University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.9

The Born-Oppenheimer approximation for molecules

Every potential-energy curve in chemistry, and every vibrational or rotational spectrum read from one, rests on this approximation: let the electrons solve their problem as if the nuclei were still, then let the nuclei move in the energy that comes out. Here it is derived, what it drops is sized, and where it fails is found.

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Connect the measurement to the mechanism.

A molecule's electrons and nuclei pull on each other with equal and opposite forces, but the nuclei are at least 1836 times heavier, so they move far more slowly and the electrons settle into a stationary state for wherever the nuclei happen to be. The Born-Oppenheimer approximation turns that into two steps. Clamp the nuclei at separation R and solve for the electrons, which gives an energy Eₙ(R) and a state φₙ(r; R) at each R: this unit's adiabatic theorem, with R as the slow parameter.

Then quantise the nuclei with Eₙ(R) as their potential energy. Written on the complete set φₙ, the molecular problem is exact; the approximation drops only the terms in which the nuclear kinetic energy differentiates φₙ with respect to R, and those couple one curve to another. Counting powers of m/M shows why they are small, since electronic, vibrational and rotational energies stand as 1 : (m/M)¹⁄² : m/M, and it turns a curve's curvature into a vibrational quantum: Unit 10's H₂⁺ curve gives 0.225 eV against a measured 0.272 eV.

The picture fails where two curves come close, because the dropped coupling grows as the inverse of their gap; there the nuclei no longer follow one potential, and the Landau-Zener formula takes over.

Simple definition: The Born-Oppenheimer approximation writes a molecular state as one electronic eigenstate φₙ(r; R), found with the nuclei clamped at R, times a nuclear wavefunction χ(R) that moves in the electronic energy Eₙ(R) as its potential.
Example: H₂⁺'s clamped-nuclei energy is lowest, 2.79 eV below H + p, at R = 2.00 a₀ = 106 pm. The protons, with μ = mₚ/2, vibrate in that well with a measured fundamental of 0.272 eV, 44 times smaller than the 11.84 eV gap to the next electronic curve.

The clamped-nuclei electronic problemHₑ(R) φₙ(rR) = Eₙ(R) φₙ(rR)

No ∂/∂R appears in Hₑ, so each R is its own eigenvalue problem. Repeated on a grid of R it traces the curves Eₙ(R), Unit 10's H₂⁺ LCAO curve among them.

Hₑ = Tₑ + every Coulomb term, nuclear repulsion included; Eₙ in J; R in m, a fixed parameter; φₙ in m^(−3/2) per electron

Exact coupled equations (derived)[−(ħ²/2μ)∂²_R + Eₘ(R) − E] χₘ = (ħ²/2μ) Σₙ (2dₘₙ ∂_R + gₘₙ) χₙ

Exact for Ψ = Σₙ χₙ(R) φₙ(r; R), since the φₙ are complete at each R. Every term on the right exists only because the nuclear kinetic energy differentiates φₙ.

dₘₙ = ⟨φₘ|∂_Rφₙ⟩ in m⁻¹; gₘₙ = ⟨φₘ|∂²_Rφₙ⟩ in m⁻²; χₘ(R) the nuclear amplitudes; μ in kg

The Born-Oppenheimer nuclear equation[−(ħ²/2μ)∇²_R + Eₙ(R)] χ(R) = E χ(R)

The coupled equations with the right side set to zero. The nuclear kinetic energy is kept in full; only its action on φₙ is dropped.

