The orbital in momentum space

A wavefunction is a function of position, and every picture in this collection so far has drawn it as one. That is a choice of variable rather than a fact about the electron. The same state is described exactly as completely by its Fourier transform — a function of momentum — and the transform loses nothing, adds nothing, and is invertible.

What changes is which statements are obvious. In position the kinetic energy is a curvature, buried inside a second derivative; in momentum it is a width, and the mean square momentum is a moment of the distribution the way the mean square radius is. Facts that require an integration by parts in one picture are read off in the other.

The transform, and what it does to the angular part #

A function with a definite angular momentum transforms in a way that keeps its shape. Writing $\psi_{nlm}(\mathbf{r}) = R_{nl}(r)\,Y_{lm}(\hat{r})$, the transform is

$$\phi_{nlm}(\mathbf{p}) = \phi_{nl}(p)\,Y_{lm}(\hat{p}), \qquad \phi_{nl}(p) = \sqrt{\tfrac{2}{\pi}} \int_0^\infty R_{nl}(r)\, j_l(pr)\, r^2 \, dr,$$

where $j_l$ is a spherical Bessel function. The angular part is untouched: a p orbital in position is a p orbital in momentum, pointing the same way. Only the radial function changes, and it changes through a kernel that carries the angular momentum as its order.

That last clause is doing real work, and it is where the check on the whole calculation lives. Hand the same radial function to the wrong kernel — $j_0$ for a 2p — and the result is still normalised, because a Hankel transform is unitary at every order and the norm cannot see which one was used. What it comes back short by is exactly the centrifugal energy: $\langle p^2 \rangle$ falls from 0.250 to 0.083, and the difference is $2\langle 1/r^2 \rangle = 1/6$ for a hydrogenic 2p. The order of the kernel is carrying a stated quantity, and the check is that number rather than a plausibility.

One case in closed form, and the reason to want it #

The 1s transform can be done by hand. With $R_{1s} = 2Z^{3/2}e^{-Zr}$ and $j_0(x) = \sin x / x$, the integral is elementary and gives

$$\phi_{1s}(p) = 4\sqrt{\tfrac{2}{\pi}}\, \frac{Z^{5/2}}{(Z^2+p^2)^2}.$$

A Lorentzian squared — which is not the transform of an exponential that a first guess supplies, because the three-dimensional transform of $e^{-Zr}$ carries the $r^2$ of the volume element with it.

The numerical transform agrees with that expression to better than a part in a million at every momentum tested. This is the same discipline the overlap integrals get: one case with an exact answer turns a quadrature from something to be trusted into something that has been tested, and the rest of the results are then worth having.

The shape says something immediately. A Lorentzian squared falls off as $p^{-4}$, so it has a long tail — the momentum distribution of a 1s has weight far out, and the reason is the cusp at the nucleus. A corner in position is a slow decay in momentum, exactly as a discontinuity in a signal is a slow decay in its spectrum. That connection is the one this picture is for.

The kinetic energy, twice #

For any hydrogenic orbital the virial theorem fixes $\langle T \rangle = -E = Z^2/2n^2$ without any integration at all, so

$$\langle p^2 \rangle = 2\langle T \rangle = \frac{Z^2}{n^2}.$$

It depends only on $n$: a 2s and a 2p have the same mean square momentum, and a 3p and a 3d have the same. Whatever the angular momentum does to the shape of the orbital, it does not change how fast the electron is going on average.

Computed by transforming and integrating $|\phi|^2 p^4$, the values come out at 0.99995, 0.24999, 0.25000, 0.11111 and 0.11111 for the 1s, 2s, 2p, 3p and 3d — against exactly 1, 1/4, 1/4, 1/9 and 1/9. One route is a quadrature over an oscillating integrand on a grid; the other is a line of algebra. They have nothing in common but the orbital.

The uncertainty product, and the function that attains it #

Multiply the two moments together and the result is dimensionless, comparable between orbitals, and bounded below. The Heisenberg relation in the form appropriate to three dimensions gives

$$\langle r^2 \rangle \langle p^2 \rangle \geq \tfrac{9}{4}.$$

The hydrogenic values are startlingly clean:

Every one is a whole number or a half. None attains the bound. The ground state is closest, at three — a third again above the floor — which is what a ground state should be, and the ordering rewards compactness at fixed $n$: the 2p beats the 2s and the 3d beats the 3p, because a node in the radial function pushes density outward without buying any momentum back.

