Physical Chemistry · Part 6 of 9

Approximation Methods

Quantum Chemistry, Part 6 · 10 sections · about 26,300 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

Beyond hydrogen, nothing is exactly solvable — so everything else in quantum chemistry is approximation. Two methods do almost all the work, and the exam reliably asks you to choose between them and to execute one by hand. This part derives both from scratch, applies them to the same problems so the comparison is concrete, and is honest about where each one fails. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 10 sections in Part 6

  • 1The variation theorem, stated and proved Free below
  • 2Trial functions, variational parameters, and the minimisation
  • 3The linear variation method, secular equations and the secular determinant
  • 4Rayleigh–Schrödinger theory: first-order energy and wavefunction
  • 5The second-order energy, its sign, and where the series fails
  • 6Degenerate perturbation theory and the Stark effect
  • 7Time-dependent perturbation theory, Fermi's golden rule and selection rules
  • 8The helium atom: both methods on the same problem
  • 9The WKB approximation
  • 10Born–Oppenheimer as an approximation, and how CSIR-NET tests all this

The variation theorem, stated and proved

Free extract

Section F.1 of Part 6, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

One inequality, three lines of proof, and it is the reason every quantum chemistry program in the world minimises something.

Here is the situation. You have a Hamiltonian Ĥ whose eigenfunctions you cannot find. You want the ground-state energy E0. You are willing to guess a function, but you have no way of checking your guess against the right answer, because you do not have the right answer. It sounds hopeless. It is not, and the escape is one of the most useful theorems in all of chemistry.

The variation theorem: for any normalisable, well-behaved function φ that satisfies the boundary conditions of the problem, the expectation value of the Hamiltonian is never less than the true ground-state energy: W[φ] = ⟨φ|Ĥ|φ⟩ / ⟨φ|φ⟩ ≥ E0. Equality holds if and only if φ is itself the ground-state eigenfunction.
W[φ] = ⟨φ|Ĥ|φ⟩ / ⟨φ|φ⟩ ≥ E0

The quantity W is called the variational energy or the Rayleigh quotient. Note the denominator: it is there so that you do not have to normalise your trial function before you start. If φ happens to be normalised, ⟨φ|φ⟩ = 1 and the denominator disappears, but in practice keeping it saves a step.

Now read the inequality slowly, because everything follows from what it says and does not say. It says W is never below E0. It does not say W is close to E0; a stupid guess gives a stupid answer. What it gives you is a direction: since no trial function can ever produce a number below the truth, lower is always better. If trial function A gives −2.75 hartree and trial function B gives −2.85 hartree, then B is the better wavefunction. Full stop. You do not need to know E0 to know that.

That is the whole practical content of the theorem, and it is why the method is called variational: you build a family of trial functions containing adjustable parameters, vary the parameters, and take the lowest energy the family can reach.

The proof, in words first

The proof depends on one idea you have already met: the eigenfunctions of a Hermitian operator form a complete set. That means any acceptable function can be written as a superposition of them — including your trial function, and including the case where you have no idea what those eigenfunctions look like. You never have to compute the expansion. You only have to know it exists.

φ = ∑n cnψn ,   Ĥψn = Enψn ,   E0 ≤ E1 ≤ E2 ≤ …

Now put that expansion into the numerator and the denominator of W. Because the ψn are orthonormal, every cross term dies and you are left with two sums:

⟨φ|Ĥ|φ⟩ = ∑n |cn|² En ,    ⟨φ|φ⟩ = ∑n |cn|²

So W is a weighted average of the exact energies, with weights |cn|² that are automatically non-negative (they are squared moduli) and automatically sum to the normalisation. And an average of a set of numbers, none of which is smaller than E0, cannot itself be smaller than E0. Done.

