Four angles the shell chooses

Three events and a ratio of two dipoles put an electric field along the z axis of a shell with three values of orbital momentum and counted the fields at which a pair of levels is brought into coincidence by it. There are three, because a field along z changes the momentum by one and leaves its projection alone: the $m = 0$ levels contribute two coupled pairs and the $m = \pm 1$ pair contributes one, with the two signs of $m$ giving the same field and so counting once.

Its closing paragraph named the loose end. Keeping the field along z keeps $m$ a good label and keeps the two halves of the shell apart; tilting it mixes them, the two coincident $m = \pm 1$ events must separate, and the count should rise towards the number of coupled pairs a five-by-five problem has.

It rises to six, which is every pair. And then, at four particular angles, it comes back down — to five at three of them and to four at forty-five degrees, where two separate collisions happen at once. Those angles are not fitted and not approximate: each is the arctangent of a ratio of two angular integrals, and the integrals are surds.

Why three was the right count before #

It is worth being clear that the axial count was not an undercount. Three is what a field along z has, and the reason two of the six pairs share a field there is the reason degeneracy is a group theorem gives: the two signs of the projection are related by a symmetry the axial field preserves, so their events are not merely close but identical, and no calculation could separate them without breaking that symmetry.

What the tilt does is remove the symmetry that made them identical. Two events where there was one is the same move made elsewhere — a perturbation that splits what a symmetry had held together — and this is that move applied to the field’s direction rather than to the atom.

What a tilt can reach #

Tilting in the xz plane leaves one symmetry standing: reflection in that plane. The functions odd under it — those built from $y$ — cannot mix with the rest, so the problem is not nine by nine but five by five, in the basis $3s$, $3p_z$, $3d_{z^2}$, $3p_x$, $3d_{xz}$.

Within that basis the two components of the field do different jobs and never the same one. The $z$ component couples three pairs — $3s$ with $3p_z$, $3p_z$ with $3d_{z^2}$, and $3p_x$ with $3d_{xz}$ — and the $x$ component couples three others: $3s$ with $3p_x$, $3p_z$ with $3d_{xz}$, and $3d_{z^2}$ with $3p_x$. Every one of the six pairs has exactly one non-zero element and the other is exactly zero, because $z$ leaves the projection of momentum alone and $x$ changes it by one.

That is the structural fact the rest follows from. A field at angle $\theta$ gives each pair an effective dipole $d_z\cos\theta + d_x\sin\theta$, and since one of the two terms is always zero, every pair’s dipole is a single number times a cosine or a single number times a sine. Nothing is a sum of two competing terms, which is why what comes next is an arctangent and not the root of a quadratic.

The exact zeros do real work here, and they are the kind the projector is unique, the basis is not relies on: a matrix element that vanishes because the integrand is odd about a plane is zero, not small, and the whole argument below needs it to be zero. If those elements were merely small, every pair would have both a cosine and a sine term, the coincidence condition would be a quadratic, and its solutions would move with the field strength.

Six curves that have to cross #

The crossover field for a pair is its energy gap divided by twice its dipole, so a pair reached by $z$ has a field going as $1/\cos\theta$ — rising as the field turns away from the axis — and a pair reached by $x$ has one going as $1/\sin\theta$, falling towards it. Along z the three $x$-pairs have no dipole at all and their fields are infinite; along x the three $z$-pairs are.

So three curves come down from infinity while three go up to it, and they are plotted on the same axes. Any rising curve and any falling curve must meet somewhere. The only question is whether the two are comparable — and they are comparable exactly when their gaps are equal, which happens whenever both pairs join the same two shells.

Where they meet #

Two pairs with the same gap coincide when their effective dipoles are equal: $|d_z|\cos\theta = |d_x|\sin\theta$, so $\tan\theta = |d_z| / |d_x|$. The angle is an arctangent of a ratio of two matrix elements and nothing else.

