

It is worth noticing which way round that is, because the intuitive ordering is the other one. The s–p gap is the larger of the two, so one might expect it to be the harder to overcome and therefore the first to matter — but a crossover is a ratio, and the s–p dipole is larger too. What decides is gap over dipole, and the p–d pair wins because its gap falls faster than its dipole does.

Coupled pairs do not cross over in order of anything obvious, and the only way to know which goes first is to compute both. The degeneracy no group predicts is the other statement about a Coulomb shell: the ordering of what happens inside one is not something the symmetry hands over.

What the levels do #

At the weak end the three levels sit where their defects put them and the field is a perturbation — the ordinary quadratic Stark effect, in which each level’s shift is a polarisability times the square of the field. At the strong end they are equally spaced — the hydrogenic pattern the degeneracy would have given, which is what the linear effect returns means.

In between they are neither, and the in-between is wide.

The exponent takes a decade #

The sharpest way to say it is with the exponent. Measure how much wider the shell is than at zero field, and take the local slope of that against the field: 2 while every shift is quadratic, 1 once every one is linear.

More than a decade of field between the two ends, which is the number to carry, and the exponent passes through 1.5 at 4.3 × 10⁻⁵ — between the two crossovers rather than at either.

That width is the answer to the second half of the question — whether the crossovers are close together or spread out. A factor of 3.66 between them and more than a decade for the exponent to travel is spread out: there is no field at which the shell can be described as having just changed behaviour, and the splitting that a symmetry statement guarantees survives in part long after it has gone in part.

So the honest answer to the question is that the linear effect returns is not an event in a shell with three levels. It is a slope, and the two crossovers are where it begins and where it has mostly finished. A reader who quoted a single field for the n = 3 shell would be quoting one of the two and hiding the other.

Why the exponent is not simply a weighted mean #

There is a tidier picture that would have been wrong, and it is worth ruling out.

If the shell were two independent two-level problems — s with p, and p with d — then the excess width would be a sum of two terms, each with its own crossover, and the exponent would be a weighted mean of two step functions: mostly 2 below both, mostly 1 above both, and a shoulder in between whose shape is set by the two terms’ relative sizes.

The p orbital is in both pairs, so they are not independent. What the diagonalisation shows is a single smooth curve with no shoulder in it — the exponent falls monotonically from 1.997 to 1.008 with no plateau anywhere — which is the signature of three levels mixing at once rather than two pairs mixing separately.

So the two crossovers are landmarks rather than causes. They say where each pair would have changed behaviour alone, and the shell’s actual behaviour is one process spread across both. That is worth having explicitly, because two crossovers invites a picture of two events and the curve says there are none.

And the higher shell goes first #

The comparison with n = 2 has a direction that is worth stating, because the naive expectation is the opposite.

Both of the higher shell’s crossovers are below the lower shell’s single one, by factors of thirty and eight. Two things push the same way: a higher shell’s defects are smaller, so its gaps are smaller; and its orbitals are larger, so its dipoles are bigger. Gap over dipole falls twice over.

The two factors are worth separating, because they are different physics. The gaps come from the quantum defects, and a defect falls with n roughly as 1/n³ once the orbital has stopped penetrating the core — so a higher shell’s levels are closer together. The dipoles come from the orbitals’ size, which grows as n², so a higher shell is more polarisable. Neither is a surprise on its own; what the ratio does is multiply them, and gap over dipole falls faster than either factor.

That is a real prediction with a chemical shape. A Rydberg series becomes hydrogenic at fields that fall steeply with n — so the field at which a spectrum stops showing quantum defects and starts showing an equally spaced Stark manifold is not one field for an atom, it is a different field for every shell, and the high ones go first.

What a two-state expression was hiding #

The two-state arithmetic was not wrong and it is worth being precise about what it was and was not.

In a two-level shell it is exact. Two levels, one coupling, one gap: the two-by-two problem has a closed-form solution and the crossover is where its two terms are equal. Nothing is approximated and no third level is being neglected, because there is not one.

In a three-level shell it is an estimate per pair. Each of the two pairs has a two-state crossover, and neither of them is where the shell changes behaviour, because at any field all three levels are mixing. The marked fields in the figures above are those two estimates, and the exponent curve is what actually happens — and the curve passes 1.5 between them rather than at either.

So what the two-state expression hides in a larger shell is not an error but a category: it answers a question about a pair, and a shell is not a pair. That is the same shape as a symmetry species that appears more than once — a description that is complete where the object has one of something and incomplete where it has two.

What is quoted, and what is computed #

Degeneracy is a group theorem everywhere except here: the degeneracy broken here is the Coulomb one, which no group predicts.

Nothing is quoted. A screened Coulomb potential with a stated screening length, a shell of three orbitals, and a uniform field.

The dipole matrix is computed by a mapped quadrature — over the whole real line rather than over a box — and the element symmetry forbids comes back at 10⁻¹⁸ rather than at something small, which is what makes two coupled pairs a count rather than an observation — the same distinction a symmetry-forbidden overlap rests on.

