University Physics V · Particle in a Box · 5.2
The Infinite-Wall Limit
The hard wall is not a place you can build; it is what a deep well looks like once its decay length shrinks below everything else you care about. This lesson takes that limit slowly, so you know what ψ(0) = 0 buys, what it costs, and where the kink at the wall came from.
Build the model
Connect the measurement to the mechanism.
Start from a real well: depth V₀, width L, a bound state at E < V₀. Outside, ψ decays as e(−κ|x|) with κ = √(2m(V₀ − E))/ħ, so the state leaks a distance 1/κ into each wall, and matching ψ and ψ′ at the edge is the same as fixing one ratio there, ψ′(0⁺)/ψ(0⁺) = κ. That is a Robin boundary condition — one member of the family that makes −(ħ²/2m)d²/dx² self-adjoint on [0, L].
Now raise V₀. κ grows, 1/κ collapses, and because the interior slope is bounded by k = √(2mE)/ħ times the amplitude, the ratio can only keep pace with a diverging κ if ψ(0) → 0. Dirichlet is the κ → ∞ endpoint of the Robin family, selected by the potential rather than assumed. What the limit costs is the derivative.
Continuity of ψ survives, because a jump in ψ would put a delta in ψ′ and make ⟨T⟩ = (ħ²/2m)∫|ψ′|² diverge; continuity of ψ′ does not, because the theorem guaranteeing it assumes a bounded V. So the box eigenfunctions carry ψ(0) = 0 with ψ′(0⁺) ≠ 0 — a kink no material supplies, bought by an idealisation that also shortens the box by 2/κ and pushes every level up.
- Simple definition
- A Dirichlet boundary condition sets the wavefunction itself to zero at a wall, and it is what the finite well's matching condition ψ′(0) = κψ(0) becomes once the barrier is high enough that the decay length 1/κ is negligible against the width.
- Example
- An electron in a 1.0 nm well 10 eV deep has 1/κ = 0.063 nm, so its true ground state is 0.298 eV while the Dirichlet box predicts 0.376 eV — a 26% overestimate from a barrier already 34 times the level's own energy.
Continuity of ψ′ is a theorem with a hypothesis, and an infinite wall is exactly the case that breaks it.
ψ′ in m(−3/2), the integral carries a length. Bounded V ⇒ the right side vanishes as b − a → 0.
The one length a wall adds. Everything the hard wall gets wrong is measured in units of δ/L.
κ in m⁻¹, δ a length. Electron with V₀ − E = 9.70 eV: κ = 16.0 nm⁻¹, δ = 0.063 nm.
The boundary condition is not extra physics bolted on: it is which self-adjoint operator you are solving.
κ = 0 is Neumann, finite κ is Robin, κ → ∞ is Dirichlet. Each choice fixes a domain for H.
Put Leff into the box formula and it recovers the finite well's level to a fraction of a percent.
k = √(2mE)/ħ. The interior sine reaches zero a distance δₑ outside each wall, not at it.
The one number that decides whether you are allowed to use a box at all.
Dimensionless. L = 1.0 nm with κ = 16.0 nm⁻¹ predicts 25%; the exact answer is 26%.
The box's ladder is unbounded only because its wall is; levels near and above V₀ are artefacts.
For L = 1.0 nm and V₀ = 10 eV: N = ⌈5.157⌉ = 6 bound states, not infinitely many.
