University Quantum Mechanics II · The Variational Principle · 10.2

One-parameter Gaussian trials tested against exact answers

The variational theorem guarantees a bound and says nothing about its size. Running one Gaussian through three wells whose answers are known calibrates the method: exact in the oscillator, 0.5 per cent high in the V-shaped well, over a third of the binding lost in the delta well, and the trial's shape decides which.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A normalised Gaussian, ψ_b(x) = (2b/π)¹⁄⁴ e^(−bx²), is the workhorse trial function because every integral it produces is a Gaussian moment. Its kinetic energy is ħ²b/2m whatever the potential, so each problem reduces to one integral, ⟨V⟩(b), and one derivative. Dimensional analysis has already fixed everything except a number: ħ, m and the potential's single strength parameter build one energy scale, so the minimum is a pure number times that scale, and testing the method means asking how close that number comes to the exact one. Three potentials with known answers make the test.

In the oscillator the family contains the exact ground state, and the bound is exactly ħω/2. In V = F|x| the exact state is an Airy function, smooth at the origin, and the bound lands 0.53 per cent high. In the delta well −αδ(x) the exact state e^(−κ|x|) has a kink at the origin that a Gaussian, flat there, cannot form, and the bound recovers only 2/π of the binding although the best Gaussian overlaps the exact state at 97 per cent. The method stops where the shape stops.

One parameter only rescales a fixed profile, so a wrong cusp or a wrong tail leaves an error that no value of b removes, and nothing inside the calculation reports how large it is. Only an exact answer, a measurement or a better trial can.

Simple definition: A one-parameter Gaussian trial is the normalised function (2b/π)¹⁄⁴ e^(−bx²) whose width parameter b is varied to minimise ⟨H⟩; the minimum is an upper bound on the ground-state energy, and comparing it with an exact answer measures how well the Gaussian shape fits.
Example: In the delta well −αδ(x) the best b is (2/π)(mα/ħ²)². For an electron with α = 0.10 eV nm it gives −41.8 meV, an upper bound on the exact −65.6 meV; the ratio, 2/π, is the price of the missing cusp.

Normalised Gaussian trialψ_b(x) = (2b/π)¹⁄⁴ e^(−bx²)

One knob, the r.m.s. width Δx = 1/(2√b). The profile itself is frozen, so an error of shape, a missing cusp or a wrong tail, survives every value of b.

b in m⁻², x in m, ψ_b in m^(−1/2); ⟨x²⟩ = 1/(4b), ⟨|x|⟩ = 1/√(2πb) and |ψ_b(0)|² = √(2b/π)

Kinetic energy of any Gaussian trial⟨T⟩ = ħ²b/(2m)

The potential never enters, so each new problem costs one integral, ⟨V⟩(b), and one derivative. It is (Δp)²/2m for a minimum-uncertainty state, Δx Δp = ħ/2.

ħ in J s, m in kg, b in m⁻², so ⟨T⟩ in J; derived from ∫|ψ_b′|² dx = 4b²⟨x²⟩ = b

Oscillator test: the family holds the answerE(b) = ħ²b/(2m) + mω²/(8b), least at b = mω/(2ħ), where E = ħω/2

E(b) − ħω/2 = (ħ²/2mb)(b − mω/2ħ)², quadratic in the error of b: a b that is 10 per cent off costs only about 0.5 per cent in energy.

ω in s⁻¹; ⟨V⟩ = ½mω²⟨x²⟩. Exact, because the minimiser is Quantum Mechanics I's ground state

Delta well test: 2/π of the bindingE(b) = ħ²b/(2m) − α√(2b/π), least at b = 2m²α²/(πħ⁴), where E = −mα²/(πħ²)

The ratio is 2/π = 0.637 for every α and m: a Gaussian is flat at x = 0 and cannot form the kink of the exact state e^(−κ|x|), κ = mα/ħ².

