University Quantum Mechanics II · The Variational Principle · 10.3

Bounds on excited states from orthogonality and symmetry

The variational method is taught as a ground-state tool, yet spectroscopy lives on excited levels. Here is when a trial energy is a genuine upper bound on one, why orthogonality to your best guess at the ground state is not enough, and how parity, angular momentum and exchange supply the exact orthogonality the proof needs.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The ground-state theorem bounds only E₀ because a trial state can always leak weight into the lowest level, and any leakage pulls ⟨H⟩ toward E₀. Forbid that and the argument moves up a rung: expand the trial on the exact eigenstates, and if its ground-state amplitude c₀ is exactly zero, ⟨H⟩ is a weighted average of E₁, E₂, … and cannot lie below E₁. The weak word is exactly, because the ground state is what we do not know.

A trial orthogonal to an approximate ground state keeps a residual c₀ about as large as that approximation's error, which adds −|c₀|²(E₁ − E₀) to ⟨H⟩: second order, the same order as the accuracy the bound was meant to certify, so the number can land on either side of E₁. Symmetry repairs this. If H commutes with parity, L² or the exchange of two electrons, every eigenstate can be chosen to lie in one symmetry sector, a trial built in another sector has c₀ = 0 by construction, and minimising there bounds that sector's lowest level without the ground state ever being known.

An odd trial bounds the first excited level of an even one-dimensional well, an l = 1 trial bounds hydrogen's 2p, and a spatially antisymmetric trial bounds orthohelium's 2³S₁. The method stops where the symmetry stops: a lopsided well, a parity-breaking Stark field, or a level sharing every conserved quantum number with one below it gets no certified bound this way and needs the Rayleigh-Ritz matrix method.

Simple definition: If a trial state has exactly zero overlap with the true ground state, its ⟨H⟩ is an upper bound on the first excited energy E₁; a symmetry of H is the usual way to guarantee that zero without knowing the ground state.
Example: For V = ½mω²x², the odd trial x e^(−bx²) is orthogonal to the even ground state for every b, so ⟨H⟩ = 3ħ²b/(2m) + 3mω²/(8b) never falls below E₁ = (3/2)ħω, and it reaches that value at b = mω/(2ħ).

Excited-state bound from exact orthogonality⟨0|ψ⟩ = 0 ⇒ E₁ ≤ ⟨ψ|H|ψ⟩/⟨ψ|ψ⟩

Orthogonal to the lowest k exact states, the same line bounds Eₖ, and the energy error stays second order in the error of the state.

|0⟩ the exact ground state, E₁ the second-lowest eigenvalue in J or eV; derived from ψ = Σ cₙ|n⟩

What approximate orthogonality costs⟨H⟩ − E₁ = Σ(n ≥ 2) |cₙ|²(Eₙ − E₁) − |c₀|²(E₁ − E₀)

The sum is never negative and the last term never positive, so unless c₀ = 0 exactly nothing fixes the sign: a 1% ground-state probability can put ⟨H⟩ 0.01(E₁ − E₀) below E₁.

ψ normalised, cₙ = ⟨n|ψ⟩ dimensionless, energies in J; exact for any trial, derived here

Parity sectors of an even potentialV(−x) = V(x) and ψ(−x) = −ψ(x) ⇒ ⟨H⟩ ≥ lowest odd level = E₁

An odd trial needs no knowledge of the ground state: its c₀ vanishes by symmetry for every value of its parameters, not only at the minimum.

Parity commutes with H, so ⟨even|odd⟩ = 0 and ⟨even|H|odd⟩ = 0; the 1D node theorem makes E₀ even and E₁ odd

Odd Gaussian trial in the oscillatorE(b) = 3ħ²b/(2m) + 3mω²/(8b), minimum (3/2)ħω at b = mω/(2ħ)

The bound is met exactly because x e^{(−mωx²/2ħ)} is the true first excited state: a trial family that contains the answer finds it.

b in m⁻², ω in s⁻¹, m in kg; derived here from Gaussian integrals

Odd Gaussian trial in V = F|x|E(b) = 3ħ²b/(2m) + F√(2/(πb)), minimum 9/(18π)¹⁄³ ε = 2.3448 ε

Derived bound, 0.29% above the exact level. At the minimum ⟨V⟩ = 2⟨T⟩, the virial theorem for a linear potential, which checks the algebra.

