The decay law, Γ, and where it fails

An unstable state cannot be an eigenstate of the full Hamiltonian; an eigenstate would sit still forever. So prepare |ψ(0)⟩ and expand it on Ĥ's continuous spectrum, |ψ(0)⟩ = ∫dE c(E)|E⟩, with the boundary condition that the decay products are purely outgoing. In that basis time evolution is trivial — each |E⟩ collects the phase e^(−iEt/ħ) — so the survival amplitude is a(t) = ⟨ψ(0)|e^(−iĤt/ħ)|ψ(0)⟩ = ∫dE ρ(E) e^(−iEt/ħ), with ρ(E) = |c(E)|² normalised to one. That is the Fock–Krylov theorem, and it is the entire apparatus of this topic: a decay law is not an extra postulate bolted onto quantum mechanics, it is the Fourier transform of a lineshape. Settle what ρ(E) looks like and the time dependence is already decided.

Weisskopf and Wigner solved the coupled amplitude equations for a discrete level emptying into a smooth, broad continuum and found a complex pole at E₀ − iΓ/2: a level shift and a width. The matching ρ is the Breit–Wigner line, ρ(E) = (Γ/2π)/[(E−E₀)² + Γ²/4], and its transform is a one-line contour integral, a(t) = e^{(−iE₀t/ħ)}e^(−Γt/2ħ). Squaring kills the phase and leaves S(t) = e^(−Γt/ħ). Read that backwards and a rate is a linewidth: λ = Γ/ħ, τ = ħ/Γ. The single reciprocal covers everything measurable. The Z boson's Γ = 2.495 GeV gives τ = 2.6 × 10⁻²⁵ s, far too short for any track; the 14.4 keV level of ⁵⁷Fe has Γ = 4.64 × 10⁻⁹ eV and τ = 142 ns, narrow enough that Mössbauer absorption resolves it. Nearly eighteen orders of magnitude, one formula.

Define the hazard rate h(t) = −(d/dt) ln S(t): the chance per unit time of going now, given survival up to now. For S = e^(−λt) it is the constant λ, and constancy is exactly memorylessness, since S(t+s)/S(s) = e^(−λt) = S(t) for every s. A nucleus that has already waited ten half-lives faces precisely the odds of one made a microsecond ago; nothing inside it is running down. Contrast a filament lamp or a person, whose hazard climbs with age — the exponential is the unique distribution with no age at all. The same distribution explains why the half-life and the mean life differ: T½ = τ ln2 = 0.693τ is the median of the decay-time density f(t) = λe^(−λt), while τ is its mean, dragged later by the tail.

Nothing so far predicts when a given nucleus goes, and nothing can. Each nucleus is an independent Bernoulli trial with survival probability p(t) = e^(−λt), so the number left is a binomial variable: ⟨N⟩ = N₀p and Var N = N₀p(1−p). The smooth curve is the mean; real data scatter about it with σ = √(N₀p(1−p)). Take N₀ = 400 and wait one half-life: ⟨N⟩ = 200 with σ = 10, a 5% run-to-run wobble the textbook curve never draws. The relative scatter √((1−p)/N₀p) grows as the source empties, which is why the last decades of a decay curve are the noisy ones and why counting statistics, not the fitting, usually limit a measured λ. So dN/dt = −λN holds exactly for the expectation value ⟨N⟩ at any N₀, and describes the count itself only as N₀ → ∞, when the relative scatter vanishes.

Expand the amplitude directly: a(t) = 1 − i⟨H⟩t/ħ − ⟨H²⟩t²/2ħ² + …, so S(t) = 1 − (ΔE)²t²/ħ² + O(t⁴) with (ΔE)² = ⟨H²⟩ − ⟨H⟩². Survival is quadratic and its slope at t = 0 is zero, which is why repeated fast measurement can freeze a decay — the quantum Zeno effect. An exponential has slope −λ at t = 0, so it cannot be right there. Why does the Lorentzian give an exact exponential anyway? Because ∫E²ρ(E)dE diverges for a Breit–Wigner: ΔE is infinite, τ_Z = ħ/ΔE collapses to zero and the quadratic window shuts. Any real ρ(E) falls faster than 1/E³ far from E₀, so ΔE is finite and the window reopens. Survival then sits above the exponential while t²/τ_Z² < t/τ, that is for t < τ_Z²/τ. For a Lorentzian cut off a distance W from E₀, (ΔE)² ≈ ΓW/π, so that window closes at τ_Z²/τ ≈ πħ/W whatever Γ is; for ⁵⁷Fe, cut off at its threshold 14.4 keV below E₀, that is about 1.4 × 10⁻¹⁹ s.

The decay products cannot carry away less than their own rest energy, so ρ(E) vanishes below a threshold Eₘᵢₙ. That single fact settles the far tail. A spectral density supported on a half-line cannot have a transform that decays exponentially for all t — the Paley–Wiener theorem — so S(t) must eventually fall more slowly than e^(−λt), typically as a power law whose exponent is set by how ρ(E) switches on at threshold. The crossover arrives only after the exponential has run some 50 to 100 lifetimes. For ⁵⁷Fe that means waiting until S ≈ e⁽⁻⁵⁰⁾ ≈ 2 × 10⁻²², whereas a sample of 10²⁰ nuclei has lost its last one at 46τ = 6.5 μs. Both failures are theorems and both are hidden: the exponential owns the nearly fourteen decades between them, which is the only stretch an experiment ever visits. Non-exponential tails have been observed in organic-molecule luminescence, never yet in nuclear decay.