University Physics V · Foundations of Quantum Mechanics · 2.8
The Born Rule & Probability
Everything before this point is linear algebra; nothing in it counts clicks. This is the postulate that cashes a state vector out as detector statistics — and the theorems that police it: a current that moves probability locally, and Gleason's proof that no rival rule was ever available.
Build the model
Connect the measurement to the mechanism.
Schrödinger's equation is deterministic: given ψ at t = 0 it fixes ψ forever. The indeterminism of quantum mechanics enters through one added postulate. Born's rule says a measurement of the observable A on the normalised state |ψ⟩ returns an eigenvalue aₙ with probability |⟨aₙ|ψ⟩|² — the squared modulus of a projection.
That one line does three jobs. It makes ⟨ψ|ψ⟩ = 1 physical rather than conventional, because the probabilities are squared expansion coefficients and Parseval makes them total one exactly when the state is normalised. It is consistent with the dynamics, because for real V the density |ψ|² obeys a continuity equation with current j = (ħ/m) Im(ψ*∂ₓψ), so probability moves locally and the norm never drifts.
And it is not adjustable: Gleason's theorem shows that in dimension three or more every additive probability assignment on projectors is Tr(ρ̂P̂), so the exponent 2 is forced, not fitted. The cost is scope. The rule delivers frequencies over an ensemble of identically prepared systems and says nothing about a single run; what — if anything — selects one outcome is the measurement problem, and it sits outside the postulate.
- Simple definition
- The Born rule assigns to each measurement outcome the squared modulus of the state's projection onto that outcome's eigenspace, turning normalised vectors in Hilbert space into probability distributions over detector results.
- Example
- For |ψ⟩ = (3|E₁⟩ + 4i|E₂⟩)/5, an energy measurement returns E₁ with probability |3/5|² = 0.36 and E₂ with |4i/5|² = 0.64 — the i drops out, and 0.36 + 0.64 = 1 because ⟨ψ|ψ⟩ = 1.
Degeneracy costs nothing: make P̂ₙ the projector onto the whole eigenspace and the rule is unchanged.
aₙ, |aₙ⟩: eigenvalue and eigenket of Hermitian Â; P̂ₙ = |aₙ⟩⟨aₙ| its projector; P is a pure number
Only the integral is a probability; the density itself may exceed 1 anywhere it likes.
⟨ψ|ψ⟩ = ∫|ψ|² dx = 1 forces [ψ] = m⁻¹ᐟ² in 1D; |ψ(x)|² is a density in m⁻¹, not a probability
For ψ = Ae^{ikx}: j = |A|²ħk/m, density times velocity — the flux behind R + T = 1 in scattering.
ħ in J s, m in kg, ψ in m⁻¹ᐟ²; j comes out in s⁻¹ — probability crossing the point x per second
Integrate over all x: d⟨ψ|ψ⟩/dt = 0, so one normalisation at t = 0 lasts forever.
Both terms in m⁻¹ s⁻¹; holds for real V — Im V ≠ 0 adds a sink term −(2 Im V/ħ)|ψ|²
The exponent 2 is a theorem in d ≥ 3, not a fit; a lone qubit is the one space that escapes.
μ: any assignment with μ ≥ 0, μ(I) = 1, additive on orthogonal projectors; ρ̂: positive, unit trace
Normalisation is physics, not housekeeping
Born's rule turns ⟨ψ|ψ⟩ = 1 from a convention into a physical requirement: the outcome probabilities are the squared moduli of expansion coefficients, and they total one exactly when the state does. Normalising fixes more than a constant. For ψ(x) = A e(−|x|/a), the integral ∫|ψ|² dx = |A|²a forces A = 1/√a, and with it the dimension of every one-dimensional wavefunction: [ψ] = m⁻¹ᐟ², so that |ψ|² dx is a pure number. Two things escape the procedure. The overall phase does — e(iα)|ψ⟩ has every squared overlap unchanged, which is why a state is a ray, not a vector. And the momentum eigenfunction e^{ikx} does — ∫|A|² dx diverges on the line, so no choice of A normalises it. Plane waves sit outside L² and take the delta convention ⟨p|p′⟩ = δ(p − p′) instead, and for them the Born rule delivers a probability density over p rather than a probability for any single sharp momentum.
