One level is not one comparison

An orbital picture is an isosurface: the set of points at which the wavefunction takes one chosen value. Two numbers can be attached to such a surface and they are not interchangeable. One is the level — the value of $|\psi|$ on the surface. The other is the enclosed fraction — how much of the probability density lies inside it.

This collection fixes the second and computes the first, as say what it encloses sets out. Almost every published plate of orbitals does the opposite, if it fixes anything at all, and the difference is not a matter of taste. It decides what the picture is a picture of.

The levels that enclose one fraction #

Take six hydrogenic orbitals and ask, for each, what level encloses ninety per cent of its density. The level is not chosen: the density inside a candidate surface is integrated, and the candidate is moved by bisection until the integral comes out at the fraction claimed.

The answers are $3.94\times10^{-2}$ for the 1s, $9.48\times10^{-3}$ for the 2p, $7.40\times10^{-3}$ for the 2s, $3.60\times10^{-3}$ for the 3d, $2.64\times10^{-3}$ for the 3s and $1.24\times10^{-3}$ for the 4s.

The largest is thirty-two times the smallest. That spread is the whole of the problem, because a plate drawn at one level has to pick one of those six numbers and use it for all six pictures.

The spread is not an artefact of choosing ninety per cent. At half the density the six levels run from $1.48\times10^{-1}$ down to $2.68\times10^{-3}$, a factor of fifty-five; at ninety-nine per cent from $8.44\times10^{-3}$ to $3.58\times10^{-4}$, a factor of twenty-four. What changes with the fraction is where each orbital’s own level sits, not how far apart they are.

Worse, the ranges overlap. The level that encloses ninety-nine per cent of a 1s, $8.44\times10^{-3}$, is larger than the level that encloses ninety per cent of a 2s, $7.40\times10^{-3}$ — so a single number cannot be found that treats both orbitals even roughly alike, in the strong sense that the interval of levels giving each of them a sensible fraction has almost nothing in it.

What one level does to the others #

Fix the level at the 1s orbital’s ninety-per-cent value, $3.9366\times10^{-2}$, and integrate again — this time asking what each of the other five encloses at that level rather than what level each needs.

The 1s encloses $90.05$ per cent, which is the check that the two calculations are the same calculation run in opposite directions. The 2p encloses $36.28$ per cent. The 2s encloses $4.44$. The 3s encloses $0.80$. The 4s encloses $0.14$.

The 3d encloses nothing whatever, and the reason is worth stating exactly, because it is not a rounding.

The maximum value a 3d wavefunction takes anywhere in space is $2.77\times10^{-2}$. The level in question is $3.94\times10^{-2}$, which is larger. There is no point in the universe at which that orbital’s wavefunction reaches the chosen value, so the surface is empty — not small, not faint, absent. A plate drawn at one level and containing a 3d orbital has either used a level low enough for the 3d, in which case the 1s is a balloon enclosing well over ninety-nine per cent, or a level suited to the 1s, in which case the 3d could not have been drawn at all and whatever is printed in its place was drawn at a different level without saying so.

There is no compromise level #

The natural response is that the level was badly chosen: pick a lower one and every orbital in the family gets a reasonable surface. The integrals say otherwise, and they say it quantitatively.

At $8.44\times10^{-3}$ — the 1s orbital’s ninety-nine-per-cent level, and about the lowest that still draws the 1s as anything but a balloon — the six enclose $99.00$, $91.60$, $87.64$, $63.94$, $9.69$ and $2.10$ per cent, in the order 1s, 2p, 2s, 3d, 3s, 4s. Three of them are within a few points of each other and three are nowhere near.

Drop further, to the 4s orbital’s own ninety-per-cent level of $1.24\times10^{-3}$, and the 4s at last shows ninety per cent. By then the 1s encloses $99.96$ per cent, the 2s $99.49$ and the 2p $99.61$. Those three surfaces are drawings of essentially the whole orbital, which is a picture with no information in it: the ninety-nine-point-nine-per-cent contour of anything is a smooth blob whose shape is set by the outermost tail rather than by the structure.

Between the two there is no better answer, because the failure is not a bad choice. It is that six functions with peak values spanning a factor of twenty and mean radii spanning a factor of sixteen have no single value of $|\psi|$ that means the same thing about all of them.

Why the discrepancy is so large #

The enclosed fraction is an integral of $|\psi|^2$ over a volume and the level is a value of $|\psi|$ at a point. Two things separate them, and both work in the same direction.

