University Quantum Mechanics I · topic UQMI-06.02
Energy quantisation as the condition for a nontrivial solution
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.
Energy quantisation: the condition for a nontrivial solution: predict a finite change
An allowed energy of the infinite well is a value of E at which ψ″ = −(2mE/ħ²)ψ has a solution that vanishes at both walls without vanishing everywhere; these are Eₙ = n²π²ħ²/(2mL²) with n = 1, 2, 3, … Use the supplied topic model Y_out = 0.0381*pi^2/w^2, where Y_out is GROUND LEVEL E₁ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Well width L, measured in nm. Keep en (Trial energy E) = 1.2 eV fixed. For this exercise use only 0.70625 nm ≤ w ≤ 1.49375 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 w = 0.8795 nm and 0.98975 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GROUND LEVEL E₁ against w
Predictions from the supplied Energy quantisation: the condition for a nontrivial solution 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): GROUND LEVEL E₁ / eV
Horizontal axis (x): w: Well width L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Well width L / nm | GROUND LEVEL E₁ / eV |
|---|---|
| 0.7062499999999999 | 0.7538896819364491 |
| 0.903125 | 0.4610297936397561 |
| 1.0999999999999999 | 0.3107701881665328 |
| 1.296875 | 0.22357770007017602 |
| 1.49375 | 0.16852676508896058 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 0.8795 nm, substitute into Y_out = 0.0381*pi^2/w^2: Y₀ = 0.486131 eV.
- At w = 0.98975 nm, the same fixed-parameter model gives Y₁ = 0.383861 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.10227 eV; the input change is Δw = 0.11025 nm.
- Average sensitivity is ΔY_out/Δw = -0.927619 eV per nm. 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 = 0.486131 eV, Y₁ = 0.383861 eV; ΔY_out = -0.10227 eV; average sensitivity = -0.927619 eV per 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.
Energy quantisation: the condition for a nontrivial solution: infer an input with uncertainty
An allowed energy of the infinite well is a value of E at which ψ″ = −(2mE/ħ²)ψ has a solution that vanishes at both walls without vanishing everywhere; these are Eₙ = n²π²ħ²/(2mL²) with n = 1, 2, 3, … Use the supplied topic model Y_out = 0.0381*pi^2/w^2, where Y_out is GROUND LEVEL E₁ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Well width L, measured in nm. Keep en (Trial energy E) = 1.2 eV fixed. For this exercise use only 0.70625 nm ≤ w ≤ 1.49375 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 = 0.26705317 eV, with a bounded reading uncertainty ±0.014779 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 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.
GROUND LEVEL E₁ against w
Predictions from the supplied Energy quantisation: the condition for a nontrivial solution 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): GROUND LEVEL E₁ / eV
Horizontal axis (x): w: Well width L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Well width L / nm | GROUND LEVEL E₁ / eV |
|---|---|
| 0.7062499999999999 | 0.7538896819364491 |
| 0.903125 | 0.4610297936397561 |
| 1.0999999999999999 | 0.3107701881665328 |
| 1.296875 | 0.22357770007017602 |
| 1.49375 | 0.16852676508896058 |
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(w) = 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(0.70625) = 0.75389 eV and Y_out(1.49375) = 0.168527 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.70625 + (0.267053 − 0.75389)(1.49375 − 0.70625)/(0.168527 − 0.75389) = 1.3612 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 w = 1.18662 nm; substituting back gives Y_out = 0.26705317 eV.
- The interpolation error in the input is |w_line − w| = 0.174576 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.25227417 eV and 0.28183217 eV. Sort the roots even if the response decreases: 1.15509 nm ≤ w ≤ 1.22089 nm.
- The lower and upper input deviations are 0.0315317 nm and 0.0342634 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 w ≈ 1.18662 nm; compatible interval [1.15509, 1.22089] nm. Endpoint interpolation gives 1.3612 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.