University Quantum Mechanics II · The Variational Principle · 10.1

The variational theorem: ⟨H⟩ is never below E₀

When a problem offers no small parameter to expand in, this is the method to reach for. Guess a state, evaluate ⟨H⟩, and the number is guaranteed to sit at or above the true ground-state energy. We prove why, pin down which guesses are allowed, and see why a rough guess earns a good energy that must not be over-read.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Perturbation theory needs a small parameter; the variational principle needs only an inequality. Take any trial state you can write down and expand it, in principle, on the eigenstates of H that you cannot write down. Its energy ⟨H⟩ = Σₙ|cₙ|²Eₙ is then an average of the true levels with non-negative weights adding to one, and an average cannot fall below its smallest entry, so ⟨H⟩ ≥ E₀ for every admissible trial. Admissible means what the proof quietly used: H must have a lowest level at all, and the trial must be normalisable with finite kinetic energy, which lets a kink in ψ through and stops a jump.

The bound is also forgiving. Write the trial as the ground state plus a wrong piece of size ε, and the cross terms vanish because the ground state is an eigenstate, so ⟨H⟩ − E₀ grows as ε²: a wrong amplitude of 10% raises the energy by only 1% of the gap to the admixed level. ⟨H⟩ is flat at every eigenstate, a minimum only at the lowest. That forgiveness is also where the method stops. The number is one-sided and carries no error bar: it certifies that E₀ lies lower, not by how much.

And because a second-order energy error hides a first-order error in the state, a good energy certifies neither the wavefunction, nor its value at any point, nor any other observable computed from it. Overlaps and lower bounds need more information: the next level, or ⟨H²⟩.

Simple definition: The variational theorem says that for any admissible trial state ψ the Rayleigh quotient ⟨ψ|H|ψ⟩/⟨ψ|ψ⟩ is at least the ground-state energy E₀, with equality only when ψ is itself a ground state.
Example: Hydrogen's model Hamiltonian has E₀ = −13.60 eV. The trial e^(−r/b) with b = 1.50a, half as wide again as the true orbital, gives ⟨H⟩ = −(8/9) × 13.60 eV = −12.09 eV: above E₀, as it must be, by 1/9 of 13.60 eV.

The bound as an average of true levels⟨H⟩ = ⟨ψ|H|ψ⟩/⟨ψ|ψ⟩ = Σₙ |cₙ|² Eₙ / Σₙ |cₙ|² ≥ E₀

Derived by inserting Σₙ|n⟩⟨n| = 1. Every weight is non-negative, so no trial undercuts E₀, and equality needs every weight above the ground level to vanish.

cₙ = ⟨n|ψ⟩ on the unknown eigenstates of H; Eₙ, E₀ and ⟨H⟩ in J; a continuum adds an integral with non-negative weights

Kinetic energy in its admissible form⟨T⟩ = (ħ²/2m) ∫ |∇ψ|² d³r / ∫ |ψ|² d³r

Derived. Finite for a kink, where ∇ψ jumps; infinite for a jump in ψ, where ∇ψ holds a delta function. Read naively, the ∇²ψ form drops a kink's delta and can undercut E₀.

m in kg, ħ = 1.055 × 10⁻³⁴ J s, ⟨T⟩ in J; one integration by parts from −(ħ²/2m)∫ψ*∇²ψ d³r

An error ε in the state costs ε² in the energy⟨H⟩ − E₀ = ε² (⟨φ|H|φ⟩ − E₀) / (1 + ε²), for |ψ⟩ = |0⟩ + ε|φ⟩

Derived: the cross terms ε⟨φ|H|0⟩ = εE₀⟨φ|0⟩ vanish. An admixture of amplitude ε = 0.1 of a level 8E₁ above the ground raises ⟨H⟩ by 0.079E₁, not 0.8E₁.

|0⟩ the normalised ground state, |φ⟩ normalised with ⟨0|φ⟩ = 0, ε dimensionless; energies in J

Stationary at every eigenstateδ⟨H⟩ = 2 Re⟨δψ|(H − ⟨H⟩)|ψ⟩ / ⟨ψ|ψ⟩

Derived. It vanishes for every δψ exactly when H|ψ⟩ = ⟨H⟩|ψ⟩: each eigenstate is a stationary point, the ground state a minimum and every excited state a saddle.

