University Physics V · Particle in a Box · 5.9

Classical Limit & Correspondence

The box is where the correspondence principle has to be stated carefully. A classical bouncer has a flat density 1/L; sin² never flattens at any n. This lesson pins down what actually converges, how fast, and why the classical limit belongs to the detector's resolution rather than to the state.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A classical particle bouncing between two walls at constant speed spends equal time in equal intervals, so its position density is flat: ρ_cl = 1/L. The box eigenstate never reproduces that curve. Write the density exactly, |ψₙ(x)|² = (1/L)[1 − cos(2nπx/L)], and the classical answer is already sitting in it at n = 1 as the mean term; all that raising n does is shorten the wavelength of the leftover cosine from L to L/n.

Its amplitude stays exactly 1/L, the density still reaches 2/L at n antinodes and exactly 0 at n − 1 interior nodes, and at x = L/2 it alternates between 2/L and 0 forever. So correspondence here cannot be pointwise convergence. What does converge is every integral of the density against a fixed function: a fixed window's probability closes on the classical value with error bounded by 1/(nπ), ⟨x²⟩ → L²/3 like 1/n², and averaging over a detector window Δ multiplies the interference term by sin(nπΔ/L)/(nπΔ/L), killing it once Δ spans a few fringes.

The cost is worth naming: the classical limit is not something the state acquires, it is a property of what you ask the state. And the parameter controlling it is not ħ, which never changes, but the round-trip action in units of Planck's constant, n = ∮p dx/h — about 3 × 10²⁷ for a gram bead crawling in a 10 cm channel, and exactly 1 for an electron in the ground state of a nanometre well.

Simple definition: The classical limit of the box is a statement about integrals, not about the wavefunction: as n grows, the probability ψₙ assigns to any fixed interval approaches the classical dwell-time value (b − a)/L, while |ψₙ(x)|² itself keeps swinging between 0 and 2/L.
Example: For [0, L/3] the exact probability is 1/3 − sin(2nπ/3)/(2nπ): 0.196 at n = 1, 0.320 at n = 10, 0.332 at n = 100 — closing on 0.333 like 1/n, while the density at x = L/2 is still 2/L for every odd n and exactly 0 for every even one.

Classical dwell-time densityρ_cl(x) = 2/(vT) = 1/L, with T = 2L/v

The target the quantum state has to reproduce. In a well with sloping walls the same rule piles density up at the turning points instead.

ρ_cl in m⁻¹, v the constant speed in m s⁻¹, T the round-trip period in s; flat only because v is constant here

Density split: classical plus interference|ψₙ(x)|² = (2/L)sin²(nπx/L) = (1/L)[1 − cos(2nπx/L)]

An identity, not an approximation: the classical density is the mean term at every n, including n = 1. Only the leftover wavelength changes.

Both terms in m⁻¹; the cosine has zero mean, fixed amplitude 1/L, and wavelength L/n

Interval probability and its 1/n errorPₙ[a, b] = (b−a)/L − [sin(2nπb/L) − sin(2nπa/L)]/(2nπ)

The exact sense of correspondence: weak convergence at rate 1/n. Fix the window first, then send n up — never the other way round.

Dimensionless; the bracket is bounded by 2, so |Pₙ − (b−a)/L| ≤ 1/(nπ) for any fixed window

Coarse-graining over a detector windowρ̄ₙ(x) = (1/L)[1 − sinc(nπΔ/L)⋅cos(2nπx/L)], sinc u = sin u / u

Says how much interference survives a real measurement: contrast dies like L/(nπΔ), so classicality is bought with resolution.

Δ in m is the probe's resolution; the factor is exactly 0 when Δ is a whole multiple of the fringe spacing L/n

Level spacing: absolute against fractionalE₍ₙ₊₁₎ − Eₙ = (2n+1)E₁ · (E₍ₙ₊₁₎ − Eₙ)/Eₙ = (2n+1)/n²

A spectrum is never continuous on its own. It is continuous relative to a resolution, and only the fractional gap decides that.

E₁ = π²ħ²/(2mL²) = h²/(8mL²) in J; the gap grows without bound while the ratio falls like 2/n

The correspondence parameter is actionn = ∮p dx / h = 2pₙL/h, with pₙ = nπħ/L

Bohr–Sommerfeld is exact for the box. Classicality means large action in units of h, not a limit in which ħ somehow gets small.