Why the variational energy can never dip below the truthEnergy →E0|c0|² = 0.62E1|c1|² = 0.22E2|c2|² = 0.10E3|c3|² = 0.04E4|c4|² = 0.02WW − E0 ≥ 0Expand the guess in the (unknown)exact eigenfunctions:  φ = ∑n cnψnThen, with φ normalised,  W = ∑n |cn|² En  ∑n |cn|² = 1W is a WEIGHTED AVERAGE of thetrue energies. An average of numbersnone of which is below E0 cannotitself be below E0.That is the entire proof. Everythingelse in F.1 is bookkeeping.The weights are unknown — you never compute them — but they exist, they are non-negative and they sum to one. That is all the proof needs.
The variation theorem, drawn. Any acceptable trial function can be written as a superposition of the exact eigenfunctions, whether or not you know what they are. Its expectation value is then a weighted average of the exact energies, with non-negative weights that sum to unity. A weighted average is always at least as large as the smallest thing being averaged, so W ≥ E0, with equality only when all the weight sits on ψ0. The weights shown here are illustrative; the logic does not depend on their values.

If you take away one image from F.1, take away that figure. The trial function secretly contains some of the ground state, some of the first excited state, some of the second, and so on. Its energy is the average of the true energies over that secret mixture. The only way to reach E0 is to have all the weight on ψ0 — that is, to have guessed the exact wavefunction.

Advanced / reference layer

The proof written out with the inequality made explicit, the equality condition, and the extensions to excited states and to states of definite symmetry.

Write φ = ∑n cnψn where {ψn} is the complete orthonormal set of eigenfunctions of Ĥ with eigenvalues E0 ≤ E1 ≤ …. Then

W − E0 = [∑n|cn|²En − E0∑n|cn|²] / ∑n|cn|² = ∑n|cn|²(En − E0) / ∑n|cn|²

Every factor on the right is non-negative: |cn|² ≥ 0 by construction, and En − E0 ≥ 0 because E0 is by definition the smallest eigenvalue. A sum of non-negative terms divided by a positive number is non-negative. Hence W ≥ E0. This is the version to reproduce in an exam: it is four lines and the marker can see the inequality being used rather than asserted.

The equality condition. W = E0 requires every term |cn|²(En − E0) to vanish. For every state above the ground state this forces cn = 0 — unless En = E0, that is unless the state is degenerate with the ground state. So the precise statement is: equality holds if and only if φ lies entirely within the ground-state eigenspace. For a non-degenerate ground state that means φ ∝ ψ0.

Three technical conditions that are usually left unsaid, and which occasionally appear as a Part-C sting:

  • The spectrum must be bounded below. If Ĥ has no lowest eigenvalue the theorem is vacuous. Real molecular Hamiltonians are bounded below; a Dirac Hamiltonian is not, which is why naive variational relativistic calculations collapse into the negative-energy continuum. This failure has a name, variational collapse, and dealing with it is a live problem in relativistic quantum chemistry.
  • φ must be in the domain of Ĥ. It must be square-integrable, continuous, with a piecewise-continuous first derivative, and it must obey the boundary conditions of the problem. A trial function for the particle in a box that does not vanish at the walls is not merely inaccurate, it is illegal, and it can return a value below E0. Students who report W < E0 have almost always violated a boundary condition or dropped a surface term when integrating by parts.
  • ⟨φ|φ⟩ must be finite and non-zero. Obvious, but it rules out trial functions that do not decay, and it is the reason a Gaussian is legal for the H atom while a constant is not.

Excited states: two ways to get at them

The theorem as stated bounds the ground state only, and this is the method's real limitation. There are two escapes.

By orthogonality. If φ is constructed to be orthogonal to the exact ground state, then c0 = 0 and the same argument gives W ≥ E1. The catch is that you need the exact ψ0 to impose the orthogonality, which you do not have. In practice one imposes orthogonality to the best available approximation to ψ0, and the bound becomes approximate.

By symmetry, which is the version that actually works. Suppose Ĥ commutes with some symmetry operator (parity, a point-group operation, L̂², Ŝ²). Then the eigenfunctions can be classified by symmetry, and a trial function of a given symmetry has zero overlap with every eigenfunction of different symmetry. The variational energy of such a φ therefore bounds the lowest eigenvalue of that symmetry.

The symmetry-restricted variation theorem: if φ belongs to a definite irreducible representation Γ (or a definite eigenvalue of some operator commuting with Ĥ), then W[φ] ≥ E0Γ, the lowest exact eigenvalue of that same symmetry. This is what makes variational calculations of excited states possible at all, and it is why quantum-chemistry programs ask you for the state symmetry before they start.

