University Quantum Mechanics II · The Variational Principle · 10.9

The hydrogen molecule ion and the chemical bond

This is where the variational method stops describing atoms and starts explaining why they stick together. Two hydrogen orbitals and three closed-form integrals give a potential curve that proves H₂⁺ binds, locates the bond, misses over a third of its strength, and shows which of those claims an upper bound can carry.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Clamp the protons a distance R apart, a step Unit 14 justifies, and the electron faces a two-centre Coulomb problem. Near either proton it should look like a hydrogen 1s electron, and inversion through the midpoint swaps the protons without changing H, so the eigenstates can be chosen even or odd; the natural trials are ψ₁ₛ(r_A) ± ψ₁ₛ(r_B), each bounding the lowest level of its own symmetry. Their energy needs three integrals, all elementary functions of R/a: the overlap S; a direct integral, the attraction between one orbital's cloud and the other proton, done by the l = 0 multipole term as for helium; and an exchange integral with no classical counterpart.

The direct term alone never binds: the cloud never attracts proton B more strongly than proton A repels it. The bond comes from the exchange term, which exists only because the electron is shared. Add e²/(4πε₀R) and the even curve dips to −15.37 eV at R = 2.49a.

The separated limit, −13.61 eV, is exact and the trial curve lies above the true one at every R, so the dip proves the clamped ion binds, by at least 1.76 eV. The exact answer, quoted, is 2.79 eV at 2.00a: the bond is found, 37% too weak and 25% too long, because two fixed-size orbitals can neither contract nor polarise as the protons close. And the bound certifies energies only, never the bond length, the curvature, or the odd state's repulsion.

Simple definition: The LCAO treatment of H₂⁺ takes the even and odd combinations ψ₁ₛ(r_A) ± ψ₁ₛ(r_B) as variational trials for one electron shared by two clamped protons, giving at every separation R an upper bound E±(R) on the lowest energy of each symmetry.
Example: At R = 2.50a = 1.32 Å the orbitals overlap by S = 0.458; the even trial gives −15.37 eV, 1.76 eV below a hydrogen atom plus a distant proton, and the odd trial gives −7.91 eV, 5.70 eV above it.

Overlap of the two 1s orbitalsS(R) = ⟨ψ_A|ψ_B⟩ = e^(−R/a) (1 + R/a + R²/(3a²))

Derived with the law of cosines, r_B² = r² + R² − 2rR cos θ. It sets the normalisation 1/√(2(1 ± S)), because ψ_A and ψ_B are not orthogonal.

R the proton separation and a = 5.29 × 10⁻¹¹ m the Bohr radius, both in m; S is dimensionless, 1 at R = 0 and 0.458 at 2.5a

Direct integral from the l = 0 multipoleD(R) = a⟨ψ_A|1/r_B|ψ_A⟩ = a/R − (1 + a/R) e^(−2R/a)

Derived: a spherical cloud keeps only the monopole of 1/|r − R|, so its charge inside radius R acts from A. It tends to a/R, a point charge, at large R.

Dimensionless; times e²/(4πε₀a) = 27.2 eV it is the attraction of proton B for cloud A; 0.391 at 2.5a

Exchange integralX(R) = a⟨ψ_A|1/r_A|ψ_B⟩ = (1 + R/a) e^(−R/a)

Derived with the same law-of-cosines substitution as S. It has no classical counterpart: it exists only because the electron is shared, and it carries the bond.

Dimensionless; the overlap density ψ_Aψ_B in one proton's field, times 27.2 eV; 0.287 at 2.5a

Electronic energy of the two trials⟨Hₑₗ⟩± = E₁ [1 + 2(D ± X)/(1 ± S)]

Derived from Hₑₗψ_A = E₁ψ_A − (e²/4πε₀r_B)ψ_A and e²/(4πε₀a) = −2E₁. As R → 0 the even value tends to 3E₁ = −40.8 eV, above He⁺'s 4E₁ = −54.4 eV.

E₁ = −13.6 eV = −2.18 × 10⁻¹⁸ J; upper signs for the even trial, lower for the odd; Hₑₗ leaves out the proton repulsion

The clamped-proton potential curveE±(R) = ⟨Hₑₗ⟩± + e²/(4πε₀R)

Derived, and an upper bound on the lowest level of each symmetry at every R. The even curve's minimum is −15.37 eV at R = 2.49a = 1.32 Å; the odd curve stays above −13.6 eV.

e²/(4πε₀R) = 27.2 eV × a/R; E± in J, quoted in eV, for protons held at separation R

What the even minimum certifiesDₑ ≥ E₁ − min E₊(R) = −13.61 eV + 15.37 eV = 1.76 eV

Derived: the true curve lies below E₊ at every R, so its minimum lies below −15.37 eV. The exact Dₑ = 2.79 eV at 2.00a is quoted, not derived here.

