Physical Chemistry · Part 8 of 9

Molecular Quantum Mechanics

Quantum Chemistry, Part 8 · 9 sections · about 23,528 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

Molecules introduce a second problem on top of the many-electron one: nuclei that move. The Born–Oppenheimer approximation separates the two, and everything chemists mean by a potential-energy surface, a bond length or a reaction path depends on it. This part builds molecular orbital theory from the one molecule that can be solved exactly, sets it honestly against valence bond theory, and ends with Hückel theory, which is the most examinable calculation in the whole subject. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 8

  • 1The molecular Hamiltonian, the BO separation and potential-energy surfaces Free below
  • 2H2+: the molecule that can be solved
  • 3LCAO in general, and where Born–Oppenheimer breaks down
  • 4Homonuclear diatomics: the correlation diagram, s–p mixing and the N2/O2 crossover
  • 5Heteronuclear diatomics, valence bond theory, and a fair comparison
  • 6Hybridisation as algebra, and MO theory for polyatomics by symmetry
  • 7The approximations, the secular determinant, and ethene
  • 8Allyl, butadiene and benzene; delocalisation energy, charge densities and bond orders
  • 9Cyclic systems, the Frost circle, (4n + 2), heteroatoms and extended Hückel

The molecular Hamiltonian, the BO separation and potential-energy surfaces

Free extract

Section H.1 of Part 8, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

What the full problem looks like, which term gets thrown away, why throwing it away is allowed, and what you get in exchange.

Writing down the exact problem

Take a molecule with n electrons at positions ri and N nuclei of charge ZA and mass MA at positions RA. Every interaction is Coulombic, and every particle has kinetic energy. In atomic units—which we use throughout, so ℏ = me = e = 4πε0 = 1—the exact non-relativistic Hamiltonian is:

Ĥ = −∑A (1/2MA)∇²A − ∑i ½∇²i − ∑i,A ZA/riA + ∑i<j 1/rij + ∑A<B ZAZB/RAB
The five terms of the molecular Hamiltonian, and which one has to goThe exact non-relativistic molecular Hamiltonian (atomic units, i = electrons, A = nuclei)T̂n−∑A (1/2MA)∇²AT̂e−∑i (1/2)∇²iV̂ne−∑i,A ZA/riAV̂ee+∑i<j 1/rijV̂nn+∑A<B ZAZB/RAB++++Ĥ =Thrown away in step 1Kept in step 1 (the electronic problem)T̂n only.Set it to zero: clamp every nucleus.Justification: MA ≥ 1836 me,so 1/2MA is at most 5.4 × 10⁻⁴of the electronic 1/2 (= 1/1836).Restored in step 2 as the nuclearSchrödinger equation on the surface.T̂e + V̂ne + V̂ee + V̂nnV̂nn is a CONSTANT once the nuclei areclamped — it shifts every electroniceigenvalue by the same amount, so it can beadded at the end without changing any ψ.The eigenvalue Eel(R) that comes outIS the potential energy surface.
Everything in molecular quantum mechanics follows from what you do with the first term. The exact Hamiltonian is not separable, because V̂ne couples electronic and nuclear coordinates in the same denominator. Born–Oppenheimer deletes T̂n, making the nuclear coordinates parameters rather than variables. V̂nn survives but is only an additive constant at fixed geometry — which is why it never appears in an orbital-shape argument, and why it makes every curve turn up at short R.

Five terms. Read them in order: nuclei moving, electrons moving, electrons pulled by nuclei, electrons pushing each other apart, nuclei pushing each other apart. Nothing is missing except relativity and the coupling of spin to orbital motion, both of which are small for light atoms.

Why can this not be separated? Because of the third term. ZA/riA = ZA/|ri − RA| contains an electronic coordinate and a nuclear coordinate inside the same modulus sign. You cannot write it as (something in r) + (something in R), so you cannot write Ψ as a product and separate. The same term is what makes chemistry: it is the electron–nucleus attraction, and without it there would be no molecules to worry about.

The one number that lets us proceed

A proton is 1836 times heavier than an electron. That is the whole argument, and it does two jobs at once.