F in N, ε = (ħ²F²/2m)¹⁄³ in J; exact E₁ = 2.3381 ε, set by the first zero of Ai at −2.3381, is quoted from a table

Exchange sector: a bound on orthoheliumΨ(r₂, r₁) = −Ψ(r₁, r₂) ⇒ E(2³S₁) ≤ ⟨Ψ|H|Ψ⟩/⟨Ψ|Ψ⟩

An antisymmetric spatial trial is orthogonal to the 1¹S ground state and to all of parahelium. For an S-state trial the bound is on the lowest ³S level.

H is spin-free and symmetric in r₁ and r₂, so spatial exchange symmetry is conserved; measured E(2³S₁) = −59.19 eV

Take away the ground state and the proof moves up a rung

The ground-state theorem is one line: with ψ = Σ cₙ|n⟩ normalised, ⟨H⟩ = Σ |cₙ|²Eₙ ≥ E₀. Nothing in that line cares which level happens to be lowest. If c₀ = 0 exactly, the sum starts at n = 1, and ⟨H⟩ − E₁ = Σ(n ≥ 2) |cₙ|²(Eₙ − E₁) ≥ 0, with equality only when ψ is the first excited state itself. Make ψ orthogonal to the lowest k exact states and the same line bounds Eₖ. The accuracy argument carries over unchanged: if ψ = |1⟩ + δ|χ⟩ with |χ⟩ orthogonal to |0⟩ and |1⟩, the energy error is of order δ². In the oscillator, mixing an amplitude 0.1 of |3⟩ into |1⟩ gives ⟨H⟩ = (1.5 + 0.01 × 3.5)/1.01 ħω = 1.520ħω, so a 10% error in the state costs 1.3% in the energy, and always from above.

Orthogonal to a guess is not orthogonal

The catch is that |0⟩ is the unknown. Aim at the oscillator's second even level, E₂ = (5/2)ħω, where parity cannot help because |2⟩ shares the ground state's parity. Take a good approximate ground state, |g⟩ = cos θ|0⟩ + sin θ|2⟩ with sin²θ = 0.01: its energy is (1/2 + 2 sin²θ)ħω = 0.52ħω, only 4% high. The trial orthogonal to it, −sin θ|0⟩ + cos θ|2⟩, has ⟨H⟩ = (5/2 − 2 sin²θ)ħω = 2.48ħω, below the very level it was meant to bound. In general ⟨H⟩ picks up −|c₀|² times the target level's height above E₀, and |c₀| is as large as the error in the ground state you orthogonalised against, so the violation is second order, the same order as the accuracy the bound was supposed to certify. The matrix method rescues this case: the ordered roots of H c = E S c bound the ordered levels, by the Hylleraas-Undheim-MacDonald theorem, quoted in the topic on linear variation.

Let a symmetry do the orthogonalising

If a Hermitian operator S commutes with H, the eigenstates of H can be chosen as eigenstates of S, and H has no matrix elements between different eigenvalues s and s′ of S. The state space splits into sectors, each with its own ladder of levels, and the ground-state theorem holds separately in each. For an even potential S is parity. The one-dimensional node theorem, quoted from Quantum Mechanics I, makes the ground state nodeless and non-degenerate, hence even, and makes the lowest odd level E₁. So any odd trial has c₀ = 0 for every value of its parameters. In three dimensions L² plays the same role: a trial f(r)Y₁ₘ(θ, φ) bounds the lowest l = 1 level. For two electrons the exchange operator separates spatially symmetric from antisymmetric states. Commuting symmetries stack, so parity, l, m and exchange can be imposed together, each choice giving a smaller sector and its own bound.

