University Physics IV · Quantum Potentials · 7.2
The Infinite Square Well
The first bound-state problem you can solve exactly, and the first place quantisation arrives as a consequence rather than a postulate. Two walls, one second-order equation, and the whole ladder of allowed energies falls out — along with the orthonormal basis you will expand every later state on.
Build the model
Connect the measurement to the mechanism.
Put a particle where the potential is zero on 0 < x < L and infinite everywhere else, and the time-independent Schrödinger equation collapses inside to ψ″ = −k²ψ with k = √(2mE)/ħ. That equation alone quantises nothing: every positive E solves it. What quantises is the pair of walls.
An infinite barrier forces ψ = 0 outside, and continuity then demands ψ(0) = ψ(L) = 0, turning a differential equation into a two-point boundary-value problem. Only sines survive, and only those fitting a whole number of half-wavelengths into the box: kₙ = nπ/L, so Eₙ = n²π²ħ²/(2mL²), with n = 1 the floor because n = 0 leaves ψ ≡ 0, which is no particle at all. Normalisation fixes the amplitude at √(2/L) for every n, and Hermiticity makes the set orthonormal, so these states are a basis rather than a list.
The cost is that infinite walls are a fiction: nothing leaks out, the well holds infinitely many bound states, the spectrum runs to infinity, and ψ′ jumps at each edge — a discontinuity only an infinite potential is permitted to produce.
- Simple definition
- The infinite square well is the model of a particle that moves freely inside a segment of length L and is forbidden outside it, so its stationary states are the standing sine waves that vanish at both walls.
- Example
- An electron in a 0.50 nm well has Eₙ = n² × 0.376 eV⋅nm² / (0.50 nm)² = n² × 1.50 eV, so the ground state sits 1.50 eV above the floor and the first gap is 4.51 eV.
Oscillatory, not exponential — and every E > 0 solves it, so the equation alone quantises nothing.
V = 0 for 0 < x < L; E in J, m in kg, ħ = 1.055×10⁻³⁴ J s, k in m⁻¹
Quantisation is a boundary-value result, not a postulate bolted on: only whole half-wavelengths fit.
n = 0 leaves ψ ≡ 0, which cannot be normalised; −n repeats the state with a sign flip
Both scalings in one line: E ∝ n² climbing the ladder, E ∝ 1/L² as the box is squeezed.
for an electron, Eₙ = n² × 0.376 eV⋅nm² / L², with L in nm
|ψₙ|² peaks at 2/L, twice the classical density 1/L, and carries n − 1 interior nodes.
ψ carries m(−1/2) in one dimension; the amplitude is the same for every n
Makes the set a basis: cₙ = ∫ψₙ*ψ dx lifts one coefficient out of the whole sum.
δₘₙ is 1 for m = n and 0 otherwise; guaranteed because Ĥ is Hermitian
Gaps widen going up yet shrink relative to Eₙ — the correspondence limit, and why E₁ is never zero.
relative spacing (2n + 1)/n² → 0, so large-n ladders look continuous
Why ψ must vanish at each wall — and why ψ′ need not
Do not start from "the particle cannot be there". Start from a finite well of depth V₀ and take the limit. Outside, ψ ∝ e(−κ|x|) with κ = √(2m(V₀ − E))/ħ, so as V₀ → ∞ the decay length 1/κ shrinks to nothing and the outside amplitude collapses. Continuity of ψ then hands you ψ(0) = ψ(L) = 0 — the boundary conditions are inherited, not assumed. What does not survive the limit is continuity of ψ′. Integrating the Schrödinger equation across a wall gives Δψ′ = (2m/ħ²)∫Vψ dx, which vanishes for finite V but not for infinite V, so the slope is allowed to jump at each edge. For ψ₁ that jump runs from 0 outside to √(2/L)(π/L) just inside. It is the first sign that this model is a limit, not a potential anyone could build.
Solving inside, and why n starts at 1
Inside, V = 0 and ψ″ = −k²ψ with k = √(2mE)/ħ, whose general solution is ψ = A sin kx + B cos kx. Impose ψ(0) = 0: cos 0 = 1, so B = 0. Impose ψ(L) = 0 with A ≠ 0: sin kL = 0, so kL = nπ, and E = ħ²k²/2m follows immediately. Now handle the integer carefully. n = 0 gives ψ ≡ 0 everywhere — a function you cannot normalise, describing a particle that is nowhere — so it is not a state and E = 0 is not in the spectrum. Negative n returns sin(−nπx/L) = −sin(nπx/L), the same function up to an overall sign, and a global phase is physically invisible, so n = −3 and n = 3 are one state, not two. The spectrum is therefore n = 1, 2, 3, … with no repeats and no rung at zero energy.
