The nodes in the other variable

An orbital has a second picture as complete as the first, and taking it adds nothing: it is the same one-electron function written in the other variable. What changes is which facts are obvious.

The most countable fact available about an orbital is its node count. An orbital with quantum numbers n and l has exactly n − l − 1 radial nodes — a count, exact, and the fastest way to catch a drawing that is wrong. It is a statement about a function of r, and nobody asks what the same function looks like in p.

The count is the same and it is an identity #

A hydrogenic radial function is r^l e^(−Zr/n) times a generalised Laguerre polynomial of degree n − l − 1 in the variable 2Zr/n. The exponential never vanishes and the power of r vanishes only at the origin, which is not a node, so the radial nodes are the zeros of that polynomial and there are n − l − 1 of them.

The momentum radial function has the same shape with different parts. It is p^l divided by (1 + n²p²/Z²)^(l+2), times a Gegenbauer polynomial of degree n − l − 1 in the variable t = (n²p²/Z² − 1)/(n²p²/Z² + 1), which runs over the open interval from minus one to one as p runs over the positive half-line. The prefactor and the denominator never vanish, so again the nodes are the polynomial’s zeros and again there are n − l − 1 of them.

So the equality of the counts is an identity rather than a coincidence. The substitution carrying one picture to the other is a rational map of degree one in $p^2$, so it carries a polynomial of degree $m$ to a polynomial of degree $m$, and a bijection of the open interval onto the positive half-line carries each zero to exactly one zero. Nothing about the physics enters; it is the arithmetic of the transform.

That is worth having because it makes the node count the one property of an orbital that reads the same in both pictures. Everything else swaps ends. A contracted orbital is small in position and wide in momentum; the nucleus is where the position function peaks and where the momentum function is flattest; a diffuse orbital has a long tail in one variable and a narrow one in the other. The count does not care.

And nothing pairs them #

Having found that both sets have the same size, the natural next move is to pair them off, and the natural guess is that the pairing is reciprocal — a node at a large radius answering a node at a small momentum.

It is not. The 4s orbital has position nodes at 1.872, 6.611 and 15.518 bohr and momentum nodes at 0.1036, 0.2500 and 0.6036 atomic units. Pairing innermost with outermost gives products of 1.130, 1.653 and 1.607; pairing innermost with innermost gives a different and equally unequal set. The 5s spreads from 1.144 to 2.082 under the better of the two orderings.

A rule would collapse one of those sets to one number on every orbital, and neither does on any orbital with more than one node.

The reason is visible in the two polynomials. Laguerre zeros and Gegenbauer zeros are the zeros of two different classical families, obeying two different recurrences with two different weight functions. There is no reason for the k-th zero of one to be related to any particular zero of the other, and there is nothing in the transform that would produce such a relation — the transform relates the functions, not their individual features.

What is exact is the product over all of them #

There is an exact relation and it is a statement about the whole set rather than about any node.

A polynomial’s roots multiply to the ratio of its constant term to its leading one, up to a sign. For the generalised Laguerre polynomial with parameter 2l + 1 and degree m, that ratio is (m + 2l + 1)!/(2l + 1)!, and the variable is 2Zr/n — so the product of the position nodes is (n/2Z)^m times (n + l)!/(2l + 1)!, using m + 2l + 1 = n + l.

For the momentum side the answer is shorter and stranger. Gegenbauer polynomials are even or odd, so their zeros come in pairs ±t and the multiset is symmetric about zero. That makes the product of (1 + t) and the product of (1 − t) identical, so the square root in the mapping contributes exactly one and the product of the momentum nodes is simply (Z/n)^m. Every factor of the radial structure cancels out of the momentum product and only the scale is left.

Multiplying the two, every power of Z and every power of n cancels, and what is left is (n + l)! divided by (2l + 1)! and by 2^(n − l − 1). For the 2s it is 1, for the 3s three halves, for the 3p two, for the 4s three, for the 4p five, for the 5p fifteen. A pure function of the quantum numbers, agreeing with the computed products to the twelve digits the arithmetic carries.

The nuclear charge is the surprising part #

Every individual node depends on Z and depends on it strongly. Raise the nuclear charge from one to ten and each position node contracts by a factor of ten while each momentum node expands by the same factor — the orbital is pulled in and its momentum distribution is pushed out, which is the whole content of what an effective charge does to an orbital and is visible in the radial distribution across the table.