One odd trial in two potentials: exact once, 0.3% high once

For ψ = x e^(−bx²), ⟨T⟩ = 3ħ²b/(2m) in any potential, three times the even Gaussian's, because the odd function bends more. The oscillator adds ⟨V⟩ = 3mω²/(8b); dE/db = 0 gives b = mω/(2ħ) and E = (3/4 + 3/4)ħω = (3/2)ħω, exact, because the family contains the true first excited state. For V = F|x|, ⟨|x|⟩ = √(2/(πb)), so E(b) = 3ħ²b/(2m) + F√(2/(πb)). In the unit ε = (ħ²F²/2m)¹⁄³, an energy since ħ²F²/m carries J³, the minimum is 9/(18π)¹⁄³ = 2.3448 ε. The exact odd level, 2.3381 ε, is set by the first zero of the Airy function and quoted from a table ahead of Unit 11, so the bound sits 0.29% high. Two checks: at the minimum ⟨V⟩ = 2⟨T⟩, as the virial theorem requires for a linear potential, and as F → 0 bound and exact level both vanish as F²⁄³.

A mirror turns the odd levels into a measured spectrum

A neutron above a horizontal mirror feels V = mgz for z ≥ 0 and a hard wall at z = 0, which forces ψ(0) = 0. On z ≥ 0 that is exactly the problem the odd states of mg|z| solve, so the bouncer's levels are the odd levels of the V-shaped well, and the odd-trial bound becomes a ground-state bound with F = mg. For a neutron ε = (ħ²mg²/2)¹⁄³ = 0.602 peV, so the trial gives E ≤ 1.411 peV against the exact 1.407 peV, and classical turning heights E/mg of 13.76 µm and 13.72 µm. Transmission of ultracold neutrons through a narrow gap above a mirror at the Institut Laue-Langevin put that height at 12.2 µm, with 1.8 µm systematic and 0.7 µm statistical uncertainty, consistent with both. The limiting-case check is g → 0: every level falls as g²⁄³, and the states spread upward without bound.

Exchange is a symmetry too: a certified bound on orthohelium

Helium's non-relativistic H is unchanged when r₁ and r₂ are swapped and contains no spin, so spatial exchange commutes with it. Unit 7 showed that the spin triplet carries an antisymmetric spatial state, while the 1¹S ground state is spatially symmetric. Any trial with Ψ(r₂, r₁) = −Ψ(r₁, r₂) is therefore orthogonal to the exact ground state and to every parahelium level, whatever those states look like, and its ⟨H⟩ bounds the lowest orthohelium level, 2³S₁, measured at −59.19 eV. The symmetric 1s2s combination has no such protection: it is orthogonal to the model state 1s² but not to the exact ground state, so its energy bounds E₀ and nothing more. Relativistic corrections are of relative order (Zα)² ≈ 2 × 10⁻⁴, about 0.01 eV here, far below the gaps simple trials leave, so a non-relativistic bound can fairly be set against the measured level.

Change one variable at a time

Make the relationship visible.

Interactive model

Width parameter β = bℓ²0.26

Even-trial probability w0.00

Power ν: 1 for F|x|, 2 for ½mω²x²1

At ν = 1 the solid curve bottoms out near β = 0.26 at 2.3448 ε, 0.29% above E₁ = 2.3381 ε, and never crosses it; raise w to 0.01 and one per cent of even weight drags ⟨H⟩ below E₁. At ν = 2 the odd curve touches E₁ = 1.5 at β = 0.50.

ODD TRIAL ⟨H⟩2.3448 ε

WITH EVEN WEIGHT w2.3448 ε

EXACT E₁ (QUOTED)2.3381 ε

MIXED ⟨H⟩ − E₁0.0067 ε

Live interpretationODD TRIAL ⟨H⟩: 2.3448 ε. WITH EVEN WEIGHT w: 2.3448 ε. EXACT E₁ (QUOTED): 2.3381 ε. MIXED ⟨H⟩ − E₁: 0.0067 ε

Catch the common trap

Explain before calculating.