From expansion coefficient to count rate
Expand the state on the eigenbasis of the observable: |ψ⟩ = Σ cₙ|aₙ⟩ with cₙ = ⟨aₙ|ψ⟩, each coefficient a projection. The rule reads off P(aₙ) = |cₙ|² = cₙ*cₙ — the conjugate is compulsory, because c² of a complex number is complex and no probability is. Take |ψ⟩ = ((1+i)|a₁⟩ + 2|a₂⟩)/√6: the weights are |1+i|² = 2 and |2|² = 4, so P(a₁) = 1/3 and P(a₂) = 2/3, and Parseval's identity guarantees Σ|cₙ|² = ⟨ψ|ψ⟩ = 1 in any orthonormal basis, not just this one. Degeneracy changes the bookkeeping, not the rule: when the eigenvalue a spans a d-dimensional eigenspace, build the projector P̂ₐ = Σⱼ|a, j⟩⟨a, j| over it and P(a) = ⟨ψ|P̂ₐ|ψ⟩ — square first and then add over the orthogonal directions, never add the amplitudes. A continuous spectrum swaps the sum for an integral, P(a ∈ Δ) = ∫Δ |⟨a|ψ⟩|² da, where |⟨a|ψ⟩|² is now a density carrying the inverse dimension of a.
The current: how the density moves without teleporting
Probability is not just conserved in total; it moves locally. Subtract ψ × (Schrödinger)* from ψ* × (Schrödinger): the V terms cancel when V is real, and what survives is ∂ₜ|ψ|² + ∂ₓj = 0 with j = (ħ/m) Im(ψ*∂ₓψ) — the exact analogue of charge conservation, with |ψ|² as the charge. The current diagnoses states at a glance. Any real ψ has j = 0, which is why the bound states of a well sit still. A plane wave Ae^{ikx} carries j = |A|²ħk/m, density times velocity. Integrated over the line, the continuity equation reduces d⟨ψ|ψ⟩/dt to a boundary term, and L² states vanish at infinity, so the norm never drifts: normalise once at t = 0 and the dynamics — unitary because H is self-adjoint — preserves it forever. The exception proves the rule: give V an imaginary part, as absorbing boundaries in numerical scattering deliberately do, and the norm decays at the local rate (2 Im V/ħ)|ψ|² — probability then leaves by construction, not by error.
Gleason: why the exponent 2 is not a choice
Why the squared modulus and not some other function of the overlap? Ask what any probability assignment on a Hilbert space must satisfy: every projector gets a number in [0,1], orthogonal outcomes add, and each complete basis totals 1. Gleason's 1957 theorem answers: in dimension three or more, the only such measures are μ(P̂) = Tr(ρ̂P̂) for some density operator ρ̂ — for a pure state, exactly |⟨a|ψ⟩|². A rival exponent dies on contact: under a |c|⁴ rule the state (1,1,1)/√3 in ℂ³ would give each basis outcome (1/3)² = 1/9, totalling 1/3 rather than 1, and no renormalisation can fix every basis at once. The theorem needs d ≥ 3 because there each direction belongs to continuum-many orthonormal bases, and additivity across all of them is crushing. A lone qubit escapes: its orthogonal projectors come only in antipodal pairs on the Bloch sphere, additivity constrains each pair separately, and explicit non-Born measures can be written down. So the exponent 2 is forced everywhere except the one space too small for the argument to bite.
Frequencies, not fates
The rule predicts the statistics of an ensemble of identically prepared systems and refuses to say anything about a single run. That makes its test statistical: prepare N copies, count outcomes, and compare against Np with the binomial spread √(Np(1−p)). With P(E₂) = 0.64 and N = 10⁴ runs, expect 6400 ± 48 — a count of 6390 confirms the rule, and asking which runs would return E₂ is not a question the theory answers. What the statistics can distinguish is a superposition from a mixture. The superposition c₁ψ₁ + c₂ψ₂ has density |c₁|²|ψ₁|² + |c₂|²|ψ₂|² + 2 Re(c₁*c₂ψ₁*ψ₂); the mixture keeps only the first two terms. That cross term is the experimental fingerprint of coherence — it is what two-slit fringes measure and what decoherence destroys. What the postulate never supplies is a definition of measurement itself, or a mechanism selecting one outcome: that is the measurement problem, and it lives outside the rule, not inside it.
The norm is your cheapest bug detector
The Born rule gives numerical work its acceptance test. Sample ψ on a grid with spacing Δx and the norm becomes Σ|ψᵢ|²Δx — one line of NumPy, np.sum(np.abs(ψ)**2)*dx — and printing it every few steps is a habit worth building. Propagate with Crank–Nicolson, whose one-step map (I + iĤΔt/2ħ)⁻¹(I − iĤΔt/2ħ) is exactly unitary for Hermitian Ĥ: the norm then holds to machine rounding over thousands of steps. Propagate with explicit Euler, ψ → (I − iĤΔt/ħ)ψ, and the norm grows every step — the map's eigenvalues 1 − iEₙΔt/ħ all have modulus √(1 + (EₙΔt/ħ)²) > 1 — so the density inflates exponentially and every probability quietly becomes a lie. The same printout flags a grid too coarse for the momenta present before any physics is misread. Conservation of Σ|ψᵢ|²Δx is the discrete shadow of the continuity equation, and a scheme that breaks it has broken the probability interpretation itself.
Change one variable at a time
Make the relationship visible.
