University Physics V · Radioactive Decay · 15.1
The Decay Law, Γ, and Where It Fails
Nobody has ever caught a nucleus ageing. This lesson shows why: Fourier-transform a Lorentzian line and the exponential drops out with λ = Γ/ħ, so the decay law is a claim about a lineshape and about an ensemble average — and it names the two clocks at which that claim stops holding.
Build the model
Connect the measurement to the mechanism.
Start from the state, not from the statistics. An unstable state |ψ(0)⟩ is not an eigenstate of the full Ĥ, so expand it on Ĥ's continuous spectrum with outgoing-wave boundary conditions; each |E⟩ then merely collects a phase e(−iEt/ħ), and the survival amplitude a(t) = ⟨ψ(0)|ψ(t)⟩ = ∫ρ(E)e(−iEt/ħ)dE is the Fourier transform of the state's own energy distribution ρ(E) = |⟨E|ψ(0)⟩|². Wigner and Weisskopf showed that a discrete level fed into a smooth, broad continuum acquires a pole at E₀ − iΓ/2, so ρ is a Breit–Wigner of full width Γ — and the transform of a Lorentzian is an exponential, S(t) = |a(t)|² = e(−Γt/ħ).
Read backwards, a rate is a linewidth: λ = Γ/ħ. Averaging that survival probability over N₀ independent nuclei gives ⟨N⟩ = N₀e(−λt), the mean of a binomial rather than the trajectory of anything. What the model costs is the Lorentzian itself: it has divergent ⟨H²⟩ and reaches to E = −∞, and no real level does either.
Restore a finite ⟨H²⟩ and survival starts quadratic; restore a decay threshold and survival must eventually beat any exponential.
- Simple definition
- The decay law says an unstable state's survival probability falls as e(−λt) with a rate λ = Γ/ħ fixed by the full width of its energy line, so the expected number of survivors in an ensemble is N₀e(−λt).
- Example
- The 14.4 keV level of ⁵⁷Fe has natural width Γ = 4.64 × 10⁻⁹ eV, so λ = Γ/ħ = 4.64 × 10⁻⁹ ÷ 6.582 × 10⁻¹⁶ = 7.05 × 10⁶ s⁻¹ and τ = 142 ns: the linewidth Mössbauer absorption resolves is the same number as the clock.
Turns a rate question into a lineshape question: the clock is the Fourier partner of the spectrum.
ρ(E) = |⟨E|ψ(0)⟩|² in eV⁻¹, with ∫ρ dE = 1; t in s
The only lineshape whose transform is exponential at every t — and it manages that because its ⟨H²⟩ diverges.
Γ is the full width at half maximum in eV; E₀ is the line centre
Γ = 1 eV means τ = 6.6 × 10⁻¹⁶ s. Quote a width and the half-life comes free.
ħ = 6.582 × 10⁻¹⁶ eV s; λ in s⁻¹, Γ in eV, τ and T½ in s
No nucleus ages: the next second looks identical after 1 ns and after ten half-lives.
S is the dimensionless survival probability; h and λ in s⁻¹
N₀ = 400 after one half-life gives 200 ± 10. The smooth curve is an average, not a path.
N₀ independent nuclei, each one Bernoulli trial; σN counted in nuclei
Zeno flatness below τZ = ħ/ΔE, a Paley–Wiener tail past roughly 50τ. Both real, both unobserved in nuclei.
(ΔE)² = ⟨H²⟩ − ⟨H⟩² taken over ρ(E), in eV²; Eₘᵢₙ is the decay threshold
The survival amplitude is a Fourier transform
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.
Wigner–Weisskopf: a Lorentzian in, an exponential out
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.
Constant hazard is the whole physical content
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 density p(t) = λe(−λt), while τ is its mean, dragged later by the tail.
One nucleus decays; N₀ nuclei make a law
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 is a statement about an expectation value, exact only as N₀ → ∞.
Short times: survival starts flat, not sloping
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, τZ = ħ/ΔE collapses to zero and the quadratic window shuts. Any real ρ(E) falls faster than 1/E² far from E₀, so ΔE is finite and the window reopens. For the ⁵⁷Fe level, with ΔE of order the 14.4 keV transition energy, τZ ≈ 4.6 × 10⁻²⁰ s, and the quadratic term only outweighs the linear one below t ≈ τZ²/τ ≈ 1.5 × 10⁻³² s.
Long times: the spectrum has a floor
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 twenty-seven 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.
Change one variable at a time
Make the relationship visible.
Drag Γ from 1 to 10 neV and watch τ fall from 658 ns to 66 ns: one Fourier pair, no new physics. Then drag t and watch the last readout refuse to move — the next 100 ns multiplies the survival by the same factor whatever t you start from, which is what memoryless means.
MEAN LIFE τ = ħ/Γ140 ns
HALF-LIFE T½ = τ ln297 ns
SURVIVAL S(t)0.368
NEXT 100 ns S(t+100)/S(t)0.490
Live interpretationMEAN LIFE τ = ħ/Γ: 140 ns. HALF-LIFE T½ = τ ln2: 97 ns. SURVIVAL S(t): 0.368. NEXT 100 ns S(t+100)/S(t): 0.490
Catch the common trap
Explain before calculating.