Concretely: the lowest triplet state of a molecule can be computed variationally even though it lies above the singlet ground state, because the spin symmetry is exact and the two spaces do not mix. The lowest 1B1u state of benzene can be bounded even though many states lie below it. But the second 1A1g state cannot be bounded this way, and that is a real restriction.

Hylleraas–Undheim and MacDonald: the bracketing theorem

For the linear variation method of F.3 there is a much stronger statement, and it is worth knowing by name because it is what licenses configuration-interaction calculations of excited states.

The Hylleraas–Undheim–MacDonald theorem: if the linear variation problem is solved in a basis of n functions, the n roots W1 ≤ W2 ≤ … ≤ Wn satisfy Wk ≥ Ek−1 for every k. That is, the k-th root is an upper bound to the k-th exact eigenvalue, not just the lowest. Adding a further basis function pushes every root down or leaves it where it was, and the new roots interlace the old ones.

This is a far stronger guarantee than the elementary theorem, and it is the reason a CI calculation can report a whole spectrum rather than one number. It is also the reason increasing a basis set never makes an excited state artificially too low.

What the theorem does not do

  • It gives no error bar. W ≥ E0 tells you the sign of the error, never its size. You can be 40% too high and the theorem is perfectly satisfied. Rigorous lower bounds exist (Temple's inequality is the classic) but they need ⟨Ĥ²⟩ and knowledge of E1, and they are almost always too loose to be useful.
  • It says nothing about other properties. A wavefunction optimised to give the best energy is not necessarily the best wavefunction for the dipole moment, the electron density at the nucleus, or anything else. F.2 explains why: the energy is correct to second order in the error of the wavefunction, whereas other expectation values are only correct to first order.
  • A lower W does not guarantee a better wavefunction in every respect, only a better energy. Two trial functions can give nearly the same energy with very different densities, which is exactly the situation that makes density-functional benchmarking difficult.

Prove that W ≥ E0, and state precisely when equality holds Medium

Set up. Let {ψn} be the complete orthonormal eigenset of Ĥ, Ĥψn = Enψn, ordered so that E0 is the smallest eigenvalue. Let φ be any acceptable trial function, not necessarily normalised.
Step 1 — expand. Completeness allows φ = ∑ncnψn with cn = ⟨ψn|φ⟩. We never need the values of cn; we need only that they exist.
Step 2 — the two integrals. Using orthonormality, ⟨φ|φ⟩ = ∑n|cn|² and ⟨φ|Ĥ|φ⟩ = ∑m∑ncm*cnEn⟨ψm|ψn⟩ = ∑n|cn|²En.
Step 3 — subtract E0. W − E0 = ∑n|cn|²(En − E0) / ∑n|cn|². Do the subtraction inside the sum; this is the step that makes the sign visible.
Step 4 — read the signs. |cn|² ≥ 0 always. En − E0 ≥ 0 because E0 is the lowest eigenvalue. The denominator is strictly positive because φ is not the zero function. Therefore W − E0 ≥ 0. □
Step 5 — equality. The sum vanishes only if every term vanishes. For each n > 0 either cn = 0 or En = E0. So equality holds exactly when φ is a member of the ground-state eigenspace — for a non-degenerate ground state, when φ is proportional to ψ0.
Marker's note. The two places marks are lost: forgetting the denominator (which makes the proof only valid for normalised φ, a weaker result), and asserting En ≥ E0 without saying it is by definition of the ground state.

A trial function returns W = −0.55 hartree for a system whose ground state you know to be −0.50 hartree. What went wrong? Easy

The theorem has been violated, so the calculation is wrong. No arithmetic slip in the integrals can produce this — a wrong number would still have to be above −0.50 if the setup were legal. Something structural is broken.
Candidate 1 — the trial function breaks a boundary condition. The commonest cause. For a box, φ(0) or φ(L) ≠ 0; for an atom, a function that does not decay at infinity. Illegal functions lie outside the space the theorem is about.
Candidate 2 — a dropped surface term. Evaluating ⟨φ|T̂|φ⟩ by parts generates a boundary term [φ*dφ/dx]. It vanishes for legal functions and does not for illegal ones. Silently discarding it manufactures energies below E0.
Candidate 3 — the normalisation integral was dropped or mis-evaluated. If ⟨φ|φ⟩ > 1 and you forgot to divide, W comes out too negative by exactly that factor.
Candidate 4 — the Hamiltonian is not the one you think. A missing term, a wrong sign on the potential, or a factor of 2 in ℏ²/2m.
The general lesson. W < E0 is a diagnostic, not a discovery. It is the most useful bug detector in computational chemistry, and it is worth building the check into every calculation you do.

