Born's rule and probability current

ρ(x, t) = ψ*ψ is real and never negative, whatever ψ is doing in the complex plane, and that is the only reason a squared modulus can serve as a probability density. Multiplying the whole state by a constant phase e^(iα) leaves ρ untouched, and so does the stationary-state factor e^(−iEt/ħ): a single energy eigenstate has a density that does not move at all. What survives squaring is relative phase between two terms, which is why a superposition of two eigenstates carries a cross term 2ψ₁ψ₂cos(ωt) and a single eigenstate carries none. Dimensions follow from the rule itself: probability is a pure number and dx is a length, so ρ has units of m⁻¹ in one dimension and ψ has units of one per root metre — the strangest-looking unit in the course, and a direct consequence of Born's rule.

The time-dependent Schrödinger equation is linear and homogeneous, so if ψ solves it then so does cψ for any complex c. Nothing in the dynamics picks c; Born's rule does, by demanding that the total probability equal 1. Try it on a shape with no walls in sight, ψ(x) = C e^(−|x|/a): the integral is |C|² × 2 × (a/2) = |C|²a, so |C| = 1/√a, and with a = 0.25 nm that is C = 2.0 nm^(−1/2) = 6.32 × 10⁴ m^(−1/2) with a peak density |ψ(0)|² = 4.0 nm⁻¹. Only the modulus of C is settled — the leftover phase is free and unmeasurable, the first sign that ψ carries more labels than physics uses. And notice what normalisation excludes: any function whose square integral diverges has no c that will work, so it is not a state at all.

Write ∂ρ/∂t = ψ*(∂ψ/∂t) + ψ(∂ψ*/∂t) and substitute the Schrödinger equation in the form ∂ψ/∂t = (iħ/2m)∂²ψ/∂x² − (i/ħ)Vψ, together with its conjugate. The potential contributes −(i/ħ)V|ψ|² from the first term and +(i/ħ)V*|ψ|² from the second, so the two cancel exactly when V is real. What remains is (iħ/2m)(ψ*ψ″ − ψψ*″), and that is a perfect x-derivative: it equals ∂/∂x[(iħ/2m)(ψ*ψ′ − ψψ*′)] = −∂j/∂x. Hence ∂ρ/∂t + ∂j/∂x = 0. Integrate over the whole line and the right-hand side becomes j(−∞) − j(+∞), which vanishes for any square-integrable ψ, so d/dt∫|ψ|²dx = 0. Normalise once and it stays normalised. Break the assumption on purpose — add an imaginary part, V → V − iΓ/2 — and the norm decays as e^(−Γt/ħ), which is how absorption and decay are modelled.

The bound states of a well are real sines, so ψ*ψ′ is already real and j is zero at every point: a stationary state is a standing wave with a frozen density, which is exactly why none of its predictions depend on t. Put two such states together with different energies and the relative phase advances at ω = (E₂ − E₁)/ħ, the density sloshes, and j stops being zero. The polar form ψ = √ρ e^(iS/ħ) says where the flow comes from — j = ρ(∂S/∂x)/m — so it is the phase, not the shape of ρ, that decides whether anything moves. The sign of j is the direction of flow, and the integrated form dP(a, b)/dt = j(a) − j(b) is the accounting you actually use: probability inside [a, b] rises only if more flows in at a than out at b. In the figure below, that is why the arrow is longest at the phase where the boundary probability is changing fastest, and zero at the two turning phases.

A plane wave ψ = A exp[i(kx − ωt)] has ρ = |A|² everywhere, so ∫ρ dx diverges and no choice of A normalises it. It is not a state; it is a limit that states approach. Its current is finite and useful all the same: j = (ħk/m)|A|² = v|A|², reading as flux equals density times speed. There are two standard repairs. Box normalisation confines the system to length L with periodic boundaries, giving A = 1/√L and a discrete set of k. Delta normalisation keeps the continuum and replaces 1 with a Dirac δ, ∫φ*ₖ φₖ′ dx = 2πδ(k − k′). For scattering problems neither is needed, because reflection and transmission are ratios in which |A|² cancels. Numbers for scale: boxing the system at L = 1.00 μm gives A = 1.00 × 10³ m^(−1/2) and neighbouring wavenumbers spaced by 2π/L = 6.28 × 10⁶ m⁻¹, fine enough that a sum over k behaves like an integral.

Nothing structural changes in 3D. The density ρ = |ψ(r, t)|² has units of m⁻³, so ψ carries m^(−3/2); the current becomes a vector J = (ħ/m)Im(ψ*∇ψ) with units m⁻² s⁻¹; and continuity reads ∂ρ/∂t + ∇⋅J = 0, whose integrated form turns the edge terms into a flux through the surface bounding the volume. What Born's rule does not give you is the individual outcome. It is a probability assignment, not a mechanism, and nothing in the deterministic linear Schrödinger equation implies it — which is why it is listed among the postulates. Nor is |ψ|² the whole distribution story: momentum has its own density, obtained from the same ψ by Fourier transform, and the two cannot be read off one classical picture.