Normalising: where √(2/L) comes from
Born's rule only means something once ∫|ψ|² dx = 1. With ψ = A sin(nπx/L), use sin²θ = (1 − cos 2θ)/2: ∫₀L A² sin²(nπx/L) dx = A²[x/2 − (L/4nπ) sin(2nπx/L)]₀L = A²L/2, since the sine vanishes at both limits. So A = √(2/L) for every n — the constant does not grow with n, even though higher states wiggle faster, because sin² averages to exactly ½ whatever the frequency. Two checks. The units are right: in one dimension |ψ|² is a probability per unit length, so ψ carries m(−1/2), and √(2/L) does. And the shape is right: |ψₙ|² peaks at 2/L, exactly twice the classical density 1/L of a particle bouncing at constant speed, because the quantum particle piles probability into n humps and empties it at the nodes between them.
Orthonormality is a theorem, not a coincidence
Compute ∫₀L ψₘψₙ dx with sin a sin b = ½[cos(a − b) − cos(a + b)]. Both cosines integrate to zero across the well when m ≠ n, and the first becomes cos 0 = 1 when m = n, returning δₘₙ. The orthogonality is not luck: Ĥ is Hermitian, so eigenfunctions belonging to different eigenvalues are automatically orthogonal, and the integral merely confirms a theorem you already had. What you gain is a basis. These sines are exactly the Fourier sine series on [0, L], which is complete, so any state of the well — a Gaussian squeezed in at t = 0, a triangle, a single half-lobe — can be written as ψ = Σ cₙψₙ with cₙ = ∫ψₙ*ψ dx, and Σ|cₙ|² = 1 makes |cₙ|² the probability of measuring Eₙ. Orthonormality is precisely what lets one overlap integral extract one coefficient.
Reading the spectrum: n², 1/L², and zero-point energy
Eₙ = n²h²/(8mL²) carries two scalings. Up the ladder E ∝ n², so gaps widen as E_{n+1} − Eₙ = (2n + 1)E₁ — unlike the harmonic oscillator's even rungs or hydrogen's converging ones, and a fingerprint you can test against a measured spectrum. Across boxes E ∝ 1/L², the price of confinement: halve a quantum dot from 5.0 nm to 2.5 nm and every level quadruples. Numbers make it concrete. For an electron Eₙ = n² × 0.376 eV⋅nm² / L², so a 0.50 nm well gives E₁ = 1.50 eV and a first gap of 4.51 eV — visible-to-ultraviolet photons. Put a proton in the same box and E₁ falls by the mass ratio to 0.82 meV. And E₁ can never be zero, because a constant ψ cannot vanish at both walls: curvature is forced, and curvature is kinetic energy.
What the model buys, and what it costs
The buy is real: an exact spectrum, an exact basis, and a first model of any tightly confined electron — a quantum dot, a nanowire segment, a π-electron on a conjugated chain. The costs are worth naming out loud. ψₙ is not a momentum eigenstate: p̂ψₙ ∝ cos(nπx/L), not sin, because a standing wave is an equal superposition of +ħkₙ and −ħkₙ. So ⟨p⟩ = 0 while ⟨p²⟩ = ħ²kₙ² = 2mEₙ, and for n = 1 the uncertainty product is Δx Δp = 0.568ħ, comfortably above the ħ/2 floor. The walls are the bigger fiction. A real well has finite depth, so the wavefunction penetrates and the effective width exceeds L — which is why the infinite well always overestimates the levels — and it holds only finitely many bound states rather than an unbounded ladder.
Change one variable at a time
Make the relationship visible.
Step n up one rung at a time and count the humps in |ψₙ|² — n of them, with n − 1 nodes between. Then narrow the well: both curves grow taller, as √(2/L) demands, while Eₙ climbs as 1/L². The dashed line sits at half the peak at every setting.
ENERGY Eₙ4.18 eV
GAP TO n+15.22 eV
WAVENUMBER kₙ10.47 nm⁻¹
INTERIOR NODES1
Live interpretationENERGY Eₙ: 4.18 eV. GAP TO n+1: 5.22 eV. WAVENUMBER kₙ: 10.47 nm⁻¹. INTERIOR NODES: 1
Catch the common trap
Explain before calculating.