Continuity of ψ′ is a theorem, and it has a hypothesis
Integrate the time-independent equation ψ″ = (2m/ħ²)(V − E)ψ across a slice [a, b]: ψ′(b) − ψ′(a) = (2m/ħ²)∫ₐᵇ (V − E)ψ dx. If |V| ≤ Vₘₐₓ on that slice, the right side is bounded by (2m/ħ²)(Vₘₐₓ + |E|)⋅max|ψ|⋅(b − a), which vanishes as b − a → 0. That is the entire proof that ψ′ is continuous, and it can fail in two ways. A delta well V = λδ(x − c) is unbounded but integrable, and leaves a prescribed jump Δψ′ = (2mλ/ħ²)ψ(c). An infinite wall is unbounded on every slice that touches it, so no bound survives at all and the jump is whatever the solution needs. Continuity of ψ itself rests on something quite different: ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx must be finite, and a step in ψ makes ψ′ a δ whose square is not integrable. A step in ψ′ only makes ψ′ a bounded step function, which integrates perfectly well. Hence the hierarchy you should carry: ψ continuous always, ψ′ continuous only where V is bounded.
One wall, one ratio: Robin becomes Dirichlet
Put the wall at x = 0, with V = V₀ for x < 0 and V = 0 inside. The exterior solution that stays normalisable is ψ = B e(κx) with κ = √(2m(V₀ − E))/ħ, so ψ′(0⁻)/ψ(0⁻) = κ. Matching that ratio, rather than ψ and ψ′ separately, removes both unknown amplitudes and leaves ψ′(0⁺) = κ ψ(0⁺): a Robin boundary condition whose parameter is κ. Neumann is κ = 0, Dirichlet is κ = ∞, and everything between is a partly transparent wall. Now raise V₀ at fixed n. The interior slope cannot run away with κ — it is at most kA with k = √(2mE)/ħ — so the only way the ratio can track a diverging κ is ψ(0) = ψ′(0)/κ → 0. Put numbers on it. An electron in a 1.0 nm well with V₀ = 10 eV has k = 2.79 nm⁻¹ and κ = 15.96 nm⁻¹, so ψ(0)/ψₘₐₓ = sin(kδₑ) = 0.17: the state is still 17% of its peak height at the wall. Deepen to V₀ = 100 eV and that falls to 0.059.
The kink at the wall, and why it is legal
In the box, ψₙ = √(2/L) sin(nπx/L), so ψₙ(0) = 0 while ψₙ′(0⁺) = √(2/L)(nπ/L); with ψ ≡ 0 outside, ψ′(0⁻) = 0. The derivative therefore jumps by 4.44 nm(−3/2) at each wall for n = 1 and L = 1.0 nm. Nothing breaks. The kink costs no energy in the form ⟨T⟩ = (ħ²/2m)∫₀L |ψ′|² dx, because ψ′ is a bounded function on the interval; and that integral really is ⟨ψ|p̂²|ψ⟩/2m here, since integrating by parts leaves the surface term ψ*ψ′|₀L, which vanishes precisely because ψ(0) = ψ(L) = 0. The Dirichlet condition is what makes the two expressions for kinetic energy agree. Where the kink does bite is anywhere you differentiate twice at a wall: ψ″ carries a delta there, so a finite-difference H on a grid converges slowly at the edges, the momentum-space wavefunction falls off only as 1/k² instead of exponentially, and ⟨p̂⁴⟩ diverges for every box eigenstate even though ⟨p̂²⟩ is perfectly finite.
The extrapolation length: the box is too short
Inside the finite well the solution is ψ ∝ sin(k(x + δₑ)), and the Robin condition k cot(kδₑ) = κ fixes the offset through tan(kδₑ) = k/κ. The interior sine reaches zero a distance δₑ outside each wall, so the ground state fits half a wavelength across L + 2δₑ rather than across L, giving k = π/(L + 2δₑ). Run the electron example. L = 1.0 nm and V₀ = 10 eV give z₀ = 8.100, and the lowest root of z tan z = √(z₀² − z²) is z₁ = 1.3974, so k = 2.795 nm⁻¹ and E₁ = 0.298 eV. Then κ = 15.96 nm⁻¹, k/κ = 0.1751, δₑ = 0.0620 nm — within 1% of the crude 1/κ = 0.0627 nm — and Leff = 1.124 nm, whose π/Leff reproduces k exactly. The Dirichlet box, using L = 1.0 nm, returns 0.376 eV: 26% high, as 4/(κL) = 0.25 predicted. The sign is structural. Leaking into the wall lengthens the wavelength, and a longer wavelength means less energy. And because κₙ falls as Eₙ rises, the fractional error 4/(κₙL) grows with n: the top of the ladder is the worst part of it.
