The projector is unique, the basis is not

Character tables and reduction does the arithmetic that every symmetry argument rests on: take a set of functions, work out the character of each operation acting on them, and reduce that character in the group’s table to get a list of whole numbers.

Those numbers answer one question completely and another not at all. How many times does each irreducible representation appear — answered exactly. Which combinations of the original functions are they — not addressed, and the reduction cannot address it, because computing a character throws away everything but a trace.

The gap matters because the combinations are what a molecular orbital diagram is drawn from, and every ligand-field argument — the splitting is a symmetry statement among them — draws one. Nobody sketches “$a_{1g} \oplus t_{1u} \oplus e_g$”; they sketch the six ligand orbitals adding up in phase, and the three that have a node through the metal, and so on. Those pictures come from somewhere, and the somewhere is a projection operator.

The operator #

For a representation $\Gamma_a$ of dimension $d_a$ in a group of order $h$:

$$P_a = \frac{d_a}{h} \sum_R \chi_a(R)^{*}\, D(R)$$

summed over every operation, not over classes — which is why character tables and reduction’s bookkeeping of classes, sufficient for a character, is not sufficient here. The characters are constant on a class and the matrices $D(R)$ are not, so the sum has to run over all $h$ of them.

The matrices are the part a character calculation throws away. A reduction needs only the trace of $D(R)$ for each operation, and the matrix itself is discarded. Getting it back costs almost nothing once the operations are in hand: for one function per atom an operation is a permutation of the atoms, and for a perpendicular $p$ function it is that permutation with a sign, which is the $zz$ element of the operation’s matrix.

The sign is why the basis has to be checked for closure rather than assumed. A perpendicular $p$ basis is only closed under a group whose operations keep $z$ along $\pm z$, and the calculation checks that rather than trusting the molecule to be planar.

Three things that make it a projector #

Something built from a formula is not a projector until it behaves like one, and three properties are checked for every representation of every group used here.

It is idempotent. $P^2 = P$, to better than $10^{-12}$ in every entry. Applying it twice is applying it once, which is what “project” means.

Its trace is the dimension of the subspace. For benzene’s six $\pi$ functions the traces come out at 1, 2, 1 and 2 for B2g, E1g, A2u and E2u — the multiplicity times the dimension in each case, and zero for the eight representations of D6h that do not appear. That is the reduction, recovered from an operator that was built without it.

They add up to the identity. Summed over all twelve representations of D6h, the projectors give the identity matrix to $10^{-12}$. Nothing in the space is missed and nothing is counted twice, which is the completeness relation and is the only one of the three that could fail for a reason other than an arithmetic slip: it would fail if the group had not closed, which is the failure point groups from coordinates had to repair before any of this was possible.

What the projector does not fix #

Hand the E1g projector a starting function and it returns something in E1g. Which something depends on what it was handed.

Apply it to the $p$ orbital on atom 1 and it gives $(0.577, 0.289, -0.289, -0.577, -0.289, 0.289)$, normalised — the combination with its largest lobes on atoms 1 and 4 and a node through the bond between 2 and 3.

Apply it to the $p$ orbital on atom 2 and it gives $(0.289, 0.577, 0.289, -0.289, -0.577, -0.289)$ — the same shape rotated by one atom.

Both are in E1g. Neither is more correct. They overlap by exactly $\tfrac12$, which is $\cos 60°$, so they are two directions in one two-dimensional space at sixty degrees to each other.

This is not a defect in the method. It is the content of the word degenerate. A two-dimensional representation means the group cannot tell two directions apart, so any orthonormal pair spanning that plane is as good as any other, and asking which pair is the pair is asking the group a question it has no way to answer.

What nothing may depend on #

The check that makes this a statement rather than a shrug: write the molecule’s own Hückel matrix in each of the two pairs and diagonalise.

Both give a $2 \times 2$ block that is the identity times $+1$ — one number, the same number, to twelve decimal places. The E1g level of benzene sits at $\alpha + \beta$ whichever basis it is written in, because a change of basis inside a degenerate subspace is a unitary transformation and every observable is invariant under one.