The four elements are closed forms — $3\sqrt6$, $3\sqrt3$, $9/2$ and $3\sqrt3/2$ — so all four ratios are surds and all four angles are exact:

• $3\sqrt6 : 3\sqrt6 = 1$, so $3s$–$3p_z$ meets $3s$–$3p_x$ at $45°$;
• $9/2 : 9/2 = 1$, so $3p_x$–$3d_{xz}$ meets $3p_z$–$3d_{xz}$ at $45°$ as well;
• $3\sqrt3 : 9/2 = 2/\sqrt3$, at $49.107°$;
• $9/2 : 3\sqrt3/2 = \sqrt3$, at $60°$;
• $3\sqrt3 : 3\sqrt3/2 = 2$, at $63.435°$.

Forty-five degrees carries two coincidences rather than one, which is why the count there is four and not five: the $s$–$p$ pair and one of the $p$–$d$ pairs both have equal partners, and both collapse at once.

None of these angles depends on the field strength, the nuclear charge, or the quantum defect that separates the levels. Change any of those and every crossover field moves; the angles at which two of them coincide do not, because the gaps cancel out of the condition. They are properties of the orbitals.

What the ratios are made of #

The four matrix elements are worth naming, because their being closed forms is what makes the angles exact rather than fitted. Between $3s$ and $3p$ the element is $3\sqrt6$; between $3p_z$ and $3d_{z^2}$ it is $3\sqrt3$; between $3p_x$ and $3d_{xz}$ it is $9/2$; and between $3d_{z^2}$ and $3p_x$ it is $3\sqrt3/2$. The quadrature reproduces all four to about a part in a million, which is the check that the integration is doing what the algebra says.

The three distinct $p$–$d$ elements are in the ratio $3\sqrt3 : 9/2 : 3\sqrt3/2$, or $2\sqrt3 : 3 : \sqrt3$ after clearing — the angular factors of a $d$ orbital seen along and across the field. Every coincidence angle is an arctangent of one of those ratios or of $1$, and there are exactly as many coincidences as there are ways of pairing a $z$-reached pair with an $x$-reached pair of the same gap: one in the $s$–$p$ family and four in the $p$–$d$ family, of which two land together at forty-five degrees.

Whether it is a meeting or a threshold #

A count of distinct values always rests on a decision about when two numbers are the same, and a reader is right to suspect that a coincidence is the threshold doing the work.

Two degrees away in either direction the count is back to six, and it stays six until the angle itself. So each coincidence is a point rather than a stretch — the two crossover fields genuinely cross, and away from the crossing they are separated by far more than the tolerance.

The check that would fail is available too, and it is the refusal: a tolerance of one, which declares every crossover equal to every other, reports a single event at every angle. That is what a threshold artefact looks like — a count that does not vary because the instrument cannot see anything — and it is not what the figures above show.

There is one subtlety worth stating rather than hiding. At the coincidence angle itself the two fields agree to machine precision however tight the tolerance is, and that is not evidence of anything: the angle is computed from the same two matrix elements that make them equal there, so the agreement is arithmetic rather than measurement. The tolerance matters for angles near the coincidence and not at one, which is why the test above steps off it rather than sitting on it.

What was computed, and how #

The matrix elements are quadratures over hydrogen-like orbitals, taken between every pair in the five-function basis for each of the three Cartesian components. Elements that vanish by symmetry come out below the quadrature’s own noise floor and are set to zero, which is this collection’s standing treatment of an exact zero — and here it does real work, since the whole argument rests on each pair having one non-zero element rather than two small ones.

The level energies are hydrogen-like with a quantum defect that separates $s$, $p$ and $d$; the crossover for a pair is the two-state estimate, gap over twice the dipole, which is the same construction used for the axial field. Using the same one is deliberate: the comparison is against the axial field’s three events, and a different definition would compare against something else.

The four coincidence angles are put into the scan rather than approximated by round numbers. That is not fussiness — $\arctan\sqrt3$ differs from a round $60°$ by about a hundred-thousandth of a degree, and at that separation the two crossovers are resolved as distinct again, so a table built on round numbers would contradict the list of coincidences printed beneath it.

The degeneracy no group predicts is the neighbouring case and the contrast is instructive. There a degeneracy appeared that the point group did not require, and the explanation was a hidden symmetry of the potential. Here the coincidences are not degeneracies of the levels at all — two events land at one field, which is a statement about where two independent near-crossings happen to fall rather than about any state being degenerate with another. The two are easy to conflate and the distinction is the last section’s question.

