University Physics V · The Quantum Wavefunction · 3.2
Born's Rule, Density & Normalisation
Everything measurable in this course exits through one gate: square the position component of the state, integrate it over the interval your detector covers, and check the total is one. Treat |ψ|² as a density with units, and half the classic wavefunction blunders become impossible to commit.
Build the model
Connect the measurement to the mechanism.
Born's rule is the bridge from the Hilbert space to the counter: it declares |ψ(x)|² = |⟨x|ψ⟩|² a probability density for position, so the theory's predictions are the integrals ∫ₐᵇ|ψ|²dx over intervals of x̂'s continuous spectrum — formally ⟨ψ|Ê([a, b])|ψ⟩ with Ê the spectral projector, since x̂ owns no eigenvectors inside L²(ℝ). Normalisation ⟨ψ|ψ⟩ = 1 is not bookkeeping but the statement that the particle is certainly somewhere, and it prices the wavefunction's units at m⁻¹ᐟ² in one dimension; Hermiticity of H then guarantees the norm never drifts, so one normalisation lasts forever.
The cost of the rule is everything it refuses to say: ψ itself is unobservable (a global phase changes nothing, which is why a state is a ray), a single point carries zero probability, the density may exceed 1 wherever it likes, and any vector with a divergent norm — a plane wave above all — supports no probability statement until a box or delta convention scaffolds it. And even then the rule predicts ensemble frequencies over identically prepared systems, never the outcome of one run.
- Simple definition
- Born's rule reads |ψ(x)|² as a probability density: integrating it across an interval gives the chance of finding the particle there, and normalisation sets the whole-line integral to one.
- Example
- For the 1.0 nm well ground state ψ₁ = √(2/L) sin(πx/L), the leftmost quarter holds P = 1/4 − 1/(2π) ≈ 0.091, not 0.25: the density vanishes at the wall and piles up at the centre.
The exit from Hilbert space: the position component of |ψ⟩, squared, predicts the statistics of an ensemble of position measurements.
P(x) carries units m⁻¹ in 1D; only its integrals are probabilities
x̂ has no eigenvectors in L², so probability attaches to intervals of its continuous spectrum — a single point gets exactly zero.
Ê([a, b]) is the spectral projector of x̂: it multiplies ψ by the interval's indicator function
"The particle is certainly somewhere." Fixes |N| but never the phase: ψ and e(iα)ψ are the same ray and the same physics.
the total is dimensionless; in 3D, ∫|ψ|² d³r = 1 gives [ψ] = m⁻³ᐟ²
Do it once: for Hermitian H, d⟨ψ|ψ⟩/dt = (i/ħ)⟨ψ|(H† − H)|ψ⟩ = 0, so unitary evolution preserves the norm forever.
eigenvalue problems return ψ only up to scale; ψ must lie in L²(ℝ)
Box normalisation buys discrete modes and a 1/Lbox density; delta normalisation trades probabilities for densities per unit k. Both are scaffolding for wave packets.
|e(ikx)|² = 1 everywhere, so ∫|ψ|² dx diverges: no plane wave is in L²(ℝ)
From projectors to a density
For an observable with discrete spectrum, Born's rule is |⟨aₙ|ψ⟩|²: square the component along the eigenvector. Position breaks that recipe — x̂'s spectrum is the whole real line and it has no eigenvectors in L²(ℝ), since ⟨x|x′⟩ = δ(x − x′) is no norm at all. The spectral theorem repairs it: to every interval [a, b] it assigns a projector Ê([a, b]), which in the position representation simply multiplies ψ by the interval's indicator function, so P(a < x < b) = ⟨ψ|Ê([a, b])|ψ⟩ = ∫ₐᵇ|ψ(x)|²dx. Read backwards, that makes |ψ(x)|² a probability density — probability per unit length, meaningful only under an integral — and it is why a continuous spectrum attaches probability to intervals while a discrete one attaches it to eigenvalues.
The units are doing physics
A probability is dimensionless, so ∫|ψ|²dx = 1 forces [|ψ|²] = m⁻¹ and [ψ] = m⁻¹ᐟ² in one dimension; in three, ∫|ψ|²d³r = 1 makes it m⁻³ᐟ². The infinite well's ground state peaks at |ψ₁(L/2)|² = 2/L — for L = 1.0 nm that is 2.0 nm⁻¹, or 2.0 × 10⁹ m⁻¹ in SI, the number swinging by nine orders of magnitude under a unit change no probability could survive. Discrete expansion coefficients behave differently: cₙ = ⟨n|ψ⟩ against an orthonormal eigenbasis is dimensionless, and |cₙ|² is already a probability with no integral needed. Track which of the two you are holding, and dimensional analysis will catch most Born-rule errors before the arithmetic starts.
