University Physics IV · topic UPIV-06.02
Admissible wavefunctions, normalisation, and the probability current
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.
Normalisation and the probability current: predict a finite change
A wavefunction is admissible when it is single-valued, continuous, and square-integrable, and normalising it means dividing by √(∫|ψ|²dx) so that the probability of finding the particle somewhere is exactly one. Use the supplied topic model Y_out = pi/max(k,0.01), where Y_out is FRINGE SPACING π / k (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. k is Wavenumber k, measured in nm⁻¹. Keep r (Reflected amplitude r = |B|/|A|) = 0.6 fixed. For this exercise use only 2.6375 nm⁻¹ ≤ k ≤ 7.3625 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 k = 3.677 nm⁻¹ and 4.3385 nm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FRINGE SPACING π / k against k
Predictions from the supplied Normalisation and the probability current 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): FRINGE SPACING π / k / nm
Horizontal axis (x): k: Wavenumber k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wavenumber k / nm⁻¹ | FRINGE SPACING π / k / nm |
|---|---|
| 2.6375 | 1.1911251767165092 |
| 3.81875 | 0.8226756539678672 |
| 5 | 0.6283185307179586 |
| 6.18125 | 0.5082455253532526 |
| 7.3625 | 0.42670188843324863 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At k = 3.677 nm⁻¹, substitute into Y_out = pi/max(k,0.01): Y₀ = 0.85439 nm.
- At k = 4.3385 nm⁻¹, the same fixed-parameter model gives Y₁ = 0.72412 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.130271 nm; the input change is Δk = 0.6615 nm⁻¹.
- Average sensitivity is ΔY_out/Δk = -0.196932 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 = 0.85439 nm, Y₁ = 0.72412 nm; ΔY_out = -0.130271 nm; average sensitivity = -0.196932 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.
Normalisation and the probability current: infer an input with uncertainty
A wavefunction is admissible when it is single-valued, continuous, and square-integrable, and normalising it means dividing by √(∫|ψ|²dx) so that the probability of finding the particle somewhere is exactly one. Use the supplied topic model Y_out = pi/max(k,0.01), where Y_out is FRINGE SPACING π / k (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. k is Wavenumber k, measured in nm⁻¹. Keep r (Reflected amplitude r = |B|/|A|) = 0.6 fixed. For this exercise use only 2.6375 nm⁻¹ ≤ k ≤ 7.3625 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 = 0.56915488 nm, with a bounded reading uncertainty ±0.021368 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 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.
FRINGE SPACING π / k against k
Predictions from the supplied Normalisation and the probability current 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): FRINGE SPACING π / k / nm
Horizontal axis (x): k: Wavenumber k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wavenumber k / nm⁻¹ | FRINGE SPACING π / k / nm |
|---|---|
| 2.6375 | 1.1911251767165092 |
| 3.81875 | 0.8226756539678672 |
| 5 | 0.6283185307179586 |
| 6.18125 | 0.5082455253532526 |
| 7.3625 | 0.42670188843324863 |
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(k) = 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(2.6375) = 1.19113 nm and Y_out(7.3625) = 0.426702 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 2.6375 + (0.569155 − 1.19113)(7.3625 − 2.6375)/(0.426702 − 1.19113) = 6.48198 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 k = 5.51975 nm⁻¹; substituting back gives Y_out = 0.56915488 nm.
- The interpolation error in the input is |k_line − k| = 0.962229 nm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.54778688 nm and 0.59052288 nm. Sort the roots even if the response decreases: 5.32002 nm⁻¹ ≤ k ≤ 5.73506 nm⁻¹.
- The lower and upper input deviations are 0.199731 nm⁻¹ and 0.215314 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 k ≈ 5.51975 nm⁻¹; compatible interval [5.32002, 5.73506] nm⁻¹. Endpoint interpolation gives 6.48198 nm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.