∮p dx is the round-trip action in J s, h = 6.626 × 10⁻³⁴ J s, and n is dimensionless

The classical target is a dwell time, and here it is flat

A classical particle in the box moves at constant speed v and turns around at each wall, so the time it spends in an interval dx is dt = dx/v on each pass, and the round trip takes T = 2L/v. The fraction of a period spent in dx is therefore 2(dx/v)/T = dx/L, and the classical position density is ρ_cl(x) = 1/L, uniform on [0, L] and independent of m, v and E. That flatness is special to the box: for the harmonic oscillator the same dwell-time rule gives ρ_cl ∝ 1/√(A² − x²), piling probability up at the turning points, and there the quantum envelope visibly matches it. Fix the target before comparing anything. ρ_cl is the density of an ensemble of identically prepared bouncers observed at random times, not a trajectory, and it sets ⟨x⟩ = L/2, ⟨x²⟩ = L²/3 and Δx = L/√12 = 0.2887L — the three numbers the quantum results must reproduce.

Split it exactly: the classical part is already there

Use sin²θ = (1 − cos 2θ)/2 and the box density becomes |ψₙ(x)|² = (2/L)sin²(nπx/L) = 1/L − (1/L)cos(2nπx/L). Read that as classical plus interference. The first term is exactly ρ_cl, and it is exactly ρ_cl already at n = 1 — nothing is approaching anything. The second term has amplitude exactly 1/L for every n; what changes with n is only its wavelength, L/n. Two consequences follow. The density reaches 2/L at the n antinodes and exactly 0 at the n − 1 interior nodes, so the supremum over x of ||ψₙ|² − ρ_cl| equals 1/L for all n: no uniform convergence. And there is no pointwise convergence either — at x = L/2 the density is 2/L when n is odd and 0 when n is even, so at that one point the sequence has no limit at all. At n = 100 there are still 99 places inside the box where the theory says the particle is never found.

Integrate and the error dies like 1/n

Integrate the same identity over a fixed window and the interference term collapses: Pₙ[a, b] = (b−a)/L − [sin(2nπb/L) − sin(2nπa/L)]/(2nπ), whose second term is bounded by 1/(nπ) whatever the window. Take [0, L/3], where Pₙ = 1/3 − sin(2nπ/3)/(2nπ): that is 0.1955 at n = 1, 0.3196 at n = 10 and 0.3320 at n = 100 against the classical 0.3333 — an error falling like 1/n, algebraically, not exponentially. Moments converge faster because they weight the cosine against a smooth function: ⟨x²⟩ = L²(1/3 − 1/(2n²π²)) is 15% below L²/3 at n = 1 but only 0.15% below at n = 10, and Δx = L√(1/12 − 1/(2n²π²)) climbs from 0.1808L to 0.2878L against the classical 0.2887L. The general statement is the Riemann–Lebesgue lemma: ∫|ψₙ|²f dx → ∫ρ_cl f dx for any fixed integrable f, because a cosine of wavelength L/n averages to nothing against a function that cannot follow it. That is weak convergence — convergence of integrals, test function fixed first, n sent up afterwards.

Coarse-graining is the integral a detector actually does

No instrument reports |ψ|² at a point; it reports a count in a bin. Average the density over a window of width Δ centred on x and the cosine picks up a sinc factor: ρ̄ₙ(x) = (1/L)[1 − sinc(nπΔ/L)cos(2nπx/L)], with sinc u = sin u/u. The surviving contrast is |sinc(nπΔ/L)| ≤ L/(nπΔ), and it is exactly zero whenever Δ is a whole number of fringe spacings L/n. So the classical limit carries a resolution criterion, not just a large-n one. An electron at n = 20 in a 5.0 nm box has fringes 0.25 nm apart; a probe averaging over Δ = 0.050 nm keeps 93.5% of the contrast, and those wiggles are physically there to be seen. Push Δ to 1.6 nm, some 6.4 fringes, and the contrast drops below 5%. The gram bead of the last section has fringes 3.3 × 10⁻²⁹ m apart, fourteen orders of magnitude below a nuclear diameter, so every conceivable probe averages and every measurement returns 1/L. The bead is classical because it is unresolvable, not because its state stopped oscillating.