μ = M_A M_B/(M_A + M_B) in kg for a diatomic; Eₙ(R) in J is the potential; E the total molecular energy in J

Non-adiabatic coupling between curves (derived)dₘₙ(R) = ⟨φₘ|∂_R Hₑ|φₙ⟩ / (Eₙ(R) − Eₘ(R)), m ≠ n

It grows as the inverse of the gap. At an avoided crossing with gap 2V and diabatic slopes differing by ΔF it peaks at ΔF/(4V), and the curves stop being separate.

dₘₙ in m⁻¹, ∂_R Hₑ in J m⁻¹, energies in J; from differentiating Hₑ φₙ = Eₙ φₙ and projecting on φₘ

The mass hierarchyEₑₗ : ħω : B ≈ 1 : (m/M)¹⁄² : m/M

With κ = (m/M)¹⁄⁴: zero-point spread ≈ κa, diagonal correction ≈ κ⁴Eₑₗ, off-diagonal level shifts ≈ κ⁶Eₑₗ, with Eₑₗ ≈ α²mc². The expansion is in κ, not α.

m the electron mass, M a nuclear mass; for H₂⁺ with μ = mₚ/2, (m/μ)¹⁄² = 0.033 and m/μ = 1.1 × 10⁻³

A vibrational quantum from the curvatureħω = ħ √(Eₙ″(Rₑ)/μ)

Unit 10's LCAO curve has Eₙ″ = 97.8 N m⁻¹ at Rₑ = 2.49 a₀, giving ħω = 0.225 eV (derived); the measured H₂⁺ fundamental is 0.272 eV.

Eₙ″ in J m⁻² = N m⁻¹, μ in kg; H₂⁺ has μ = mₚ/2 = 8.363 × 10⁻²⁸ kg

Clamp the nuclei and the electrons return a potential

With the centre of mass removed, a diatomic molecule's Hamiltonian is H = −(ħ²/2μ)∇²_R + Hₑ(r; R), where μ = M_A M_B/(M_A + M_B) and Hₑ holds the rest: the electrons' kinetic energy and every Coulomb term, the nuclear repulsion e²/(4πε₀R) included. Mass-polarisation terms of relative order m/M are set aside. Hₑ contains R but no derivative with respect to R, so at each fixed R it is an ordinary electronic problem, Hₑ(R)φₙ(r; R) = Eₙ(R)φₙ(r; R), and solving it on a grid of R traces out the curves Eₙ(R). This is topic 14.02's adiabatic set-up with the nuclear separation as the slow parameter. In H₂⁺ the gap to the next electronic curve at Rₑ is 11.84 eV (quoted), an electronic period h/ΔE of 0.35 fs, while one vibration, at the measured 0.272 eV, takes 15.2 fs. Unit 10's LCAO calculation was this first step done variationally: its minimum, 1.76 eV below H + p at 2.49 a₀, is a clamped-nuclei energy, against the exact 2.79 eV at 2.00 a₀ (quoted).

Expand on the electronic states, then drop one class of terms

At each R the φₙ(r; R) are complete in the electron coordinates, so any molecular state can be written exactly as Ψ = Σₙ χₙ(R)φₙ(r; R), the Born-Huang expansion. The nuclear Laplacian acts on both factors, ∂²_R(χφ) = φ∂²_Rχ + 2(∂_Rχ)(∂_Rφ) + χ∂²_Rφ, and projecting on φₘ gives, for the single coordinate R, [−(ħ²/2μ)∂²_R + Eₘ(R) − E]χₘ = (ħ²/2μ)Σₙ(2dₘₙ∂_R + gₘₙ)χₙ, with dₘₙ = ⟨φₘ|∂_Rφₙ⟩ and gₘₙ = ⟨φₘ|∂²_Rφₙ⟩ (derived). Setting the right side to zero is the Born-Oppenheimer approximation. Keeping its n = m terms is the adiabatic approximation: for a real φₙ, dₙₙ = ½∂_R⟨φₙ|φₙ⟩ = 0, while gₙₙ = −⟨∂_Rφₙ|∂_Rφₙ⟩ adds (ħ²/2μ)⟨∂_Rφₙ|∂_Rφₙ⟩ to the potential, a small positive shift that, unlike Eₙ(R), depends on the nuclear mass. The off-diagonal dₘₙ and gₘₙ are the non-adiabatic couplings.

