The degeneracy time reversal forces

A coincidence the group did not ask for sorts every degeneracy in this collection’s band models into three bins — forced by the unitary operations, forced by an antiunitary one, or accidental — and then admits that three rows do not fit in any of them. They are zone-boundary wavevectors of pg and pgg, the levels there double robustly, the characters of the little group do not require it, and the essay names what is missing: Herring’s criterion, the bookkeeping for antiunitary operations, which the collection had not built. Six further rows are worse off still, marked only projective, because the factor system there cannot be rephased away and the ordinary character machinery does not apply at all.

Nine open rows. This essay closes them, and the whole of the argument is one exact number.

Time reversal, in a model with no spin #

A Hamiltonian whose hoppings are real numbers is unchanged by complex conjugation in the site basis. That is not a symmetry of the crystal in any sense a group of motions recognises — conjugation is antiunitary, it takes a scalar to its conjugate rather than leaving it alone, and no set of matrices contains it.

It is nevertheless a symmetry of the operator, and its consequence for a Bloch state is immediate: conjugating a state at wavevector k gives a state at −k. So on its own, conjugation is a symmetry of a wavevector only where −k and k are the same modulo the reciprocal lattice — the centre of the zone and its three half-lattice points.

Composed with a crystal operation it does better. If some element a of the group has a rotation part carrying k to −k, then

Θ = D(a) · K

is an antiunitary operator that fixes k, where D(a) is a’s Bloch operator and K is conjugation. Every element with that property gives one, and the elements with it form a coset of the little group — the antiunitary coset. It is empty at the corner of p3’s zone, where −K is the other corner and nothing in a three-fold rotation reaches it, and it is not empty at the corner of p6’s, where the six-fold contains a half turn. The two groups have the same little group there and behave completely differently, and that difference is not in any character table.

That is worth dwelling on, because it is the one place where a question about an operator turns into a question about the star of a wavevector. Time reversal relates k to −k always. Whether it says anything at k depends on whether the crystal can bring −k back, and that is orbit arithmetic: −k has to lie in the star. Nineteen of the hundred and two rows in the census have no antiunitary operator at all for that reason, and every one of them is a group without a half-turn at a wavevector that is not its own negative.

The same accounting appears in a completely different corner of this subject as Friedel’s law — a diffraction pattern is centrosymmetric whether or not the crystal is, because scattering is time-reversal symmetric. It is the same operation making the same identification of k with −k, and it is the reason the two subjects keep producing the same sentence with different nouns in it.

Θ² is exact, and it does not depend on the gauge #

Here is the number the essay rests on.

Θ² = D(a)·conj(D(a)), which is an ordinary linear operator rather than an antiunitary one. Two things are true of it and both are load-bearing.

It is gauge-free. Multiplying D(a) by a phase multiplies Θ² by that phase times its own conjugate, which is one. So the ambiguity that makes projective representations awkward to compute with — the phases of the operators are conventions, and the factor system moves when they move — simply does not touch Θ². That is not a workaround; the quantity is invariant.

It is exact. The Bloch operators here are monomial matrices whose entries are twelfth roots of unity: one non-zero entry per row, and its value an exponent modulo twelve. Conjugation negates an exponent. The product of two monomial matrices is monomial. So “Θ² = −1” is the entirely finite statement that the permutation is the identity and every exponent is six, decided in integers with no tolerance anywhere.

Why it is −1 there is one line, and it is the line a glide sticks two levels together already computes. A glide applied twice is a lattice translation. A lattice translation acts on a Bloch state as the phase of that translation. At the edge of the zone that phase is −1.

Kramers’ theorem, arriving without spin #

An antiunitary operator that commutes with a Hamiltonian and squares to −1 forces every eigenvalue to be at least doubly degenerate. The proof is three lines: if Θ² = −1 then a state and its image under Θ are orthogonal, because their inner product equals minus itself; they have the same energy, because Θ commutes with H; so every level is at least a pair.

That is Kramers’ theorem, and it is usually introduced as a statement about half-integer spin, where Θ² = −1 because rotating a spinor through 2π returns minus the spinor. There is no spin in any model in this collection. The −1 comes from the lattice instead, through a glide whose square is a half translation and a wavevector at which a half translation is a sign.