Dₑ the clamped-ion binding energy, in J or eV; E₁ is the exact energy of an H atom and a distant proton

Clamp the protons, then let symmetry choose the trial

With the protons held at A and B, a distance R apart, the electron's Hamiltonian is Hₑₗ = −(ħ²/2m)∇² − e²/(4πε₀r_A) − e²/(4πε₀r_B), and the clamped ion adds the constant e²/(4πε₀R). Clamping is the Born–Oppenheimer step Unit 14 justifies; here it is assumed. Inversion through the midpoint swaps r_A and r_B and commutes with H, so the eigenstates can be chosen even or odd, and an even trial is orthogonal to every odd one. The variational theorem therefore holds in each symmetry class separately, the move that bounded excited states earlier in this unit. Far from B the electron should be a 1s electron on A, so the trials are ψ± = [ψ₁ₛ(r_A) ± ψ₁ₛ(r_B)]/√(2(1 ± S)), with ψ₁ₛ(r) = e^(−r/a)/√(πa³) from Unit 2. For the overlap S, write r_B² = r² + R² − 2rR cos θ: then d(cos θ) = −r_B dr_B/(rR), and the angular integral becomes an integral over r_B from |r − R| to r + R, giving S = e^(−R/a)(1 + R/a + R²/(3a²)).

The direct integral is the shell theorem again

The electron on A feels proton B through ⟨ψ_A|1/r_B|ψ_A⟩. Expand 1/r_B = Σₗ (sˡ/tˡ⁺¹) Pₗ(cos θ), with s and t the smaller and larger of r and R, as for helium: |ψ_A|² is spherical about A, so every term with l ≥ 1 averages to zero and only the monopole, 1/max(r, R), survives. Charge within R of A acts as if it sat at A; charge farther out contributes its own 1/r. Two radial integrals give D = a/R − (1 + a/R)e^(−2R/a), which tends to a/R at large R and to 1 as R → 0. Now keep the electron on A alone, a trial with no sharing: H_AA = E₁ − 27.2 eV × D + 27.2 eV × a/R = E₁ + 27.2 eV × (1 + a/R)e^(−2R/a). The correction is positive at every R. Proton B sits inside A's cloud: the charge within R of A pulls on it as if from A, the charge farther out pulls more weakly, and proton A repels it with full strength, so a rigid 1s cloud that is neither shared nor polarised never binds. At R = 2.5a it leaves the pair 0.26 eV above E₁.

The exchange integral is where the bond comes from

Sharing adds the off-diagonal element. Because Hₑₗψ_B = E₁ψ_B − (e²/4πε₀r_A)ψ_B, the cross element needs only S and the exchange integral X = a⟨ψ_A|1/r_A|ψ_B⟩ = (1 + R/a)e^(−R/a), done by the same law-of-cosines substitution as S. The orbitals ψ_A and ψ_B form a two-function Rayleigh–Ritz basis, and the proton-swap symmetry makes ψ_A ± ψ_B its eigenvectors, with roots E± = (H_AA ± H_AB)/(1 ± S). Measured from the unshared energy they are E± − H_AA = ∓27.2 eV × (X − SD)/(1 ± S). X − SD is the part of the exchange integral not already counted by weighting the direct term with the overlap, and it is positive at every R: 0.108 at R = 2.5a, where it lowers the even state by 2.02 eV and raises the odd one by 5.44 eV. The odd shift is larger by (1 + S)/(1 − S), because the odd state's normalisation 1/√(2(1 − S)) is the larger. Only the even state gains enough to beat the 0.26 eV the unshared electron already lost.