First-order change in the Rayleigh quotient when ψ becomes ψ + δψ; in J

Eckart's overlap bound1 − |⟨0|ψ⟩|² ≤ (⟨H⟩ − E₀) / (E′ − E₀)

Derived from the same sum with Eₙ ≥ E′ above the ground. It turns an energy into a guaranteed overlap only if E₀ and E′ are known, and says nothing about ψ at a point.

ψ normalised; E′ the lowest excited level ψ can contain; both sides dimensionless

Temple's lower boundE₀ ≥ ⟨H⟩ − (ΔH)² / (E′ − ⟨H⟩), where (ΔH)² = ⟨H²⟩ − ⟨H⟩²

Derived from Σₙ|cₙ|²(Eₙ − E₀)(Eₙ − E′) ≥ 0. The upper bound is free; bracketing E₀ from below costs ⟨H²⟩ and a lower bound on the next level.

Needs ⟨H⟩ below E′, or below any lower bound used in its place; energies in J, (ΔH)² in J²

An average cannot fall below its smallest entry

H is Hermitian, so its eigenstates |n⟩ form a complete orthonormal set even when nobody can write one down. Any normalised trial can be expanded as |ψ⟩ = Σₙ cₙ|n⟩ with Σₙ|cₙ|² = 1, and then ⟨H⟩ = Σₙ|cₙ|²Eₙ: a weighted average of the true levels, with weights that are non-negative and add to one. Subtract E₀ = E₀Σₙ|cₙ|² and ⟨H⟩ − E₀ = Σₙ|cₙ|²(Eₙ − E₀), each term a weight times a gap that cannot be negative. A continuum, such as hydrogen's above 0 eV, adds an integral whose weights and gaps are just as non-negative. Equality needs every weight above the ground level to vanish. The averaging is unforgiving of stray weight high up. A box trial with weights 0.96, 0.03 and 0.01 on ψ₁, ψ₃ and ψ₅ has ⟨H⟩ = (0.96 + 0.03 × 9 + 0.01 × 25)E₁ = 1.48E₁: 96% of the probability in the right state, yet 48% too high, because the stray 4% sits 8E₁ and 24E₁ up. Notice what the proof never used: the eigenstates themselves. They only have to exist.

What the proof assumed: a floor and a finite kinetic energy

Two assumptions hide in that algebra. First, E₀ must exist, so H must be bounded below. Hydrogen is, at −13.6 eV. Hydrogen in a uniform field F is not: the Stark term eFz, with F along z, falls without limit as z → −∞, a trial parked there has ⟨H⟩ as negative as you please, and the field-free ground state becomes a long-lived resonance. Second, ⟨H⟩ must be finite, which needs a normalisable trial with finite kinetic energy. The safe form is ⟨T⟩ = (ħ²/2m)∫|∇ψ|² ÷ ∫|ψ|², one integration by parts from the ∇²ψ form. A kink, where ψ is continuous and its slope jumps, keeps ∫|ψ′|² finite, so kinked trials are admissible. A jump in ψ makes ψ′ a delta function whose square has an infinite integral: the bound survives and says nothing. The ∇²ψ form is the trap. A tent in a box of width L, rising linearly to the centre and back, has ψ″ = 0 away from the kink, so read naively ⟨T⟩ = 0, below E₁, which the theorem forbids. The kink's delta function in ψ″ carries it all: ∫|ψ′|² gives 12ħ²/(2mL²) = 1.216E₁.

An error of order ε costs order ε²

Write the trial as the ground state plus a wrong piece, |ψ⟩ = |0⟩ + ε|φ⟩, with |φ⟩ normalised and orthogonal to |0⟩. Then ⟨ψ|ψ⟩ = 1 + ε² and ⟨ψ|H|ψ⟩ = E₀ + ε⟨0|H|φ⟩ + ε⟨φ|H|0⟩ + ε²⟨φ|H|φ⟩. The cross terms vanish, and not by luck: H|0⟩ = E₀|0⟩, so ⟨φ|H|0⟩ = E₀⟨φ|0⟩ = 0. What is left is ⟨H⟩ − E₀ = ε²(⟨φ|H|φ⟩ − E₀)/(1 + ε²), quadratic in the state error, with the energy of the wrong piece above E₀ as its prefactor. The parabola 1 − u² in a box, with u measured from the centre in half-widths, shows it at work. Its components on the exact states are c₁ = 0.9993, c₃ = −0.0370 and c₅ = +0.0080, falling as 1/n³. The largest wrong amplitude, 3.7%, costs 0.0370² × 8E₁ = 0.0110E₁, c₅ adds 0.0015E₁, and the rest bring ⟨H⟩ to 10/π² E₁ = 1.0132E₁. A trial drawn with no knowledge of sines lands within 1.3% of the energy. Limiting case: as ε → 0 the excess vanishes with zero slope.