Normalise once; Hermiticity keeps it
Solving an eigenvalue problem returns ψ only up to scale, so fix it. A Gaussian ψ(x) = N e(−x²/4σ²) has ∫|N|²e(−x²/2σ²)dx = |N|²σ√(2π), and ⟨ψ|ψ⟩ = 1 gives N = (2πσ²)(−1/4) — about 0.63 nm⁻¹ᐟ² for σ = 1.0 nm. Only the modulus is fixed: N e(iα) works equally well for any real α, which is the precise sense in which a state is a ray rather than a vector. And the job is done once. The Schrödinger equation gives d⟨ψ|ψ⟩/dt = (i/ħ)⟨ψ|(H† − H)|ψ⟩, which vanishes exactly because H is Hermitian — self-adjointness is not a technicality but the guarantee that today's normalisation still holds at every later time.
Integrals, not point values
The n = 2 well state ψ₂ = √(2/L) sin(2πx/L) puts P(0 < x < L/8) = 1/8 − 1/(4π) ≈ 0.045 in the leftmost eighth, against the classical bouncer's 0.125 — the density rises from zero at a hard wall, so wall-hugging intervals are quantum-poor. Ask instead for the probability at exactly x = L/4 and the honest answer is zero: an integral over a width-zero interval vanishes, node or antinode alike. What a point value does buy is a ratio and an estimate. The density at L/4 is 2/L while at L/8 it is 1/L, so equal narrow windows around those points collect probability 2:1, and a window of width 0.010L at L/4 holds ≈ (2/L)(0.010L) = 0.020.
Plane waves and the two rescues
e(ikx) solves the free eigenvalue problem, yet |e(ikx)|² = 1 makes ⟨ψ|ψ⟩ = ∫1 dx diverge: momentum eigenfunctions are not vectors of L²(ℝ), and Born's rule says nothing about them directly. Two conventions rescue the calculation. Box normalisation confines the wave to a length Lbox with periodic boundary conditions, giving ψₖ = e(ikx)/√Lbox, discrete k = 2πn/Lbox and a uniform density 1/Lbox, with Lbox cancelling from any physical answer at the end. Delta normalisation keeps the line infinite and sets ⟨k|k′⟩ = δ(k − k′) via ψₖ = e(ikx)/√(2π), trading probabilities for densities per unit k. Both are scaffolding: every preparable state is a normalisable wave packet, and a "plane wave" is shorthand for a packet much longer than anything else in the problem.
Put it on a grid
Discretise ψ on N points with spacing dx and every statement above becomes NumPy. The norm is dx*np.sum(np.abs(ψ)**2) — a Riemann sum standing in for ⟨ψ|ψ⟩ — and ψ /= np.√(norm) normalises; an interval probability is the same sum over a slice. The Born rule itself can be simulated: np.random.choice(x, p=np.abs(ψ)**2*dx) draws positions one at a time, and a histogram of 10⁴ draws hugs |ψ|² with the 1/√N scatter of any counting experiment. Two checks keep the grid honest: the summed norm should approach 1 as dx shrinks (trapezoid error ~ dx² for smooth ψ), and the box edges must sit where ψ has already decayed, or the grid silently renormalises probability that belongs outside.
Change one variable at a time
Make the relationship visible.
Set w = 0.50 and sweep φ from 0° to 180°: P(0 < x < L/2) swings from 0.924 to 0.076 while the total area never moves — the coherence term, not the populations, carries the probability. Then slide c and watch P track the interval your detector actually covers.
P(0 < x < cL)0.924
COHERENCE TERM0.424
MIXTURE P (NO COHERENCE)0.500
NORM ⟨ψ|ψ⟩1.000
Live interpretationP(0 < x < cL): 0.924. COHERENCE TERM: 0.424. MIXTURE P (NO COHERENCE): 0.500. NORM ⟨ψ|ψ⟩: 1.000
Catch the common trap
Explain before calculating.
A normalised state of an electron on a line has probability density |ψ(x₀)|² = 3.0 nm⁻¹ at the point x₀, where the density is smooth. Which conclusion is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle on a line is prepared in ψ(x) = N e(−|x|/a) with a = 2.0 nm. Find N with its units, and the probability of finding the particle within one decay length of the origin, P(−a < x < a).