α: well strength in J m, not the fine-structure constant; the exact E₀ = −mα²/(2ħ²) is derived in Quantum Mechanics I

Linear well test: half a per cent highE(b) = ħ²b/(2m) + F/√(2πb), least at 1.0242 ε, with ε = (ħ²F²/2m)¹⁄³

Only 0.53 per cent above the exact level: the true state is smooth at x = 0, where most of its weight sits, and a Gaussian copies that shape closely.

F in N, ε in J; 1.0242 is (3/2)π^(−1/3). The exact 1.0188 ε, from the first zero of Ai′, is quoted, not derived

Virial relation at a scale optimum2⟨T⟩ = n⟨V⟩ for V ∝ |x|ⁿ, so Eₘᵢₙ = (1 + 2/n)⟨T⟩

It checks the algebra, not the accuracy: the delta well's best Gaussian obeys ⟨V⟩ = −2⟨T⟩ exactly while missing 36 per cent of the binding.

n = 2 for the oscillator, 1 for F|x|, −1 for −αδ(x) since δ(λx) = δ(x)/|λ|; true at the minimum of any one-scale trial

One kinetic integral serves every potential

Write ψ_b = (2b/π)¹⁄⁴ e^(−bx²). Its density is a normal distribution of variance 1/(4b), so ⟨x²⟩ = 1/(4b), and the three tests need only this and two more moments, ⟨|x|⟩ and |ψ_b(0)|². The kinetic energy follows by integrating by parts, with no boundary term because ψ_b vanishes at infinity: ⟨T⟩ = (ħ²/2m)∫|ψ_b′|² dx, and ψ_b′ = −2bxψ_b gives ∫|ψ_b′|² dx = 4b²⟨x²⟩ = b. So ⟨T⟩ = ħ²b/2m (derived), in J² s² m⁻² kg⁻¹ = J. Quantum Mechanics I gives the same result in one line: the Gaussian is a minimum-uncertainty state, Δp = ħ/(2Δx) = ħ√b, and ⟨T⟩ = (Δp)²/2m. The potential never appeared, so every problem reads E(b) = ħ²b/2m + ⟨V⟩(b). Dimensions fix what is left. A potential with one strength parameter builds, with ħ and m, exactly one length L and one energy Eₛ, so E(b) = Eₛ f(bL²) and the whole calculation produces one pure number. The tests ask how close that number comes to the exact one.

The oscillator: a test the Gaussian cannot fail

With ⟨V⟩ = ½mω²⟨x²⟩ = mω²/(8b), E(b) = ħ²b/2m + mω²/8b. Setting dE/db = ħ²/2m − mω²/8b² to zero gives b = mω/(2ħ), where E = ħω/4 + ħω/4 = ħω/2: Quantum Mechanics I's ground-state energy, derived here without the ladder operators. It had to be exact, because e^{(−mωx²/2ħ)} is the ground state and the family contains it. Completing the square shows the second-order behaviour directly: E(b) − ħω/2 = (ħ²/2mb)(b − mω/2ħ)², so a b that is 10 per cent too large raises E by 0.45 per cent, and one 10 per cent too small by 0.56 per cent. Both limits behave: as b → 0 the potential term diverges, because a wide state climbs the parabola, and as b → ∞ the kinetic term does. At the minimum ⟨T⟩ = ⟨V⟩ = ħω/4, the oscillator's virial theorem. What this test certifies is the algebra and nothing more: a family that contains the exact state must return it.

The delta well: the cusp a Gaussian cannot make

For V = −αδ(x), with α a strength in J m, ⟨V⟩ = −α|ψ_b(0)|² = −α√(2b/π), so E(b) = ħ²b/2m − α√(2b/π). For small b the √b term wins and E(b) goes negative, which already proves that the well binds. In s = √b the energy is a parabola; its minimum lies at b = (2/π)κ², with κ = mα/ħ², where E = −mα²/(πħ²) (derived), an energy because mα²/ħ² carries kg J² m² J⁻² s⁻² = J. The exact state, √κ e^(−κ|x|), has E₀ = −mα²/(2ħ²) (quoted from Quantum Mechanics I), so the Gaussian recovers 2/π = 63.7 per cent of the binding for every α and m. The reason is the cusp. Integrating the Schrödinger equation across x = 0 makes ψ′ jump by −(2mα/ħ²)ψ(0), and the exact state meets that condition with a kink, while a Gaussian has zero slope there. The potential rewards density at a single point, and a Gaussian can raise ψ(0)² only by narrowing everywhere, paying kinetic energy across its whole width. At the optimum it reaches a density of 2κ/π at the origin against the exact κ.