Stationary at every eigenstate, a minimum only at the bottom

The cancellation of cross terms needed |0⟩ to be an eigenstate, not the lowest one. Vary any trial by δψ and the Rayleigh quotient changes, to first order, by δ⟨H⟩ = 2Re⟨δψ|(H − ⟨H⟩)|ψ⟩/⟨ψ|ψ⟩. That vanishes for every δψ exactly when H|ψ⟩ = ⟨H⟩|ψ⟩, so the eigenvalue equation and the stationarity of ⟨H⟩ are one statement, and every eigenstate is a flat point. Only the ground state is a minimum. Start at the box's ψ₃, with E₃ = 9E₁. Mixing in ψ₅ with amplitude 0.1 raises ⟨H⟩ to (9 + 0.01 × 25)/1.01 E₁ = 9.158E₁; mixing in ψ₁ with amplitude 0.1 lowers it to (9 + 0.01)/1.01 E₁ = 8.921E₁. That is a saddle, uphill one way and downhill the other. Two consequences follow. An unconstrained minimisation slides off any excited state towards the ground state, which is why the theorem bounds E₀ and nothing above it. And a single trial is guaranteed to bound an excited level only when it is kept exactly out of the downhill directions, for instance by a symmetry the lower states lack, as a later lesson in this unit shows; orthogonality to an approximate ground state is not enough.

What a variational number certifies, and what it cannot

The theorem gives one-sided information: E₀ ≤ ⟨H⟩, and nothing about how far below. Helium shows it against experiment, with numbers quoted here and derived later in this unit: first-order perturbation theory, itself an upper bound, gives −74.8 eV, an effective nuclear charge −77.5 eV, and the measured energy is −79.0 eV. Neither calculation could see the missing 1.5 eV. Nor does a good energy certify the state. Eckart's inequality turns the energy into a guaranteed overlap, but only if E₀ and the next level E′ are known: for the box parabola, whose symmetry keeps out ψ₂ so that E′ = 9E₁, it allows an infidelity of 0.0132/8 = 0.0017 against a true 0.0014. Local values fare worse. The same trial puts 6.25% too little density at the centre while its energy is 1.3% high, because a density is first order in ε. A lower bound costs more again. In units of ħ²/(2mL²) the parabola gives ⟨H⟩ = 10 and ⟨H²⟩ = 120, and Temple's inequality with E′ = 9π² places the ground level between 9.746 and 10.000; the exact π² = 9.870 lies inside.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle θ6 °

Admixed level n (odd, same symmetry)3

At n = 3 and θ = 6°, ε = 0.105: the centre density is 21% low and ⟨H⟩ is 8.7% above E₁. Halve θ to 3° and the density error only halves, to 10.5%, while the energy excess falls fourfold, to 2.2%. Then drive θ to 90°, where the trial is ψₙ and the solid curve is flat again.

⟨H⟩ ÷ E₁1.087

WRONG AMPLITUDE ε = sin θ0.105

ENERGY EXCESS ÷ E₁0.0874

CENTRE DENSITY ÷ EXACT0.792

Live interpretation⟨H⟩ ÷ E₁: 1.087. WRONG AMPLITUDE ε = sin θ: 0.105. ENERGY EXCESS ÷ E₁: 0.0874. CENTRE DENSITY ÷ EXACT: 0.792

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that an energy within 1% means a wavefunction within 1%, and that of two trials the one with the lower energy is the better function. Neither follows. The energy error is second order in the state error, so it hides a first-order error in everything else, and ⟨H⟩ weighs each error only by its energy cost, so quite different wrong functions can cost the same. In a box of width L, with u measured from the centre in half-widths, the admissible trials 1 − u² and 1 − |u|³⁄² both give ⟨H⟩ = 10ħ²/(2mL²) = 1.0132E₁ exactly, yet one puts 6.25% too little probability density at the centre and the other 11.1% too much. Lower is better only for the energy itself: 1 − |u|^(1.55) cuts the energy error to 0.89%, while its central density error grows to 8.8%. Judge a wavefunction by the quantity you need, computed and checked in its own right.

