The wave function, Born's rule and expectation values

Fire electrons at a double slit one at a time and each arrives as one dot at one place; the fringes exist only in the accumulated histogram. Born's rule says what ψ(x, t) predicts about that histogram: the probability of detecting the particle between x and x + dx is |ψ(x, t)|² dx, with |ψ|² = ψ*ψ. Over a finite interval, integrate. Over one much narrower than the distance on which ψ changes, P ≈ |ψ(x₀)|² Δx: a detector pixel 1.0 pm wide at the centre of a Gaussian packet whose |ψ|² has standard deviation σ = 50 pm, so that |ψ(0)|² = 1/(σ√(2π)), fires on 0.0080 of runs. Because |ψ|² is a probability per unit length, a one-dimensional ψ carries m⁻¹⁄², a unit check worth running on every normalisation constant. ψ has to be complex: a matter wave of definite momentum, e^{(i(kx − ωt))}, has a uniform |ψ|² and carries its momentum entirely in the phase, where the density cannot see it. Multiply ψ by a global factor e^(iα) and no prediction changes.

The Schrödinger equation is linear, so if ψ solves it so do 3ψ and −iψ; the physics picks the multiple by demanding ∫ |ψ|² dx = 1 over all space. Take ψ = A e^(-|x|/a). Then ∫ |ψ|² dx = 2|A|² ∫₀^∞ e^(−2x/a) dx = |A|² a, so |A| = 1/√a, in m⁻¹⁄² as it must be. For a = 0.10 nm the particle is found within 0.10 nm of the origin with probability 1 − e⁽⁻²⁾ = 0.865. Normalisation fixes |A| and leaves its phase free, which costs nothing because a global phase is unobservable. You normalise once: the Schrödinger equation conserves ∫ |ψ|² dx in time, so probability can flow from place to place but cannot appear or vanish. A function whose integral diverges cannot be normalised. A pure plane wave e^(ikx) is the standard case; it is the limit of physical, normalisable wave packets and is used for ratios of probabilities, never for probabilities themselves.

Because |ψ|² is a probability density, any function of position has a mean over repeated runs, ⟨f(x)⟩ = ∫ f(x) |ψ|² dx, computed exactly as the mean of a histogram, and a spread Δx = √(⟨x²⟩ − ⟨x⟩²). For the ground state of a box, ψ₁ = √(2/L) sin(πx/L), symmetry gives ⟨x⟩ = L/2 and the integral gives ⟨x²⟩ = L²(1/3 − 1/(2π²)), so Δx = 0.181L — narrower than the 0.289L of a uniform spread, because |ψ₁|² piles up in the middle. These are statements about ensembles: after N runs the sample mean scatters about ⟨x⟩ by Δx/√N, which for L = 1.0 nm and N = 100 is 0.018 nm. The same integral supplies the correspondence check. In the nth standing wave ⟨x²⟩ = L²(1/3 − 1/(2n²π²)), which tends to the classical L²/3 of a particle bouncing at constant speed as n grows; |ψₙ|², averaged over its rapid wiggles, tends to the uniform 1/L.

Position enters an average as a multiplier; momentum enters as a derivative. Acting on a plane wave e^(ikx), −iħ ∂/∂x returns ħk times the same function, and ħk = h/λ is de Broglie's momentum, so ⟨p⟩ = ∫ ψ*(−iħ ∂ψ/∂x) dx extends that to any state. Two results are worth knowing on sight. If ψ is real and normalisable, ∫ ψ ∂ψ/∂x dx = [ψ²/2] between −∞ and +∞ = 0, so ⟨p⟩ = 0: a real standing wave is an equal mix of leftward and rightward motion. Put a phase ramp on it, ψ = R(x) e^(ik₀x) with R real, and ⟨p⟩ = ħk₀; k₀ = 5.0 × 10⁹ m⁻¹ gives 5.3 × 10⁻²⁵ kg m s⁻¹. For the spread, ⟨p²⟩ = ħ² ∫ |∂ψ/∂x|² dx: a real ψ squeezed into a narrower region must change more steeply, so its Δp grows, which is the uncertainty trade-off seen from inside one state. Ehrenfest's theorem, m d⟨x⟩/dt = ⟨p⟩ with d⟨p⟩/dt = −⟨dV/dx⟩, is the correspondence limit: while the packet is narrow compared with the scale on which V changes, the averages move as a classical particle does.

Because the Schrödinger equation is linear, c₁ψ₁ + c₂ψ₂ is an allowed state whenever ψ₁ and ψ₂ are. Its density is |c₁|²|ψ₁|² + |c₂|²|ψ₂|² + 2 Re(c₁* c₂ ψ₁* ψ₂), and the last term, absent from any statistical mixture of the two, depends on the relative phase of c₁ and c₂. Take the two lowest box states in equal weight, ψ = (ψ₁ + e^(iφ)ψ₂)/√2. The probability of finding the particle in the left half is 1/2 + (4/(3π)) cos φ: 0.924 at φ = 0, 0.500 at φ = 90° and 0.076 at φ = 180°. Those three states have identical energy content, since |c₁|² and |c₂|² are unchanged; only a phase separates them. In time the relative phase advances at (E₂ − E₁)/ħ, so the density sloshes from one half of the box to the other with period h/(E₂ − E₁) — 3.7 fs for an electron in a 1.0 nm box — while either stationary state on its own keeps a static density.

To predict a measurement of a quantity A, expand ψ in A's eigenstates, ψ = Σ cₙψₙ with cₙ = ∫ ψₙ*ψ dx. Each run returns one eigenvalue aₙ, with probability |cₙ|². Take the normalised parabola ψ = √(30/L⁵) x(L − x) in a box and measure energy. Its overlap with the ground state is c₁ = 4√60/π³ = 0.9993, so E₁ appears in 99.86% of runs; E₂ never, because c₂ = 0 by symmetry; 9E₁ in 0.14%; higher odd levels share the last 0.0075%. The mean is ⟨E⟩ = 10E₁/π² = 1.013E₁, an average rather than an outcome, and never below E₁. If ψ is an eigenstate of A, one |cₙ| is 1 and every run agrees; otherwise identical preparations give different results and the theory fixes only their frequencies. Immediately after a run returns aₙ, a repeat returns aₙ again, because the state is now ψₙ. That is how the model treats measurement: as an ideal projection. How an apparatus comes to show one definite result is the measurement problem, which this calculus takes as given rather than explains.