The consequence is a doubling that is real, robust, and completely invisible to the little group’s character table — because the character table describes unitary operations and the operation responsible is not one.

Whether Θ actually commutes with the Hamiltonian is not assumed. The entries of H(k) are cyclotomic integers, conjugation negates their exponents, and the comparison is entry by entry with no tolerance: eighty-three wavevectors carry an antiunitary operator and every one of them commutes.

It is worth pausing on how little the group has to be doing for this. The little group at pg’s zone boundary has order two: the identity and a glide. A group of order two has two one-dimensional representations and nothing else, so the largest degeneracy its characters permit is one. The measurement is a pair, twice over. There is no room whatever in the unitary account for what is observed, which is what makes the row an admission rather than an imprecision.

Three cases, and which of them was missing #

Herring’s criterion sorts a representation of the little group into three cases, and it is worth stating what each of them is before saying which the collection already had.

Case (a). The representation is real, and time reversal adds nothing. This is most of the census: sixty-nine rows.

Case (b). The representation is equivalent to its conjugate, but only through a matrix that cannot be made symmetric. The level doubles. In this subject case (b) is exactly Θ² = −1, and it is what was missing.

Case ©. The representation is not equivalent to its conjugate at all, so it and its conjugate become one level of twice the dimension. This the collection had: it is the Frobenius–Schur indicator being zero, and it is what doubles the levels of p3, p4 and p6 at the centre of their zones.

The three are not shades of one thing. Case © is a statement about two different representations being forced to the same energy — the levels were always going to be there, and time reversal decides they coincide. Case (b) is a statement about one representation being unable to occur singly at all. A level in case (b) has no one-dimensional version, in the same way that a two-dimensional representation has no one-dimensional version, except that the obstruction is antiunitary and therefore in no table.

Nine rows are case (b), and every one of them is a non-symmorphic group — pg, pmg, pgg, p4g — at a zone boundary. That is not a coincidence and it is checked as a refusal: a symmorphic group has no operation whose square is a half translation, so it has no route to a −1.

What the older test got wrong #

Building case (b) turned up an error in case ©, and it is the kind that only shows when a second computation is put beside the first.

The collection’s existing test for the conjugate pairing asks two things: is −k in the star of k, and is the indicator zero. Both are necessary. Together they are not sufficient, because time reversal does not simply conjugate a representation. Θ carries D to

D′(g) = conj( D(a⁻¹ g a) ),

and the relabelling by a can undo the conjugation. When it does, D′ is D again and nothing happens; when it does not, D′ is the conjugate representation and the two stick.

p3m1 at the corner of its zone is the case. The little group is a three-fold with two complex representations, −K is in the star, the indicator is zero — and the levels stay single, because the mirror carrying K to −K also exchanges the two rotations and the conjugation is undone. The older prediction says the levels pair; the spectrum says they do not; and the spectrum is right.

One row out of a hundred and two. It is the only place in the census where the two tests disagree, which is exactly why nothing found it before: a test that is wrong once in a hundred and two is a test that looks correct.

The control, and why it needs two halves #

A degeneracy attributed to time reversal has to be destroyed by breaking time reversal and by nothing else, and the perturbation that does it has to be chosen with some care.

Take an arbitrary Hermitian operator. Split it into a part that commutes with Θ and a part that anticommutes with it — every operator splits that way, and both halves are Hermitian. Then average each half over the little group, which lands both in the commutant: they commute with every unitary operation of the group, so neither can lift a degeneracy those operations force.

What is left is a clean pair of experiments. The even half preserves everything, so nothing should move. The odd half preserves the unitary symmetry and breaks time reversal, so anything time reversal was holding together should fall apart.

The even half lifts nothing at any of the fourteen wavevectors, which is what makes the odd half’s result mean anything. The odd half splits nine of them.

The five it does not split are the informative ones. At pmg’s two boundary points, two of pgg’s and one of p4g’s, the doublet survives a perturbation that breaks time reversal completely — so time reversal was not what was holding it. Those are exactly the wavevectors whose factor system cannot be rephased away, where the little group has a two-dimensional representation of its own and the levels were already required to be pairs. Kramers applies there too; it is simply adding nothing.

