A transition rate emerges in a restricted continuum window
Derive the golden rule from a finite-time peak and check density, normalization, and depletion assumptions.
On this page
The question for today
What you’ll understand
- Explain how a finite-time sinc-squared peak produces a continuum rate.
- Use a density of final states with consistent energy normalization.
- Check correlation, recurrence, and depletion time scales.
A single weakly coupled final state does not usually acquire probability at a constant rate forever. Its finite-time transition amplitude is coherent and oscillatory. A nearly continuous family of final states changes the result: as the resonance peak narrows, more finely spaced states contribute a total probability proportional to time. Fermi’s golden rule summarizes this intermediate regime. Its familiar formula must be accompanied by a time window, a density-of-states convention, and a weak-depletion check.
A finite observation gives a finite energy window
Switch on a weak time-independent perturbation at time zero and start in one unperturbed discrete state. The first-order probability for a specified final state is a sinc-squared function of its energy difference, multiplied by duration squared. The central peak has energy width of order ħ/T, not an infinitely sharp conservation law at finite T. The pulse onset and observation time supply this finite width. Its height grows as T² while its integrated area over final energy grows only as T.
A normalized version of this peak approaches a delta distribution as T increases. That statement means its integral against sufficiently smooth functions approaches their value at the resonant energy. It does not mean an individual finite-time probability becomes a literal delta function with an ordinary finite height. The replacement is useful only when the density of final states and squared matrix element vary slowly across the peak. A nearby continuum threshold, narrow resonance, or sharp selection feature can invalidate that simplification.
Integrating final states produces the rate
Replace the sum over closely spaced final states by an energy integral weighted by their density ρ(E). If both the density and matrix elements are smooth across the finite-time peak, evaluate their product at the initial energy and integrate the peak. The total leading probability becomes ΓT, with Γ equal to 2π over ħ times the squared coupling times the final-state density. For a static perturbation the resonant unperturbed final energy equals the initial one. A harmonic drive instead shifts that energy by its supplied or removed quantum.
The matrix-element convention must match the density convention. In a large normalization box, a final extended state often contributes a coupling proportional to inverse square root of volume, while the number of final states grows with volume. Their factors cancel in the physical rate. Counting the volume in the density but forgetting it in the matrix element yields a spurious size dependence. For a real cosine drive, use the relevant positive- or negative-frequency Fourier-component matrix element, rather than silently inserting the full peak amplitude.
Long compared with memory is still short compared with depletion
Let Λ be an energy scale over which the product of density and coupling changes appreciably. The continuum peak should be narrow relative to Λ, requiring T much longer than ħ/Λ. For a finite dense spectrum, it should still cover many levels, requiring T much shorter than ħ divided by their typical spacing. Finally ΓT must remain small for the first-order calculation. These can be compatible when the coupling is weak; sending T to infinity at fixed coupling is not justified by the same derivation.
At sufficiently short times a finite-bandwidth system starts with quadratic transition probability, and at very long times depletion, recurrences, or more detailed spectral features matter. Exponential survival can follow from additional controlled resummation or reservoir assumptions, not merely by extending ΓT beyond one. The probability lab illustrates independent preparations: Reset and set both controls to p=0.05 and n=10, predicting mean count 0.5; Reset and set p=0.05 and n=20, predicting mean one. It computes exact binomial counts, not a continuum spectrum, transition trajectory, or golden-rule rate.
Work it through
Check a numerical rate and its window
Use scaled ħ=1 energy-unit times time-unit, a static matrix element |V|=0.01 energy unit, and density ρ=50 per energy unit. Take T=0.5 time unit, a smooth spectral variation scale Λ=100 energy units, and typical final-level spacing 0.02 energy unit. Estimate Γ and total probability, then check the stated window.
Insert the coupling and density: Γ=2π×(0.01)²×50=0.01π per time unit, approximately 0.0314. The energy factors cancel to give inverse time.
Multiply by T to obtain probability 0.005π≈0.0157. This is well below one, supporting the weak-depletion requirement.
The memory scale is ħ/Λ=0.01 time unit and the spacing scale is ħ/0.02=50 time units. The chosen 0.5 lies between them, so the peak is narrow on the supplied smooth scale yet broad across many final levels.
The estimate is Γ≈0.0314 per time unit and P≈0.0157, with the supplied scale hierarchy compatible with the golden-rule approximation.
The living notebook
Explore the model
Interactive lab · probability
Count weak transitions across independent preparations
Reset and set BOTH controls to p=0.05 and n=10; predict mean np=0.5. Reset and set p=0.05 and n=20; predict mean one. The exact binomial distribution describes independent trials with an assigned p, not dynamical transitions.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA probability per preparation and an event-count distribution are distinct from the physical calculation of a rate.
Pause & reason
Check your understanding
Your turn
Try it yourself
In a box-normalized calculation, enlarge the normalization volume by a factor of eight without changing the physical system. The final-state density increases eightfold while each matrix element scales as volume^(−1/2). Find the change in its magnitude, squared magnitude, and golden-rule rate.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Apply the volume scaling to the amplitude before squaring, then combine it with the density.
Compare with the solution
The matrix-element magnitude is divided by √8, so its square is divided by eight. Multiplying that square by the density, which grows by eight, leaves the rate unchanged. A rate that grew eightfold in this relabeling would indicate inconsistent normalization rather than a real enhancement of the physical transition.
Keep the idea
Take it with you
- A continuum rate arises after integrating finite-time coherent probabilities.
- Density of states and matrix elements must use matching normalization conventions.
- The golden rule has a window between memory, discrete recurrence, and depletion scales.
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