Misconception check

Putting a node in the trial makes it an excited-state trial. The first excited state of a one-dimensional well has one node and the second has two, so it is tempting to draw a function with the right nodes and minimise. But the bound rests on c₀ = 0, not on nodes, and minimisation actively hunts for overlap with the ground state. Take the oscillator's second even level, E₂ = (5/2)ħω, and the two-node trial (x² − c)e^(−x²/2), with x in units of √(ħ/mω). At c = 1/2 it is exactly |2⟩. Move the nodes out to x = ±1, c = 1, and ⟨H⟩ = (11/6)ħω = 1.83ħω, already 0.67ħω below the level it was meant to bound; let c grow and the nodes run off to infinity while ⟨H⟩ sinks to (1/2)ħω. The trial with nodes has simply become the ground state. Only an exact symmetry, or the ordered roots of a matrix method, turns a node pattern into a bound.

V(x) is an even potential, and a student wants a certified upper bound on its first excited energy E₁. Which trial state's ⟨H⟩ is guaranteed to lie at or above E₁?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron sits in an infinite square well 1.00 nm wide, with walls at x = ±a and a = 0.500 nm. Without using the known sine solutions, bound its first excited energy from above with the parameter-free trial ψ = x(a² − x²), then compare with the exact value.

The well is symmetric, so parity commutes with H and the ground state is even. ψ = x(a² − x²) is odd and vanishes at both walls, so it is admissible and ⟨0|ψ⟩ = 0 exactly: its ⟨H⟩ bounds the first excited level, n = 2 in the well's numbering.
Inside the well V = 0, so ⟨H⟩ = (ħ²/2m)∫ψ′² dx / ∫ψ² dx over −a to a. With ψ′ = a² − 3x², ∫(a² − 3x²)² dx = 8a⁵/5 and ∫x²(a² − x²)² dx = 16a⁷/105.
The ratio is (8/5)(105/16)/a² = 10.5/a², so ⟨H⟩ = 10.5 ħ²/(2ma²). Dimensions: ħ²/(ma²) is (J s)²/(kg m²) = J.
Numbers: ħ²/(2mₑa²) = (1.0546 × 10⁻³⁴)² / (2 × 9.109 × 10⁻³¹ × (5.00 × 10⁻¹⁰)²) J = 2.442 × 10⁻²⁰ J = 0.1524 eV, so ⟨H⟩ = 10.5 × 0.1524 eV = 1.600 eV.
Exact, from Quantum Mechanics I: E₂ = 4π²ħ²/(2m(2a)²) = π²ħ²/(2ma²) = 9.870 × 0.1524 eV = 1.504 eV. The bound lies above it, as it must, by the factor 10.5/π² = 1.064, which is the same for any particle and any width.

AnswerE₂ ≤ 1.60 eV against the exact 1.50 eV: a certified bound 6.4% high, guaranteed because an odd trial cannot overlap the even ground state.

MediumTake the trial ψ = z e^(−βr²) = r cos θ e^(−βr²) for hydrogen with a fixed proton, in atomic units (ħ = mₑ = e²/4πε₀ = 1, lengths in a₀, energies in Eₕ = 27.211 eV). Say which level its ⟨H⟩ bounds, minimise it, and compare with the measured energy of that level.

