University Physics III · topic UPIII-09.05
Wave packets and the uncertainty principle
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.
Wave packets and the uncertainty principle: predict a finite change
A wave packet is a superposition of plane waves spanning a band of wavenumbers, localised because the components cancel away from its centre; its rms widths obey Δx Δk ≥ ½, and with p = ħk, Δx Δp ≥ ħ/2. Use the supplied topic model Y_out = 1/(2*s), where Y_out is RMS SPREAD Δk (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Initial rms width σ₀, measured in nm. Keep k (Central wavenumber k₀) = 8 nm⁻¹; t (Time since preparation t) = 0 fs fixed. For this exercise use only 0.516875 nm ≤ s ≤ 1.38313 nm 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 s = 0.70745 nm and 0.828725 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RMS SPREAD Δk against s
Predictions from the supplied Wave packets and the uncertainty principle 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): RMS SPREAD Δk / nm⁻¹
Horizontal axis (x): s: Initial rms width σ₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Initial rms width σ₀ / nm | RMS SPREAD Δk / nm⁻¹ |
|---|---|
| 0.516875 | 0.9673518742442564 |
| 0.7334375 | 0.6817213463996592 |
| 0.9500000000000001 | 0.5263157894736842 |
| 1.1665625000000002 | 0.42860969729440124 |
| 1.3831250000000002 | 0.3615002259376412 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 0.70745 nm, substitute into Y_out = 1/(2*s): Y₀ = 0.706764 nm⁻¹.
- At s = 0.828725 nm, the same fixed-parameter model gives Y₁ = 0.603336 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.103427 nm⁻¹; the input change is Δs = 0.121275 nm.
- Average sensitivity is ΔY_out/Δs = -0.852833 nm⁻¹ per nm. 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 = 0.706764 nm⁻¹, Y₁ = 0.603336 nm⁻¹; ΔY_out = -0.103427 nm⁻¹; average sensitivity = -0.852833 nm⁻¹ per nm.
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.
Wave packets and the uncertainty principle: infer an input with uncertainty
A wave packet is a superposition of plane waves spanning a band of wavenumbers, localised because the components cancel away from its centre; its rms widths obey Δx Δk ≥ ½, and with p = ħk, Δx Δp ≥ ħ/2. Use the supplied topic model Y_out = 1/(2*s), where Y_out is RMS SPREAD Δk (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Initial rms width σ₀, measured in nm. Keep k (Central wavenumber k₀) = 8 nm⁻¹; t (Time since preparation t) = 0 fs fixed. For this exercise use only 0.516875 nm ≤ s ≤ 1.38313 nm 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 = 0.4783373 nm⁻¹, with a bounded reading uncertainty ±0.017526 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000086625 nm, 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.
RMS SPREAD Δk against s
Predictions from the supplied Wave packets and the uncertainty principle 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): RMS SPREAD Δk / nm⁻¹
Horizontal axis (x): s: Initial rms width σ₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Initial rms width σ₀ / nm | RMS SPREAD Δk / nm⁻¹ |
|---|---|
| 0.516875 | 0.9673518742442564 |
| 0.7334375 | 0.6817213463996592 |
| 0.9500000000000001 | 0.5263157894736842 |
| 1.1665625000000002 | 0.42860969729440124 |
| 1.3831250000000002 | 0.3615002259376412 |
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(s) = 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(0.516875) = 0.967352 nm⁻¹ and Y_out(1.38313) = 0.3615 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.516875 + (0.478337 − 0.967352)(1.38313 − 0.516875)/(0.3615 − 0.967352) = 1.21607 nm. 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 s = 1.04529 nm; substituting back gives Y_out = 0.4783373 nm⁻¹.
- The interpolation error in the input is |s_line − s| = 0.170783 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.4608113 nm⁻¹ and 0.4958633 nm⁻¹. Sort the roots even if the response decreases: 1.00834 nm ≤ s ≤ 1.08504 nm.
- The lower and upper input deviations are 0.0369451 nm and 0.0397553 nm. 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 s ≈ 1.04529 nm; compatible interval [1.00834, 1.08504] nm. Endpoint interpolation gives 1.21607 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.