University Quantum Mechanics I · topic UQMI-01.09
The requirements on psi(x,t), and the reason for each
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.
The requirements on ψ(x, t), and the reason for each: predict a finite change
An admissible wavefunction is single-valued, continuous and square-integrable, and an energy eigenstate also has ψ′ continuous wherever V is finite; the first and third conditions come from reading |ψ|² as a probability density, the second and fourth from the Schrödinger equation. Use the supplied topic model Y_out = -sqrt(en/0.0381)*sqrt(1-en/v0), where Y_out is SHARED SLOPE ψ′(0)/A (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. v0 is Step height V₀, measured in eV. Keep en (Electron energy E) = 2 eV fixed. For this exercise use only 13.7531 eV ≤ v0 ≤ 89.7469 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 v0 = 30.4718 eV and 41.1109 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SHARED SLOPE ψ′(0)/A against v0
Predictions from the supplied The requirements on ψ(x, t), and the reason for each 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): SHARED SLOPE ψ′(0)/A / nm⁻¹
Horizontal axis (x): v0: Step height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Step height V₀ / eV | SHARED SLOPE ψ′(0)/A / nm⁻¹ |
|---|---|
| 13.753124999999999 | -6.697743204690018 |
| 32.7515625 | -7.020533106534718 |
| 51.74999999999999 | -7.103851519031988 |
| 70.7484375 | -7.142092977328008 |
| 89.74687499999999 | -7.164051045000124 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v0 = 30.4718 eV, substitute into Y_out = -sqrt(en/0.0381)*sqrt(1-en/v0): Y₀ = -7.00343 nm⁻¹.
- At v0 = 41.1109 eV, the same fixed-parameter model gives Y₁ = -7.0668 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0633703 nm⁻¹; the input change is Δv0 = 10.6391 eV.
- Average sensitivity is ΔY_out/Δv0 = -0.00595634 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 = -7.00343 nm⁻¹, Y₁ = -7.0668 nm⁻¹; ΔY_out = -0.0633703 nm⁻¹; average sensitivity = -0.00595634 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.
The requirements on ψ(x, t), and the reason for each: infer an input with uncertainty
An admissible wavefunction is single-valued, continuous and square-integrable, and an energy eigenstate also has ψ′ continuous wherever V is finite; the first and third conditions come from reading |ψ|² as a probability density, the second and fourth from the Schrödinger equation. Use the supplied topic model Y_out = -sqrt(en/0.0381)*sqrt(1-en/v0), where Y_out is SHARED SLOPE ψ′(0)/A (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. v0 is Step height V₀, measured in eV. Keep en (Electron energy E) = 2 eV fixed. For this exercise use only 13.7531 eV ≤ v0 ≤ 89.7469 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 = -7.1236816 nm⁻¹, with a bounded reading uncertainty ±0.0060554 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 v0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00759937 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.
SHARED SLOPE ψ′(0)/A against v0
Predictions from the supplied The requirements on ψ(x, t), and the reason for each 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): SHARED SLOPE ψ′(0)/A / nm⁻¹
Horizontal axis (x): v0: Step height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Step height V₀ / eV | SHARED SLOPE ψ′(0)/A / nm⁻¹ |
|---|---|
| 13.753124999999999 | -6.697743204690018 |
| 32.7515625 | -7.020533106534718 |
| 51.74999999999999 | -7.103851519031988 |
| 70.7484375 | -7.142092977328008 |
| 89.74687499999999 | -7.164051045000124 |
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(v0) = 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(13.7531) = -6.69774 nm⁻¹ and Y_out(89.7469) = -7.16405 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v0_line = 13.7531 + (-7.12368 − -6.69774)(89.7469 − 13.7531)/(-7.16405 − -6.69774) = 83.1679 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 v0 = 60.1093 eV; substituting back gives Y_out = -7.1236816 nm⁻¹.
- The interpolation error in the input is |v0_line − v0| = 23.0586 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -7.129737 nm⁻¹ and -7.1176262 nm⁻¹. Sort the roots even if the response decreases: 57.2811 eV ≤ v0 ≤ 63.2341 eV.
- The lower and upper input deviations are 2.82821 eV and 3.12479 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 v0 ≈ 60.1093 eV; compatible interval [57.2811, 63.2341] eV. Endpoint interpolation gives 83.1679 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.