The quantum defects come from the closed form for a screened potential, evaluated at three angular momenta rather than two. The levels at each field are eigenvalues of the three-by-three Hamiltonian, diagonalised: the two-state expression is what this replaces, and it appears here only as the estimate of where each crossover is.

What this cannot say #

One electron, and no spin. The shell here is a one-electron problem in a screened potential, so there is no fine structure, no exchange and no configuration interaction. A real n = 3 shell of an alkali atom has spin–orbit splitting of the p and d levels that is comparable with the defect differences at the top of the series, and that would add more gaps and therefore more coupled pairs — pushing further in the direction already found here.

A screened Coulomb potential is not an atom. It is a standard model and it reproduces the qualitative fact — an s is shifted more than a p, which is more than a d — without reproducing any real atom’s defects. Sodium’s are 1.35, 0.85 and 0.01, which are far larger and far more unequal, and the crossovers would be further apart still.

The m = 0 subshell is not the shell. A full n = 3 shell has nine orbitals and a field mixes within each m separately, so the m = ±1 and m = ±2 subshells have their own level sets with their own crossovers. Those would add more events rather than change the ones here, and the count of distinct crossovers for a whole shell is larger than two.

One screening length. Every number here is at λ = 0.01, and how the answer moves with the screening is a subject of its own. The two crossovers move with it too, and whether their ratio of 3.66 is stable across screenings is a sweep not run here — it is one loop, and it would say whether two events, a factor of three apart is a property of the shell or of one screening.

And the crossover is an estimate even here. Gap over twice the dipole is the two-state expression, and the whole point here is that the shell is not two states — so the marked fields are where a pair would cross over if it were alone. What the diagonalisation establishes is the exponent curve, and the two estimates are shown against it rather than instead of it.

What the two shells establish together #

Set beside each other, the two-level shell and the three-level shell make a pair of statements about broken symmetry that neither makes alone.

A broken symmetry has a scale, which is the two-level result: there is a field at which the breaking stops mattering, it is computable, and it is gap over twice the dipole. That is a statement about a pair of levels and it is exact for one.

A shell does not have one scale, which is the three-level result: with three levels there are two such fields, they are far apart, and the shell’s actual behaviour is a slope between them rather than an event at either.

Together they say that how badly is a symmetry broken is a question with a number in it only when the object has two of something. How much symmetry is left asks the same question about a distorted molecule and gets a single number for a different reason — there the measure is defined as a distance rather than found as a crossing.

Counting the events before computing them #

Two crossovers where the shell below had one is a measured result, and the count follows from an argument that needs no calculation — which makes it a prediction for every shell rather than an observation about two.

A crossover happens where a coupling overcomes a gap. So the number of crossovers is the number of gaps, and the number of gaps in a set of levels is one less than the number of levels.

The shell below has two levels in the manifold being followed, one gap and one crossover. The next has three levels, two gaps and two crossovers. The pattern continues: a manifold with $k$ levels has $k-1$ crossovers, and $k$ grows with the shell.

That is a count rather than a curve, and it says what to expect without any of the arithmetic performed here. What the arithmetic supplies is where they fall — the factor of 3.66 between the two, and the decade of field the exponent takes to travel between them — and neither of those is available from counting.

The count also says why the events crowd together as the shell rises. More levels in the same shell means more gaps, and the gaps are the differences between levels that are themselves converging, so the crossovers arrive closer together in field. A high enough shell has many crossovers within a narrow range, which is the same crowding the scaling of a single crossover predicts and is why the transition from quadratic to linear behaviour looks like a single event when it is not.

What the calculation requires #

Three different defects, required as three distinct values rather than as an ordering, because the count is what makes two gaps.

Two pairs the field couples, so two crossovers — and the requirement is about the pairs the dipole matrix connects rather than about neighbouring levels, which is the distinction that keeps the n = 2 shell at one crossover instead of three.

They are a factor of more than three apart, which is the answer to close together or spread out.

And the exponent goes from above 1.7 to below 1.3 across the range, checked at both ends. The refusal is the n = 2 shell: it has one coupled pair and must give one crossover, and a procedure that counted neighbouring levels would report three there and two of them would be between levels that do not interact at all.

Still open: a real atom #

The obvious open question is a real atom. Sodium’s n = 3 defects are 1.35, 0.85 and 0.01, which are two orders of magnitude larger than this model’s and much more unequal — so its two crossovers should be at higher fields and much further apart, and both are quoted quantities rather than computed ones. Running the same three-by-three problem with measured defects and the same computed dipoles would turn the qualitative statement here into a prediction about a spectrum somebody has measured.

The nearer question is the rest of the shell. The m = ±1 subshell of n = 3 is a two-level ladder — 3p and 3d — with its own gap and its own dipole, and it is one diagonalisation away. Whether its crossover falls between the two here or outside them decides how many distinct events a whole shell has, and the answer is either four or three, which is a small enough number to be worth having exactly.