University Quantum Mechanics I · topic UQMI-05.08
Quadratic dispersion and why a free packet spreads
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.
Quadratic dispersion and why a free packet spreads: predict a finite change
Wave-packet spreading is the growth of a free packet's position spread σₓ with time. It happens because ω = ħk²/2m is quadratic, so each wavenumber moves at its own group velocity ħk/m and the packet's components drift apart. Use the supplied topic model Y_out = sqrt(s^2 + 0.1157676*beta*t + 0.003350535*(1+beta^2)*t^2/s^2), where Y_out is WIDTH σₓ AT 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 Time since preparation t, measured in fs. Keep s (Initial width σ₀) = 1.2 nm; beta (Chirp β, phase βx²/(4σ₀²)) = -2 fixed. For this exercise use only 3.1875 fs ≤ t ≤ 9.4875 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 = 4.5735 fs and 5.4555 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.
WIDTH σₓ AT t against t
Predictions from the supplied Quadratic dispersion and why a free packet spreads 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): WIDTH σₓ AT t / nm
Horizontal axis (x): t: Time since preparation t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since preparation t / fs | WIDTH σₓ AT t / nm |
|---|---|
| 3.1875 | 0.9056394409190901 |
| 4.762499999999999 | 0.7753609490322425 |
| 6.3374999999999995 | 0.6632530483955991 |
| 7.9125 | 0.5799506976578847 |
| 9.487499999999999 | 0.5389797935058578 |
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 = 4.5735 fs, substitute into Y_out = sqrt(s^2 + 0.1157676*beta*t + 0.003350535*(1+beta^2)*t^2/s^2): Y₀ = 0.790201 nm.
- At t = 5.4555 fs, the same fixed-parameter model gives Y₁ = 0.723264 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0669367 nm; the input change is Δt = 0.882 fs.
- Average sensitivity is ΔY_out/Δt = -0.075892 nm per fs. 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 = 0.790201 nm, Y₁ = 0.723264 nm; ΔY_out = -0.0669367 nm; average sensitivity = -0.075892 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.
Quadratic dispersion and why a free packet spreads: infer an input with uncertainty
Wave-packet spreading is the growth of a free packet's position spread σₓ with time. It happens because ω = ħk²/2m is quadratic, so each wavenumber moves at its own group velocity ħk/m and the packet's components drift apart. Use the supplied topic model Y_out = sqrt(s^2 + 0.1157676*beta*t + 0.003350535*(1+beta^2)*t^2/s^2), where Y_out is WIDTH σₓ AT 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 Time since preparation t, measured in fs. Keep s (Initial width σ₀) = 1.2 nm; beta (Chirp β, phase βx²/(4σ₀²)) = -2 fixed. For this exercise use only 3.1875 fs ≤ t ≤ 9.4875 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 = 0.62227529 nm, with a bounded reading uncertainty ±0.012494 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 0.00063 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.
WIDTH σₓ AT t against t
Predictions from the supplied Quadratic dispersion and why a free packet spreads 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): WIDTH σₓ AT t / nm
Horizontal axis (x): t: Time since preparation t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since preparation t / fs | WIDTH σₓ AT t / nm |
|---|---|
| 3.1875 | 0.9056394409190901 |
| 4.762499999999999 | 0.7753609490322425 |
| 6.3374999999999995 | 0.6632530483955991 |
| 7.9125 | 0.5799506976578847 |
| 9.487499999999999 | 0.5389797935058578 |
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(3.1875) = 0.905639 nm and Y_out(9.4875) = 0.53898 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 3.1875 + (0.622275 − 0.905639)(9.4875 − 3.1875)/(0.53898 − 0.905639) = 8.0563 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 = 7.0305 fs; substituting back gives Y_out = 0.62227529 nm.
- The interpolation error in the input is |t_line − t| = 1.0258 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.60978129 nm and 0.63476929 nm. Sort the roots even if the response decreases: 6.80786 fs ≤ t ≤ 7.26657 fs.
- The lower and upper input deviations are 0.222639 fs and 0.236072 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 ≈ 7.0305 fs; compatible interval [6.80786, 7.26657] fs. Endpoint interpolation gives 8.0563 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.