The V-shaped well: half a per cent from a smooth state

For V = F|x|, with F a force in N, ⟨V⟩ = F⟨|x|⟩ = F/√(2πb), so E(b) = ħ²b/2m + F/√(2πb). The minimum lies at b = (2π)^(−1/3)(mF/ħ²)²⁄³, where ⟨V⟩ = 2⟨T⟩ and E = (3/2)π^(−1/3) ε = 1.0242 ε, with ε = (ħ²F²/2m)¹⁄³ the only energy that ħ, m and F can build: its units, (J² s² N² kg⁻¹)¹⁄³, reduce to J (derived). The exact answer is quoted. For x above 0 the Schrödinger equation is Airy's, an even ground state needs Ai′ = 0 at the origin, and the first zero of Ai′, at −1.01879, gives E₀ = 1.01879 ε; Airy functions return in Unit 11 at WKB turning points. The Gaussian is 0.53 per cent high. Here the exact state has no cusp, because a continuous potential leaves ψ″ continuous; its tail, exp(−(2/3)(x/ℓ)³⁄²) with ℓ = (ħ²/2mF)¹⁄³, falls between an exponential and a Gaussian; and the best Gaussian overlaps it at 99.89 per cent, a number computed numerically. Good shape, good bound.

The same cusp in three dimensions: hydrogen

The Coulomb potential also forces a cusp: the exact 1s state e^(−r/a), derived in Unit 2, has a kink at the nucleus. A Gaussian e^(−br²) factorises into three Cartesian Gaussians, so ⟨T⟩ = 3ħ²b/2μ, and ⟨1/r⟩ = 2√(2b/π). Minimising E(b) = 3ħ²b/2μ − 2(e²/4πε₀)√(2b/π) gives b = 8/(9πa²), with a = 4πε₀ħ²/(μe²) = 52.95 pm the reduced-mass Bohr radius, and E = −(8/3π) × 13.598 eV = −11.54 eV (derived). The measured ionisation energy is 13.598 eV (tabulated), so the Gaussian finds 84.9 per cent of the binding. Both energies are a pure number times ½μc²α², order α²: a trial function can change the coefficient, never the power of the fine-structure constant. The cusp costs less here than in the delta well because ⟨1/r⟩ = ∫4πr|ψ|² dr draws on the whole cloud, with the region near the nucleus suppressed by the factor r, rather than on one point. Quantum chemistry builds its basis sets from Gaussians because many-centre integrals stay analytic, and repairs the cusp by adding more of them: the Rayleigh–Ritz method of topic 10.08.

What a variational number can and cannot certify

Line the results up: the oscillator exact, the linear well 0.53 per cent high, hydrogen 15 per cent short of the binding, the delta well 36 per cent short. Each came from a clean minimum, and each satisfies 2⟨T⟩ = n⟨V⟩ exactly, because the minimum of any one-scale family does, whatever its shape. Overlap is no better guide. The best Gaussian overlaps the delta well's exact state at |⟨ψ₀|ψ_b⟩|² = π e^(π/4) [erfc(√(π/8))]² = 0.972, derived with erfc from a table, yet misses over a third of the binding, because the 2.8 per cent it lacks lies in the continuum at high momentum, where each unit of weight costs about 12.8 |E₀|. The energy error is second order in the state error, with a coefficient set by what the error is made of. Swap the shapes and the failures swap: e^(−b|x|) is exact for the delta well and gives ħω/√2 for the oscillator, 41 per cent high. So a variational number certifies E₀ ≤ Eₘᵢₙ and no more. Every gap quoted here came from an exact answer or a measurement; without one, the honest test is to enlarge the trial family and watch the bound fall.