A level's energy distribution is measured. It is Breit–Wigner of width Γ across the resolved region, it is cut off at a threshold a few MeV below E₀, and far above E₀ it falls as 1/E⁴ rather than 1/E². What does that measured ρ(E) imply about the survival probability S(t)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe Z boson shows up in e⁺e⁻ annihilation as a Breit–Wigner resonance of full width Γ = 2.495 GeV. Taking ħ = 6.582 × 10⁻¹⁶ eV s, find its decay constant, mean life and half-life, then estimate how far it travels while moving at essentially c.
- The rate is the width in units of ħ: λ = Γ/ħ. Put Γ in eV first, Γ = 2.495 GeV = 2.495 × 10⁹ eV.
- λ = 2.495 × 10⁹ ÷ 6.582 × 10⁻¹⁶ = 3.79 × 10²⁴ s⁻¹.
- τ = 1/λ = 2.64 × 10⁻²⁵ s, and T½ = τ ln2 = 0.6931 × 2.64 × 10⁻²⁵ = 1.83 × 10⁻²⁵ s.
- Flight distance: cτ = 2.998 × 10⁸ × 2.64 × 10⁻²⁵ = 7.9 × 10⁻¹⁷ m, about a tenth of a proton radius. No detector can time that, which is why the Z's lifetime is only ever reported as the width of a resonance curve.
Answerλ = 3.79 × 10²⁴ s⁻¹, τ = 2.64 × 10⁻²⁵ s, T½ = 1.83 × 10⁻²⁵ s, and cτ ≈ 7.9 × 10⁻¹⁷ m — a tenth of a proton radius, so the lifetime is measured as a linewidth because there is no track to time.
MediumA source contains N₀ = 400 atoms of a nuclide with T½ = 20 min. Find the expected number surviving at 20 min and at 60 min with the standard deviation of each, then find the probability that an atom still present at 60 min survives the following 20 min.
- λ = ln2/T½ = 0.6931/20 min = 0.03466 min⁻¹, so the single-atom survival probability is p(t) = e(−λt): p(20) = 0.500 and p(60) = e(−2.079) = 0.125.
- The atoms are independent and identical, so N(t) is binomial: N(t) ~ Bin(400, p(t)), with ⟨N⟩ = 400p and Var N = 400p(1−p).
- At 20 min: ⟨N⟩ = 400 × 0.500 = 200 and σ = √(400 × 0.500 × 0.500) = √100 = 10, a 5.0% spread the smooth curve N₀e(−λt) does not show.
- At 60 min: ⟨N⟩ = 400 × 0.125 = 50 and σ = √(400 × 0.125 × 0.875) = √43.75 = 6.6, now 13% — the relative scatter √((1−p)/N₀p) grows as the source empties.
- Conditional survival: the hazard rate is the constant λ, so P(alive at 80 | alive at 60) = e(−λ × 20) = 0.500, identical to a freshly made atom. Having lasted an hour buys nothing.
Answer⟨N⟩ = 200 ± 10 at 20 min and 50 ± 6.6 at 60 min; an atom alive at 60 min still has probability 0.500 of surviving the next 20 min, exactly like a new one.
HardThe 14.4 keV level of ⁵⁷Fe has mean life τ = 141.8 ns. Find its natural width Γ and fractional width Γ/E₀. Then bound both failure regimes: estimate the Zeno time from ΔE ≈ E₀, find where the quadratic term stops outweighing the linear one, and find when the exponential would empty a 10²⁰-nucleus sample.
- Γ = ħ/τ = 6.582 × 10⁻¹⁶ eV s ÷ 1.418 × 10⁻⁷ s = 4.64 × 10⁻⁹ eV, so Γ/E₀ = 4.64 × 10⁻⁹ ÷ 1.441 × 10⁴ = 3.2 × 10⁻¹³ — the resolution that makes Mössbauer spectroscopy possible.
- The Zeno time uses the spread of the whole energy distribution, not the width of the line. Taking ΔE ≈ E₀ = 14.41 keV as the scale on which the Lorentzian must be cut off: τZ = ħ/ΔE = 6.582 × 10⁻¹⁶ ÷ 1.441 × 10⁴ = 4.57 × 10⁻²⁰ s.
- The quadratic term beats the linear one only while (t/τZ)² > t/τ, that is t < t₁ = τZ²/τ = (4.57 × 10⁻²⁰)² ÷ 1.418 × 10⁻⁷ = 1.5 × 10⁻³² s.
- At the far end, the exponential predicts the last of N₀ = 10²⁰ nuclei at N₀e(−t/τ) = 1, so t = τ ln(10²⁰) = 141.8 ns × 46.05 = 6.5 μs. Any Paley–Wiener tail sets in later than that, below one count.
- So the exponential is unchallenged from 1.5 × 10⁻³² s to 6.5 × 10⁻⁶ s: log₁₀(6.5 × 10⁻⁶ ÷ 1.5 × 10⁻³²) = 26.6, nearly twenty-seven decades. Both breakdowns are real theorems and neither is reachable here.
AnswerΓ = 4.64 × 10⁻⁹ eV and Γ/E₀ = 3.2 × 10⁻¹³; τZ ≈ 4.6 × 10⁻²⁰ s with the quadratic regime ending near 1.5 × 10⁻³² s, and the exponential's last nucleus leaving at 46τ = 6.5 μs — a 27-decade window in which the exponential is untouchable.