Where the model stops #

The tilt is in one plane. A field out of the xz plane would break the reflection symmetry that keeps the $y$-built functions out, and the problem would be nine by nine with more pairs and presumably more coincidences. Whether the extra angles are also arctangents of ratios is not answered here, though the argument that makes them so — one component per pair — is a consequence of the selection rule rather than of the plane, so it would be surprising if it failed.

The two-state crossover is also an estimate. It is where two levels would cross if the other three were absent, and in a five-level problem with a field strong enough to matter they are not absent. What the figures count is the number of distinct such estimates, which is the right quantity for comparing against the axial count and is not the same as the number of avoided crossings a full diagonalisation would show.

And the shell is hydrogen-like with a defect standing in for everything a real atom does. Nothing here needs the defect’s value — the angles are independent of it — but the fields do, and they should be read as a scale rather than as a prediction for any element.

The symmetry that is not a rotation is worth reading beside this one, because it is the case where an operation nobody had listed turned out to be doing the work. Here nothing hidden is at play: every coincidence follows from selection rules anybody would write down, and the only reason they are not obvious is that nobody solves for where two independent crossings coincide.

The generalisation #

The useful part is not about Stark effects. It is that a count of distinct events is not a monotone function of how much symmetry has been broken.

The intuition being corrected is a natural one: symmetry makes things coincide, breaking it separates them, so the further it is broken the more distinct things appear. The first two steps are right. The third does not follow, because a partial breaking replaces one exact relation with another — here, the requirement that two dipoles be equal — and that relation has its own solutions.

What is left when a symmetry is broken is a smaller symmetry, and the smaller one has its own degeneracies at its own special configurations. Those configurations are usually not the ones the original symmetry singled out, which is why they are easy to miss: nobody looks for a coincidence at $63.435°$.

One table, three groups makes the companion point about descriptions: the same object admits several symmetry labels depending on what is being asked, and which one is in force decides what may coincide. A tilted field is exactly a case where the label changes continuously — full axial symmetry at nought degrees, a single reflection plane in between, and axial symmetry again about a different axis at ninety.

The practical form is a warning about scans. A sweep at round values of a parameter will pass straight through a coincidence that sits at an arctangent of a surd and report nothing, and the coincidence will not show up as an anomaly because the count on either side is the same. If a count is being reported against a continuous parameter, the places where it can drop should be solved for rather than sampled.

One more thing the sweep settles cheaply. The count at nought and at ninety degrees is the same — three — but the three fields are not the same three: along z they are set by the $z$-reached pairs and along x by the $x$-reached ones, and the numbers differ. So a measurement that reported only how many events a shell has would find the two orientations indistinguishable, and one that reported where they are would not. That is the same distinction why a character table stops draws between what counting can settle and what it cannot.

Who found it, and when #

The Stark effect on hydrogen-like shells is a century old, and that a field at a general orientation couples states differing in the projection of angular momentum is the elementary selection rule. Nothing about the physics here is new.

What is new here is the counting: that the number of distinct two-state crossovers in this shell is three along an axis, six at a general tilt, and reduced at four angles which can be written down exactly as arctangents of ratios of the shell’s own angular integrals. It is a small statement and its interest is in the shape rather than the subject.

Still open: fields out of the plane, and the avoided crossings #

The obvious open question is out of the plane. A field with a component along $y$ admits the three functions excluded here, giving a nine-by-nine problem with far more coupled pairs, and the count becomes a function of two angles rather than one. The coincidences would then be curves on a sphere rather than points on a line, and whether those curves are the same four arctangents extended, or something with more structure, is a question the same calculation answers directly.

The nearer question is the avoided crossing. Everything counted here is a two-state estimate, and at each of the four coincidence angles two such estimates land on one field — which means four levels are involved at once and the two-state picture is at its least trustworthy exactly where the interesting thing happens. Diagonalising the full five-by-five at and around each coincidence would say whether the two crossings that meet there produce one avoided crossing or two, and that is the difference between a real degeneracy of the tilted problem and two independent near-misses that happen to occur at the same field. The matrices are already built; it is one sweep of an eigenvalue problem that is five levels deep.