Momentum agrees exactly where position agrees only weakly

Correspondence is not one statement; it is one per observable. In momentum, ψₙ is a superposition of two plane waves, ψₙ = (1/2i)√(2/L)(e^(ikₙx) − e^(−ikₙx)) with kₙ = nπ/L, cut off outside [0, L]. Its Fourier transform is therefore a pair of diffraction peaks centred on p = ±nπħ/L, and the main lobe's half-width comes from the length-L window alone: 2πħ/L, fixed as n grows. Relative to the peak position that is 2/n, so the momentum distribution collapses onto the two classical values ±p in the ordinary, strong sense. Better still, ⟨p⟩ = 0 and ⟨p²⟩ = (nπħ/L)² = 2mEₙ hold exactly at every n, matching the classical bouncer's p² = 2mE with no error whatever, while ⟨x²⟩ only approaches its classical value. One caveat from the operator's domain: p̂ is symmetric but not self-adjoint on Dirichlet functions, so 'the momentum distribution' here means the transform of ψₙ extended by zero, and the peak width is that window's diffraction pattern.

The knob is action in units of h, not a shrinking ħ

The classical limit is often written ħ → 0, which is a category error: ħ is a constant of nature and nothing in a laboratory can vary it. What varies is a dimensionless ratio, and for the box that ratio is exact — pₙ = nπħ/L makes the round-trip action ∮p dx = 2pₙL = nh, so n is literally the action measured in units of Planck's constant, and Bohr–Sommerfeld quantisation is not an approximation here. Two numbers show the range. A 1.0 g bead at 1.0 cm s⁻¹ in a 10.0 cm channel has p = 1.0 × 10⁻⁵ kg m s⁻¹, hence n = 2pL/h = 3.0 × 10²⁷; its fractional level spacing is (2n+1)/n² ≈ 2/n = 6.6 × 10⁻²⁸, so the gap is 3.3 × 10⁻³⁵ J, about 2.1 × 10⁻¹⁶ eV. An electron in a 1.0 nm box has E₁ = h²/(8mL²) = 0.376 eV and a first gap of 3E₁ = 1.13 eV, roughly 44 times kT at 300 K, and is quantum by any thermal standard. Same equation, same ħ. And note which spacing vanishes: E₍ₙ₊₁₎ − Eₙ = (2n+1)E₁ grows without bound, and only its ratio to Eₙ falls, like 2/n.

Change one variable at a time

Make the relationship visible.

Interactive model

Quantum number n3

Detector window Δ / L0.10 L

Set n = 5, then drag Δ from 0.02 L up to 0.20 L: the averaged curve collapses onto the classical 1/L exactly when the window spans one fringe, while the solid |ψ₅|² behind it still runs from 0 to 2/L.

SURVIVING CONTRAST0.858

FRINGE SPACING L/n0.333 L

WINDOW IN FRINGES0.30 fringes

FRACTIONAL GAP (2n+1)/n²0.778

Live interpretationSURVIVING CONTRAST: 0.858. FRINGE SPACING L/n: 0.333 L. WINDOW IN FRINGES: 0.30 fringes. FRACTIONAL GAP (2n+1)/n²: 0.778

Catch the common trap

Explain before calculating.

Misconception check

The standard slip is that the wiggles die out — that as n grows |ψₙ|² flattens onto the classical 1/L. It does not. Written as (1/L)[1 − cos(2nπx/L)], the interference term keeps amplitude exactly 1/L at every n: the maximum is 2/L, the minimum is exactly 0, and at x = L/2 the density is 2/L for odd n and 0 for even n however large n gets. Only the fringe spacing L/n shrinks. The second slip is to repair this with "take ħ → 0". ħ is a constant of nature and cannot be taken anywhere; the dimensionless quantity that grows is the action in units of h, n = ∮p dx/h. The third is reading the spectrum backwards: the gap E₍ₙ₊₁₎ − Eₙ = (2n+1)E₁ grows without limit, and it is the fractional gap that vanishes. Correspondence here is a statement about integrals over fixed windows, not about pointwise limits.

For the infinite-square-well eigenstates ψₙ on [0, L], which statement correctly describes the n → ∞ limit?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA particle is in the box eigenstate ψₙ on [0, L]. Compute the probability of finding it in the left third, [0, L/3], for n = 1 and for n = 10, and compare each with the classical prediction.