The row that makes the point sharpest is p4g at the corner, where the perturbation splits two of the four doublets and leaves two alone. One wavevector, two causes, separated by an experiment rather than by an argument about which explanation is prettier.

What the criterion refuses #

Two of those deserve naming. A case-(a) wavevector must have no time-reversal-odd invariant available — sixty-nine rows, and at every one of them the odd half of the perturbation comes out zero, which is a much stronger statement than “nothing was lifted”. There is nothing to lift with. And the new test must disagree with the old one somewhere, because a criterion that reproduces what was already computed is a longer way of computing it. It disagrees at p3m1’s corner and nowhere else, which is the least it could do and still be worth building.

Leaving the old verdict where it is #

One decision about the machinery is worth stating, because the tidier alternative would have been wrong.

The earlier census still returns unaccounted at those three rows and projective at the six. Nothing in this essay changes it. That census is the unitary instrument — it decomposes a Bloch character against a little group’s characters and reports where the decomposition fails — and its answer is still correct: the unitary operations do not account for the doubling. Rewriting its verdict to say antiunitary would make the earlier essay’s prose false, and would also merge two instruments that answer different questions into one that answers neither cleanly.

So the criterion is a second layer, run beside the first, and the interesting output is the pair of verdicts rather than either alone. A row where both say the same thing is a row nothing was learnt at; the nine rows where they differ are the content.

Where the exactness stops #

Computed here: the antiunitary coset at every special wavevector of all seventeen plane groups; the exact square of every operator in it as an exponent modulo twelve; an exact check that each commutes with the Hamiltonian; the action of conjugation on every irreducible representation of every little group; Herring’s three cases and the corrected degeneracies; and a two-part perturbation applied at every wavevector where time reversal does anything.

Two dimensions, and models rather than materials. Every group here is a plane group and every Hamiltonian is one orbital per site with weights chosen for convenience. What the model has to be is a legitimate operator with the right symmetry and real hoppings, and it is; what it is not is anybody’s material, and the level orderings it produces mean nothing at all. Only the degeneracies are being read.

A real Hamiltonian is a choice about the physics. Time reversal is a symmetry of this operator because its hoppings are real, which is the spinless, field-free case. A magnetic field, a spin–orbit term or a current-carrying ground state each break it, and everything on this page then goes away — which is the content of the control, run deliberately rather than suffered.

And spin would change the arithmetic, not the argument. With half-integer spin, Θ² picks up an extra factor of −1 from the rotation of the spinor, so the cases exchange: what is (a) here becomes (b) and what is (b) becomes (a). Every step of the reasoning survives and every verdict flips, which is a good reason to have computed Θ² rather than assumed a sign.

Who found it, and why it took a Bell Labs metallurgist #

Conyers Herring published the criterion in 1937, at twenty-three, in two papers that also gave the accidental degeneracies of band structures their first serious treatment. Wigner had established the general theory of antiunitary operations two years earlier; what Herring supplied was the space-group bookkeeping — the sum over the antiunitary coset, and the recognition that the three cases must be worked out at each wavevector separately because the coset changes from one to the next.

The reason it was needed then is worth remembering. Band structures in the 1930s were computed at a handful of points by hand, and a degeneracy that appeared in the numbers had to be explained or it meant the calculation was wrong. A doubling nobody could account for at the edge of a zone was, in that setting, an alarm. Knowing that it was required — and required by an operation that is in no space group — is what let the arithmetic be trusted.

It has not become less useful. The whole apparatus of symmetry indicators for topological materials rests on knowing which representations are joined at which wavevectors, and time reversal is exactly the operation that joins them.

Where the ladder goes next #

Back, to the census this fills in: a coincidence the group did not ask for, where the method of perturbing what symmetry leaves alone is set out, and where the three unaccounted rows are named as unaccounted.

Down, to the exact fact case (b) rests on: a glide sticks two levels together, and the factor system that no rephasing removes.

Sideways, to what happens on the way out of these points, where the little group shrinks and the levels are free to split again: which levels join which, and, at the points themselves, where two levels must meet — the unitary half of the same question, and the half that was never in doubt.