University Quantum Mechanics I · topic UQMI-05.07
Phase velocity, group velocity, and which one is the particle
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.
Phase velocity, group velocity, and which is the particle: predict a finite change
The group velocity v_g = dω/dk, evaluated at the peak k₀ of φ(k), is the speed of a narrow-band packet's envelope |ψ|²; for a free particle it is ħk₀/m = p₀/m, twice the phase velocity ω/k at which the individual crests move. Use the supplied topic model Y_out = s0^2/0.0578838, where Y_out is SPREADING TIME (fs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s0 is Initial packet width, measured in nm. Keep t (Time t) = 6 fs; v0 (Energy-zero shift V₀) = 0 eV fixed. For this exercise use only 0.516875 nm ≤ s0 ≤ 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 s0 = 0.70745 nm and 0.828725 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPREADING TIME against s0
Predictions from the supplied Phase velocity, group velocity, and which is the particle 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): SPREADING TIME / fs
Horizontal axis (x): s0: Initial packet width / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s0: Initial packet width / nm | SPREADING TIME / fs |
|---|---|
| 0.516875 | 4.615449670287714 |
| 0.7334375 | 9.293283550946033 |
| 0.9500000000000001 | 15.59158175517157 |
| 1.1665625000000002 | 23.510344282964326 |
| 1.3831250000000002 | 33.04957113432429 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s0 = 0.70745 nm, substitute into Y_out = s0^2/0.0578838: Y₀ = 8.64638 fs.
- At s0 = 0.828725 nm, the same fixed-parameter model gives Y₁ = 11.8649 fs.
- Signed change: ΔY_out = Y₁ − Y₀ = 3.21851 fs; the input change is Δs0 = 0.121275 nm.
- Average sensitivity is ΔY_out/Δs0 = 26.5389 fs per nm. 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 = 8.64638 fs, Y₁ = 11.8649 fs; ΔY_out = 3.21851 fs; average sensitivity = 26.5389 fs 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.
Phase velocity, group velocity, and which is the particle: infer an input with uncertainty
The group velocity v_g = dω/dk, evaluated at the peak k₀ of φ(k), is the speed of a narrow-band packet's envelope |ψ|²; for a free particle it is ħk₀/m = p₀/m, twice the phase velocity ω/k at which the individual crests move. Use the supplied topic model Y_out = s0^2/0.0578838, where Y_out is SPREADING TIME (fs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s0 is Initial packet width, measured in nm. Keep t (Time t) = 6 fs; v0 (Energy-zero shift V₀) = 0 eV fixed. For this exercise use only 0.516875 nm ≤ s0 ≤ 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 = 18.876196 fs, with a bounded reading uncertainty ±2.126 fs. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s0 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.
SPREADING TIME against s0
Predictions from the supplied Phase velocity, group velocity, and which is the particle 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): SPREADING TIME / fs
Horizontal axis (x): s0: Initial packet width / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s0: Initial packet width / nm | SPREADING TIME / fs |
|---|---|
| 0.516875 | 4.615449670287714 |
| 0.7334375 | 9.293283550946033 |
| 0.9500000000000001 | 15.59158175517157 |
| 1.1665625000000002 | 23.510344282964326 |
| 1.3831250000000002 | 33.04957113432429 |
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(s0) = 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) = 4.61545 fs and Y_out(1.38313) = 33.0496 fs. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s0_line = 0.516875 + (18.8762 − 4.61545)(1.38313 − 0.516875)/(33.0496 − 4.61545) = 0.951331 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 s0 = 1.04529 nm; substituting back gives Y_out = 18.876196 fs.
- The interpolation error in the input is |s0_line − s0| = 0.0939566 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 16.750196 fs and 21.002196 fs. Sort the roots even if the response decreases: 0.984665 nm ≤ s0 ≤ 1.10258 nm.
- The lower and upper input deviations are 0.0606226 nm and 0.0572944 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 s0 ≈ 1.04529 nm; compatible interval [0.984665, 1.10258] nm. Endpoint interpolation gives 0.951331 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.