Change one variable at a time

Make the relationship visible.

Interactive model

Gaussian width parameter b1.00

Drag b down from 1.00. The ½x² curve touches its dashed line at b = 0.50, where the trial is the exact state; the |x| curve bottoms out at 0.54, 0.53 per cent high; the −δ(x) curve bottoms out at 0.64, still 36 per cent of |E₀| above: the cusp no Gaussian can make.

⟨T⟩ = b/2, ALL THREE0.500

½x²: E ABOVE EXACT25.0 % of |E₀|

|x|: E ABOVE EXACT11.17 % of |E₀|

−δ(x): E ABOVE EXACT40.4 % of |E₀|

Live interpretation⟨T⟩ = b/2, ALL THREE: 0.500. ½x²: E ABOVE EXACT: 25.0 % of |E₀|. |x|: E ABOVE EXACT: 11.17 % of |E₀|. −δ(x): E ABOVE EXACT: 40.4 % of |E₀|

Catch the common trap

Explain before calculating.

For V = −αδ(x) the best Gaussian gives −mα²/(πħ²), against the exact −mα²/(2ħ²), yet its overlap with the exact ground state is |⟨ψ₀|ψ_b⟩|² = 0.97. Which account of the 36 per cent shortfall in binding is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron in GaAs, effective mass m = 0.067 mₑ, is held against a sheet of ionised donors whose field is 5.0 × 10⁶ V m⁻¹ on either side, so its potential energy is V = F|x| with F = eℰ. Use the Gaussian trial to find the best b, the trial's r.m.s. width and the upper bound on the ground-state energy, then compare with the exact 1.0188 (ħ²F²/2m)¹⁄³.

Inputs: F = (1.602 × 10⁻¹⁹ C)(5.0 × 10⁶ V m⁻¹) = 8.01 × 10⁻¹³ N, and m = 0.067 × 9.109 × 10⁻³¹ kg = 6.10 × 10⁻³² kg.
Energy scale: ε = (ħ²F²/2m)¹⁄³ = [(1.055 × 10⁻³⁴ J s)² (8.01 × 10⁻¹³ N)² / (1.221 × 10⁻³¹ kg)]¹⁄³ = (5.85 × 10⁻⁶² J³)¹⁄³ = 3.88 × 10⁻²¹ J = 24.2 meV.
Best width: mF/ħ² = 4.40 × 10²⁴ m⁻³, so b = (2π)^(−1/3)(mF/ħ²)²⁄³ = 0.5419 × 2.68 × 10¹⁶ m⁻² = 1.45 × 10¹⁶ m⁻². The r.m.s. width is Δx = 1/(2√b) = 1/(2 × 1.206 × 10⁸ m⁻¹) = 4.15 nm.
At that b, ⟨T⟩ = ħ²b/2m = 8.27 meV and ⟨V⟩ = F/√(2πb) = 16.54 meV, twice ⟨T⟩ as the virial relation for n = 1 requires, so Eₘᵢₙ = 24.81 meV, which is 1.0242 ε.
The exact level, quoted from the first zero of Ai′, is 1.0188 ε = 24.68 meV: the bound sits 0.13 meV, or 0.53 per cent, above it, and on the right side.

Answerb = 1.45 × 10¹⁶ m⁻² (r.m.s. width 4.15 nm); E₀ ≤ 24.81 meV, against the exact 24.68 meV, a bound 0.53 per cent high.

MediumA 1.0 nm GaAs layer lies 0.30 eV below its AlGaAs surroundings; for an electron of mass 0.067 mₑ, model it as V = −αδ(x) with α = 0.30 eV nm. Find the exact energy and the best Gaussian bound. Then, given erfc(√(π/8)) = 0.37550, find the Gaussian's overlap with the exact state and the average excitation energy of the part of the trial that is not the ground state.