Count powers of m/M to size what is dropped

Let the electrons set an energy ε ≈ ħ²/(ma²) over a length a near a Bohr radius; for a = a₀, ε = α²mc² = 27.2 eV. A curve changes by about ε when R changes by about a, so its curvature is k ≈ ε/a² and the nuclei vibrate with ħω = ħ√(k/M) ≈ ε(m/M)¹⁄². Their zero-point spread is ΔR ≈ √(ħ/(Mω)) ≈ a(m/M)¹⁄⁴, and rotation costs ħ²/(MR²) ≈ ε(m/M). With κ = (m/M)¹⁄⁴, electronic, vibrational and rotational energies go as κ⁰, κ² and κ⁴ (estimates, derived). Now the dropped terms. φₙ changes over a, so ∂_Rφₙ ≈ φₙ/a, while χ changes over ΔR ≈ κa. The diagonal term, (ħ²/2M)⟨∂_Rφₙ|∂_Rφₙ⟩ ≈ ħ²/(Ma²) = εκ⁴, is the size of a rotational energy. The off-diagonal term (ħ²/M)dₘₙ∂_Rχ ≈ εκ⁴/κ = εκ³, but it connects curves about ε apart, so it moves levels only in second order, by (εκ³)²/ε = εκ⁶ = ε(m/M)³⁄². Every one of these energies is α²mc² times a power of m/M: the hierarchy is in m/M, a small number unrelated to α.

A vibrational quantum from the curvature of the H₂⁺ curve

Near its minimum a curve is a parabola, Eₙ(R) ≈ Eₙ(Rₑ) + ½k(R − Rₑ)² with k = Eₙ″(Rₑ), and the nuclear equation becomes Quantum Mechanics I's oscillator, so ħω = ħ√(k/μ). The two protons of H₂⁺ give μ = mₚ/2 = 8.363 × 10⁻²⁸ kg. Differentiating Unit 10's closed-form LCAO curve twice at its minimum, Rₑ = 2.49 a₀, gives k = 1.710 eV a₀⁻² = 97.8 N m⁻¹, and ħω = 1.0546 × 10⁻³⁴ J s × √(97.8 N m⁻¹ ÷ 8.363 × 10⁻²⁸ kg) = 3.61 × 10⁻²⁰ J = 0.225 eV (derived from that curve; √(N m⁻¹ kg⁻¹) = s⁻¹). The measured fundamental, v = 0 → 1, is 0.272 eV. Two things account for the difference. The LCAO well is too shallow and too wide, 1.76 eV deep against 2.79 eV, and its curvature is 39 per cent low: the exact curve's harmonic constant, ωₑ = 0.288 eV (quoted), corresponds to k = 2.80 eV a₀⁻². And the true well softens above its bottom, so the first spacing is ωₑ − 2ωₑxₑ = 0.288 − 0.016 = 0.272 eV (table). The ratio 0.272/11.84 = 0.023 sits beside (m/μ)¹⁄² = 0.033, as the counting predicts.

Isotopes share one curve: testing the separation

Hₑ contains the nuclear charges but not the nuclear masses, so every isotopologue shares one set of curves, with the same Rₑ, Dₑ and k; mass enters only through μ in the nuclear equation. The prediction is sharp: ħω ∝ μ^(−1/2), B = ħ²/(2μRₑ²) ∝ μ⁽⁻¹⁾ and ΔR ∝ μ^(−1/4). For H₂ and D₂, μ(D₂)/μ(H₂) = m_d/mₚ = 1.99901. The tabulated rotational constants, 60.853 and 30.443 cm⁻¹, stand in the ratio 1.99892, and the vibrational ones, 4401.21 and 3115.50 cm⁻¹, in the ratio 1.41268 against √1.99901 = 1.41386. The misses, 5 × 10⁻⁵ and 8 × 10⁻⁴, are no larger than m/μ = 1.1 × 10⁻³, the fractional size of the dropped terms. Precision calculations on H₂ split its measured dissociation energy, 36 118.07 cm⁻¹, into pieces (quoted): the clamped-nuclei curve gives 36 112.59, the adiabatic correction adds 5.77 and the non-adiabatic couplings 0.43, while relativity removes 0.53 and QED 0.19. As fractions of the whole, 1.6 × 10⁻⁴ and 1.2 × 10⁻⁵ sit beside m/M = 5.4 × 10⁻⁴ and (m/M)³⁄² = 1.3 × 10⁻⁵, while relativity's 1.5 × 10⁻⁵ is set by α² = 5.3 × 10⁻⁵ instead.