Why a factor of 1836 is enoughThe one number that does all the workWhat follows from itmp / me = 1836.153At the SAME kinetic energy,v ∝ 1/√m, so  ve / vp = √(mp/me) = 42.85The natural expansion parameter ofthe Born–Oppenheimer series isκ = (me/M)1/4 = 0.1528so successive corrections fall offroughly by a factor of 7 each time.Electronic motion is ~43× faster.• Nuclei look STATIONARY to the electrons:  solve for ψel at fixed R.• Electrons look INSTANTANEOUS to the  nuclei: they feel only the averaged  electronic energy Eel(R).• Typical electronic excitation ~ 5 eV;  typical vibrational quantum ~ 0.1–0.4 eV;  typical rotational quantum ~ 10⁻³ eV.  Three separated energy scales — and BO is  good exactly when they stay separated.The approximation is not that the nuclei do not move. It is that at each instant the electrons have already finished adjusting.
The whole of Born–Oppenheimer rests on one mass ratio, and the numbers here are computed at build time from the CODATA masses. Note the fourth root: the small parameter of the systematic expansion is not me/M but (me/M)1/4, about 0.15 for hydrogen and 0.08 for carbon. That is also why deuteration measurably changes bond lengths and zero-point energies: the approximation is good, but it is not mass-independent.

Job one: the nuclear kinetic-energy term is tiny. Look at −(1/2MA)∇²A next to −½∇²i. The prefactors differ by a factor of at least 1836. If the two Laplacians were of comparable size, the nuclear term would contribute at the 0.05% level. Delete it.

Job two: the electrons keep up. At the same kinetic energy a particle's speed goes as 1/√m, so electrons move about 43 times faster than protons. From the electron's point of view the nuclei are frozen scenery; from the nuclei's point of view the electron cloud has already relaxed to its new equilibrium before they have moved appreciably. This is the adiabatic picture, and it is the physical content of the approximation.

Note: Be careful how you state it. The approximation is not ‘the nuclei do not move’. Of course they move — vibration and rotation are nuclear motion, and Part 9 quantises them. The approximation is that at each instant the electrons have already finished adjusting to wherever the nuclei happen to be. An examiner who sets ‘state the Born–Oppenheimer approximation’ is testing exactly this distinction.

The two-step procedure

Having deleted T̂n, what is left is the electronic Hamiltonian, in which R appears only as a set of numbers:

Ĥel(R) = −∑i ½∇²i − ∑i,A ZA/riA + ∑i<j 1/rij
Ĥel(R) ψel(r; R) = Eel(R) ψel(r; R)

Solve that at one geometry and you get a number, Eel(R). Add the nucleus–nucleus repulsion, which is just a constant at fixed geometry:

U(R) = Eel(R) + ∑A<B ZAZB/RAB

Now repeat at every geometry. The function U(R) is the potential-energy surface, and step two is to put it into a Schrödinger equation for the nuclei alone:

[−∑A (1/2MA)∇²A + U(R)] χ(R) = Etotal χ(R)

Its solutions are the vibrational and rotational states of the molecule. Part 9 does exactly this for a diatomic and recovers the harmonic oscillator and the rigid rotor. Note what has happened to the total energy: it is not the electronic energy, and it is not the vibrational energy; it is a single eigenvalue of the second equation, in which the first equation's answer has been buried in the potential.

Potential-energy surface (PES): the total energy of a molecule — electronic energy plus nuclear repulsion — as a function of nuclear geometry, with the electrons in a specified state (usually the ground state). It has 3N − 6 dimensions for a non-linear molecule of N atoms, 3N − 5 if linear.
One curve, and everything a chemist reads off it050100150200250300−5−4−3−2−1012internuclear separation R / pmE(R) − E(∞) / eVdissociation limit: two free atomsRe = equilibrium bond length(the MINIMUM, not an average)Dev = 0v = 1v = 2v = 3v = 4zero-point energy ½ℏωD0 = De − ZPE— and D0 is what a thermochemist measuresV̂nn = ZAZB/R wins as R → 0— every curve goes to +∞the curvature at the bottom, k = d²E/dR², is the force constant— and ν = (1/2π)√(k/μ) is the infrared stretching frequency
The potential-energy curve is not an extra piece of theory; it IS the electronic eigenvalue Eel(R) plotted against R, and it exists only because Born–Oppenheimer made R a parameter. Read off it: the bond length (the minimum), the depth De (theoretical — measured from the bottom of a well nothing ever sits in), the true dissociation energy D0 (from v = 0, what calorimetry gives), the force constant (the curvature) and hence the vibrational frequency, and the anharmonicity (converging levels). Drawn from a Morse function with illustrative parameters: the shape, not the numbers, is the point.