Where the limit is not uniform, and which levels are real
The convergence is genuine but partial. As V₀ → ∞ the finite-well eigenfunctions converge to the box ones in the L² norm and uniformly on any closed interval inside (0, L). They do not converge in derivative at the walls: every finite well has ψ′ continuous there and the box never does, so the limit of the derivatives is not the derivative of the limit. The spectrum breaks a second way. A well of depth V₀ holds only ⌈(L/π)√(2mV₀)/ħ⌉ bound states — six for L = 1.0 nm and V₀ = 10 eV, seventeen at V₀ = 100 eV — while the box has an infinite ladder. So the two limits do not commute. Fix n and send V₀ → ∞, and the finite-well level converges to n²π²ħ²/(2mL²); fix V₀ and send n → ∞, and the state leaves the well for the scattering continuum instead. Whenever you quote a box result, quote the n you trust with it: the condition is n ≪ (L/π)√(2mV₀)/ħ, not merely V₀ ≫ E₁.
Deciding by δ/L: metals, dyes, and quantum wells
The decision is one ratio, δ/L, and it takes a line to compute. A conduction electron held in by a work function Φ ≈ 4.5 eV has δ = √(0.0381/4.5) nm = 0.092 nm; in a 10 nm metal grain that is 0.9% per wall and a 3.7% shift in energy, so a hard wall is honest there. The π-electrons of a cyanine dye sit in a box about 1 nm long behind a barrier of a few eV, so δ/L is nearer 0.06 — which is why a box fit to the absorption maxima always returns a length longer than the bond geometry allows: the extrapolation length is hiding inside the fitted L. A GaAs/AlGaAs quantum well is the outright failure case. With a barrier ΔEc = 0.30 eV and m* = 0.067 mₑ, so that ħ²/2m* = 0.569 eV nm², the extrapolation length is δₑ = 1.40 nm, and a 10 nm well leaks 14% of its width into each barrier. The box predicts 56.1 meV where the true ground state is 34.2 meV, and the well holds three bound states, not infinitely many. Device modelling never uses a hard wall.
Change one variable at a time
Make the relationship visible.
Pull δ down to 0.02 nm: the tails vanish and ψ(0)/peak falls toward zero, but the slope readout stays near π — a vanishing value beside a surviving slope is exactly the kink. Watch the barrier depth climb past 90 eV to buy it.
ψ(0) / PEAK0.299
L⋅ψ′(0⁺) / PEAK2.42
BARRIER DEPTH V₀2.7 eV
E₁ / E₁(BOX)0.650
Live interpretationψ(0) / PEAK: 0.299. L⋅ψ′(0⁺) / PEAK: 2.42. BARRIER DEPTH V₀: 2.7 eV. E₁ / E₁(BOX): 0.650
Catch the common trap
Explain before calculating.
A finite well of width L holds a bound state at energy E below its depth V₀. Holding L and the quantum number fixed, V₀ is raised without bound. What happens at the left wall as that limit is taken?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is bound in a well 1.0 nm wide and 10 eV deep, with its ground state at E ≈ 0.30 eV. Take ħ²/2m = 0.0381 eV nm². Find the decay constant κ outside the wall, the penetration depth δ = 1/κ, and estimate the fraction by which a Dirichlet box overestimates E₁.
- Outside the wall the bound state decays as e(−κ|x|) with κ = √(2m(V₀ − E))/ħ. In the given units that is κ = √((V₀ − E)/0.0381) nm⁻¹.