The infinite-well eigenfunctions are ψₙ(x) = √(2/L) sin(nπx/L) on 0 ≤ x ≤ L, with Eₙ = n²π²ħ²/(2mL²). Which single statement about this set is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is confined to a one-dimensional infinite well of width L = 0.50 nm. Find E₁ and E₃ in eV, then find the wavelength of the photon emitted when it drops from n = 3 to n = 1. Use h²/8mₑ = 0.376 eV⋅nm² and hc = 1240 eV⋅nm.
- Eₙ = n²h²/(8mₑL²). With the shortcut constant, Eₙ = n² × 0.376 eV⋅nm² / (0.50 nm)² = n² × 0.376/0.25 eV = n² × 1.50 eV.
- Ground state: E₁ = 1.50 eV. Third level: E₃ = 9 × 1.50 = 13.5 eV — an n² ladder, not an even one.
- Photon energy: ΔE = E₃ − E₁ = (9 − 1)E₁ = 8 × 1.504 = 12.0 eV.
- Wavelength: λ = hc/ΔE = 1240 eV⋅nm ÷ 12.03 eV = 103 nm, in the far ultraviolet.
AnswerE₁ = 1.50 eV, E₃ = 13.5 eV, and the 3 → 1 photon has λ = 103 nm.
MediumA particle is in the n = 2 state of an infinite well of width L. (a) Show that the normalisation constant is √(2/L). (b) Find the probability of finding it in the left quarter of the well, 0 ≤ x ≤ L/4. (c) Repeat (b) for n = 1 and account for the difference.
- (a) Require ∫₀L A² sin²(2πx/L) dx = 1. With sin²θ = (1 − cos 2θ)/2 the integral is A²[x/2 − (L/8π) sin(4πx/L)]₀L = A²L/2, since the sine vanishes at both limits. So A²L/2 = 1 and A = √(2/L).
- (b) P = ∫₀(L/4) (2/L) sin²(2πx/L) dx = (1/L)∫₀(L/4) [1 − cos(4πx/L)] dx = (1/L)[x − (L/4π) sin(4πx/L)]₀(L/4).
- At x = L/4 the sine is sin π = 0, so P = (1/L)(L/4) = 1/4 exactly — that quarter holds exactly one complete hump of |ψ₂|², so it matches the classical 25%.
- (c) The same integral for n = 1 is (1/L)[x − (L/2π) sin(2πx/L)]₀(L/4) = 1/4 − (1/2π) sin(π/2) = 0.250 − 0.159 = 0.0908.
- The n = 1 density is a single hump centred on L/2, so the outer quarter sits on its rising flank and collects only 9.1%. Higher n spreads the density more evenly — the correspondence limit in miniature.
AnswerA = √(2/L); the probability is 1/4 exactly for n = 2, but 1/4 − 1/(2π) = 0.091 for n = 1.
HardA short nanowire segment is modelled as a one-dimensional infinite well holding one electron of mass mₑ. Its n = 1 → n = 2 absorption sits at 620 nm. (a) Find L. (b) How many levels lie below 20 eV? (c) Give the fractional spacing (E_{n+1} − Eₙ)/Eₙ at the top of that set and at n = 100, and say what it means. Use h²/8mₑ = 0.376 eV⋅nm² and hc = 1240 eV⋅nm.
- Photon energy: ΔE = hc/λ = 1240 eV⋅nm ÷ 620 nm = 2.00 eV.
- (a) The transition energy is E₂ − E₁ = (4 − 1)E₁ = 3E₁, so E₁ = 2.00/3 = 0.667 eV. Then L² = 0.376 eV⋅nm² ÷ 0.667 eV = 0.564 nm², giving L = 0.751 nm.
- (b) Eₙ = n² × 0.667 eV ≤ 20 eV needs n² ≤ 30.0, so n ≤ 5.48: five levels, n = 1 to 5, with E₅ = 16.7 eV inside and E₆ = 24.0 eV just over the cut.
- (c) The fractional spacing is (E_{n+1} − Eₙ)/Eₙ = (2n + 1)/n². At n = 5 that is 11/25 = 0.44; at n = 100 it is 201/10⁴ = 0.020.
- Absolute gaps widen without limit, but relative gaps fall as 2/n, so at large n no spectrometer resolves the rungs and the ladder looks continuous. That is the correspondence limit, and why a 1.0 g bead at 1.0 mm s⁻¹ in a 10 cm box (n ≈ 3 × 10²⁶) never shows quantisation.
AnswerL = 0.751 nm; five levels lie below 20 eV (n = 1–5); the fractional spacing falls from 0.44 at n = 5 to 0.020 at n = 100.