A diffuse orbital never reaches a high value. Normalisation, the condition that makes what an orbital is a probability amplitude at all, fixes the integral of $|\psi|^2$ over all space at one, so a function spread over a large volume must be small everywhere. The peak $|\psi|$ falls from $5.61\times10^{-1}$ for a 1s to $1.99\times10^{-1}$ for a 2s to $7.02\times10^{-2}$ for a 4s. A level chosen for the compact member of a family is above the whole of the diffuse one.

The volume enclosed grows as the cube of the radius. Even where a level does cut a diffuse orbital, the surface it cuts sits far out on a tail that carries very little density, because the density at that radius is small and the volume element grows more slowly than the density falls.

The two effects multiply, and thirty-two fold in the level becomes a factor of hundreds in the enclosed fraction.

What the fraction does not change #

The obvious worry, given all of that, is that the level chosen decides not only how big the picture is but what shape it has. It does not, and the reason is exact.

A radial node is a sphere on which the wavefunction is exactly zero — the objects nodes counts, and the reason a 2s is a shell inside a shell rather than a ball. Every contour is drawn at a positive value of $|\psi|$, so no contour can cross a node: the surface must close on each side of it. That fixes the topology of the picture completely. The number of closed shells is $n - l$, one more than the number of radial nodes, at every level a contour can be drawn at.

Rendering the surfaces at fractions from thirty per cent to ninety-nine and counting the closed shells gives two for a 2s at every fraction, three for a 3s, four for a 4s, one for every p and d orbital drawn here. The count is $n - l$ each time, and the only thing the fraction moves is how far out the outermost shell reaches.

This is a more useful result than it sounds, because it says which complaints about orbital pictures are worth making. That surface is too big is a complaint about a choice somebody made and did not state. That surface should be two shells and is drawn as one is a complaint about an error, and the two are often confused.

A 2s at fifty per cent of its density has |ψ| = 2.05×10⁻² and reaches 5.88 bohr, with the inner shell a small closed surface within it. That number is the level, and it is what a figure at a fixed isovalue is holding constant.

The same orbital at ninety-nine per cent has |ψ| = 1.84×10⁻³ and reaches 12.73 bohr — a factor of 2.2 in radius. Both pictures are correct, and neither caption would distinguish them if it said only “the 2s orbital”.

What a sphere does instead #

There is a way of drawing a family of orbitals on a common footing that is not a common level and is not a contour at all: a sphere holding a stated fraction.

How big is an orbital computes that radius among its four measures. For these orbitals the ninety-per-cent spheres sit at $2.66$, $9.12$, $7.99$, $19.44$, $15.80$ and $33.83$ bohr for the 1s, 2s, 2p, 3s, 3d and 4s, and they are honest comparisons: each holds the same amount of its own density, so the radii can be put beside one another.

What a sphere throws away is the shape, which for anything but an s orbital is most of the content — and the shape is what carries the sign structure that complex harmonics against real ones is about. A 2p and a 2s have ninety-per-cent contours that reach the same distance — both $9.13$ bohr — while their ninety-per-cent spheres differ by more than a bohr, because the 2p’s density is gathered along one axis and the sphere has to be large enough to hold the tail in that direction while enclosing empty space in the other two. Neither number is wrong. They answer different questions, and so does the contour.

That is three distinct notions of size now on the table — the contour at a stated fraction, the sphere at a stated fraction, and the level — and only the last of the three fails to be a notion of size at all.

The one comparison a level does support #

There is a question a common level answers well, and it is not the question a plate of orbitals is usually printed to answer.

Fixing the level and asking what each orbital encloses measures how sharply each function is peaked — how much of its density sits in the region where it is large. That is a real property and it is the property that governs, for instance, whether an orbital penetrates near a nucleus. A 2p enclosing $36$ per cent at the 1s ninety-per-cent level while a 2s encloses $4.4$ says something true and interesting: at that distance scale the 2p is the more concentrated of the two, even though its total spread is smaller in mean radius by a bohr.

The trouble is that no caption ever claims this. The claim made is about size, and size is exactly what a common level does not measure.

Four measures of size for the same six orbitals — mean radius, root-mean-square radius, most probable radius and the radius holding ninety per cent — order them differently. The level is a fifth quantity, and it is the only one of the five that is not a length at all.

Where this leaves the pictures already drawn #

Every orbital picture here states the fraction it encloses, and the level behind it is found by integration and then checked by integrating again at that level.