⚠ Common mistakes & exam traps

  • Saying ‘the variation theorem gives the ground-state energy’. It gives an upper bound to it. The difference is the whole theorem.
  • Forgetting the denominator ⟨φ|φ⟩. Legal only if you have already normalised φ. In an exam, write the quotient; it costs one line and it is sometimes worth a mark on its own.
  • Believing a lower energy means a better wavefunction for every purpose. It means a better energy. Dipole moments, densities and spin properties from a variationally optimised function can be worse than those from a cruder one.
  • Applying the plain theorem to an excited state. Without an orthogonality or symmetry constraint there is no bound at all: your trial function will simply collapse onto the ground state when you minimise.
  • Using a trial function that violates the boundary conditions and then being pleased with an energy below E0. That is a broken calculation, not a breakthrough.
  • Assuming the theorem needs the exact eigenfunctions to be known. It needs them to exist and to be complete. This is the conceptual point most often missed in Part-C answers.
Easy
A student computes W = −13.4 eV for the hydrogen atom ground state using a trial function with one parameter, then a second student gets −13.55 eV with a two-parameter function. Which wavefunction is better, and can either be exact?
Show solution
The second: it lies lower, and by the variation theorem lower is always closer to the true E0 = −13.6057 eV. Neither is exact, because both lie above the true value. Note that you could answer ‘the second is better’ even without knowing the exact energy — that is the point of the theorem.
Med
Explain why a trial function of odd parity gives a variational bound on the lowest odd-parity state of a symmetric potential, even though that state is not the ground state.
Show solution
For a symmetric potential the parity operator commutes with Ĥ, so eigenfunctions have definite parity and functions of opposite parity are orthogonal. An odd trial function therefore has cn = 0 for every even eigenstate, and the expansion W = ∑|cn|²En/∑|cn|² runs only over odd states. Since the lowest of those is Eodd,0, we get W ≥ Eodd,0. This is the symmetry-restricted variation theorem, and it generalises to any operator commuting with Ĥ.
Hard
Trial function φ is exactly ψ0 + 0.1ψ1 for a system with E0 = −10 eV and E1 = −4 eV. Compute W and comment on how the error in W compares with the error in the wavefunction.
Show solution
⟨φ|φ⟩ = 1 + 0.01 = 1.01. ⟨φ|Ĥ|φ⟩ = 1×(−10) + 0.01×(−4) = −10.04 eV. So W = −10.04/1.01 = −9.9406 eV, which is above E0 as required. The error in the energy is 0.0594 eV, i.e. 0.59%. The wavefunction, however, is contaminated by 10% of the wrong state in amplitude. The energy error is second order in the amplitude error (0.1² = 0.01 times the level spacing of 6 eV gives 0.06 eV, which is what we got). This is the key practical fact about the variation method: energies are far more accurate than the wavefunctions that produce them.
Med
Can the variation method ever give a lower bound to E0?
Show solution
Not in its standard form — the inequality only runs one way. Lower bounds require extra information. Temple's inequality gives E0 ≥ W − (⟨Ĥ²⟩ − W²)/(E1 − W), but it needs ⟨Ĥ²⟩ (expensive, and very sensitive to the quality of φ near singularities) and a lower bound on E1 that you usually do not have. In practice, quantum chemistry lives entirely with upper bounds and estimates the remaining error empirically.

Read the rest of Part 6

The remaining 9 sections of this part — Trial functions, variational parameters, and the minimisation, The linear variation method, secular equations and the secular determinant, Rayleigh–Schrödinger theory: first-order energy… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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