Normalisation and the probability current

The Schrödinger equation is a linear PDE, and like any such equation it has far more solutions than physics wants. Born's rule is the filter. If |ψ(x, t)|²dx is the probability of finding the particle in dx, then ∫|ψ|²dx over all space must converge — ψ must be square-integrable. e^(−|x|/L) qualifies; e^(+x²/L²) and the constant function do not, and neither is a state however neatly it solves the equation. Two further requirements have nothing to do with convergence, and they come from different places. ψ must be single-valued, or |ψ|² is not even a function of position — a demand that bites on an angular coordinate, where ψ(φ + 2π) has to equal ψ(φ). And ψ must be continuous everywhere, with ψ′ continuous wherever V is finite, because ψ″ appears in the equation: a kink in ψ makes ψ″ a delta function, and the only thing that can balance a delta on one side is a delta on the other. Where V is not finite the licence is withdrawn — at the walls of an infinite square well, ψ′ jumps, and it is allowed to.

Square-integrability makes normalisation possible; it does not perform it. If ∫|ψ|²dx = I, then ψ_N = ψ/√I integrates to 1. Take the trial function ψ = A x(a − x) on 0 < x < a, zero outside. Then ∫₀ᵃ x²(a − x)²dx = a⁵/3 − a⁵/2 + a⁵/5 = a⁵/30, so A = √(30/a⁵); with a = 1.00 nm that is A = 5.48 nm^(−5/2). Notice that A itself does not carry m^(−1/2) here: the factor x(a − x) already supplies two powers of length, so A has to carry m^(−5/2) to leave ψ at m^(−1/2). The dimension of a normalisation constant is whatever the rest of the trial function leaves over. Normalisation leaves exactly one thing undetermined, an overall factor e^(iα) with α real. That global phase cancels out of |ψ|², out of j, and out of every expectation value, so it is not a physical degree of freedom — two states differing only by it are the same state. A relative phase is a different animal: in ψ = c₁ψ₁ + c₂ψ₂ the difference arg c₂ − arg c₁ survives in the cross term and is measurable.

Normalising at t = 0 is worth nothing unless the norm stays at 1, so test it. With ρ = ψ*ψ, ∂ρ/∂t = ψ*ψ̇ + ψψ̇*, and the equation supplies both pieces: ψ̇ = (iħ/2m)ψ″ − (i/ħ)Vψ, and its conjugate ψ̇* = −(iħ/2m)ψ*″ + (i/ħ)V*ψ*. Substituting gives ∂ρ/∂t = (iħ/2m)(ψ*ψ″ − ψψ*″) + (i/ħ)(V* − V)|ψ|². Take V real and the second term dies outright. The first is already a perfect derivative, because ψ*ψ″ − ψψ*″ = ∂ₓ(ψ*ψ′ − ψψ*′). So ∂ρ/∂t + ∂j/∂x = 0 with j = (ħ/2mi)(ψ*ψ′ − ψψ*′) = (ħ/m)Im(ψ*ψ′), the last form following because ψ*ψ′ − ψψ*′ = 2i Im(ψ*ψ′). Beyond taking V real, nothing was assumed: not the shape of that potential, not the form of ψ. This is a local conservation law of the same type as charge conservation in electromagnetism: probability moves through the intervening space rather than disappearing here and reappearing there.

j has units of probability per second — a rate at which probability crosses a point — and its sign gives the direction. Three readings are worth committing to memory. For a plane wave ψ = A e^{(i(kx − ωt))}, ψ*ψ′ = ik|A|², so j = (ħk/m)|A|² = vρ: density times velocity, exactly the classical flux. For any real ψ, ψ*ψ′ is real, so Im(ψ*ψ′) = 0 and j vanishes identically — every non-degenerate bound state of a real one-dimensional potential can be chosen real, so bound states carry no current, though they are certainly not at rest, since ⟨p⟩ = 0 while ⟨p²⟩ > 0. And in any stationary state ρ is time-independent, so ∂ρ/∂t = 0 forces dj/dx = 0: the current is the same number at every x. That last fact is the whole basis of scattering coefficients, R = |j_refl|/j_inc and T = jₜᵣₐₙₛ/j_inc, and it is why R + T = 1 is a statement about currents rather than about amplitudes.

Integrate the continuity relation over all space and you get dN/dt = −[j] from −∞ to +∞ plus (2/ħ)∫Im(V)|ψ|²dx, with N = ∫|ψ|²dx. Two conditions have to hold for N to stay at 1. First, the boundary terms must vanish: a normalisable state takes ψ and ψ′ to zero at infinity and j goes with them, but nothing about a raw solution of the equation guarantees this in advance — it is a condition you impose, not one you inherit. Second, V must be real, or Im(V) acts as a source or a sink. Physicists exploit that deliberately. Setting V = V₀ − iΓ/2 gives dN/dt = −(Γ/ħ)N, so N(t) = e^(−t/τ) with τ = ħ/Γ: a state that leaks away with a lifetime, and an energy width tied to that lifetime by Γτ = ħ. Short life, broad line — that is where a natural linewidth comes from, and the Hard example puts numbers on it. This is a modelling device, not new physics — the missing probability has gone into channels the Hamiltonian omits, and the price is that Ĥ is no longer Hermitian.

ψ = Ae^(ikx) has |ψ|² = |A|² everywhere, so ∫|ψ|²dx = |A|² times the length of all space. The momentum eigenstates therefore sit outside the space of physical states, and there are three standard repairs. Box normalisation: confine the particle to a length L with periodic boundaries, take A = 1/√L, allow only k = 2πn/L, and let L → ∞ at the end of the calculation. Delta normalisation: keep the continuum and accept that ∫e^(−ik′x)e^(ikx)dx = 2πδ(k − k′), so that ψₖ = e^(ikx)/√(2π) obeys ∫ψ*_(k′)ψₖ dx = δ(k − k′) — normalised to a Dirac δ rather than to a Kronecker δ, with sums over states becoming integrals. Or build a packet: superpose a spread of k with a square-integrable weight, and the result is normalisable, with a finite Δx and Δp. For scattering none of this is required, because a ratio of currents stays finite even when neither state is normalisable — which is why transmission is always quoted as a current ratio.