A student bounds hydrogen's ground state using the Hamiltonian p²/2μ − e²/(4πε₀r), whose lowest eigenvalue is −13.598 eV. One trial gives ⟨H⟩ = −13.05 eV; a more elaborate trial gives ⟨H⟩ = −13.75 eV. What should the student conclude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyNitrogen's vibration is close to harmonic, with ħω = 0.2924 eV from its measured harmonic wavenumber of 2358.6 cm⁻¹. The unnormalised trial ψ = ψ₀ + 0.10ψ₂, built from the oscillator's own eigenstates, is offered as the ground state. Find ⟨H⟩ and its excess over E₀, then ⟨V⟩ and ⟨T⟩ separately, and explain why the energy comes out so much better than either part.

The eigenstates are orthonormal, so with ε = 0.10, ⟨ψ|ψ⟩ = 1 + ε² = 1.01 and ⟨ψ|H|ψ⟩ = E₀ + ε²E₂, the cross terms carrying ⟨0|2⟩ = 0. With E₀ = ħω/2 and E₂ = 5ħω/2, quoted from Quantum Mechanics I: ⟨H⟩ = (0.5 + 0.01 × 2.5)ħω/1.01 = 0.51980ħω = 0.1520 eV.
E₀ = 0.5 × 0.2924 eV = 0.1462 eV, so the excess is 5.79 meV, or 4.0% of E₀. It equals ε²(E₂ − E₀)/(1 + ε²) = 0.01 × 2ħω/1.01 = 0.0198ħω, the ε² law: halving ε to 0.05 would cut it to 1.46 meV.
Potential energy: x² = (ħ/2mω)(a + a†)² gives ⟨0|x²|0⟩ = ħ/(2mω), ⟨2|x²|2⟩ = 5ħ/(2mω) and ⟨0|x²|2⟩ = √2 ħ/(2mω). So ⟨V⟩ = ½mω²⟨x²⟩ = (ħω/4)(1 + 2√2ε + 5ε²)/(1 + ε²) = (ħω/4)(1.3328/1.01) = 0.32991ħω = 0.09647 eV, 32% above the exact ħω/4 = 0.07310 eV.
Kinetic energy: p² = −(mħω/2)(a − a†)² flips the sign of the off-diagonal element, ⟨0|p²|2⟩ = −√2 mħω/2, so ⟨T⟩ = (ħω/4)(1 − 2√2ε + 5ε²)/(1 + ε²) = (ħω/4)(0.7672/1.01) = 0.18989ħω = 0.05552 eV, 24% below ħω/4. Check: 0.09647 + 0.05552 = 0.1520 eV.
The first-order pieces, ±2√2ε × ħω/4 = ±0.0707ħω = ±20.7 meV, are equal and opposite, because ψ₀ is an eigenstate and ⟨H⟩ is stationary there; only the ε² remainder survives. Dimensions: every term is a multiple of ħω. Limit: ε → 0 returns ⟨T⟩ = ⟨V⟩ = ħω/4.

Answer⟨H⟩ = 0.1520 eV, 5.8 meV (4.0%) above E₀ = 0.1462 eV, while ⟨V⟩ = 0.0965 eV is 32% high and ⟨T⟩ = 0.0555 eV is 24% low: their first-order errors of ±20.7 meV cancel in the sum.

MediumA particle is confined to an infinite well of width L. Measure u from the centre in half-widths, so the walls sit at u = ±1, and try ψ = 1 − |u|^p. For which p is this trial admissible? Find ⟨H⟩/E₁ as a function of p, with E₁ = π²ħ²/(2mL²), evaluate it at p = 1 and p = 2, find the best p, and check every value against the theorem.

