The symmetry that is not a rotation

Hydrogen’s shells have a degeneracy that no group of rotations predicts, and the usual account of it stops short. The classical demonstration is easy — an orbit closes for an inverse-square force and precesses for anything else — and the quantum object behind it is harder: an operator that commutes with the Hamiltonian and connects orbitals of different l.

It is constructed here, and the material it is built from is ordinary dipole integrals between hydrogenic functions.

What has to be produced #

A symmetry of a quantum system is an operator that commutes with the Hamiltonian. Rotations supply three of them — the components of the angular momentum — and they explain a threefold degeneracy among the p orbitals and say nothing whatever about why the s orbital of the same shell is at the same energy.

So a fourth operator is needed, and it has to do something the angular momentum cannot: connect states of different l. Angular momentum cannot, by construction. Every one of its components leaves l alone, which is precisely why the degeneracy it explains is the degeneracy within one l.

The classical statement of what that operator should be is old.

The vector that points along the major axis of a closed orbit is conserved for an inverse-square force and for no other. It carries several names — Laplace’s, Runge’s, Lenz’s — and its quantum form is

$$\mathbf{M} = \tfrac{1}{2}(\mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - \hat{\mathbf{r}}$$

which needs momentum operators. The construction here avoids them. What it uses is every matrix element of position between hydrogenic functions, by quadrature, to about a part in a hundred million.

That turns out to be enough.

The projection that makes it computable #

Inside one degenerate shell every vector operator is proportional to every other. This is a standard theorem about vector operators and it is worth stating plainly, because it is what converts an impossible calculation into an easy one: within a fixed n, the conserved vector cannot do anything the position operator does not already do, and can only differ from it by a number.

The number is fixed by requiring the algebra to close, and it is

$$A_i = \frac{2}{3n}\, x_i \qquad \text{(inside the } n \text{ shell)}$$

where $\mathbf{A} = n\mathbf{M}$ is the conventionally scaled version. That is written down rather than derived — and then checked in three independent ways, each of which could have failed.

The single non-zero element is $\langle 2s | z | 2p_z \rangle$, which comes out at −3.000000 against a closed form of exactly −3. Everything below is built from that number and its two partners in x and y.

The three identities, as residuals #

The rotation generators in this basis are written down: a rotation about z carries the p orbital along x into the one along y and leaves the s orbital alone. Real spherical harmonics make them real antisymmetric matrices, $G = -i\mathbf{L}$, and the identities to be satisfied are

$$[G_i, G_j] = \varepsilon_{ijk} G_k, \qquad [G_i, A_j] = \varepsilon_{ijk} A_k, \qquad [A_i, A_j] = -\varepsilon_{ijk} G_k$$

The first says the rotations close. The second says the constructed object is a vector under them. The third is the one that matters: the commutator of two components of A is not a new kind of object but a rotation, so the six operators together close into an algebra of their own.

The algebra those six operators close into is the algebra of rotations in four dimensions, not three. That is the whole answer to the question of the accidental degeneracy: the symmetry hydrogen has is a rotational symmetry, of a space the atom does not sit in.

And the size of the object it makes is the last line of the figure. The combination $A^2 + L^2$ comes out at 3.0000000 times the identity, against $n^2 - 1 = 3$ — which says that all four states of the shell are one irreducible object of the larger group, of dimension $n^2$. A fourfold degeneracy from a group whose irreducible representations have dimensions 1, 3, 5 requires a coincidence; a fourfold degeneracy from a group whose representation here is four-dimensional requires nothing at all.

The same matrix is an observable #

None of the above would be worth much if it were only bookkeeping. It is not: the matrix whose eigenvalues were just taken is the matrix a uniform electric field adds to the Hamiltonian, so its eigenvalues are a measurement.

A field along z adds $Fz$ to the energy. Inside a degenerate shell, first-order perturbation theory says the shifts are the eigenvalues of that operator restricted to the shell — and those are the numbers above.

The n = 2 shell splits into three levels, at $-3eFa_0$, zero twice, and $+3eFa_0$. The shift is proportional to the field, not to its square, which is what makes hydrogen’s Stark effect different in kind from every other atom’s.

The eigenvectors are worth as much as the eigenvalues. Each extreme state of the n = 2 shell is an equal mixture of 2s and 2pz — which is to say, an sp hybrid — and it carries a computed dipole of $3ea_0$, or 7.6 debye. That is larger than the dipole of any molecule in this collection, and it belongs to a single atom of hydrogen.