That is the same argument as hybrids are a basis, arriving from the other end. There the freedom was in mixing $s$ and $p$ on one atom and the invariance was the density; here it is in mixing two functions of one representation and the invariance is a level. The general statement is one line and it is the whole reason a basis can be chosen for convenience: a rotation of a basis is not a change to the object it describes.

The same operator on methane, where it is actually used #

Benzene’s π system is the easy case because the functions are all equivalent. The place a projector earns its keep is a molecule with several kinds of atom, and methane is the one already reduced twice.

Its four hydrogen 1s functions span $a_1 \oplus t_2$, which is the reduction hybridisation does not explain turns into a prediction about a photoelectron spectrum. The projector turns the same two symbols into four combinations: one totally symmetric sum, and three functions with a node through the carbon.

Here the freedom has three dimensions rather than two, and the choice usually made is the one aligned with the Cartesian axes, because the $t_2$ functions transform like $x$, $y$ and $z$ and those are the axes of the table. That is a convention imported from a coordinate system, not a fact about methane — and it is the same convention that makes the four sp³ hybrids of the localisation transformation look canonical when they are a rotation of the same subspace.

Which is why the pictures in textbooks disagree #

Open three books at benzene’s $\pi$ system and the two E1g orbitals will be drawn differently: one book puts a node through two atoms, another through two bonds, a third draws them as complex functions with no node marked at all.

None of them is wrong, and the disagreement is not sloppiness. It is the choice the projector leaves open, made three times by three authors who had no reason to make it the same way. What is common to all three is the pair, and what differs is how the pair is coordinatised.

The same thing happens one level down. Real and complex harmonics is the same choice for a $p$ shell: $p_x$ and $p_y$ are a real basis for a two-dimensional space that the complex functions $m = \pm 1$ span just as well, and which pair is drawn is a decision about what looks like a picture rather than about what the atom has. There the freedom has a physical consequence when a magnetic field is applied, because the field picks a direction and lifts the degeneracy; here the equivalent statement is that a distortion picks a direction, and the vibration that lowers the symmetry is what happens when it does.

Where the choice stops being free #

A degenerate subspace has a rotational freedom only while the degeneracy is exact, and three things remove it.

A lower symmetry. In D2h — benzene stretched, or a para-disubstituted ring — E1g splits into B2g and B3g, and each of those is one-dimensional. Then the projector fixes the combination completely, and the two functions that were arbitrary become determined: they are the pair aligned with the surviving twofold axes, which is why a substituted benzene’s orbitals look like the two “canonical” pictures and an unsubstituted one’s do not.

A perturbation with a direction. Any operator that does not commute with the full group picks a basis for its own reasons, and the right one is then the one that diagonalises it. That is first-order degenerate perturbation theory, and its whole content is the instruction to choose the basis the perturbation prefers.

A second copy of the same representation. If a basis contains two independent sets of functions spanning the same representation — the $\sigma$ and $\pi$ combinations of a set of ligands, say, both giving $t_{1u}$ — then the projector returns a space of dimension twice the representation’s, and the two copies mix by an amount no symmetry argument supplies. That is the case where the freedom stops being harmless: the two combinations are genuinely different orbitals, one is lower than the other, and only an energy can say which mixture each level is. Symmetry has done everything it can and the remaining question is a two-by-two diagonalisation, which is exactly the shape of the argument in back-bonding is two interactions.

A convention. Where nothing physical chooses, somebody chooses anyway — as real and complex harmonics records for the p shell — and the choice usually follows the axis convention of the character table. That is legitimate as long as it is called what it is.

The projector in the calculations that use it #

Symmetry-adapted combinations are not an aesthetic preference; they are what makes a large calculation tractable, and it is worth saying how.

A Hamiltonian commutes with every operation of the molecule’s group. Written in a symmetry-adapted basis it is therefore block diagonal: every matrix element between functions of different representations vanishes exactly, by the vanishing-integral theorem selection rules are one theorem states once. So a matrix of dimension $n$ becomes a set of smaller matrices, one per representation, and the eigenvalue problem is solved in pieces.