The product does not move at all. It cannot: the position product carries Z^(−m) and the momentum product carries Z^(+m), and m is the same number because the two polynomials have the same degree — which is the identity the first section established, doing work in a second place.

So the count being equal is what makes the product be charge-free. Two facts that look like separate curiosities are the same fact, and the second is the reason the first is worth more than a tally.

The orbital this matters for is not hydrogen’s. A 4s electron in potassium does not see a charge of nineteen and does not see one either; Slater’s rules give it about 2.2, and that number is a fit rather than a derivation. The product identity is indifferent to which value is right, which makes it one of very few exact statements available about an orbital whose effective charge is a guess.

The one-node case, and why it is the wrong case to argue from #

Three of the nine orbitals here have exactly one node in each picture, and for them the identity reads as a pairing rule.

The 2s has a position node at 2.000 bohr and a momentum node at 0.500, and their product is exactly 1. The 3p has 6.000 and 0.3333, product exactly 2. The 4d has 12.000 and 0.2500, product exactly 3. In every case the answer is l + 1, and it follows from the identity in a line: with m = 1 the closed form is (n+l)!/((2l+1)! · 2), and n + l = 2l + 2 when m = 1, so the factorial ratio is (2l+2)(2l+1)!/((2l+1)!·2) = l + 1.

Which is exactly why an argument from these three would have been worthless. With one node of each kind there is one pair to form, so the product of the paired nodes is constant across the orbital and the product of all the nodes is a function of the quantum numbers are the same sentence — and the first is the claim the previous section refuted. The three nice cases cannot tell the true statement from the false one, and they are the cases whose numbers are nicest to quote.

That is the reason six of the nine orbitals in the test have two or more nodes, and the reason a check confirms that they do. A test set chosen for the elegance of its answers is a test set chosen to agree with the simplest hypothesis about them.

What a measurement would show #

The momentum picture has one place where it is not an alternative way of writing something: a Compton experiment measures the distribution of one component of the electron’s momentum, and getting that from a position-space wavefunction means transforming first.

So the momentum nodes are in principle observable — and what is observable is not a node.

A Compton profile is one half of the integral of the momentum density times p, taken from the measured momentum q up to infinity — an integral of a squared function over everything above the momentum being measured. It is strictly positive: the integrand is a square and the range is not empty. So the profile never vanishes anywhere, and a radial node in the momentum wavefunction is not a zero of the thing an experiment returns.

What it is is a horizontal tangent. Differentiating under the integral gives minus one half of the momentum density at q times q, which vanishes exactly where the momentum wavefunction does. One momentum node, one stationary point in the profile; n − l − 1 of them in general. On the 2s the tangent sits at q = 0.500, where the profile is still at 24 per cent of its value at zero — a flat place in a curve, not a dip to nothing.

That is the sharpest available statement of what this picture can and cannot show. The node is there, it is at a computable place, and the measurement renders it as an absence of slope rather than an absence of density.

What was computed, and how #

Both node sets are zeros of polynomials, found by scanning for sign changes and bisecting on the polynomial itself rather than on the wavefunction. That distinction is the reason the numbers are exact to twelve digits: a scan over a quadrature of the momentum function finds its nodes to three or four figures, which is the resolution of the scan, and the polynomial has no such limit.

The grid used for the scan has an odd number of points, and that is a repair rather than a detail. A Gegenbauer polynomial of odd degree has a zero at t = 0 exactly, and an even grid over a symmetric interval puts a sample exactly on it — where the product of neighbouring values is zero rather than negative, so a sign-change scan steps over the one root it should be surest of. The first run reported the 2s and the 3p as having no momentum node at all.

Fourteen things are checked. Per orbital: that the two counts are both n − l − 1, and that the product of all the nodes equals the closed form. Then the closed forms against the numerical transform on the 3s, both at its momentum nodes and at its position nodes, since a closed form that agreed with itself and not with the function would be a closed form for something else. Then the charge sweep: that the product is the same at Z = 1 and Z = 10 while every position node has moved in by ten and every momentum node out by ten. Then that the test contains several orbitals with more than one node, and that on each of them neither pairing gives a constant. Then the one-node orbitals, where the identity collapses to r · p = l + 1 exactly. And finally the profile: that it is positive throughout, that it has exactly one stationary point for the 2s, and that the stationary point is at the momentum the closed form predicts.