A normalised Gaussian $e^{-\alpha r^2}$ has $\langle r^2 \rangle = 3/4\alpha$ and $\langle p^2 \rangle = 3\alpha$, so its product is $9/4$ for every $\alpha$. Exactly the bound, everywhere.

That is worth sitting with. A Gaussian is the wrong shape for an orbital at both ends of the range — no cusp, too fast a decay — and it is the shape the entire subject uses, for reasons of integral arithmetic. It is also, uniquely, the shape that saturates the uncertainty relation. The property that makes it optimal is not the property anybody chose it for, and it is not a property that helps: being minimum-uncertainty is exactly what makes a Gaussian unable to be simultaneously peaked at the nucleus and slow to die away, which is what an orbital needs to be.

The cusp and the tail are the same statement #

The two things a Gaussian gets wrong about an orbital look like two separate defects in position — a missing corner at the nucleus, and a decay that is too fast far out. In momentum they are one defect, and seeing that is the clearest thing this change of variable does.

A function’s behaviour at large momentum is controlled by its least smooth feature. An exponential has a corner at the origin, so its transform decays as a power — $p^{-4}$ for the 1s. A Gaussian is infinitely smooth everywhere, so its transform is another Gaussian and decays faster than any power. The missing cusp in position is the missing tail in momentum; they are not two failures but one, viewed from two sides.

And the consequence is quantitative rather than aesthetic. The kinetic energy is $\tfrac{1}{2}\langle p^2 \rangle$, so it is an integral of the momentum distribution weighted by $p^4$ — dominated by exactly the far tail a Gaussian basis does not have. A fitted function reproduces the kinetic energy anyway, because the variational principle drives the total energy down and the exponents adjust until the two errors balance. What comes out right is a sum; what does not is either term separately.

This is the sharp form of a statement made loosely elsewhere: an energy can be excellent while the function is wrong, and here it is visible as a cancellation rather than as a coincidence.

The atom, rather than the orbital #

A single orbital in momentum has the same angular shape it has in position, which means a 2p momentum distribution is a dumbbell pointing along the same axis. That raises the question a filled shell always raises, and it has the same answer.

Summing $|Y_{lm}|^2$ over a complete shell gives a constant, in either variable, so a closed shell has an isotropic momentum distribution. A filled shell has no shape is therefore a statement about the momentum density as well, and it is the reason a Compton profile of a closed-shell atom is a single curve with no orientation dependence — which is what makes the measurement usable on a gas or a powder at all.

An open shell is a different matter, and it is where the technique earns its keep experimentally: a directional Compton profile of an oriented crystal does depend on direction, and the dependence is a picture of which orbitals are occupied. That is a measurement of an anisotropy in the wavefunction, taken directly, and there are not many of those.

What a virial check is worth #

The agreement between the transformed $\langle p^2 \rangle$ and $2\langle T \rangle$ deserves more than a line, because it is a particular kind of check and a powerful one.

The two routes share nothing. One integrates an oscillating Bessel kernel against a radial function over sixty bohr, then integrates the result against $p^4$ over forty atomic units of momentum, on grids chosen by hand. The other is the statement that for a potential going as $1/r$ the kinetic and total energies are related by a factor of $-1$, which follows from a scaling argument and involves no integration whatever. A quadrature error would break the agreement; an error in the kernel would break it; a wrong normalisation would break it. Nothing plausible leaves it standing at three parts in a thousand for five different orbitals.

This is worth contrasting with a check that would have looked similar and proved nothing. Requiring the momentum function to be normalised is such a check: it passes for the wrong kernel, as the refusal above records, because unitarity does not care which order was used. The distinction between the two is the whole difference between a test and a formality, and it is not always obvious in advance which one is being written.

What a measurement returns #

The momentum picture has one place where it is not an alternative way of writing something down. A Compton scattering experiment measures the distribution of one component of the electron momentum, through the Doppler broadening of the scattered photon’s energy, and what it returns is

$$J(q) = \tfrac{1}{2}\int_{|q|}^{\infty} |\phi(p)|^2\, p \, dp,$$

which integrates to one over the whole line. For the hydrogenic 1s this has the closed form $8Z^5/3\pi(Z^2+q^2)^3$, and $J(0) = 8/3\pi = 0.8488$.

This is the reason the momentum picture is not merely elegant. There exists an experiment whose direct output is $|\phi(p)|^2$, integrated in a stated way. It is a much cruder probe than a spectrum — it gives a single smooth curve rather than a set of lines — but what it constrains is the wavefunction rather than an energy difference, and there are very few measurements of which that is true. A photoelectron spectrum measures energies and is read for what it says about orbitals only through a model; a Compton profile is a moment of the density, in momentum, with no model between.

