Chapter 1 — The Single Bond: H₂
The simplest molecule. Two protons, two electrons, one bond. Classic chemistry gets it right. C/T gets it right more cheaply — and tells you exactly when the classic picture breaks down.
The classic picture
Two hydrogen atoms, each with one electron in a 1s orbital, approach each other. Their orbitals overlap and split into two combinations:
- σ (bonding): electrons between the nuclei, lowering the energy
- σ* (antibonding): electrons outside the nuclei, raising the energy
At the equilibrium bond length (0.74 Å), both electrons go into σ. Bond order = 1. Done.
| Quantity | Value |
|---|---|
| Bond length | 0.74 Å |
| Bond energy | 432 kJ/mol |
| Bond order | 1 |
| Orbital picture | σ² (both electrons in bonding MO) |
This is what every first-year chemistry course teaches, and it is correct.
The C/T picture
In C/T language we ask: which orbitals are C-boxes (frozen — NOON near 0 or 2) and which are T-arrows (active — NOON between 0.02 and 1.98)?
At equilibrium (R = 0.74 Å):
| Orbital | NOON | C/T classification |
|---|---|---|
| σ (bonding) | 1.975 | C-box (frozen, doubly occupied) |
| σ* (antibonding) | 0.025 | C-box (frozen, unoccupied) |
Both orbitals are C-boxes. There are no T-arrows. The molecule is H⁰.
| This means: the Lewis diagram is exact. No correlation energy is needed. Hartree-Fock gives the right answer. The wavefunction is a single Slater determinant | σσ⟩ and nothing else matters. |
The C/T method gives the same answer as the classic method — from a single number (the NOON) rather than from a diagram.
What C/T adds: the bond-breaking story
Now stretch the bond. What happens to the NOONs?
| R (Å) | n(σ) | n(σ*) | C/T tier | Classic description |
|---|---|---|---|---|
| 0.50 | 1.990 | 0.010 | H⁰ | Compressed bond, still H⁰ |
| 0.74 | 1.975 | 0.025 | H⁰ | Equilibrium — Lewis exact |
| 1.00 | 1.939 | 0.061 | H⁰ | Still H⁰, but weakening |
| 0.67* | 1.980 | 0.020 | H⁰→H¹ snap | C/T threshold crossed |
| 1.50 | 1.747 | 0.253 | H¹ | Significant correlation |
| 2.00 | 1.424 | 0.576 | H¹/H² | Strong correlation |
| 3.50 | 1.027 | 0.973 | H² | Near-dissociation: two radicals |
* The snap is at R* ≈ 0.67 Å on the compression side (shorter than equilibrium in STO-3G). On stretching, the snap occurs later, around R ≈ 1.0–1.5 Å depending on the basis set.
The key point: the C/T method tells you exactly when the Lewis diagram breaks down. There is no ambiguity, no judgement call about “when does this molecule become multi-reference.” The NOON crossing 0.02 is the answer.
The Bloch sphere: H₂ as a point on S²
For H₂ in a minimal basis (STO-3G), there are only two orbitals: σ and σ*. The active subspace is at most 1-dimensional, sitting inside a 2-dimensional one-particle space. The space of all 1-dimensional subspaces of ℂ² is CP¹ ≅ S² — the Bloch sphere.
The C/T compression map sends each bond length R to a point on S²:
\[\theta(R) = 2\arcsin\!\sqrt{\frac{n_{\sigma^*}(R)}{2}}\]At equilibrium: θ ≈ 8.2° (near the north pole — mostly C-box).
At dissociation: θ → 90° (equator — equal occupation, H²).
The bond-stretching trajectory is a geodesic on S² — a great-circle arc along the meridian. This is not a model: it is the exact FCI trajectory, confirmed numerically.
The snap at R* is the point where the geodesic crosses the Schubert variety — the locus on S² where the C/T threshold is reached. Before the crossing: H⁰ (Lewis diagram valid). After: H¹ or H² (correlation required).
What Jaynes says
The maximum-entropy state consistent with the NOONs (n_σ, n_σ*) is the free-fermion Gibbs state:
\[\rho_{\text{MaxEnt}} = \frac{e^{-\lambda_\sigma \hat{n}_\sigma - \lambda_{\sigma^\ast} \hat{n}_{\sigma^\ast}}}{Z}, \qquad \lambda_i = \log\!\left(\frac{2}{n_i} - 1\right)\]| At equilibrium: λ_σ = −3.66, λ_σ* = +3.66 (σ strongly favoured, σ* strongly disfavoured). The MaxEnt state has 97.5% of its weight in | σσ⟩ — consistent with the Lewis diagram. |
| At dissociation: λ_σ ≈ λ_σ* ≈ 0 (both equally probable). The MaxEnt state is maximally mixed between | σσ⟩, | σ*σ*⟩, | σσ*⟩ — consistent with two independent radicals. |
The MaxEnt Lagrange multipliers reproduce the FCI NOONs exactly (confirmed numerically: reconstruction error = 0.000000 at every geometry). This is Proposition 1 of Paper 594: MaxEnt is not an approximation here, it is exact at the level of the 1-RDM.
The sceptic test
“You’ve just redescribed the σ/σ* picture in different language. What have you actually gained?”
Three things:
1. Automaticity. The classic picture requires you to know that σ and σ* are the right orbitals to draw. For H₂ this is obvious. For a transition metal complex with 50 d-orbital combinations, it is not. The C/T method computes the active orbitals from the 1-RDM eigenvalues — no chemical intuition required.
2. The breakdown criterion. The classic picture has no built-in way to tell you when it stops being valid. “Multi-reference character” is a judgement call. The C/T method gives a sharp criterion: the molecule is H⁰ if and only if all NOONs are outside (0.02, 1.98). For H₂ this happens at R* ≈ 0.67 Å (inward) or around 1.0–1.5 Å (outward, basis-dependent). No guesswork.
3. The energy decomposition. Once you know the H^k tier, you know what kind of calculation is needed:
- H⁰: Hartree-Fock is exact. No correlation calculation needed.
- H¹: CASSCF with the T-arrow orbitals is sufficient.
- H²: Multi-reference methods (CASPT2, MRCI) are needed. The Maslov index tells you how many non-trivial correlation insertions are required.
For H₂ at equilibrium: H⁰ → use HF → one Slater determinant → done. The C/T method confirms what every chemist knows, but now from a theorem rather than from tradition.
Numbers
From experiment x594a (FCI/STO-3G):
| Quantity | Value |
|---|---|
| Equilibrium bond length | R_eq = 0.74 Å (literature) |
| NOON snap geometry | R* = 0.67 Å (STO-3G) |
| Bloch-sphere angle at snap | θ* = 12.3° |
| Fubini-Study distance R_eq → R* | 0.036 rad |
| MaxEnt reconstruction error | 0.000000 (exact) |
In one sentence
H₂ at equilibrium is H⁰: both orbitals are frozen (C-boxes), the Lewis diagram is exact, and no correlation is needed. The C/T method confirms this from the 1-RDM eigenvalues alone — and tells you precisely where it stops being true.