Where two curves approach, the separation fails

Differentiate Hₑφₙ = Eₙφₙ with respect to R and project on φₘ, m ≠ n: ⟨φₘ|∂_R Hₑ|φₙ⟩ + Eₘdₘₙ = Eₙdₘₙ, so dₘₙ = ⟨φₘ|∂_R Hₑ|φₙ⟩/(Eₙ − Eₘ) (derived). Where two curves of one symmetry approach, with diabatic slopes differing by ΔF and minimum gap 2V, it peaks at ΔF/(4V) over a width 2V/ΔF, and nuclei passing at speed v stay on one curve only if ħv|dₘₙ| ≪ 2V, that is ħvΔF ≪ 8V²: topic 14.02's condition with d/dt = v∂_R. Otherwise the passage is a Landau-Zener problem, the diabatic character surviving with probability exp(−2πV²/(ħvΔF)) (topic 14.04, quoted), which is how ionic and covalent curves trade an electron in alkali-halide collisions. A diatomic's same-symmetry curves only avoid each other (the von Neumann-Wigner rule, quoted), but with two or more nuclear coordinates, two surfaces can touch at a conical intersection. Carrying the nuclei round one reverses the sign of the real electronic state, a Berry phase of π sourced by the degeneracy (topic 14.07), and the nuclear equation then carries the Berry connection as a vector potential, (P − ħAₙ)²/2μ with Aₙ = i⟨φₙ|∇_Rφₙ⟩.

Change one variable at a time

Make the relationship visible.

Interactive model

Nuclear reduced mass μ/mₑ918

At 918 mₑ, H₂⁺, D₀ = 2.651 eV matches the measured 2.651 eV. Drag to 2748 mₑ, T₂⁺, which triples μ to within 0.3 per cent: the curve stays put, ħω falls by √3 to 166 meV, B drops to a third and the cloud narrows by 3¹⁄⁴. At 100 mₑ, about two positive muons, ħω is 872 meV and the well holds only six levels.

VIBRATIONAL QUANTUM ħω288 meV

ROTATIONAL CONSTANT B3.705 meV

ZERO-POINT SPREAD ΔR12.0 pm

DISSOCIATION ENERGY D₀2.651 eV

Live interpretationVIBRATIONAL QUANTUM ħω: 288 meV. ROTATIONAL CONSTANT B: 3.705 meV. ZERO-POINT SPREAD ΔR: 12.0 pm. DISSOCIATION ENERGY D₀: 2.651 eV

Catch the common trap

Explain before calculating.

Misconception check

The usual slip is to read the approximation as neglecting the nuclear kinetic energy, or as holding the nuclei still. It does neither. Clamping the nuclei is only the first step, used to compute Eₙ(R); in the second, −(ħ²/2μ)∇²_R is kept in full, and it produces every vibrational and rotational level. The nuclei are not still: in H₂⁺ their zero-point spread is 12 pm, 11 per cent of the 106 pm bond, and zero-point energy is why D₂ is bound 0.078 eV more strongly than H₂ on the very same curve. What is dropped is narrower: the part of the nuclear kinetic energy that differentiates the electronic state, ⟨φₘ|∂_Rφₙ⟩ and ⟨φₘ|∂²_Rφₙ⟩. So the test is not whether the nuclei are slow in some absolute sense but whether ħv|dₘₙ| is small against the gap to the nearest other curve, and near a narrow avoided crossing that fails even for heavy nuclei.

