University Physics V · topic UPV-06.01
Bound and unbound states: two sectors, two normalisations
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.
Bound states and the continuum: predict a finite change
A bound state is a square-integrable eigenvector of Ĥ with energy below the potential's value at infinity, normalised by ⟨n|m⟩ = δ_nm; an unbound state has energy above that floor, is not square-integrable, and is normalised by ⟨k|k′⟩ = δ(k − k′). Use the supplied topic model Y_out = 1-exp(-2*13.123*a*w), where Y_out is BOUND WEIGHT IN ±X (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Well strength α, measured in eV nm. Keep k (Continuum wavenumber k) = 8 nm⁻¹; w (Window half-width X) = 0.5 nm fixed. For this exercise use only 0.197813 eV nm ≤ a ≤ 0.552187 eV 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 a = 0.275775 eV nm and 0.325387 eV nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BOUND WEIGHT IN ±X against a
Predictions from the supplied Bound states and the continuum 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): BOUND WEIGHT IN ±X
Horizontal axis (x): a: Well strength α / eV nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Well strength α / eV nm | BOUND WEIGHT IN ±X |
|---|---|
| 0.1978125 | 0.9254207855670302 |
| 0.28640625 | 0.9766815227754335 |
| 0.375 | 0.9927090760581645 |
| 0.46359374999999997 | 0.9977203669256057 |
| 0.5521874999999999 | 0.9992872334157742 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 0.275775 eV nm, substitute into Y_out = 1-exp(-2*13.123*a*w): Y₀ = 0.97319.
- At a = 0.325387 eV nm, the same fixed-parameter model gives Y₁ = 0.986019.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0128287; the input change is Δa = 0.0496125 eV nm.
- Average sensitivity is ΔY_out/Δa = 0.258577 1 per eV nm. Its positive sign means that this response increases 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 = 0.97319, Y₁ = 0.986019; ΔY_out = 0.0128287; average sensitivity = 0.258577 1 per eV 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.
Bound states and the continuum: infer an input with uncertainty
A bound state is a square-integrable eigenvector of Ĥ with energy below the potential's value at infinity, normalised by ⟨n|m⟩ = δ_nm; an unbound state has energy above that floor, is not square-integrable, and is normalised by ⟨k|k′⟩ = δ(k − k′). Use the supplied topic model Y_out = 1-exp(-2*13.123*a*w), where Y_out is BOUND WEIGHT IN ±X (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Well strength α, measured in eV nm. Keep k (Continuum wavenumber k) = 8 nm⁻¹; w (Window half-width X) = 0.5 nm fixed. For this exercise use only 0.197813 eV nm ≤ a ≤ 0.552187 eV 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 = 0.99562862, with a bounded reading uncertainty ±0.00054879. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000354375 eV 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.
BOUND WEIGHT IN ±X against a
Predictions from the supplied Bound states and the continuum 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): BOUND WEIGHT IN ±X
Horizontal axis (x): a: Well strength α / eV nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Well strength α / eV nm | BOUND WEIGHT IN ±X |
|---|---|
| 0.1978125 | 0.9254207855670302 |
| 0.28640625 | 0.9766815227754335 |
| 0.375 | 0.9927090760581645 |
| 0.46359374999999997 | 0.9977203669256057 |
| 0.5521874999999999 | 0.9992872334157742 |
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(a) = 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(0.197813) = 0.925421 and Y_out(0.552187) = 0.999287. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.197813 + (0.995629 − 0.925421)(0.552187 − 0.197813)/(0.999287 − 0.925421) = 0.534635 eV 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 a = 0.413981 eV nm; substituting back gives Y_out = 0.99562862.
- The interpolation error in the input is |a_line − a| = 0.120654 eV nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99507983 and 0.99617741. Sort the roots even if the response decreases: 0.404969 eV nm ≤ a ≤ 0.424204 eV nm.
- The lower and upper input deviations are 0.00901199 eV nm and 0.0102226 eV 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 a ≈ 0.413981 eV nm; compatible interval [0.404969, 0.424204] eV nm. Endpoint interpolation gives 0.534635 eV nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.