University Quantum Mechanics I · topic UQMI-06.01
The well, the boundary conditions, and what they cost
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 well, the boundary conditions, and what they cost: predict a finite change
The infinite square well is the potential V = 0 for 0 < x < L and V = ∞ elsewhere; its states obey ψ(0) = ψ(L) = 0, with ψ continuous at each wall and ψ′ free to jump there. Use the supplied topic model Y_out = 0.0381*(pi*f/w)^2/cos(pi*f/2)^2, where Y_out is WALL HEIGHT V₀ NEEDED (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Inside wavenumber k, in units of π/L, measured in π/L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.552063 π/L ≤ f ≤ 0.937937 π/L 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 f = 0.636955 π/L and 0.690978 π/L. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WALL HEIGHT V₀ NEEDED against f
Predictions from the supplied The well, the boundary conditions, and what they cost 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): WALL HEIGHT V₀ NEEDED / eV
Horizontal axis (x): f: Inside wavenumber k, in units of π/L / π/L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Inside wavenumber k, in units of π/L / π/L | WALL HEIGHT V₀ NEEDED / eV |
|---|---|
| 0.5520625 | 0.2737902767957964 |
| 0.64853125 | 0.574982552224795 |
| 0.745 | 1.3726794046089223 |
| 0.84146875 | 4.383558838666285 |
| 0.9379375 | 34.91809082913382 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = 0.636955 π/L, substitute into Y_out = 0.0381*(pi*f/w)^2/cos(pi*f/2)^2: Y₀ = 0.523457 eV.
- At f = 0.690978 π/L, the same fixed-parameter model gives Y₁ = 0.824735 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.301278 eV; the input change is Δf = 0.0540225 π/L.
- Average sensitivity is ΔY_out/Δf = 5.5769 eV per π/L. 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 = 0.523457 eV, Y₁ = 0.824735 eV; ΔY_out = 0.301278 eV; average sensitivity = 5.5769 eV per π/L.
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 well, the boundary conditions, and what they cost: infer an input with uncertainty
The infinite square well is the potential V = 0 for 0 < x < L and V = ∞ elsewhere; its states obey ψ(0) = ψ(L) = 0, with ψ continuous at each wall and ψ′ free to jump there. Use the supplied topic model Y_out = 0.0381*(pi*f/w)^2/cos(pi*f/2)^2, where Y_out is WALL HEIGHT V₀ NEEDED (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Inside wavenumber k, in units of π/L, measured in π/L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.552063 π/L ≤ f ≤ 0.937937 π/L 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 = 2.1711374 eV, with a bounded reading uncertainty ±0.2846 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000385875 π/L, 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.
WALL HEIGHT V₀ NEEDED against f
Predictions from the supplied The well, the boundary conditions, and what they cost 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): WALL HEIGHT V₀ NEEDED / eV
Horizontal axis (x): f: Inside wavenumber k, in units of π/L / π/L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Inside wavenumber k, in units of π/L / π/L | WALL HEIGHT V₀ NEEDED / eV |
|---|---|
| 0.5520625 | 0.2737902767957964 |
| 0.64853125 | 0.574982552224795 |
| 0.745 | 1.3726794046089223 |
| 0.84146875 | 4.383558838666285 |
| 0.9379375 | 34.91809082913382 |
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(f) = 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(0.552063) = 0.27379 eV and Y_out(0.937937) = 34.9181 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.552063 + (2.17114 − 0.27379)(0.937937 − 0.552063)/(34.9181 − 0.27379) = 0.573196 π/L. 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 f = 0.787446 π/L; substituting back gives Y_out = 2.1711374 eV.
- The interpolation error in the input is |f_line − f| = 0.214251 π/L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8865374 eV and 2.4557374 eV. Sort the roots even if the response decreases: 0.775059 π/L ≤ f ≤ 0.797861 π/L.
- The lower and upper input deviations are 0.0123875 π/L and 0.0104144 π/L. 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 f ≈ 0.787446 π/L; compatible interval [0.775059, 0.797861] π/L. Endpoint interpolation gives 0.573196 π/L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.