The two mixtures are easiest to picture as directions. An sp hybrid pair points in opposite directions, and that is exactly what the two extreme Stark states are: one with its density pushed along the field, one against it. The states labelled by l have no dipole at all, and they are not the states a field picks out.

The two states the field does not move #

Two of the four levels sit at zero, and they are the p orbitals perpendicular to the field. They are worth a paragraph because their being unmoved is not an approximation.

A field along z can only connect states whose angular parts differ in the way z does — one unit of angular momentum, no change in the component along the field. The 2px and 2py have a component along z of ±1 in the complex labelling, and there is no state in the shell for them to be connected to: the 2s has none of the right kind and the 2pz is the one the field has already used. So their rows of the matrix are empty, and empty rows give eigenvalue zero exactly rather than approximately.

That produces an observable feature. A hydrogen line split by a field shows a central unshifted component flanked by shifted ones, and the central component is these states. Counting the components of a Stark pattern is therefore a count of which pairs of parabolic states a transition can connect, which is a selection rule in a basis that is not the usual one — and the reason the pattern was solvable in 1916 with no matrix mechanics available is that the parabolic separation hands the same answer without any of this.

Why the same problem separates two ways #

There is a fact about the hydrogen atom that is usually mentioned as a curiosity and is really the same statement as everything above: the Schrödinger equation for a 1/r potential separates in two coordinate systems, spherical and parabolic, and no other central potential separates in more than one.

A separation constant is a conserved quantity. Spherical coordinates produce the square of the angular momentum and its component along an axis; parabolic coordinates produce the component of the angular momentum along the axis and — the constant that appears in no other problem — the component of the conserved vector along it. Two separations, two sets of labels, one set of states.

The eigenvectors computed above are the transformation between them, written out. Each is $(2s \pm 2p_z)/\sqrt{2}$, which is a state with a definite value of the conserved vector’s z component and no definite l at all. Going the other way, a 2s orbital is an equal mixture of the two states a field would produce, which is why it does not sit still when one is switched on.

So the question which are the real orbitals of a hydrogen shell has the same answer as the question about hybrids: both bases describe the same four-dimensional space, and which one is natural depends on what has been done to the atom. In a field, the parabolic ones are the states with definite energies; with no field, every basis is as good as every other, and the l-labelled one is chosen for convenience rather than because the atom prefers it.

Why no other atom does this #

Every other atom’s shell is split before the field arrives, because its potential is not a 1/r potential — the electron is screened by the others, and screening acts hardest where the orbital penetrates closest, which depends on l.

The same two moments for the n = 3 shell say it once more and one subshell wider: ⟨1/r⟩ is identical across s, p and d, and ⟨1/r²⟩ runs five to one across them. So any correction to the potential falling off faster than 1/r splits the shell, and splits it in the order the periodic table shows — which is the whole of why the accidental degeneracy is a hydrogen fact rather than an atomic one.

Once the shell is split by an energy Δ, a field of strength F no longer acts within a degenerate space, and the two-state problem gives shifts of $\sqrt{(\Delta/2)^2 + (Fd)^2}$ — quadratic in the field while $Fd$ is small compared with Δ, and linear only when it is large.

So the linear Stark effect is hydrogen’s alone, and it is the symmetry showing itself. Stark found it in 1913 and it was one of the results that made the old quantum theory look temporarily complete; Schwarzschild and Epstein derived it that same decade from the parabolic separation of the same problem — which is the other face of the vector constructed above, since the parabolic quantum numbers are precisely the eigenvalues labelled k here.

What this construction cannot say #

Three limits, and the third is the one that stops the argument being about the real atom.

It is a projection, not a derivation. What is built above is the conserved vector restricted to one shell, identified by a theorem and then verified. Constructing the full operator would need momentum operators, which are not used here, and the commutator with the Hamiltonian — the property that makes it conserved at all — is not computed here. Inside a shell it is trivially true, because every operator commutes with a Hamiltonian that is a multiple of the identity there.

One electron. The whole argument is about a hydrogenic Hamiltonian. Add a second electron and the potential each one feels is not 1/r, the shell splits, and the symmetry goes — which is not a technicality but the reason the periodic table has the shape it does.

No relativity, and this is the sharp one. Real hydrogen’s 2s and 2p are not degenerate. The Lamb shift separates them by 1058 megahertz — an energy of 1.6 × 10⁻⁷ hartree, from an effect that has nothing to do with the potential’s shape — and the same two-state arithmetic applies to it.

That number is the honest statement of what the model is worth. The degeneracy this essay builds an operator for is exact for the Hamiltonian it is built from, approximate for the atom in the laboratory, and recovered in the laboratory by any field above about a hundred volts per centimetre.