Sweep φ through 360° — time evolution does exactly this, at rate (E₂ − E₁)/ħ: the density sloshes and the ⟨x⟩ dot moves, yet P(E₁) and P(E₂) never flinch, because energy-basis Born weights carry no phase. Then set θ to 0° or 90° and the cross term dies with only one coefficient left.
P(E₁) = cos²θ0.59
P(E₂) = sin²θ0.41
P(E₁) + P(E₂)1.00
⟨x⟩ / L0.323
Live interpretationP(E₁) = cos²θ: 0.59. P(E₂) = sin²θ: 0.41. P(E₁) + P(E₂): 1.00. ⟨x⟩ / L: 0.323
Catch the common trap
Explain before calculating.
A system is prepared in the normalised state |ψ⟩ = (|a₁⟩ + 2i|a₂⟩)/√5, where |a₁⟩ and |a₂⟩ are orthonormal eigenkets of a Hermitian operator belonging to distinct eigenvalues a₁ and a₂. What is the probability that a single measurement returns a₂?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA three-level system is prepared as |ψ⟩ = N(|1⟩ + 2|2⟩ − 2i|3⟩), with (|1⟩, |2⟩, |3⟩) orthonormal. Fix N, then find the probability that a measurement in this basis returns outcome 2.
- Born probabilities are the squared moduli of the coefficients, so they can only total one if ⟨ψ|ψ⟩ = 1. Sum the squared moduli: |1|² + |2|² + |−2i|² = 1 + 4 + 4 = 9.
- ⟨ψ|ψ⟩ = |N|² × 9 = 1 gives |N| = 1/3. Any phase on N is global and unobservable, so take N = 1/3.
- P(2) = |⟨2|ψ⟩|² = |2/3|² = 4/9 ≈ 0.44. Note that outcome 3 gets |−2i/3|² = 4/9 as well — the i and the sign are invisible to a measurement in this basis.
- Parseval check: 1/9 + 4/9 + 4/9 = 1, so no probability has leaked.
AnswerN = 1/3; P(2) = 4/9 ≈ 0.44 (with P(1) = 1/9 and P(3) = 4/9, summing to 1).
MediumA particle in an infinite square well of width L occupies the ground state ψ₁(x) = √(2/L) sin(πx/L). Find the probability of finding it in the leftmost quarter, 0 < x < L/4, and compare with the classical uniform answer.
- The Born rule for position integrates the density: P(0 < x < L/4) = ∫₀^{L/4} (2/L) sin²(πx/L) dx — never a value of |ψ|², always its integral.
- Use sin²u = (1 − cos 2u)/2: the integrand becomes (1/L)(1 − cos(2πx/L)).
- Integrate: (1/L)[x − (L/2π) sin(2πx/L)] from 0 to L/4 = 1/4 − (1/2π) sin(π/2) = 1/4 − 1/(2π).
- Numerically: 0.2500 − 0.1592 = 0.0908, about 9.1% — far below the classical uniform 25%, because the ground-state density is pushed away from the walls where ψ must vanish.
AnswerP = 1/4 − 1/(2π) ≈ 0.091, versus 0.25 for a uniform classical density.
HardFar to the left of a barrier, a stationary electron scattering state is ψ(x) = Ae^{ikx} + Be(−ikx) with k = 5.00 × 10⁹ m⁻¹, |A|² = 1.00 × 10⁶ m⁻¹ and reflectivity R = |B|²/|A|² = 0.36. Find the probability current, and reconcile it with the interference fringes in |ψ|².
- Differentiate: ∂ₓψ = ik(Ae^{ikx} − Be(−ikx)), so ψ*∂ₓψ = ik(|A|² − |B|²) + ik(B*Ae^{2ikx} − A*Be(−2ikx)).
- The bracket B*Ae^{2ikx} − A*Be(−2ikx) has the form z − z*, purely imaginary; multiplied by ik it becomes purely real and contributes nothing to the Im. So j = (ħk/m)(|A|² − |B|²) exactly — the cross terms cancel in the current even though they survive in the density.
- Speed: ħk/m = (1.055 × 10⁻³⁴ × 5.00 × 10⁹)/(9.11 × 10⁻³¹) = 5.79 × 10⁵ m s⁻¹ (a 0.95 eV electron).
- Current: j = 5.79 × 10⁵ × (1.00 − 0.36) × 10⁶ = 3.7 × 10¹¹ s⁻¹ in +x — incident flux v|A|² minus reflected flux vR|A|².
- Consistency: |ψ|² carries standing-wave fringes, but the state is stationary, so ∂ₜ|ψ|² = 0 and continuity forces ∂ₓj = 0 — the current must be uniform in x, which is exactly what the cancelling cross terms deliver. Dividing by v|A|² gives 1 − R = T: R + T = 1 is probability conservation wearing scattering clothes.
Answerj = (ħk/m)(|A|² − |B|²) ≈ 3.7 × 10¹¹ s⁻¹, uniform in x; the fringes live in the density, not the current.