University Quantum Mechanics II · topic UQMII-12.07
A level decaying into a quasi-continuum, solved numerically
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.
A level decaying into a quasi-continuum, solved numerically: predict a finite change
A quasi-continuum is a band of discrete levels spaced closely enough that a level coupled to them decays exponentially at the golden-rule rate Γ = 2π|V|²ρ/ħ for as long as the spacing cannot be resolved: up to the recurrence time hρ. Use the supplied topic model Y_out = 0.658212*d/(2*pi*v^2), where Y_out is LIFETIME 1/Γ (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Coupling |V| to each band level, measured in µeV. Keep d (Band level spacing 1/ρ) = 0.5 µeV; f (Offset f of E₀ from a band level) = 0 spacings fixed. For this exercise use only 0.0871875 µeV ≤ v ≤ 0.362812 µeV 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 v = 0.147825 µeV and 0.186412 µeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LIFETIME 1/Γ against v
Predictions from the supplied A level decaying into a quasi-continuum, solved numerically 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): LIFETIME 1/Γ / ns
Horizontal axis (x): v: Coupling |V| to each band level / µeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Coupling |V| to each band level / µeV | LIFETIME 1/Γ / ns |
|---|---|
| 0.0871875 | 6.890448352671489 |
| 0.15609375 | 2.149734880908141 |
| 0.225 | 1.0346438854558282 |
| 0.29390625000000004 | 0.6063708126569722 |
| 0.3628125 | 0.3979160427208282 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 0.147825 µeV, substitute into Y_out = 0.658212*d/(2*pi*v^2): Y₀ = 2.39696 ns.
- At v = 0.186412 µeV, the same fixed-parameter model gives Y₁ = 1.50732 ns.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.889635 ns; the input change is Δv = 0.0385875 µeV.
- Average sensitivity is ΔY_out/Δv = -23.055 ns per µeV. 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 = 2.39696 ns, Y₁ = 1.50732 ns; ΔY_out = -0.889635 ns; average sensitivity = -23.055 ns per µeV.
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.
A level decaying into a quasi-continuum, solved numerically: infer an input with uncertainty
A quasi-continuum is a band of discrete levels spaced closely enough that a level coupled to them decays exponentially at the golden-rule rate Γ = 2π|V|²ρ/ħ for as long as the spacing cannot be resolved: up to the recurrence time hρ. Use the supplied topic model Y_out = 0.658212*d/(2*pi*v^2), where Y_out is LIFETIME 1/Γ (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Coupling |V| to each band level, measured in µeV. Keep d (Band level spacing 1/ρ) = 0.5 µeV; f (Offset f of E₀ from a band level) = 0 spacings fixed. For this exercise use only 0.0871875 µeV ≤ v ≤ 0.362812 µeV 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 = 0.80350857 ns, with a bounded reading uncertainty ±0.060839 ns. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000275625 µeV, 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.
LIFETIME 1/Γ against v
Predictions from the supplied A level decaying into a quasi-continuum, solved numerically 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): LIFETIME 1/Γ / ns
Horizontal axis (x): v: Coupling |V| to each band level / µeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Coupling |V| to each band level / µeV | LIFETIME 1/Γ / ns |
|---|---|
| 0.0871875 | 6.890448352671489 |
| 0.15609375 | 2.149734880908141 |
| 0.225 | 1.0346438854558282 |
| 0.29390625000000004 | 0.6063708126569722 |
| 0.3628125 | 0.3979160427208282 |
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(v) = 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(0.0871875) = 6.89045 ns and Y_out(0.362812) = 0.397916 ns. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 0.0871875 + (0.803509 − 6.89045)(0.362812 − 0.0871875)/(0.397916 − 6.89045) = 0.345594 µeV. 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 v = 0.255319 µeV; substituting back gives Y_out = 0.80350857 ns.
- The interpolation error in the input is |v_line − v| = 0.0902753 µeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.74266957 ns and 0.86434757 ns. Sort the roots even if the response decreases: 0.246169 µeV ≤ v ≤ 0.265571 µeV.
- The lower and upper input deviations are 0.00914952 µeV and 0.0102519 µeV. 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 v ≈ 0.255319 µeV; compatible interval [0.246169, 0.265571] µeV. Endpoint interpolation gives 0.345594 µeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.