University Physics IV · topic UPIV-05.09
Energy-time uncertainty and the classical limit
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.
Energy–time uncertainty and the classical limit: predict a finite change
The energy–time uncertainty relation says a state whose energy is spread by σ_E cannot change appreciably in less than a time of order ħ/σ_E — Δt is a property of the state's own evolution, never the reading of a clock operator. Use the supplied topic model Y_out = 159.155*(1+c*tr)/max(tr,0.1), where Y_out is LINE FWHM Δν (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. tr is Radiative lifetime τ_rad, measured in ns. Keep c (Collisional quench rate) = 0 1/ns fixed. For this exercise use only 1.15938 ns ≤ tr ≤ 2.34063 ns 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 tr = 1.41925 ns and 1.58462 ns. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LINE FWHM Δν against tr
Predictions from the supplied Energy–time uncertainty and the classical limit 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): LINE FWHM Δν / MHz
Horizontal axis (x): tr: Radiative lifetime τ_rad / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tr: Radiative lifetime τ_rad / ns | LINE FWHM Δν / MHz |
|---|---|
| 1.159375 | 137.2765498652291 |
| 1.4546875000000001 | 109.40837808807733 |
| 1.75 | 90.94571428571429 |
| 2.0453125 | 77.81451489686783 |
| 2.340625 | 67.99679572763685 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of tr separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At tr = 1.41925 ns, substitute into Y_out = 159.155*(1+c*tr)/max(tr,0.1): Y₀ = 112.14 MHz.
- At tr = 1.58462 ns, the same fixed-parameter model gives Y₁ = 100.437 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -11.7032 MHz; the input change is Δtr = 0.165375 ns.
- Average sensitivity is ΔY_out/Δtr = -70.7677 MHz per ns. 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 = 112.14 MHz, Y₁ = 100.437 MHz; ΔY_out = -11.7032 MHz; average sensitivity = -70.7677 MHz per ns.
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.
Energy–time uncertainty and the classical limit: infer an input with uncertainty
The energy–time uncertainty relation says a state whose energy is spread by σ_E cannot change appreciably in less than a time of order ħ/σ_E — Δt is a property of the state's own evolution, never the reading of a clock operator. Use the supplied topic model Y_out = 159.155*(1+c*tr)/max(tr,0.1), where Y_out is LINE FWHM Δν (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. tr is Radiative lifetime τ_rad, measured in ns. Keep c (Collisional quench rate) = 0 1/ns fixed. For this exercise use only 1.15938 ns ≤ tr ≤ 2.34063 ns 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 = 84.659729 MHz, with a bounded reading uncertainty ±2.4994 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tr by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125 ns, 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.
LINE FWHM Δν against tr
Predictions from the supplied Energy–time uncertainty and the classical limit 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): LINE FWHM Δν / MHz
Horizontal axis (x): tr: Radiative lifetime τ_rad / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tr: Radiative lifetime τ_rad / ns | LINE FWHM Δν / MHz |
|---|---|
| 1.159375 | 137.2765498652291 |
| 1.4546875000000001 | 109.40837808807733 |
| 1.75 | 90.94571428571429 |
| 2.0453125 | 77.81451489686783 |
| 2.340625 | 67.99679572763685 |
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(tr) = 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(1.15938) = 137.277 MHz and Y_out(2.34063) = 67.9968 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tr_line = 1.15938 + (84.6597 − 137.277)(2.34063 − 1.15938)/(67.9968 − 137.277) = 2.05651 ns. 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 tr = 1.87994 ns; substituting back gives Y_out = 84.659729 MHz.
- The interpolation error in the input is |tr_line − tr| = 0.176577 ns. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 82.160329 MHz and 87.159129 MHz. Sort the roots even if the response decreases: 1.82603 ns ≤ tr ≤ 1.93713 ns.
- The lower and upper input deviations are 0.0539096 ns and 0.0571896 ns. 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 tr ≈ 1.87994 ns; compatible interval [1.82603, 1.93713] ns. Endpoint interpolation gives 2.05651 ns.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.