Why the dip is a proof, and how to check it

Add the proton repulsion and minimise: E₊(R) = E₁ + 27.2 eV × [a/R − (D + X)/(1 + S)] has its minimum at R = 2.49a = 1.32 Å, where E₊ = −15.37 eV. Two facts turn that number into a theorem. At each fixed R the trial bounds the lowest even level from above, so the true curve reaches −15.37 eV or lower at 2.49a and its own minimum lies lower still. And the separated limit is not an estimate: as R → ∞ the ion becomes a hydrogen atom and a bare proton at exactly E₁ = −13.61 eV. So the clamped ion binds, with Dₑ ≥ 1.76 eV. Check it every way available. Dimensions: every term is e²/(4πε₀a) = 27.2 eV times a function of R/a. Limits: E₊ → E₁ as R → ∞, and as R → 0 the electronic part tends to 3E₁ = −40.8 eV, above the −54.4 eV of He⁺, as a bound must be. Order in α: the whole curve is α²mc² times a function of R/a, so relativistic corrections, smaller by α², are about 10⁻³ eV and cannot touch a 1 eV discrepancy.

Against exact and measured values, and why it falls short

The clamped two-centre problem separates in spheroidal coordinates and has been solved exactly; its results are quoted, not derived: a minimum at R = 2.00a = 1.06 Å with Dₑ = 2.79 eV. Measurement agrees through tabulated energies: D₀(H₂) = 4.48 eV plus the ionisation energy of H, 13.60 eV, minus that of H₂, 15.43 eV, gives D₀ = 2.65 eV for H₂⁺, and adding the zero-point energy ½ħω ≈ 0.14 eV, from the measured vibrational frequency (Unit 14), restores Dₑ ≈ 2.79 eV. The trial finds 63% of the binding and puts the bond 25% too far out, and it shows why. For an exact state at a minimum of E(R) the virial theorem, quoted, requires 2⟨T⟩ + ⟨V⟩ = 0; the trial gives 2 × 10.41 eV − 25.78 eV = −4.96 eV, so its kinetic energy is too small and its orbitals too diffuse. It cannot polarise either: ψ_A ± ψ_B has no p-like distortion, so it misses even the attraction between a proton and the atom it polarises, −(9/4) × 27.2 eV × (a/R)⁴, quoted from hydrogen's polarisability 4.5 × 4πε₀a³, which rules at large R. This curve is the potential Unit 14 turns into a vibrational frequency.

Change one variable at a time

Make the relationship visible.

Interactive model

Weight c of ψ_B in ψ_A + cψ_B1.0

Proton separation R2.5 a

Set c = 0: the electron stays on one proton and the curve never dips below −13.6 eV, so this unshared trial proves no bond. Raise c to +1 and a well 1.76 eV deep opens at 2.49a; at c = −1 every R lies above the line, which an upper bound cannot turn into a proof of repulsion.

OVERLAP S0.458

TRIAL ENERGY E(R)-15.37 eV

EVEN TRIAL E₊(R)-15.37 eV

BINDING PROVED, AT LEAST1.76 eV

Live interpretationOVERLAP S: 0.458. TRIAL ENERGY E(R): −15.37 eV. EVEN TRIAL E₊(R): −15.37 eV. BINDING PROVED, AT LEAST: 1.76 eV

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading treats the variational curves as blurred copies of the true ones, so that every feature on them is bounded too: the odd curve, above −13.6 eV at every R, is taken to prove the antibonding state repulsive, and the even minimum at 2.49a to bound the bond length. Neither follows. The theorem bounds one number at each R, the energy, and only from above. A bound lying above the dissociation limit is consistent with a true curve on either side of that limit, so the odd trial proves nothing about binding; the exact odd curve, quoted, is repulsive at bonding distances yet has a well about 1.6 meV deep near 12.5a. The position and curvature of a minimum are not energies at all: the exact bond, at 2.00a, is shorter than the trial's, but no theorem put it there. The certified reading is Dₑ ≥ 1.76 eV.

The even trial for H₂⁺ has its minimum, −15.37 eV, at R = 2.49a, and a hydrogen atom plus a distant proton has −13.61 eV. What does this calculation, on its own, establish about the clamped ion?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyEvaluate the overlap, direct and exchange integrals at R = 2.00a, the exact bond length, and use them to find the even trial's energy there, proton repulsion included. The exact clamped-ion energy at that separation is −16.40 eV (quoted). Is the comparison consistent with the variational theorem?