- Impose ⟨ψ|ψ⟩ = 1: ∫|N|² e(−2|x|/a) dx over the whole line = 2|N|² ∫₀^∞ e(−2x/a) dx = 2|N|²(a/2) = |N|² a.
- So |N| = a⁻¹ᐟ² — the inverse square root of a length, as any 1D wavefunction must carry. With a = 2.0 nm, N = (2.0 nm)⁻¹ᐟ² ≈ 0.71 nm⁻¹ᐟ², times an arbitrary global phase.
- P(−a < x < a) = (1/a) ∫₋ₐ^ₐ e(−2|x|/a) dx = 2(1/a)(a/2)(1 − e⁻²) = 1 − e⁻².
- Numerically 1 − 0.135 = 0.865: dimensionless, as every probability must be — though the density at the origin, |ψ(0)|² = 1/a = 0.50 nm⁻¹, is not.
AnswerN = a⁻¹ᐟ² ≈ 0.71 nm⁻¹ᐟ² (global phase free); P(−a < x < a) = 1 − e⁻² ≈ 0.865.
MediumAn electron occupies the n = 2 eigenstate of an infinite well of width L = 1.0 nm, ψ₂(x) = √(2/L) sin(2πx/L). Find P(0 < x < L/8), compare it with the classical uniform answer, and give the probability density at the midpoint x = L/2.
- P(0 < x < L/8) = (2/L) ∫₀(L/8) sin²(2πx/L) dx; with sin² = (1 − cos)/2 the antiderivative is x/L − sin(4πx/L)/(4π).
- At x = L/8 this is 1/8 − sin(π/2)/(4π) = 0.1250 − 0.0796 = 0.0454; the lower limit contributes zero.
- A classical particle rattling in the box spends time uniformly, giving 1/8 = 0.125 — about 2.8 times more. The quantum density rises from zero at a hard wall, so wall-hugging intervals are probability-poor.
- At the midpoint, sin(2π · ½) = sin π = 0, so |ψ₂(L/2)|² = 0 exactly: a node. Its content is not "zero probability at that point" — every exact point has that — but that intervals straddling L/2 collect less probability than their width suggests.
AnswerP(0 < x < L/8) = 1/8 − 1/(4π) ≈ 0.045 versus 0.125 classically; |ψ₂(L/2)|² = 0 — a node, though the probability at any exact point is zero regardless.
HardIn the same L = 1.0 nm well, prepare the superposition ψ = (ψ₁ + e(iφ) ψ₂)/√2 with a real relative phase φ. Show that ψ is normalised for every φ, find P(0 < x < L/2) as a function of φ, evaluate it at φ = 0 and φ = π, and state what a 50/50 statistical mixture of the two eigenstates gives for the same interval.
- ⟨ψ|ψ⟩ = (⟨1|1⟩ + ⟨2|2⟩ + e(iφ)⟨1|2⟩ + e(−iφ)⟨2|1⟩)/2 = (1 + 1 + 0 + 0)/2 = 1: orthonormality kills the cross terms over the full box, whatever φ is.
- Over the half box they survive. Since ψ₁ and ψ₂ are real, |ψ|² = (ψ₁² + ψ₂²)/2 + ψ₁ψ₂ cos φ.
- ∫₀(L/2) ψ₁² dx = ∫₀(L/2) ψ₂² dx = 1/2 by symmetry about L/2, so the diagonal part contributes (1/2 + 1/2)/2 = 1/2 regardless of φ.
- Cross integral: (2/L)∫₀(L/2) sin(πx/L) sin(2πx/L) dx = (4/L)∫₀(L/2) sin²(πx/L) cos(πx/L) dx = (4/L)(L/π)[sin³(πx/L)/3]₀(L/2) = 4/(3π) ≈ 0.4244.
- So P(0 < x < L/2) = 1/2 + (4/(3π)) cos φ: 0.924 at φ = 0, 0.076 at φ = π. The relative phase steers probability between the halves while the total stays 1.
- A mixture carries no coherence: ρ = (|1⟩⟨1| + |2⟩⟨2|)/2 gives P = (1/2)(1/2) + (1/2)(1/2) = 1/2 for every φ. That 4/(3π) interference term is the entire experimental difference between a superposition and ignorance of which eigenstate was prepared.
Answer⟨ψ|ψ⟩ = 1 for every φ; P(0 < x < L/2) = 1/2 + (4/(3π)) cos φ — 0.924 at φ = 0, 0.076 at φ = π. The 50/50 mixture gives exactly 1/2.