For a diatomic there is one internuclear distance and the surface is a curve. Everything a chemist says about the bond is a statement about that curve. The equilibrium bond length Re is where dU/dR = 0. The force constant k = (d²U/dR²) at Re gives the stretching frequency. The well depth De is the energy from the bottom of the well to the dissociation asymptote — but nothing ever sits at the bottom of the well, because of zero-point energy, so what a calorimeter measures is D0 = De − ½ℏω.

⚠ Common mistakes & exam traps

  • Confusing De and D0. De is the well depth, a theoretical quantity. D0 is measured from the v = 0 level and is what experiment gives. D0 < De, always, and the difference is the zero-point energy. Isotopic substitution changes D0 but not De, which is a clean test of whether a question is about the surface or about the nuclear motion on it.
  • Saying the PES is ‘the electronic energy’. It is the electronic energy plus the nuclear repulsion. Leave out the 1/R and your curve does not turn up at short distance and has no minimum.
  • Thinking the reaction coordinate is a physical coordinate. It is a path constructed on the surface after the fact, usually the steepest-descent path from the saddle point. It is not a normal mode and not a bond length.
  • Assuming one PES per molecule. There is one per electronic state. Photochemistry is the study of what happens on the excited-state surfaces and how molecules get between them.
Advanced / reference layer

The formal expansion, the adiabatic versus the Born–Oppenheimer wavefunction, the diagonal correction, and the coupling terms that were quietly dropped.

The exact expansion, and what is thrown away

The derivation above was a physical argument, not a proof. Here is the proper version. The electronic eigenfunctions ψk(r; R), at each fixed R, form a complete orthonormal set in the electronic coordinates. So the exact total wavefunction can be expanded in them with no approximation whatever:

Ψ(r, R) = ∑k χk(R) ψk(r; R)

Substituting into the full Schrödinger equation, multiplying by ψj*(r; R) and integrating over the electronic coordinates gives a set of coupled equations for the nuclear functions χj(R). The nuclear Laplacian acting on a product generates the cross terms, and after the dust settles:

[T̂n + Uj(R) + Λjj(R)] χj + ∑k≠j Λjk χk = E χj
Λjk = −∑A (1/MA) ⟨ψj|∇A|ψk⟩ · ∇A − ∑A (1/2MA) ⟨ψj|∇²A|ψk⟩

Three levels of approximation now present themselves, and the names are worth getting right because examiners use them precisely:

LevelWhat is keptNameComment
Exactall ΛjkBorn–Huang / coupled channelsno approximation; the electronic states are coupled by nuclear motion
Drop k ≠ jΛjj onlyadiabatic approximationone surface, but corrected: Uj + Λjj. Λjj is the diagonal Born–Oppenheimer correction (DBOC), of order me/M.
Drop all ΛnothingBorn–Oppenheimer approximationthe surface is Uj(R) alone; the mass of the nuclei enters only through T̂n
In careless usage ‘adiabatic’ and ‘Born–Oppenheimer’ are used interchangeably. Strictly, the adiabatic approximation retains the diagonal correction and the BO approximation does not; both keep the molecule on a single surface, and both fail in the same places.

The first term of Λjk is the dangerous one. Using the Hellmann–Feynman theorem, the off-diagonal derivative coupling can be written

⟨ψj|∇A|ψk⟩ = ⟨ψj|(∇AĤel)|ψk⟩ / (Ek − Ej)

There is the whole story in one denominator. The neglected coupling is inversely proportional to the energy gap between electronic states. When surfaces are far apart, the coupling is negligible and Born–Oppenheimer is superb — often better than chemical accuracy. When two surfaces approach, the coupling diverges and the approximation collapses. Everything in H.3's discussion of conical intersections is a consequence of this single expression.