With R/a = 2.00 and e⁻² = 0.13534: S = e⁻²(1 + 2 + 4/3) = 0.13534 × 4.3333 = 0.5865.
D = 1/2 − (1 + 1/2)e⁻⁴ = 0.5000 − 1.5 × 0.018316 = 0.4725, and X = (1 + 2)e⁻² = 0.4060.
Electronic energy: (D + X)/(1 + S) = 0.8785/1.5865 = 0.5537, so ⟨Hₑₗ⟩₊ = −13.606 eV × (1 + 2 × 0.5537) = −13.606 eV × 2.1074 = −28.673 eV.
Proton repulsion: e²/(4πε₀R) = 27.211 eV/2.00 = 13.606 eV, so E₊(2.00a) = −28.673 eV + 13.606 eV = −15.067 eV, or −15.07 eV.
The exact energy at the same R is 1.33 eV lower. The trial lies above it, as the theorem requires at every fixed R; it also lies 0.30 eV above the trial's own minimum, −15.37 eV at 2.49a.

AnswerE₊(2.00a) = −15.07 eV, 1.33 eV above the exact −16.40 eV: consistent with the bound, which holds at each R separately.

MediumAt R = 4.00a (2.12 Å), find the energy with the electron on proton A alone, then the even and odd energies, with the proton repulsion included throughout. Show that the odd state is pushed up (1 + S)/(1 − S) times as far as the even state is pulled down.

With e⁻⁴ = 0.018316 and e⁻⁸ = 3.3546 × 10⁻⁴: S = e⁻⁴(1 + 4 + 16/3) = 0.18926, D = 1/4 − (5/4)e⁻⁸ = 0.24958, and X = 5e⁻⁴ = 0.091578.
Unshared: H_AA = E₁ + 27.211 eV × (1 + 1/4)e⁻⁸ = −13.606 eV + 0.0114 eV = −13.5946 eV, just above E₁, as the direct term always leaves it.
Sharing term: X − SD = 0.091578 − 0.18926 × 0.24958 = 0.044342, worth 27.211 eV × 0.044342 = 1.2066 eV.
Even: E₊ = H_AA − 1.2066 eV/(1 + S) = −13.5946 eV − 1.0146 eV = −14.6092 eV. Odd: E₋ = H_AA + 1.2066 eV/(1 − S) = −13.5946 eV + 1.4883 eV = −12.1063 eV.
Ratio of the shifts: 1.4883/1.0146 = 1.467, and (1 + S)/(1 − S) = 1.18926/0.81074 = 1.467.

AnswerH_AA = −13.595 eV, E₊ = −14.61 eV and E₋ = −12.11 eV; the odd shift, +1.49 eV, is (1 + S)/(1 − S) = 1.47 times the size of the even shift, −1.01 eV.

HardAt the even trial's minimum, R = 2.49a, the kinetic energy is ⟨T⟩ = |E₁|(1 − S + 2X)/(1 + S) = 10.41 eV and the potential energy, electron–proton attraction plus proton repulsion, is ⟨V⟩ = −25.78 eV. Test this against the virial theorem, then shrink both orbitals to e^(−ζr/a) while holding ζR = 2.49a fixed. Find the best ζ, the new R and energy, and what the improved number proves.

⟨T⟩ + ⟨V⟩ = −15.37 eV, the minimum of E₊. The virial theorem (quoted) requires 2⟨T⟩ + ⟨V⟩ = 0 for an exact state at a minimum of E(R); here 20.82 eV − 25.78 eV = −4.96 eV, so the orbitals are too diffuse.
Shrinking every length by 1/ζ, R included, is the change of variable r → r/ζ in every integral. Kinetic energy goes as 1/length² and every Coulomb term, e²/(4πε₀R) included, as 1/length, so E(ζ) = ζ² × 10.41 eV − ζ × 25.78 eV.
dE/dζ = 0 gives ζ = −⟨V⟩/(2⟨T⟩) = 25.78/20.82 = 1.238, so the protons now sit at R = 2.49a/1.238 = 2.01a = 1.06 Å.
E = −⟨V⟩²/(4⟨T⟩) = −(25.78)²/41.64 = −15.96 eV, and 2ζ²⟨T⟩ + ζ⟨V⟩ = 0: the scaled trial satisfies the virial theorem by construction. Re-minimising over R as well changes E by under 1 meV.
The scaled state is an admissible trial at R = 2.01a, so the true curve reaches −15.96 eV or lower there: Dₑ ≥ 15.96 eV − 13.61 eV = 2.35 eV. Against the exact 2.79 eV (quoted) the gap falls from 1.03 eV to 0.44 eV; most of the rest is polarisation, which no rescaling of s orbitals supplies.

Answerζ = 1.24, R = 2.01a (1.06 Å) and E = −15.96 eV, proving Dₑ ≥ 2.35 eV against the exact 2.79 eV.