The refusal is the one-node case, and it is the case worth explaining. For an orbital with a single node of each kind, a pairing rule and the product identity are the same statement — there is only one pair to form. So an orbital with two or more nodes has to be in the test, or the general claim rests entirely on the case that cannot distinguish it from the claim it refutes. Six of the nine orbitals here have two or more.

Where this stops #

Everything above is hydrogenic and exact, which means it is about a model. A real many-electron orbital is not a Laguerre polynomial times an exponential, its radial function comes from a self-consistent calculation, and its nodes are wherever that calculation puts them. The count usually survives — a 4s orbital in an atom still has three radial nodes — but the product identity has no reason to, because it is derived from the polynomial’s coefficients and a numerical orbital has no polynomial.

The transform is exact for a hydrogenic function and a numerical one has to be transformed numerically. The transform used here is a spherical Bessel quadrature, which reproduces the 1s closed form to a part in a million and is the thing the nodes above were checked against. On a numerical orbital it would be the only route, and the nodes would come out to the accuracy of the quadrature rather than to twelve digits.

And the Compton statement is about a one-electron orbital, not about an atom. A measured profile is a sum over every occupied orbital, and a stationary point from one of them sits on top of a strictly decreasing contribution from the rest. Whether a shell’s momentum node survives as a visible feature of the total is a question about weights, and this argument has not asked it.

The generalisation #

The habit is to ask, of any exact count, whether it is exact in more than one representation — and if it is, to look for what the two representations’ versions multiply to.

A count that survives a transform is usually a count of something topological, and the quantity it belongs to is usually not the count itself but something the transform preserves. Here the count survives because the two polynomials have equal degree, and what that equality buys is the cancellation of Z between two products with opposite scaling. The identity was not looked for; it fell out of asking what the count’s survival was a consequence of.

The corollary is about what a measurement does to a zero. A wavefunction’s node is a zero of an amplitude, and almost every measurement returns a density or an integral of one — so the node’s signature in the measurement is generically a stationary point rather than a vanishing. That is a general fact about squaring and integrating and it applies wherever a feature of an amplitude is being looked for in data. Looking for the zero and failing to find it is available, and the flat place is where the zero went.

Who found it, and when #

The hydrogenic momentum wavefunctions are Podolsky and Pauling, 1929, and the Gegenbauer form is theirs; Fock’s 1935 four-dimensional argument is why the variable t is the natural one. The Compton profile’s relation to the momentum density is DuMond, 1929. The node counts and the polynomials are textbook. What is computed here is the product identity, its independence of Z, the refusal of every one-to-one pairing, and the stationary point of a profile at a node.

The number worth carrying is not three. It is that the 4s’s six nodes move by a factor of ten each across the first ten elements and the product of all six does not move at all.

Still open: whether the identity survives a real atom #

The obvious open question is the many-electron case. The product identity rests on two polynomials’ coefficients, and a Hartree–Fock 4s orbital has no polynomial — but it does have nodes, in both variables, and their product is a number somebody could compute. Whether it lands near (n+l)!/((2l+1)! 2^(n−l−1)) or nowhere near it would say whether the identity is a fact about the Coulomb potential or a fact about hydrogenic functions specifically, and those are very different results. The transform already in hand would do it; what it needs is a radial function that is not hydrogenic, which is a different calculation from any of the nine here.

The nearer question is the screened case, which is cheaper and sharper. A hydrogenic function with an effective charge is still hydrogenic, so the identity holds by construction and says nothing — but the whole point of an effective charge is that different shells of the same atom get different ones, and Slater’s rules give the 2s and the 2p the same value when the two are separated by several electronvolts. The product identity is one of the few exact statements about an orbital that does not depend on which charge is right, so a measured product, from a profile with enough resolution to locate a stationary point, would be a constraint on n and l that no choice of screening constant could adjust. Whether any real spectrum has that resolution is the question, and it is one for an experiment rather than for this arithmetic.