University Quantum Mechanics I · topic UQMI-05.06
Building a wave packet and evolving it
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.
Building a wave packet and evolving it: predict a finite change
To evolve a free particle, write ψ(x,0) as a superposition of plane waves with weight φ(k), multiply each weight by e^(−iħk²t/(2m)), and superpose again: ψ(x,t) = (1/√(2π)) ∫ φ(k) e^(i(kx − ħk²t/(2m))) dk. Use the supplied topic model Y_out = 2*pi/(7*dk), where Y_out is GRID SPACING Δx = L/7 (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dk is Component spacing δk, measured in nm⁻¹. Keep t (Time t) = 8 fs; k0 (Mean wavenumber k₀) = 5 nm⁻¹ fixed. For this exercise use only 0.208437 nm⁻¹ ≤ dk ≤ 0.641562 nm⁻¹ 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 dk = 0.303725 nm⁻¹ and 0.364362 nm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GRID SPACING Δx = L/7 against dk
Predictions from the supplied Building a wave packet and evolving it 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): GRID SPACING Δx = L/7 / nm
Horizontal axis (x): dk: Component spacing δk / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dk: Component spacing δk / nm⁻¹ | GRID SPACING Δx = L/7 / nm |
|---|---|
| 0.2084375 | 4.306316766539875 |
| 0.31671875 | 2.8340535602191386 |
| 0.42499999999999993 | 2.111995061236836 |
| 0.5332812499999999 | 1.6831604355593888 |
| 0.6415624999999999 | 1.3990809952664867 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dk separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dk = 0.303725 nm⁻¹, substitute into Y_out = 2*pi/(7*dk): Y₀ = 2.9553 nm.
- At dk = 0.364362 nm⁻¹, the same fixed-parameter model gives Y₁ = 2.46347 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.491823 nm; the input change is Δdk = 0.0606375 nm⁻¹.
- Average sensitivity is ΔY_out/Δdk = -8.11087 nm per nm⁻¹. 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 = 2.9553 nm, Y₁ = 2.46347 nm; ΔY_out = -0.491823 nm; average sensitivity = -8.11087 nm per nm⁻¹.
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.
Building a wave packet and evolving it: infer an input with uncertainty
To evolve a free particle, write ψ(x,0) as a superposition of plane waves with weight φ(k), multiply each weight by e^(−iħk²t/(2m)), and superpose again: ψ(x,t) = (1/√(2π)) ∫ φ(k) e^(i(kx − ħk²t/(2m))) dk. Use the supplied topic model Y_out = 2*pi/(7*dk), where Y_out is GRID SPACING Δx = L/7 (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dk is Component spacing δk, measured in nm⁻¹. Keep t (Time t) = 8 fs; k0 (Mean wavenumber k₀) = 5 nm⁻¹ fixed. For this exercise use only 0.208437 nm⁻¹ ≤ dk ≤ 0.641562 nm⁻¹ 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 = 1.8991003 nm, with a bounded reading uncertainty ±0.075003 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dk by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000433125 nm⁻¹, 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.
GRID SPACING Δx = L/7 against dk
Predictions from the supplied Building a wave packet and evolving it 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): GRID SPACING Δx = L/7 / nm
Horizontal axis (x): dk: Component spacing δk / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dk: Component spacing δk / nm⁻¹ | GRID SPACING Δx = L/7 / nm |
|---|---|
| 0.2084375 | 4.306316766539875 |
| 0.31671875 | 2.8340535602191386 |
| 0.42499999999999993 | 2.111995061236836 |
| 0.5332812499999999 | 1.6831604355593888 |
| 0.6415624999999999 | 1.3990809952664867 |
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(dk) = 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(0.208437) = 4.30632 nm and Y_out(0.641562) = 1.39908 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dk_line = 0.208437 + (1.8991 − 4.30632)(0.641562 − 0.208437)/(1.39908 − 4.30632) = 0.567069 nm⁻¹. 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 dk = 0.472644 nm⁻¹; substituting back gives Y_out = 1.8991003 nm.
- The interpolation error in the input is |dk_line − dk| = 0.094425 nm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8240973 nm and 1.9741033 nm. Sort the roots even if the response decreases: 0.454686 nm⁻¹ ≤ dk ≤ 0.492078 nm⁻¹.
- The lower and upper input deviations are 0.0179574 nm⁻¹ and 0.0194341 nm⁻¹. 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 dk ≈ 0.472644 nm⁻¹; compatible interval [0.454686, 0.492078] nm⁻¹. Endpoint interpolation gives 0.567069 nm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.