- V₀ − E = 10 − 0.30 = 9.70 eV, so κ = √(9.70/0.0381) = √254.6 = 16.0 nm⁻¹.
- δ = 1/κ = 0.0627 nm ≈ 0.63 Å — about a third of a C–C bond, and 6.3% of the box on each side.
- The box is short by 2δ, and E ∝ (L + 2δ)⁻², so (Ebox − E)/E ≈ 4δ/L = 4/(κL) = 4/16.0 = 0.25.
Answerκ = 16.0 nm⁻¹ and δ = 0.063 nm, so the hard wall runs about 25% high. The exact transcendental solution gives 26%, so the linear estimate is already reliable at this depth.
MediumFor the same well, the ground-state root of z tan z = √(z₀² − z²), with z₀ = (L/2)√(2mV₀)/ħ, is z₁ = 1.3974. Find E₁, then the extrapolation length δₑ from tan(kδₑ) = k/κ, the effective width, and the size of the Dirichlet box's error.
- z₀ = 0.5 × √(10/0.0381) = 0.5 × 16.20 = 8.100, so the quoted root is the lowest even state of the well.
- k = 2z₁/L = 2(1.3974)/1.0 = 2.795 nm⁻¹, hence E₁ = 0.0381 k² = 0.0381 × 7.811 = 0.2976 eV.
- κ = √((10 − 0.2976)/0.0381) = √254.7 = 15.96 nm⁻¹, so the Robin parameter gives k/κ = 0.1751.
- tan(kδₑ) = 0.1751 ⇒ kδₑ = 0.1734 rad ⇒ δₑ = 0.1734/2.795 = 0.0620 nm, within 1% of the crude 1/κ = 0.0627 nm.
- Leff = L + 2δₑ = 1.124 nm, and π/Leff = 2.795 nm⁻¹ returns k exactly: the ground state fits half a wavelength across the extended box, not the real one.
- The Dirichlet box gives E₁ = π²(0.0381)/1.0² = 0.3760 eV, which is 0.3760/0.2976 = 1.263 times the truth.
AnswerE₁ = 0.298 eV against the box's 0.376 eV. An extrapolation length of 0.062 nm per wall widens the box to 1.124 nm and cuts the energy by 26%, even though V₀ is 34 times E₁.
HardA GaAs quantum well 10 nm wide sits between AlGaAs barriers 0.30 eV high, with electron effective mass m* = 0.067 mₑ. Find ħ²/2m*, the Dirichlet estimate of E₁, the number of bound states, and the true E₁ given that the lowest root of z tan z = √(z₀² − z²) is z₁ = 1.226. Then judge the hard-wall model.
- ħ²/2m* = 0.0381/0.067 = 0.569 eV nm², about fifteen times the free-electron value, so every energy scale in this well is stretched down.
- Dirichlet box: E₁ = π²(0.569)/10² = 9.8696 × 0.00569 = 0.0561 eV = 56.1 meV.
- z₀ = (L/2)√(V₀/0.569) = 5√(0.5276) = 5 × 0.7264 = 3.632, so the well holds ⌈2z₀/π⌉ = ⌈2.312⌉ = 3 bound states, not an infinite ladder.
- k = 2z₁/L = 2(1.226)/10 = 0.2453 nm⁻¹, so the true E₁ = 0.569 × 0.06017 = 0.0342 eV = 34.2 meV.
- Leff = π/k = 12.81 nm, so δₑ = (12.81 − 10)/2 = 1.40 nm per wall — 14% of the well on each side, and 4δₑ/L = 0.56 is no longer a small correction.
- The box therefore overestimates by 56.1/34.2 = 1.64, and the exact ratio (Leff/L)² = 1.281² = 1.641 confirms it.
AnswerE₁ = 34.2 meV against the box's 56.1 meV, a 64% overestimate, with only three bound states and δₑ = 1.40 nm per wall. Hard walls are honest only when δ/L is small; here it is 0.14.