
Physics V course sequence · Interactive lesson
Born's rule, density and normalisation
Everything measurable in this course exits through one gate: take the squared modulus of 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.
UPV-03.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Build the model
Probability lives in the integral of |ψ|², never at a point.
The key idea
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 complete physical model and assumptions
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.
1From projectors to a density
For an observable with a non-degenerate discrete spectrum, Born's rule is |⟨aₙ|ψ⟩|²: the squared modulus of 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.
2The 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.
3Normalise 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.
4Integrals, not point values
Take the n = 2 well state ψ₂ = √(2/L) sin(2πx/L) and ask for the probability at exactly x = L/4, its antinode: the honest answer is zero, because 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. The estimate is only as good as the window is narrow against the scale on which the density changes: stretch it to a sizeable fraction of the L/2 spacing between nodes and only the integral will do.
5Plane 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.
6Put it on a grid
Discretise ψ on a grid of spacing dx and every statement above becomes a few lines of NumPy. The norm is the Riemann sum dx Σⱼ|ψⱼ|², standing in for ⟨ψ|ψ⟩; dividing ψ by the square root of that sum normalises it, and an interval probability is the same sum over a slice of the grid. The Born rule itself can be simulated: numpy.random.choice, given the grid positions and the weights |ψⱼ|²dx, draws positions one at a time, and a histogram of 10⁴ draws hugs |ψ|² with the relative scatter 1/√(counts per bin) of any counting experiment. Two checks keep the grid honest: the summed norm of an analytically normalised ψ 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.
Key equations
Read the units, assumptions and limits before substituting.
Born's rule as a density
P(x) = |⟨x|ψ⟩|² = |ψ(x)|²
The exit from Hilbert space: the squared modulus of the position component of |ψ⟩ predicts the statistics of an ensemble of position measurements.
P(x) carries units m⁻¹ in 1D; only its integrals are probabilitiesInterval probability
P(a < x < b) = ⟨ψ|Ê([a, b])|ψ⟩ = ∫ₐᵇ |ψ(x)|² dx
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 functionNormalisation fixes the units
⟨ψ|ψ⟩ = ∫|ψ(x)|² dx = 1 ⇒ [ψ] = m⁻¹ᐟ²
"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⁻³ᐟ²Normalising a raw solution
ψ → ψ / √⟨ψ|ψ⟩, legal iff 0 < ∫|ψ|² dx < ∞
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)Plane-wave conventions
ψₖ = e(ikx)/√Lbox (k = 2πn/Lbox) or ⟨k|k′⟩ = δ(k − k′)
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)Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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, moves the probability between the halves. Then slide c and watch P track the interval your detector actually covers.
Live results
- P(0 < x < cL)
- 0.924
- COHERENCE TERM
- 0.424
- MIXTURE P (NO COHERENCE)
- 0.500
- NORM ⟨ψ|ψ⟩
- 1.000
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Weight w of state |2⟩: 0.50; Relative phase φ: 0 °; Interval edge c, fraction of L: 0.50
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The claim "the particle is at x with probability |ψ(x)|²" is wrong twice over. |ψ(x)|² carries units of m⁻¹, and nothing bounds it by 1: squeeze a normalised Gaussian to σ = 0.05 nm and its peak density is 1/(σ√(2π)) ≈ 8.0 nm⁻¹ — a perfectly good density and an absurd probability. And the probability of any exact position is strictly zero, because an integral over a width-zero interval vanishes whatever the density does. Normalisation constrains the area under |ψ|², never its height. The repair is to let the density do only a density's job: multiply by a small width for a nearby-interval estimate, integrate for anything exact, and reserve the word probability for the dimensionless results.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A 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).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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) ∫₋ₐa 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.
Answer. N = a⁻¹ᐟ² ≈ 0.71 nm⁻¹ᐟ² (global phase free); P(−a < x < a) = 1 − e⁻² ≈ 0.865.
MediumExample 2
An 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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.
Answer. P(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.
HardExample 3
In 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- ⟨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)) cos φ interference term is, for this interval, 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.
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.