Every positive p gives a continuous ψ that vanishes at both walls. With dx = (L/2) du, ∫ψ² dx = L[1 − 2/(p + 1) + 1/(2p + 1)] = 2p²L/[(p + 1)(2p + 1)].
Slope: dψ/dx = −(2p/L)|u|^{(p − 1)} sign u, so ∫|ψ′|² dx = (4p²/L²)(L/2)∫|u|^{(2p − 2)} du from −1 to 1 = 4p²/[(2p − 1)L]. The integral converges only if the power 2p − 2 exceeds −1, that is only for p above ½. For p ≤ ½ the trial is continuous, yet its slope diverges so fast at the centre that ⟨T⟩ is infinite: continuity is not the test, a finite ∫|ψ′|² is.
Dividing, ⟨H⟩ = (ħ²/2mL²) × 2(p + 1)(2p + 1)/(2p − 1), so ⟨H⟩/E₁ = 2(p + 1)(2p + 1)/[π²(2p − 1)]. Dimensions: ħ²/(mL²) is an energy, and the ratio is a pure number.
p = 1 is a kinked tent: 12/π² = 1.216. p = 2 is a smooth parabola: 10/π² = 1.0132. Limits: as p → ½ the factor 1/(2p − 1) diverges; as p → ∞ the trial tends to a flat top that jumps to zero at the walls, and ⟨H⟩ grows in proportion to p. Both inadmissible limits cost infinite energy, as they must.
Minimise: setting the derivative to zero gives 4p² − 4p − 5 = 0, so p = (1 + √6)/2 = 1.725 and ⟨H⟩/E₁ = (5 + 2√6)/π² = 9.899/9.870 = 1.0030. For an electron in a 1.00 nm well, E₁ = 0.3760 eV and the bound is 0.3771 eV.
Every value exceeds 1, as the theorem demands. At the best p the density at the centre, (p + 1)(2p + 1)/(4p²) times the exact value, is 1.019: a 1.9% error in a local property sits behind a 0.30% error in the energy.

AnswerAdmissible only for p above ½. ⟨H⟩/E₁ = 2(p + 1)(2p + 1)/[π²(2p − 1)]: 1.216 at p = 1, 1.0132 at p = 2, and a minimum of 1.0030 at p = 1.725, all above E₁.

HardHydrogen's model Hamiltonian p²/2μ − e²/(4πε₀r) has E₀ = −R_H = −13.598 eV and a 2s level at −R_H/4. Try e^(−ζr/a) with ζ = 1.2, a 20% error in the decay constant, where a is the reduced-mass Bohr radius. Find ⟨H⟩, the overlap with the true 1s state, ⟨r⟩ and Temple's lower bound, and set the result against hydrogen's measured ionisation energy, 13.598434 eV.

Quote Unit 2's hydrogenic integrals with Z replaced by ζ: ⟨T⟩ = ζ²R_H and ⟨a/r⟩ = ζ, so ⟨V⟩ = −2ζR_H. Then ⟨H⟩ = (ζ² − 2ζ)R_H = [(ζ − 1)² − 1]R_H = −0.96 × 13.598 eV = −13.05 eV, above E₀ as required. Dimensions: ζ and a/r are pure numbers, so every term is a multiple of R_H.
The excess is exactly (ζ − 1)²R_H = 0.04 × 13.598 eV = 0.544 eV. There is no term linear in ζ − 1, because the true state is a stationary point: a 20% error in the decay constant costs 4% of R_H.
Overlap: ⟨1s|ψ⟩ = 8ζ³⁄²/(1 + ζ)³ = 0.9876, so 1 − |⟨1s|ψ⟩|² = 0.0246. The trial is spherical, so the lowest excited level it can contain is 2s: Eckart allows at most (⟨H⟩ − E₀)/(E′ − E₀) = 0.04/0.75 = 0.053, satisfied and loose by about a factor of two.
A property that is not stationary: ⟨r⟩ = 3a/(2ζ) = 1.25a against the true 1.50a, 17% low. That error is first order in ζ − 1, about four times the 4% energy error.
Lower bound. With H = R_H(−a²∇² − 2a/r), Hψ − ⟨H⟩ψ = 2R_H(ζ − 1)(a/r − ζ)ψ, and ⟨a²/r²⟩ = 2ζ² gives (ΔH)² = 4R_H²(ζ − 1)²ζ², so ΔH = 2ζ|ζ − 1|R_H = 0.48R_H = 6.53 eV. Temple with E′ = −R_H/4 = −3.400 eV: E₀ ≥ −13.054 eV − (6.527 eV)²/(9.655 eV) = −13.054 eV − 4.413 eV = −17.467 eV.
The measured 13.598 eV lies inside the bracket from 13.05 eV to 17.47 eV. It exceeds the model's R_H = 13.598287 eV by 0.147 meV: the fine-structure shift of 1s, −α²R_H/4 = −0.181 meV, derived in Unit 8, partly undone by the +0.034 meV Lamb shift, of order α³ ln(1/α) relative to R_H, which Unit 8 quotes from QED.

Answer⟨H⟩ = −13.05 eV, 0.544 eV above E₀; overlap 0.988, infidelity 0.025 within Eckart's 0.053; ⟨r⟩ 17% low; Temple gives E₀ ≥ −17.47 eV, so −17.47 eV ≤ E₀ ≤ −13.05 eV, containing the measured −13.598 eV.