Exact state, from Quantum Mechanics I: κ = mα/ħ² = (6.10 × 10⁻³² kg)(4.81 × 10⁻²⁹ J m)/(1.112 × 10⁻⁶⁸ J² s²) = 2.64 × 10⁸ m⁻¹, so 1/κ = 3.79 nm, nearly four layer widths, which is what makes the delta model fair. E₀ = −ħ²κ²/2m = −39.6 meV.
Gaussian: b = (2/π)κ² = 4.43 × 10¹⁶ m⁻², an r.m.s. width of 2.38 nm, and Eₘᵢₙ = (2/π)E₀ = −25.2 meV, which is 14.4 meV above E₀.
Overlap: ⟨ψ₀|ψ_b⟩ = 2√κ (2b/π)¹⁄⁴ ∫₀^∞ e^{(−κx − bx²)} dx. Completing the square at b = (2/π)κ² reduces it to √π e^(π/8) erfc(√(π/8)) = 1.77245 × 1.48097 × 0.37550 = 0.98567, a probability of 0.9715.
Expand the trial on the exact states: ⟨H⟩ − E₀ = Σ|cₙ|²(Eₙ − E₀). Only the weight w = 1 − 0.9715 = 0.0285 is outside the ground state, and all of it is continuum with E ≥ 0, so its average excitation is 14.4 meV / 0.0285 = 0.50 eV above E₀.
That is 12.8 |E₀|: continuum states near 0.47 eV, with wavenumbers about 3.4κ. A kink is built from high Fourier components and a Gaussian lacks them, so a 3 per cent error in the state becomes a 36 per cent error in the binding.

AnswerE₀ = −39.6 meV; the Gaussian gives −25.2 meV. The overlap probability is 0.9715, and the missing 2.85 per cent averages about 0.50 eV above E₀, some 12.8 |E₀|.

HardThe best single Gaussian for hydrogen's ground state is e^(−br²) with b = 8/(9πa²), a = 52.95 pm; it bounds the energy at −11.54 eV. The hyperfine splitting of an s state is proportional to |ψ(0)|². Predict the 21 cm line's frequency from this Gaussian, given the measured 1420.406 MHz, and compare its ⟨r⟩ and ⟨r²⟩ with the exact 3a/2 and 3a². What does an energy-optimised trial certify about other properties?

Density at the nucleus: |ψ_b(0)|² = (2b/π)³⁄² = [16/(9π²a²)]³⁄² = 64/(27π³a³) = 0.0764/a³. The exact 1s state, derived in Unit 2, has |ψ(0)|² = 1/(πa³) = 0.3183/a³.
Their ratio is 64/(27π²) = 0.240, with no α in it: both densities scale as a⁻³, that is as (μcα/ħ)³, so the Gaussian keeps the order in α and fails only in the coefficient.
The Fermi contact splitting, quoted from Unit 9, is proportional to |ψ(0)|² with everything else fixed, so the Gaussian predicts 0.240 × 1420.406 MHz = 341 MHz, a wavelength of 87.9 cm instead of 21.1 cm: 76 per cent too small.
Moments: ⟨r⟩ = √(2/(πb)) = √(9a²/4) = 1.5a = 79.4 pm, identical to the exact 3a/2. ⟨r²⟩ = 3/(4b) = 27πa²/32 = 2.65a², 12 per cent below 3a². ⟨1/r⟩ = 8/(3πa), 15 per cent low, the energy's shortfall again through the virial theorem.
The errors rank 0 per cent in ⟨r⟩, 12 in ⟨r²⟩, 15 in the energy and 76 in |ψ(0)|². Minimising tunes the trial to the energy, a whole-cloud average; it certifies nothing about a property weighted toward the cusp, which needs a trial with the right cusp.

AnswerThe Gaussian predicts the hyperfine line at 341 MHz (87.9 cm) against 1420.406 MHz measured, 24 per cent of the true splitting, while getting ⟨r⟩ = 1.5a exactly and ⟨r²⟩ = 2.65a², 12 per cent low.