Where the two pictures disagree about “size” #

The word size has been used in this collection to mean the radius enclosing a stated fraction of the density, and there are several other things it could mean. The momentum picture adds one more, and it runs backwards.

A 1s is the smallest orbital by every position-space measure and the largest by every momentum-space one. That is not a paradox and it is not a curiosity: it is the content of the uncertainty relation, which says the two cannot both be made small and does not say they are the same quantity read twice. Any statement of the form this orbital is bigger than that one is a statement about a chosen measure in a chosen variable, and it has now been shown to be ambiguous in two independent ways — which measure, and which variable.

The one quantity that is easier in momentum space #

Nothing is added by changing variable, which is true and leaves open the question of what the second picture is for. There is an answer, and it is a quantity usually computed in the other variable without noticing how awkward it is there.

The kinetic energy is a second derivative in position space — an operator that has to be applied to the wavefunction, integrated, and watched for cancellations. In momentum space it is $p^2/2m$: a number times the momentum density, averaged. So the kinetic energy is a width of the momentum distribution, and nothing else.

That makes the uncertainty relation do chemical work. Squeeze a function in position and its momentum distribution must broaden, so its kinetic energy must rise. Every statement about an orbital contracting is therefore also a statement about its kinetic energy going up, and the two cannot be separated.

Which produces a result that is exact and unwelcome. The virial theorem says that at an equilibrium geometry the total energy equals minus the kinetic energy, so a bond that lowers the total energy by some amount raises the kinetic energy by exactly that amount — and lowers the potential energy by twice it.

Forming a bond increases the kinetic energy of the electrons. Not as a correction: by precisely the binding energy.

And a variational calculation measures the mechanism. Letting a hydrogen molecule ion choose the size of its atomic functions gives an exponent of 1.238 rather than 1 — the orbital contracts by a quarter when the bond forms. A contracted orbital is a broadened momentum distribution, which is a raised kinetic energy, which is what the virial theorem requires.

So the two results are one statement in two variables. In position space a bond forms by the orbitals shrinking; in momentum space it forms by their momentum distributions spreading; and the virial theorem fixes the exchange rate between the two at exactly one to one against the binding energy.

It also disposes of a picture that is drawn constantly and is wrong in a specific way. Bonding is often described as electrons spreading out over two atoms and thereby lowering their kinetic energy, on the particle-in-a-box reasoning that a larger box has lower levels. The spreading is real — the orbital does cover both nuclei — and the kinetic energy goes the other way, because the contraction perpendicular to the bond and towards each nucleus outweighs the extension along it.

Which of the two effects wins is not something a picture can settle, and the virial theorem settles it without any picture at all: at equilibrium the sign is fixed, the magnitude is fixed, and both are fixed by the binding energy alone. That is a rare thing in this subject — a qualitative question with a quantitative answer that needs no calculation of the molecule.

What this cannot say #

The usual limits, and one that is particular to this picture.

One electron. Every function transformed here is hydrogenic. A many-electron momentum density is not the sum of one-electron ones any more than a many-electron position density is, and the orbital approximation is doing the same work here as everywhere else.

No relativity. The momentum distribution has a long tail, and it is precisely the tail that relativistic corrections act on — the region where the electron’s speed is a significant fraction of the speed of light. For hydrogen this is negligible; for a heavy element it is the reason the 6s orbital contracts and gold is the colour it is. The transform above is exact for the equation it was given and that equation is not the right one for a heavy atom.

The nodes do not survive as nodes. A momentum function has nodes too, but not at the transformed positions of the position-space nodes: the transform mixes the whole function. What a radial node does to a picture is a statement about one variable, and asking where the 2s node “goes” in momentum has no answer of that form.

Still open: momentum space and a lattice #

The transform gives a way to ask what a basis set does that the energy cannot see. A contraction freezes the outer shape of a fitted function, and the outer shape in position is the near origin in momentum — the region a Compton profile measures most accurately. So a basis that gives an excellent energy and a poor tail should give a visibly poor Compton profile, and that is a check on a basis set which uses a measurement rather than a comparison against a better calculation.

More immediately, the momentum picture is the natural home for a question usually asked in position and answered awkwardly: what an electron in a solid is doing. A band is a function of wavevector, which is momentum, and the extended structures here have been built as finite matrices with no periodicity assumed, precisely to avoid needing that variable. The transform above is the bridge between the two descriptions, and building it properly means committing to a lattice — which is a different subject with a different set of rules, and one deliberately left aside here.