University Physics V · topic UPV-04.05
The time-independent equation in the position basis
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.
Position-basis Schrödinger equation: predict a finite change
In the position basis the energy eigenvalue problem becomes the ordinary differential equation −(ħ²/2m)ψ″(x) + V(x)ψ(x) = Eψ(x), in which the second derivative of ψ at each point is fixed by ψ itself, the eigenvalue E, and the local value of V. Use the supplied topic model Y_out = 1/max(sqrt(dv/0.0381),0.01), where Y_out is PENETRATION 1/κ (nm). 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. dv is Step height above E, V₀ − E, measured in eV. Keep en (Energy above the floor E) = 1 eV fixed. For this exercise use only 1.08438 eV ≤ dv ≤ 5.41562 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 dv = 2.03725 eV and 2.64362 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PENETRATION 1/κ against dv
Predictions from the supplied Position-basis Schrödinger equation 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): PENETRATION 1/κ / nm
Horizontal axis (x): dv: Step height above E, V₀ − E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dv: Step height above E, V₀ − E / eV | PENETRATION 1/κ / nm |
|---|---|
| 1.084375 | 0.18744451628649733 |
| 2.1671875 | 0.13259106051875483 |
| 3.2499999999999996 | 0.1082731588302333 |
| 4.332812499999999 | 0.0937729416574408 |
| 5.415624999999999 | 0.08387609359493044 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dv separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dv = 2.03725 eV, substitute into Y_out = 1/max(sqrt(dv/0.0381),0.01): Y₀ = 0.136754 nm.
- At dv = 2.64362 eV, the same fixed-parameter model gives Y₁ = 0.12005 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.016704 nm; the input change is Δdv = 0.606375 eV.
- Average sensitivity is ΔY_out/Δdv = -0.0275473 nm 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 = 0.136754 nm, Y₁ = 0.12005 nm; ΔY_out = -0.016704 nm; average sensitivity = -0.0275473 nm 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.
Position-basis Schrödinger equation: infer an input with uncertainty
In the position basis the energy eigenvalue problem becomes the ordinary differential equation −(ħ²/2m)ψ″(x) + V(x)ψ(x) = Eψ(x), in which the second derivative of ψ at each point is fixed by ψ itself, the eigenvalue E, and the local value of V. Use the supplied topic model Y_out = 1/max(sqrt(dv/0.0381),0.01), where Y_out is PENETRATION 1/κ (nm). 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. dv is Step height above E, V₀ − E, measured in eV. Keep en (Energy above the floor E) = 1 eV fixed. For this exercise use only 1.08438 eV ≤ dv ≤ 5.41562 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 = 0.101115 nm, with a bounded reading uncertainty ±0.0025858 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000433125 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.
PENETRATION 1/κ against dv
Predictions from the supplied Position-basis Schrödinger equation 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): PENETRATION 1/κ / nm
Horizontal axis (x): dv: Step height above E, V₀ − E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dv: Step height above E, V₀ − E / eV | PENETRATION 1/κ / nm |
|---|---|
| 1.084375 | 0.18744451628649733 |
| 2.1671875 | 0.13259106051875483 |
| 3.2499999999999996 | 0.1082731588302333 |
| 4.332812499999999 | 0.0937729416574408 |
| 5.415624999999999 | 0.08387609359493044 |
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(dv) = 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(1.08438) = 0.187445 nm and Y_out(5.41562) = 0.0838761 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dv_line = 1.08438 + (0.101115 − 0.187445)(5.41562 − 1.08438)/(0.0838761 − 0.187445) = 4.69469 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 dv = 3.72644 eV; substituting back gives Y_out = 0.101115 nm.
- The interpolation error in the input is |dv_line − dv| = 0.968254 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0985292 nm and 0.1037008 nm. Sort the roots even if the response decreases: 3.54292 eV ≤ dv ≤ 3.9246 eV.
- The lower and upper input deviations are 0.183522 eV and 0.19816 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 dv ≈ 3.72644 eV; compatible interval [3.54292, 3.9246] eV. Endpoint interpolation gives 4.69469 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.