University Physics II · topic UPII-15.06
Wavefunctions, probability, and uncertainty
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.
Wavefunctions, Probability & Uncertainty: predict a finite change
A wavefunction is a complex amplitude Ψ(x, t) whose modulus squared |Ψ(x, t)|² is the probability per unit length of finding the particle at x, so every probability is an area under |Ψ|² and never a value of Ψ itself. Use the supplied topic model Y_out = 0.39894/sigma, where Y_out is PEAK DENSITY |Ψ|² (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sigma is Position spread Δx, measured in nm. Keep x0 (Packet centre x₀) = 0 nm fixed. For this exercise use only 0.12125 nm ≤ sigma ≤ 0.27875 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 sigma = 0.1559 nm and 0.17795 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsigma across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PEAK DENSITY |Ψ|² against sigma
Predictions from the supplied Wavefunctions, Probability & Uncertainty 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): PEAK DENSITY |Ψ|² / nm⁻¹
Horizontal axis (x): sigma: Position spread Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sigma: Position spread Δx / nm | PEAK DENSITY |Ψ|² / nm⁻¹ |
|---|---|
| 0.12125 | 3.2902268041237117 |
| 0.160625 | 2.483673151750973 |
| 0.19999999999999996 | 1.9947000000000006 |
| 0.23937499999999995 | 1.6665900783289822 |
| 0.27874999999999994 | 1.4311748878923771 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sigma separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sigma = 0.1559 nm, substitute into Y_out = 0.39894/sigma: Y₀ = 2.55895 nm⁻¹.
- At sigma = 0.17795 nm, the same fixed-parameter model gives Y₁ = 2.24187 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.317082 nm⁻¹; the input change is Δsigma = 0.02205 nm.
- Average sensitivity is ΔY_out/Δsigma = -14.3802 nm⁻¹ per nm. 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 = 2.55895 nm⁻¹, Y₁ = 2.24187 nm⁻¹; ΔY_out = -0.317082 nm⁻¹; average sensitivity = -14.3802 nm⁻¹ 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.
Wavefunctions, Probability & Uncertainty: infer an input with uncertainty
A wavefunction is a complex amplitude Ψ(x, t) whose modulus squared |Ψ(x, t)|² is the probability per unit length of finding the particle at x, so every probability is an area under |Ψ|² and never a value of Ψ itself. Use the supplied topic model Y_out = 0.39894/sigma, where Y_out is PEAK DENSITY |Ψ|² (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sigma is Position spread Δx, measured in nm. Keep x0 (Packet centre x₀) = 0 nm fixed. For this exercise use only 0.12125 nm ≤ sigma ≤ 0.27875 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 = 1.8356839 nm⁻¹, with a bounded reading uncertainty ±0.060676 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sigma by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00001575 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.
PEAK DENSITY |Ψ|² against sigma
Predictions from the supplied Wavefunctions, Probability & Uncertainty 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): PEAK DENSITY |Ψ|² / nm⁻¹
Horizontal axis (x): sigma: Position spread Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sigma: Position spread Δx / nm | PEAK DENSITY |Ψ|² / nm⁻¹ |
|---|---|
| 0.12125 | 3.2902268041237117 |
| 0.160625 | 2.483673151750973 |
| 0.19999999999999996 | 1.9947000000000006 |
| 0.23937499999999995 | 1.6665900783289822 |
| 0.27874999999999994 | 1.4311748878923771 |
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(sigma) = 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(0.12125) = 3.29023 nm⁻¹ and Y_out(0.27875) = 1.43117 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sigma_line = 0.12125 + (1.83568 − 3.29023)(0.27875 − 0.12125)/(1.43117 − 3.29023) = 0.24448 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 sigma = 0.217325 nm; substituting back gives Y_out = 1.8356839 nm⁻¹.
- The interpolation error in the input is |sigma_line − sigma| = 0.0271548 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.7750079 nm⁻¹ and 1.8963599 nm⁻¹. Sort the roots even if the response decreases: 0.210371 nm ≤ sigma ≤ 0.224754 nm.
- The lower and upper input deviations are 0.00695354 nm and 0.00742893 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 sigma ≈ 0.217325 nm; compatible interval [0.210371, 0.224754] nm. Endpoint interpolation gives 0.24448 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.