University Physics IV · topic UPIV-07.01
Piecewise-constant potentials and the matching conditions
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.
Piecewise potentials and matching: predict a finite change
A piecewise-constant potential is flat inside each of a few regions and jumps between them, so the Schrödinger equation is solved separately in every region and the pieces are joined by demanding that ψ and ψ′ agree at each boundary. Use the supplied topic model Y_out = 5.1227*sqrt(en)*cos(2.56135*sqrt(en))+5.1227*sqrt(max(v0-en,0.02))*sin(2.56135*sqrt(en)), where Y_out is ψ′ MISMATCH AT STEP (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; unmarked trigonometric functions use radians. en is Energy E, measured in eV. Keep v0 (Step height V₀) = 6 eV fixed. For this exercise use only 0.67625 eV ≤ en ≤ 2.3825 eV 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 en = 1.05163 eV and 1.2905 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ψ′ MISMATCH AT STEP against en
Predictions from the supplied Piecewise potentials and matching 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): ψ′ MISMATCH AT STEP / nm⁻¹
Horizontal axis (x): en: Energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Energy E / eV | ψ′ MISMATCH AT STEP / nm⁻¹ |
|---|---|
| 0.67625 | 8.015408474004285 |
| 1.1028125 | 0.10937867828723302 |
| 1.529375 | -6.61431321527364 |
| 1.9559374999999999 | -10.873470809858201 |
| 2.3825 | -12.510709565756605 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 1.05163 eV, substitute into Y_out = 5.1227*sqrt(en)*cos(2.56135*sqrt(en))+5.1227*sqrt(max(v0-en,0.02))*sin(2.56135*sqrt(en)): Y₀ = 1.04026 nm⁻¹.
- At en = 1.2905 eV, the same fixed-parameter model gives Y₁ = -3.10871 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.14896 nm⁻¹; the input change is Δen = 0.238875 eV.
- Average sensitivity is ΔY_out/Δen = -17.3688 nm⁻¹ per eV. 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 = 1.04026 nm⁻¹, Y₁ = -3.10871 nm⁻¹; ΔY_out = -4.14896 nm⁻¹; average sensitivity = -17.3688 nm⁻¹ per eV.
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.
Piecewise potentials and matching: infer an input with uncertainty
A piecewise-constant potential is flat inside each of a few regions and jumps between them, so the Schrödinger equation is solved separately in every region and the pieces are joined by demanding that ψ and ψ′ agree at each boundary. Use the supplied topic model Y_out = 5.1227*sqrt(en)*cos(2.56135*sqrt(en))+5.1227*sqrt(max(v0-en,0.02))*sin(2.56135*sqrt(en)), where Y_out is ψ′ MISMATCH AT STEP (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; unmarked trigonometric functions use radians. en is Energy E, measured in eV. Keep v0 (Step height V₀) = 6 eV fixed. For this exercise use only 0.67625 eV ≤ en ≤ 2.3825 eV 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 = -8.8173564 nm⁻¹, with a bounded reading uncertainty ±0.554 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 en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000170625 eV, 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.
ψ′ MISMATCH AT STEP against en
Predictions from the supplied Piecewise potentials and matching 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): ψ′ MISMATCH AT STEP / nm⁻¹
Horizontal axis (x): en: Energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Energy E / eV | ψ′ MISMATCH AT STEP / nm⁻¹ |
|---|---|
| 0.67625 | 8.015408474004285 |
| 1.1028125 | 0.10937867828723302 |
| 1.529375 | -6.61431321527364 |
| 1.9559374999999999 | -10.873470809858201 |
| 2.3825 | -12.510709565756605 |
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(en) = 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(0.67625) = 8.01541 nm⁻¹ and Y_out(2.3825) = -12.5107 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 0.67625 + (-8.81736 − 8.01541)(2.3825 − 0.67625)/(-12.5107 − 8.01541) = 2.07549 eV. 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 en = 1.71706 eV; substituting back gives Y_out = -8.8173564 nm⁻¹.
- The interpolation error in the input is |en_line − en| = 0.358425 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -9.3713564 nm⁻¹ and -8.2633564 nm⁻¹. Sort the roots even if the response decreases: 1.66553 eV ≤ en ≤ 1.77268 eV.
- The lower and upper input deviations are 0.0515336 eV and 0.0556148 eV. 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 en ≈ 1.71706 eV; compatible interval [1.66553, 1.77268] eV. Endpoint interpolation gives 2.07549 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.