University Physics IV · topic UPIV-05.07
Free evolution: wave packets, group velocity, and spreading
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.
Free evolution, group velocity and spreading: predict a finite change
A free wave packet is a superposition of plane waves in which time only multiplies each component by e^(−iħk²t/2m): the envelope travels at the group velocity ħk₀/m while its width grows from the spread of speeds it already contained. Use the supplied topic model Y_out = s*sqrt(1+(t/(17.276*s*s))^2), where Y_out is RMS WIDTH σ(t) (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. t is Elapsed time t, measured in fs. Keep s (Initial rms width σ₀) = 2 nm; k (Central wavenumber k₀) = 2 nm⁻¹ fixed. For this exercise use only 6.375 fs ≤ t ≤ 53.625 fs 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 = 16.77 fs and 23.385 fs. 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.
RMS WIDTH σ(t) against t
Predictions from the supplied Free evolution, group velocity and spreading 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 WIDTH σ(t) / nm
Horizontal axis (x): t: Elapsed time t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Elapsed time t / fs | RMS WIDTH σ(t) / nm |
|---|---|
| 6.375 | 2.008492448587008 |
| 18.1875 | 2.0681093926947525 |
| 29.999999999999996 | 2.1803369970618047 |
| 41.81249999999999 | 2.3376100611187764 |
| 53.62499999999999 | 2.5315470746983904 |
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 = 16.77 fs, substitute into Y_out = s*sqrt(1+(t/(17.276*s*s))^2): Y₀ = 2.05805 nm.
- At t = 23.385 fs, the same fixed-parameter model gives Y₁ = 2.11141 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0533633 nm; the input change is Δt = 6.615 fs.
- Average sensitivity is ΔY_out/Δt = 0.00806702 nm per fs. 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 = 2.05805 nm, Y₁ = 2.11141 nm; ΔY_out = 0.0533633 nm; average sensitivity = 0.00806702 nm per fs.
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.
Free evolution, group velocity and spreading: infer an input with uncertainty
A free wave packet is a superposition of plane waves in which time only multiplies each component by e^(−iħk²t/2m): the envelope travels at the group velocity ħk₀/m while its width grows from the spread of speeds it already contained. Use the supplied topic model Y_out = s*sqrt(1+(t/(17.276*s*s))^2), where Y_out is RMS WIDTH σ(t) (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. t is Elapsed time t, measured in fs. Keep s (Initial rms width σ₀) = 2 nm; k (Central wavenumber k₀) = 2 nm⁻¹ fixed. For this exercise use only 6.375 fs ≤ t ≤ 53.625 fs 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 = 2.244485 nm, with a bounded reading uncertainty ±0.035399 nm. 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 0.004725 fs, 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 WIDTH σ(t) against t
Predictions from the supplied Free evolution, group velocity and spreading 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 WIDTH σ(t) / nm
Horizontal axis (x): t: Elapsed time t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Elapsed time t / fs | RMS WIDTH σ(t) / nm |
|---|---|
| 6.375 | 2.008492448587008 |
| 18.1875 | 2.0681093926947525 |
| 29.999999999999996 | 2.1803369970618047 |
| 41.81249999999999 | 2.3376100611187764 |
| 53.62499999999999 | 2.5315470746983904 |
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(6.375) = 2.00849 nm and Y_out(53.625) = 2.53155 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 6.375 + (2.24449 − 2.00849)(53.625 − 6.375)/(2.53155 − 2.00849) = 27.6933 fs. 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 = 35.1975 fs; substituting back gives Y_out = 2.244485 nm.
- The interpolation error in the input is |t_line − t| = 7.50417 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.209086 nm and 2.279884 nm. Sort the roots even if the response decreases: 32.4138 fs ≤ t ≤ 37.8162 fs.
- The lower and upper input deviations are 2.78373 fs and 2.61873 fs. 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 ≈ 35.1975 fs; compatible interval [32.4138, 37.8162] fs. Endpoint interpolation gives 27.6933 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.