In the Born-Oppenheimer approximation, H₂⁺ has Rₑ = 2.00 a₀, a harmonic vibrational quantum ħω = 0.288 eV and a rotational constant B = 3.70 meV. Replacing both protons by deuterons doubles the reduced mass to within 0.05 per cent. What does the approximation predict for D₂⁺?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn H₂⁺ at Rₑ = 2.00 a₀ the gap to the first excited electronic curve is 11.84 eV (quoted), the measured vibrational fundamental is 0.2717 eV, and the rotational constant is B = ħ²/(2μRₑ²) with μ = mₚ/2 = 918.1 mₑ. Find the periods h/ΔE of the electronic, vibrational and rotational (J = 0 → 1) motion, and compare their ratios with the estimate (μ/mₑ)¹⁄².

Electronic: h/ΔE = 4.1357 × 10⁻¹⁵ eV s ÷ 11.84 eV = 3.49 × 10⁻¹⁶ s = 0.349 fs.
Vibrational: 4.1357 × 10⁻¹⁵ eV s ÷ 0.2717 eV = 1.522 × 10⁻¹⁴ s = 15.2 fs.
Rotational: ħ²/(mₑa₀²) = 27.211 eV, so B = 27.211 eV × (mₑ/μ)/(2 × 2.00²) = 27.211 eV/(8 × 918.1) = 3.705 meV. The J = 0 → 1 spacing is 2B = 7.410 meV, a period of 4.1357 × 10⁻¹⁵ eV s ÷ 7.410 × 10⁻³ eV = 5.58 × 10⁻¹³ s = 558 fs.
Ratios: 15.2 fs/0.349 fs = 43.6 and 558 fs/15.2 fs = 36.7. The counting gives ħω/Eₑₗ ≈ (mₑ/μ)¹⁄² and B/ħω ≈ (mₑ/μ)¹⁄², so each step should be about √918.1 = 30.3 times slower.
Checks: B = ħ²/(2μR²) has units (J s)²/(kg m²) = J. The estimate and the numbers agree within a factor of 1.5, which is all a power count promises, and the electron completes about 44 cycles per vibration, so it can follow the protons adiabatically.

Answer0.349 fs, 15.2 fs and 558 fs. Each step down the hierarchy is 37 to 44 times slower, against the estimate √918 = 30: the same order, as the counting promises.

MediumTabulated constants for H₂ are ωₑ = 4401.21 cm⁻¹, ωₑxₑ = 121.33 cm⁻¹ and Bₑ = 60.853 cm⁻¹; the nuclear masses are mₚ = 1.007276 u and m_d = 2.013553 u. Assuming D₂ moves on H₂'s Born-Oppenheimer curve, predict D₂'s ωₑ and Bₑ, then use G(0) = ωₑ/2 − ωₑxₑ/4 to predict D₀(D₂) − D₀(H₂). The measured D₀ are 36 118.07 cm⁻¹ for H₂ and 36 748.36 cm⁻¹ for D₂.

One curve means one Rₑ, Dₑ and k; only μ changes, by μ(D₂)/μ(H₂) = (m_d/2)/(mₚ/2) = 2.013553/1.007276 = 1.999008.
ωₑ ∝ μ^(−1/2): 4401.21/√1.999008 = 4401.21/1.413863 = 3112.90 cm⁻¹, against the tabulated 3115.50 cm⁻¹, 0.08 per cent away. Bₑ ∝ 1/μ: 60.853/1.999008 = 30.442 cm⁻¹, against the tabulated 30.443 cm⁻¹.
Anharmonicity goes as 1/μ (for a Morse well ωₑxₑ = (ħωₑ)²/(4Dₑ), with Dₑ fixed), so D₂ has 121.33/1.999008 = 60.70 cm⁻¹.
Zero-point energies: G(0) = 2200.61 − 30.33 = 2170.27 cm⁻¹ for H₂ and 1556.45 − 15.17 = 1541.28 cm⁻¹ for D₂.
Both are measured up from the same well bottom, so D₀ = Dₑ − G(0) and D₀(D₂) − D₀(H₂) = 2170.27 − 1541.28 = 629.00 cm⁻¹ = 0.0780 eV. Measured: 36 748.36 − 36 118.07 = 630.29 cm⁻¹. The 1.3 cm⁻¹ left over, 0.2 per cent, is the size of the higher vibrational terms and mass-dependent corrections this model omits.