The same vector, and the measurement that broke it #

The three matrices built here out of computed integrals have a classical ancestor, and the ancestor’s failure is one of the most famous measurements in physics — which makes the symmetry’s fragility observable on a scale of astronomical units as well as of ångström.

Classically the same conserved quantity is a vector pointing along the long axis of an orbit. It is conserved for a $1/r$ force and for nothing else, and what its conservation says is that the orbit’s orientation never changes: the ellipse closes, the point of closest approach returns to the same place every revolution, and the orbit is a fixed curve rather than a moving one.

Change the force away from $1/r$ by any amount, and the vector stops being conserved. It rotates slowly, the orbit’s long axis turns with it, and the point of closest approach advances a little every revolution.

That advance is measured. Mercury’s perihelion moves, and after every Newtonian effect of the other planets is subtracted, forty-three seconds of arc per century remain — which is the general-relativistic correction to the $1/r$ potential, a departure of about one part in ten million, breaking a conservation law that would otherwise hold exactly.

So the same symmetry that gives hydrogen four states at one energy gives a planet an orbit that closes, and both are broken by the same kind of thing: a potential that is not quite $1/r$.

The two breakings even scale alike. In the atom the departure comes from the other electrons screening the nucleus, the degeneracy lifts, and the size of the lifting is the quantum defect. In the solar system the departure comes from relativity, the orbit precesses, and the size of the precession is the correction to the force. Neither is a small correction to a symmetric answer; both are the appearance of a term that the symmetric answer had no room for at all.

That is worth carrying because it says what kind of object the extra symmetry is. It is not a property of atoms, or of quantum mechanics, or of electrons. It is a property of the inverse-square law, and it holds wherever that law holds exactly and nowhere else — in a hydrogen atom, in a two-body orbit, and in no chemistry beyond the first.

The quantum and classical versions differ in one respect worth naming, because it explains why the quantum case needed matrices at all. Classically the vector points somewhere, and its three components are three numbers that can be written down. In the quantum problem the components do not commute with one another — their commutators close back into the rotations, which is what the construction here checks — so there is no state in which all three have definite values, and the vector does not point anywhere.

What survives is the algebra rather than the direction. The three operators together with the three rotations form a closed set of six, and it is the closure that produces the degeneracy: a set of six operators acting within one shell, with the shell as one irreducible object rather than as four separate orbitals that happen to coincide.

So the quantum statement is stronger than the classical one and less picturable. The four states of the second shell are not four things at one energy; they are one thing with four components, and the operators that connect them are the reason the coincidence is exact rather than approximate.

That also settles what the Casimir coming out at exactly $n^2 - 1$ is doing. It is the label of the object — the quantity that says which irreducible set of the enlarged algebra a shell is — and its being an exact integer for every $n$ is the statement that the shell really is one object, computed here from integrals rather than assumed from a table.

What was computed, and what was assumed #

Assumed: the proportionality between the conserved vector and the position operator inside a shell, and the form of the rotation generators in a real basis.

Computed: every matrix element of x, y and z within the shell, by the mapped quadrature rule that integrates over the whole real line; the products and commutators of those matrices; their eigenvalues and eigenvectors; and the radial moments that say what a non-Coulomb correction would do.

Checked: the two non-zero elements against closed forms of −3, −3√6 and −3√3; the eight vanishing elements against zero, at 10⁻¹⁸; three commutator identities; the vanishing of A·L; the Casimir against $n^2 - 1$; the Stark eigenvalues against whole numbers times $(3/2)n$; and the sum of the shifts against zero, since a field spreads a shell without moving its centre.

The last of those is the cheapest check and the one most likely to catch an error of sign, because it fails if any single element is entered with the wrong one.

Still open: beyond one shell, and a slightly broken symmetry #

The construction stops at a shell, and the interesting questions are all outside one.

The operator connects states of different l within a shell. The full one also connects different n, which is what makes it a spectrum-generating algebra rather than a symmetry — the whole hydrogen spectrum as one object rather than a tower of shells. Nothing here reaches that, and reaching it needs the momentum operators.

The nearer question is what happens to this symmetry when it is broken slightly rather than badly. A screened potential splits a shell by an amount that can be computed from ⟨1/r²⟩, and the field at which the linear effect returns follows from the splitting by the two-state arithmetic above. That gives a curve of crossover field against screening, which would say — for a real atom rather than for hydrogen — how nearly the symmetry survives. It is one figure and it is not drawn here.