University Physics V · topic UPV-12.08
Atomic transitions, the golden rule, and lifetimes
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.
Transition rates, the golden rule and lifetimes: predict a finite change
Fermi's golden rule is the constant transition rate W = (2π/ħ)|⟨f|V̂|i⟩|²ρ(E_f) that first-order perturbation theory yields once the final states form a continuum dense enough to smear out the drive's energy uncertainty ħ/t. Use the supplied topic model Y_out = 1/max(t*w,0.02), where Y_out is LOBE ÷ BAND WIDTH (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Final-state band half-width w, measured in GHz. Keep t (Drive time t) = 1.5 ns fixed. For this exercise use only 0.550625 GHz ≤ w ≤ 3.14937 GHz 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 w = 1.12235 GHz and 1.48617 GHz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LOBE ÷ BAND WIDTH against w
Predictions from the supplied Transition rates, the golden rule and lifetimes 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): LOBE ÷ BAND WIDTH
Horizontal axis (x): w: Final-state band half-width w / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Final-state band half-width w / GHz | LOBE ÷ BAND WIDTH |
|---|---|
| 0.5506249999999999 | 1.2107453651153994 |
| 1.2003125 | 0.5554109172958431 |
| 1.8499999999999999 | 0.3603603603603604 |
| 2.4996875 | 0.2667000041671876 |
| 3.149375 | 0.2116822120791162 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 1.12235 GHz, substitute into Y_out = 1/max(t*w,0.02): Y₀ = 0.593992.
- At w = 1.48617 GHz, the same fixed-parameter model gives Y₁ = 0.448579.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.145413; the input change is Δw = 0.363825 GHz.
- Average sensitivity is ΔY_out/Δw = -0.399678 1 per GHz. Its negative sign means that this response decreases 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 = 0.593992, Y₁ = 0.448579; ΔY_out = -0.145413; average sensitivity = -0.399678 1 per GHz.
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.
Transition rates, the golden rule and lifetimes: infer an input with uncertainty
Fermi's golden rule is the constant transition rate W = (2π/ħ)|⟨f|V̂|i⟩|²ρ(E_f) that first-order perturbation theory yields once the final states form a continuum dense enough to smear out the drive's energy uncertainty ħ/t. Use the supplied topic model Y_out = 1/max(t*w,0.02), where Y_out is LOBE ÷ BAND WIDTH (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Final-state band half-width w, measured in GHz. Keep t (Drive time t) = 1.5 ns fixed. For this exercise use only 0.550625 GHz ≤ w ≤ 3.14937 GHz 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 = 0.31212996, with a bounded reading uncertainty ±0.015067. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000259875 GHz, 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.
LOBE ÷ BAND WIDTH against w
Predictions from the supplied Transition rates, the golden rule and lifetimes 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): LOBE ÷ BAND WIDTH
Horizontal axis (x): w: Final-state band half-width w / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Final-state band half-width w / GHz | LOBE ÷ BAND WIDTH |
|---|---|
| 0.5506249999999999 | 1.2107453651153994 |
| 1.2003125 | 0.5554109172958431 |
| 1.8499999999999999 | 0.3603603603603604 |
| 2.4996875 | 0.2667000041671876 |
| 3.149375 | 0.2116822120791162 |
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(w) = 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(0.550625) = 1.21075 and Y_out(3.14937) = 0.211682. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.550625 + (0.31213 − 1.21075)(3.14937 − 0.550625)/(0.211682 − 1.21075) = 2.88809 GHz. 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 w = 2.13586 GHz; substituting back gives Y_out = 0.31212996.
- The interpolation error in the input is |w_line − w| = 0.752229 GHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.29706296 and 0.32719696. Sort the roots even if the response decreases: 2.03751 GHz ≤ w ≤ 2.24419 GHz.
- The lower and upper input deviations are 0.0983537 GHz and 0.108331 GHz. 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 w ≈ 2.13586 GHz; compatible interval [2.03751, 2.24419] GHz. Endpoint interpolation gives 2.88809 GHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.