For benzene’s six π functions that is four blocks of sizes 1, 2, 1 and 2 rather than one of size 6. For an octahedral complex’s ligand set it is 1, 2 and 3 rather than 6. The saving grows quickly with size, and for a large highly symmetric system it is the difference between a calculation that runs and one that does not. Diagonalisation costs grow as the cube of the matrix size, so replacing one six-by-six problem with blocks of one, two, one and two cuts the work by a factor of twelve before any physics has been used.

Two consequences follow that are relevant here.

The blocks are what is unique. Which functions span each block is the choice; the block structure itself is not, and neither are the eigenvalues that come out of each block.

And the degenerate blocks are the ones that repay attention. A block of dimension 2 arising from one representation appearing once is a scalar — the E1g case above, where both bases gave the same $1 \times 1$ answer twice. A block of dimension 4 from a two-dimensional representation appearing twice is a genuine $2 \times 2$ problem whose off-diagonal element depends on the physics and not on the group, which is the case the last section of this essay leaves open.

Where the model stops #

Nothing here is an energy. A projector is built from the group and the basis, and it knows nothing about which combination is lower. The E1g pair of benzene sits above the A2u orbital because the Hückel matrix says so, not because a projector does, and Hückel theory and what it gets right is where that matrix comes from.

The basis has to be closed. These projectors act on a set of functions the group permutes among themselves. A basis the group takes outside itself is not one to project in, which is why the perpendicular-$p$ case is checked rather than assumed.

Real characters only. Groups with complex characters — the $C_n$ groups with $n \geq 3$ — have pairs of one-dimensional representations that are complex conjugates, and the projector onto either returns a complex function. Every group used here avoids the case, and the honest form of that is that it has not come up rather than that it does not exist.

There is one more place the freedom shows and it is worth a note. A projector built from a group is a projector for any basis that group permutes, so the same four matrices computed here for benzene’s π functions apply unchanged to its six C–H bonds, its six ring C–C bonds, or any other set of six objects the group moves the same way. The multiplicities differ because the characters differ; the arbitrariness does not, and neither does the block structure.

That is why a symmetry-adapted picture drawn for one basis so often looks like a picture drawn for another. It is not a coincidence and it is not a deep fact — it is one group acting on several sets of six things in the same pattern, and the projector is a function of the pattern.

The invariance test the freedom implies #

The projector fixing a subspace and not a basis in it is the same freedom met whenever a degenerate set is drawn, and it comes with a test that decides in one line whether any given quantity is affected.

A quantity summed over a complete degenerate set is invariant. Charge densities, bond orders, total energies, the trace of anything — all are sums over the whole set, and a rotation within the set leaves every one of them exactly unchanged.

A quantity attached to one member is not. The shape of a particular orbital, its individual contribution to a bond, the amplitude at a chosen atom — all depend on which pair the projector happened to return, and a different first function gives a different answer.

So the test is: is this a sum over the whole set? If it is, quote it. If it is not, it is a property of a choice.

That disposes of a class of statements that are made constantly and are not about anything. This degenerate orbital is concentrated on these atoms is such a statement, and a rotation within the pair moves the concentration elsewhere without changing a single observable.

It also explains why good figures draw degenerate sets together wherever they can. One member of a pair is a picture of a choice; both members side by side are a picture of the subspace, which is the object the projector actually determined.

What the projector adds #

The symmetry argument so far generates a group from coordinates, reduces bases in it, shows that degeneracy is fixed before any energy is computed, that a selection rule is one theorem, and that a character table is square because both of its dimensions are forced.

Every one of those is a statement about how many. This essay is the first about which: the operator that turns a multiplicity into a set of functions, the three properties that make it one, and the fact that for a degenerate representation it stops one step short and hands the choice back. What it fixes is a subspace; what it leaves open is a basis for it; and the reason nobody notices is that every number anyone computes from either basis is the same.

The open question is the case where the freedom is not free — where two representations of the same symmetry appear more than once in a basis, and the projector returns a subspace containing genuinely different orbitals that mix by an amount only an energy can settle.