Answerωₑ(D₂) = 3112.9 cm⁻¹ and Bₑ(D₂) = 30.442 cm⁻¹, against tabulated 3115.50 and 30.443 cm⁻¹; D₀(D₂) − D₀(H₂) = 629.0 cm⁻¹ = 0.0780 eV, against the measured 630.3 cm⁻¹.

HardModel lithium fluoride with a covalent curve Li + F, flat at E = 0, and an ionic curve Li⁺ + F⁻ at ΔE − e²/(4πε₀R), where ΔE = IE(Li) − EA(F) = 5.3917 − 3.4011 eV (table values), coupled by V = 0.030 eV (a model value). Find where the curves cross and the peak non-adiabatic coupling d₁₂, test the condition ħv|d₁₂| ≪ 2V for nuclei passing at 2.0 km s⁻¹, and find the Landau-Zener probability of staying diabatic.

ΔE = 1.9906 eV. The curves cross where e²/(4πε₀R_c) = ΔE: R_c = 1.43996 eV nm ÷ 1.9906 eV = 0.7234 nm. Their slopes there differ by ΔF = e²/(4πε₀R_c²) = ΔE/R_c = 2.752 eV nm⁻¹.
Near R_c the electronic Hamiltonian in the diabatic basis has diagonal ±ΔF(R − R_c)/2 and off-diagonal V, Quantum Mechanics I's avoided crossing with gap 2V. Its eigenvectors are rotated by θ(R) with tan 2θ = 2V/[ΔF(R − R_c)], so |d₁₂| = |dθ/dR| = VΔF/[ΔF²(R − R_c)² + 4V²].
At R_c this peaks at ΔF/(4V) = 2.752/0.120 = 22.9 nm⁻¹, matching ⟨φ₁|∂_R Hₑ|φ₂⟩/(E₂ − E₁) = (ΔF/2)/(2V), over a width 2V/ΔF = 0.0218 nm = 21.8 pm. At 0.20 nm from R_c it has fallen to 0.269 nm⁻¹.
At v = 2.0 km s⁻¹, about 0.11 eV of relative kinetic energy for ⁷Li¹⁹F, ħv = 6.582 × 10⁻¹⁶ eV s × 2.0 × 10¹² nm s⁻¹ = 1.316 × 10⁻³ eV nm. At the crossing ħv|d₁₂| = 0.0302 eV, half the gap 2V = 0.060 eV, so the condition fails. At 0.20 nm away it is 3.5 × 10⁻⁴ eV, 0.6 per cent of the gap, and the separation holds.
Landau-Zener (topic 14.04, quoted): P = exp[−2πV²/(ħvΔF)] = exp[−2π × 9.0 × 10⁻⁴ eV²/(1.316 × 10⁻³ eV nm × 2.752 eV nm⁻¹)] = exp(−1.561) = 0.210. Dimensions: eV² over (eV nm × eV nm⁻¹) is a pure number.
Limits: as v → 0, P → 0, the Born-Oppenheimer limit; as V → 0, P → 1 and the electron never transfers. The two tests are one: ħvΔF/(8V²) = 0.503 is the step-4 ratio, and the exponent is π/(4 × 0.503) = 1.561.

AnswerR_c = 0.723 nm; |d₁₂| peaks at 22.9 nm⁻¹ over about 22 pm; ħv|d₁₂| = 0.030 eV is half the 0.060 eV gap, so the separation fails there, and P(stay diabatic) = e^(−1.56) = 0.21.