What this essay adds is the reason the alternative convention is not merely less informative but wrong in a specific way. A plate at one level is not an under-labelled comparison; it is a picture of a different quantity with a size caption attached.

Four orbitals at one shared fraction and one scale is the comparison this essay is arguing for: the sizes on the page are the sizes, because every surface encloses the same amount of its own density. Without that the pictures are four contours at four unrelated values.

The habit worth keeping is therefore narrow and cheap. A picture that draws a contour should say what the contour was solved for — a level or an enclosed fraction — in one form, beside the picture, and the statement should be the number the contour was actually computed from rather than a description written afterwards. A caption naming a fraction the drawing never used is worse than no caption, because it lets the picture and its label drift apart in exactly the way this essay is about.

The practical test for any published plate is short. If the caption states a level, the pictures are comparable to each other and are not a comparison of size. If it states a fraction, the pictures are a comparison of size and the levels behind them differ, as they must. If it states neither, the plate cannot be read at all, and this is the commonest case.

The 2p and the 2s, which is where this goes next #

One row of the table is worth pulling out, because it is not a curiosity about drawing conventions.

At the 1s orbital’s ninety-per-cent level the 2p encloses $36.3$ per cent and the 2s encloses $4.4$ — the 2p is eight times better represented at that level than the 2s is. Yet the 2s is the larger orbital by mean radius, $6.0$ bohr against $5.0$, and it is the one that lies lower in energy in every many-electron atom.

The resolution is that the 2s has an inner lobe and the 2p has none. A 2s puts a small part of its density very close to the nucleus and the rest of it further out than the 2p; a 2p puts all of its density in one shell. So a level chosen to catch the region near the nucleus catches a little of the 2s’s inner lobe and none of its main shell, while the same level cuts through the middle of the 2p.

That difference is called penetration, and it is what what an electron actually feels is about from the direction of the nuclear charge. It is also the natural next question: the same integrals run at the effective charges a screening model supplies, where the answer turns out to be one the screening model itself cannot see.

Where the model stops #

Everything here is hydrogenic: one electron, one nucleus, no repulsion. Three limits are worth naming so that the numbers are not read further than they go.

The levels are specific to $Z = 1$. Every orbital drawn here belongs to a hydrogen atom, and screening changes both the level and the enclosed fraction — by a great deal, as the radial distribution across the periodic table shows. What does not change is the argument, since it depends only on normalisation and on the volume element.

A many-electron orbital is not an observable. The density is; the individual orbital is a piece of an approximation, as orbitals are not where the electron is and where the electron is both insist. That does not make the level-against-fraction distinction less sharp — it makes it a distinction about the picture rather than about the atom.

The surface is not a boundary. Ten per cent of the density is outside a ninety-per-cent contour and it does not stop anywhere. A larger fraction pushes the surface out and never closes it off.

Who chose the convention, and when #

The isosurface picture is much younger than the wavefunctions it draws. Schrödinger’s 1926 papers contain no such plates and the early textbook figures were plots of the radial function or of the angular part — one-dimensional pictures of a separable problem, which is what could be drawn by hand.

The familiar three-dimensional lobes arrive with computer graphics in the 1960s, and the ninety-per-cent caption arrives with them, apparently as a convention rather than as a computed result. It is easy to see why, and the same economics shaped the rest of the subject — the aufbau order is not a property of the atom records another rule that survived because computing the alternative was expensive. At the time, integrating a density inside a candidate isosurface and bisecting on the answer was a genuine calculation, while choosing a level that made a pleasing picture was free. The caption then travelled from figure to figure independently of what had been plotted, which is how two books came to print visibly different pictures under the same words.

None of that is a reason to keep the convention now. The integral is a few lines and it runs in milliseconds; the reason to state the fraction is that a picture with a stated fraction can be wrong, and one without cannot.

Still open: nuclear charges other than one #

The open question is the one deliberately not touched here: what happens to these surfaces when the nuclear charge is not one. The level, the enclosed fraction and the size all move, and they do not move together — which is the same lesson as this one, arriving from the direction of the physics rather than of the drawing.

Two neighbouring arguments are worth following instead. How big is an orbital answers the size question with four different measures and shows they disagree by more than a factor of two; this essay adds a fifth candidate and disqualifies it. And orbitals are not where the electron is makes the prior point that these surfaces are drawings of an approximation, which is what stops the present argument from being pedantry about labels: a picture of an approximate object still has to say which picture it is.