H commutes with L², and ψ ∝ Y₁₀ is an l = 1 state, so it has zero overlap with every s state, the 1s ground state included. Its minimum therefore bounds the lowest l = 1 level, 2p.
Kinetic energy: in Cartesian form ψ = z e^{(−β(x² + y² + z²))} separates. The z factor is an odd one-dimensional Gaussian with ⟨T⟩ = 3β/2, and the x and y factors add β/2 each, so ⟨T⟩ = 5β/2.
Potential: with u = rR = r²e^(−βr²), ⟨1/r⟩ = ∫r³e^(−2βr²)dr / ∫r⁴e^(−2βr²)dr = (4/3)√(2β/π), so E(β) = 5β/2 − (4/3)√(2β/π).
Setting dE/dβ = 0 gives 5/2 = (2/3)√(2/π) β^(−1/2), so √β = (4/15)√(2/π) = 0.2128 and β = 0.04527 a₀⁻². At the minimum ⟨V⟩ = −2⟨T⟩, the Coulomb virial theorem, so E = −5β/2 = −16/(45π) Eₕ = −0.11318 Eₕ.
E(2p) ≤ −0.11318 × 27.211 eV = −3.080 eV. Measured: Lyman-α at 121.57 nm carries 10.199 eV, so 2p lies at −13.598 + 10.199 = −3.399 eV, agreeing with Unit 2's derived −Eₕ/8 = −3.401 eV to within the 0.05% reduced-mass correction. The bound is 0.32 eV, 9.4%, above.
The same Gaussian without the z factor has ⟨T⟩ = 3β/2 and ⟨1/r⟩ = 2√(2β/π), so E(β) = 3β/2 − 2√(2β/π), minimised at β = 8/(9π) to bound 1s at −4/(3π) Eₕ = −11.55 eV, 15% high. The 2p does better because its true wavefunction grows as r at the nucleus, with no cusp for a Gaussian to miss; the exponential trial z e^(−r/2a₀) would return −3.40 eV exactly. Checks: every term is a multiple of Eₕ = α²mₑc², so the bound is order α² like the level it bounds, and in SI it scales as mₑ, as a gross-structure energy must.

AnswerE(2p) ≤ −3.08 eV, 0.32 eV or 9.4% above the measured −3.40 eV, certified because an l = 1 trial cannot overlap any s state.

HardBound helium's lowest orthohelium level with the spatially antisymmetric trial Ψ = [φ₁ₛ(r₁)φ₂ₛ(r₂) − φ₂ₛ(r₁)φ₁ₛ(r₂)]/√2, built from hydrogenic Z = 2 orbitals. State what the number certifies, compare it with the measured 2³S₁ energy, and explain why the symmetric partner certifies nothing about 2¹S₀.

H = h₁ + h₂ + e²/(4πε₀|r₁ − r₂|) is unchanged by swapping r₁ and r₂ and contains no spin, so exchange commutes with it. The 1¹S ground state is spatially symmetric and Ψ is antisymmetric, so their overlap is exactly zero, as is Ψ's overlap with every parahelium state. Ψ is also an S state, so its ⟨H⟩ bounds the lowest ³S level.
The orbitals are orthonormal eigenstates of the Z = 2 hydrogenic h, so Ψ is normalised and ⟨h₁ + h₂⟩ = −Z²(13.606 eV)(1 + 1/4) = −68.03 eV.
The repulsion contributes J − K. For two s orbitals only the monopole term of the multipole expansion of 1/|r₁ − r₂| survives the angular integrals, and the radial integrals give J = (17/81)ZEₕ = 11.42 eV and K = (16/729)ZEₕ = 1.19 eV, quoted from the standard table of hydrogenic integrals; Unit 7 left them unevaluated.
⟨H⟩ = −68.03 + 11.42 − 1.19 = −57.80 eV, so E(2³S₁) ≤ −57.80 eV. Measured: He⁺ lies at −54.418 eV and 2³S₁ is bound 4.768 eV below it, at −59.19 eV. The bound holds with 1.39 eV, 2.3%, to spare, the price of letting the 2s electron feel the full Z = 2 instead of a charge screened to about 1.
The symmetric partner gives −68.03 + 11.42 + 1.19 = −55.41 eV. It is orthogonal to the model state 1s², but the exact ground state is not 1s², so c₀ ≠ 0 and −55.41 eV certifies only E₀ ≤ −55.41 eV, true since E₀ = −79.0 eV. That it also lies above the measured 2¹S₀ at −58.39 eV is luck the theorem does not underwrite.
Checks: every term is a multiple of Eₕ = α²mₑc², so the result is order α²; relativistic corrections are smaller by (Zα)² ≈ 2 × 10⁻⁴, about 0.01 eV, a hundred times below the 1.39 eV gap. As Z grows, J and K grow as Z while ⟨h₁ + h₂⟩ grows as Z², so the bound tightens in relative terms.

AnswerE(2³S₁) ≤ −57.80 eV, 1.39 eV (2.3%) above the measured −59.19 eV and certified by exchange antisymmetry; the symmetric value −55.41 eV certifies only that E₀ ≤ −55.41 eV.