How good is it, quantitatively?

The systematic expansion is not in me/M but in κ = (me/M)1/4, because the amplitude of nuclear vibration scales as (me/M)1/4 in units of the electronic length: the curvature of the well is set by electronic energies while the mass is nuclear. Successive orders in κ are the electronic energy (κ⁰, 1–100 eV), the harmonic vibrational energy (κ², 0.1–0.5 eV), the anharmonic and vibration–rotation corrections (κ³), and the rotational energy together with the diagonal BO correction (κ⁴); odd powers vanish at a stationary point. For a proton κ = 0.1528, for carbon about 0.0822. The clean separation of electronic, vibrational and rotational spectroscopy into three energy regimes is a direct consequence of this ordering.

Note: The Hellmann–Feynman theorem lives entirely inside this framework: for an exact eigenfunction, dE/dλ = ⟨ψ|∂Ĥ/∂λ|ψ⟩. Taking λ to be a nuclear coordinate, the force on a nucleus is the ordinary electrostatic force exerted by the electron density and the other nuclei — which is the licence for molecular dynamics and for analytic-gradient geometry optimisation. The theorem needs a basis that does not move with the atoms; real atom-centred bases do move, and the extra term is the Pulay force.

Estimate the fractional error made by neglecting the nuclear kinetic energy in H2, and comment Medium

Set-up. In a bound Coulomb system a particle's kinetic energy is of the order of its confinement energy, ℏ²/(2mΔx²).
Electron. Δxe ≈ a₀, so Te ≈ ℏ²/(2mea₀²) = 1 hartree = 27.21 eV.
Nucleus. The v = 0 vibrational amplitude of H2 is roughly 0.15 a₀, and the reduced mass is mp/2. So Tn ≈ (me/(mp/2)) × (1/0.15²) × ½ hartree = 0.00109 × 44.4 × 0.5 = 0.0242 hartree ≈ 0.659 eV.
Compare. That is the right order for the H2 zero-point energy (about 0.27 eV) — roughly 1% of the electronic energy scale and about 6% of the bond energy. Neglecting T̂n in the electronic step is therefore safe to a fraction of a percent, but you cannot then forget nuclear motion altogether: the zero-point energy is chemically significant.
The moral. Born–Oppenheimer is an excellent approximation for the shape of the surface and a poor excuse for ignoring what happens on it. Hydrogen is the worst case; heavier nuclear energies shrink as 1/M.
Easy
Why does the nucleus–nucleus repulsion term not affect the shapes of the molecular orbitals?
Show solution
At a fixed nuclear geometry Vnn is a constant, not an operator on the electronic coordinates. Adding a constant c to a Hamiltonian leaves every eigenfunction unchanged and shifts every eigenvalue by c. So it changes the energy of the surface but not the orbitals at that geometry. It does of course change the shape of the surface across geometries, which is why the curve turns up at small R.
Med
A student says the BO approximation must be worse for I2 than for H2 because iodine has far more electrons. Is that right?
Show solution
No — and the reasoning is backwards. The quality of the BO separation is governed by me/M, the electron-to-nuclear mass ratio, and by the electronic energy gaps. Iodine nuclei are about 127 times heavier than a proton, so κ = (me/M)1/4 is smaller by a factor of 1271/4 ≈ 3.4 and the separation is better. What is worse for iodine is the neglected relativistic and spin–orbit physics, which has nothing to do with BO.
Hard
Explain, using the energy denominator in the derivative coupling, why Born–Oppenheimer is usually excellent for ground-state thermal chemistry and usually inadequate for photochemistry.
Show solution
The non-adiabatic coupling carries a factor 1/(Ek − Ej). In thermal ground-state chemistry the molecule stays on S0, and the gap to S1 is typically 3–6 eV — enormous compared with the vibrational quanta driving the nuclear motion — so the coupling is negligible. In photochemistry the molecule is placed on an excited surface which, along the relaxation path, approaches or touches another surface: the gap goes to zero, the coupling diverges, and population transfers between surfaces on a femtosecond timescale. The single-surface picture is then not a small error but the wrong picture entirely.
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