Use the exact identity |ψₙ(x)|² = (2/L)sin²(nπx/L) = (1/L)[1 − cos(2nπx/L)]. The first term integrates to the classical answer; only the cosine can spoil it.
Integrate over [0, L/3]: Pₙ = 1/3 − (1/(2nπ))[sin(2nπ/3) − sin 0] = 1/3 − sin(2nπ/3)/(2nπ).
n = 1: sin(2π/3) = +0.8660, so the correction is 0.8660/(2π) = 0.1378 and P₁ = 0.3333 − 0.1378 = 0.1955.
n = 10: 20π/3 exceeds 2π/3 by 6π, so sin(20π/3) = 0.8660 again; the correction is 0.8660/(20π) = 0.01378 and P₁₀ = 0.3333 − 0.0138 = 0.3196.
The error fell by exactly the factor 10, i.e. like 1/n, as the bound |Pₙ − (b−a)/L| ≤ 1/(nπ) requires. It is algebraic, not exponential, and for this window it never becomes exactly zero.

AnswerP₁ = 0.196 and P₁₀ = 0.320, against the classical 0.333. The approach is algebraic in 1/n, and at n = 10 the density still falls to exactly zero at nine interior nodes.

MediumAn electron sits in the n = 20 state of a 5.0 nm box. (a) What is the spatial period of |ψ₂₀|²? (b) A probe averages the density over a window Δ = 0.050 nm — what fraction of the interference contrast survives? (c) How large must Δ be to push the surviving contrast below 5%?

|ψₙ|² = (1/L)[1 − cos(2nπx/L)], so the density's spatial period is L/n = 5.0 nm / 20 = 0.25 nm. That is half the de Broglie wavelength 2L/n = 0.50 nm, because squaring the wave doubles the spatial frequency.
Averaging over a window of width Δ centred on x multiplies the cosine term by sinc u = sin u / u with u = nπΔ/L, leaving ρ̄ = (1/L)[1 − sinc(u)cos(2nπx/L)].
(b) u = 20π(0.050)/5.0 = 0.2π = 0.6283, so sinc u = 0.5878/0.6283 = 0.935. Averaging over a fifth of a fringe removes almost nothing: the coarse-grained density still runs from 0.065/L to 1.935/L.
(c) |sinc u| ≤ 1/u, so u ≥ 20 guarantees a contrast under 0.05. Then Δ ≥ 20L/(nπ) = 20(5.0 nm)/(20π) = 1.6 nm, about 6.4 fringe spacings.
In between there are exact zeros: whenever Δ = kL/n = 0.25k nm the sinc vanishes identically and the averaged density is flat at 1/L, for k = 1, 2, 3, …

Answer(a) 0.25 nm. (b) 93.5% of the contrast survives Δ = 0.050 nm. (c) Δ ≈ 1.6 nm, about 6.4 fringes, drives it below 5%. Classicality here is a property of the probe, not of the state.

HardA 1.0 g bead slides at 1.0 cm s⁻¹ inside a rigid 10.0 cm channel, modelled as an infinite square well. Find its quantum number, the fractional and absolute level spacing there, and the spatial period of |ψₙ|². Then say what those numbers do and do not establish about the correspondence principle.

Momentum: p = mv = (1.0 × 10⁻³ kg)(1.0 × 10⁻² m s⁻¹) = 1.0 × 10⁻⁵ kg m s⁻¹, and E = p²/2m = (1.0 × 10⁻⁵)²/(2.0 × 10⁻³) = 5.0 × 10⁻⁸ J.
Round-trip action: ∮p dx = 2pL = 2(1.0 × 10⁻⁵)(0.100) = 2.0 × 10⁻⁶ J s, so n = ∮p dx/h = 2.0 × 10⁻⁶ / 6.626 × 10⁻³⁴ = 3.0 × 10²⁷.
Fractional spacing (2n+1)/n² ≈ 2/n = 6.6 × 10⁻²⁸. Absolute spacing ΔE = (2/n)E = 6.6 × 10⁻²⁸ × 5.0 × 10⁻⁸ = 3.3 × 10⁻³⁵ J = 2.1 × 10⁻¹⁶ eV.
Density fringes: L/n = 0.100 m / 3.0 × 10²⁷ = 3.3 × 10⁻²⁹ m, about fourteen orders of magnitude below a nuclear diameter.
Nothing about the state has gone classical: |ψₙ|² still runs from 0 to 2/L and still has n − 1 ≈ 3 × 10²⁷ interior nodes. What has happened is that no instrument can resolve the level gap or the fringes, so every measurement returns the coarse-grained 1/L.

Answern ≈ 3.0 × 10²⁷, ΔE/E ≈ 6.6 × 10⁻²⁸ so ΔE ≈ 3.3 × 10⁻³⁵ J ≈ 2.1 × 10⁻¹⁶ eV, and the fringes lie 3.3 × 10⁻²⁹ m apart. The bead is classical because it is unresolvable, not because sin² stopped oscillating.