University Quantum Mechanics II · topic UQMII-12.06
Fermi's golden rule derived from a sum over a continuum
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Fermi's golden rule derived from a sum over a continuum: predict a finite change
Fermi's golden rule is the constant rate Γ = (2π/ħ)|V_fi|²ρ at which a weak, steady perturbation drains a state into a continuum of final states with ρ states per unit energy, read at E_f = E_i. Use the supplied topic model Y_out = 1-exp(-w*t/0.6582)*(cos(d*t/0.6582)-(d/w)*sin(d*t/0.6582)), where Y_out is dP/dt ÷ GOLDEN RULE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; unmarked trigonometric functions use radians. t is Time since switch-on t, measured in ps. Keep w (Band half-width w) = 1 meV; d (Band centre minus E_i, d) = 0 meV fixed. For this exercise use only 1.0675 ps ≤ t ≤ 4.5325 ps on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at t = 1.8298 ps and 2.3149 ps. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
dP/dt ÷ GOLDEN RULE against t
Predictions from the supplied Fermi's golden rule derived from a sum over a continuum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): dP/dt ÷ GOLDEN RULE
Horizontal axis (x): t: Time since switch-on t / ps
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since switch-on t / ps | dP/dt ÷ GOLDEN RULE |
|---|---|
| 1.0675 | 0.8024665738786187 |
| 1.9337499999999999 | 0.9470250859517759 |
| 2.8 | 0.9857930803230626 |
| 3.66625 | 0.996189959524555 |
| 4.5325 | 0.9989782156333231 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 1.8298 ps, substitute into Y_out = 1-exp(-w*t/0.6582)*(cos(d*t/0.6582)-(d/w)*sin(d*t/0.6582)): Y₀ = 0.937962.
- At t = 2.3149 ps, the same fixed-parameter model gives Y₁ = 0.970312.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0323502; the input change is Δt = 0.4851 ps.
- Average sensitivity is ΔY_out/Δt = 0.0666878 1 per ps. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.937962, Y₁ = 0.970312; ΔY_out = 0.0323502; average sensitivity = 0.0666878 1 per ps.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Fermi's golden rule derived from a sum over a continuum: infer an input with uncertainty
Fermi's golden rule is the constant rate Γ = (2π/ħ)|V_fi|²ρ at which a weak, steady perturbation drains a state into a continuum of final states with ρ states per unit energy, read at E_f = E_i. Use the supplied topic model Y_out = 1-exp(-w*t/0.6582)*(cos(d*t/0.6582)-(d/w)*sin(d*t/0.6582)), where Y_out is dP/dt ÷ GOLDEN RULE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; unmarked trigonometric functions use radians. t is Time since switch-on t, measured in ps. Keep w (Band half-width w) = 1 meV; d (Band centre minus E_i, d) = 0 meV fixed. For this exercise use only 1.0675 ps ≤ t ≤ 4.5325 ps on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.99203824, with a bounded reading uncertainty ±0.001041. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0003465 ps, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
dP/dt ÷ GOLDEN RULE against t
Predictions from the supplied Fermi's golden rule derived from a sum over a continuum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): dP/dt ÷ GOLDEN RULE
Horizontal axis (x): t: Time since switch-on t / ps
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since switch-on t / ps | dP/dt ÷ GOLDEN RULE |
|---|---|
| 1.0675 | 0.8024665738786187 |
| 1.9337499999999999 | 0.9470250859517759 |
| 2.8 | 0.9857930803230626 |
| 3.66625 | 0.996189959524555 |
| 4.5325 | 0.9989782156333231 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.0675) = 0.802467 and Y_out(4.5325) = 0.998978. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 1.0675 + (0.992038 − 0.802467)(4.5325 − 1.0675)/(0.998978 − 0.802467) = 4.41013 ps. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 3.18115 ps; substituting back gives Y_out = 0.99203824.
- The interpolation error in the input is |t_line − t| = 1.22898 ps. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99099724 and 0.99307924. Sort the roots even if the response decreases: 3.10027 ps ≤ t ≤ 3.27338 ps.
- The lower and upper input deviations are 0.0808804 ps and 0.0922299 ps. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 3.18115 ps; compatible interval [3.10027, 3.27338] ps. Endpoint interpolation gives 4.41013 ps.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.