University Quantum Mechanics I · topic UQMI-07.08
Comparison with the infinite well
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.
Comparison with the infinite well: predict a finite change
Set beside the infinite well of the same width and floor, every finite-well level Eₙ lies below its counterpart Eₙ(∞) = n²π²ħ²/(2mL²) and above Eₙ₋₁(∞), because the tails let the well act about 2/κ wider. Use the supplied topic model Y_out = 0.0381*(pi/w)^2, where Y_out is INFINITE-WELL 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 v (Well depth V₀) = 1.2 eV fixed. For this exercise use only 0.659375 nm ≤ w ≤ 1.84062 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 = 0.91925 nm and 1.08462 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.
INFINITE-WELL E₁ against w
Predictions from the supplied Comparison with the infinite well 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): INFINITE-WELL 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 | INFINITE-WELL E₁ / eV |
|---|---|
| 0.659375 | 0.8648877921561974 |
| 0.9546875 | 0.41257437320253687 |
| 1.25 | 0.2406604337161629 |
| 1.5453125 | 0.15746791815976166 |
| 1.840625 | 0.11099261617078836 |
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 = 0.91925 nm, substitute into Y_out = 0.0381*(pi/w)^2: Y₀ = 0.444997 eV.
- At w = 1.08462 nm, the same fixed-parameter model gives Y₁ = 0.319643 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.125354 eV; the input change is Δw = 0.165375 nm.
- Average sensitivity is ΔY_out/Δw = -0.757999 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 = 0.444997 eV, Y₁ = 0.319643 eV; ΔY_out = -0.125354 eV; average sensitivity = -0.757999 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.
Comparison with the infinite well: infer an input with uncertainty
Set beside the infinite well of the same width and floor, every finite-well level Eₙ lies below its counterpart Eₙ(∞) = n²π²ħ²/(2mL²) and above Eₙ₋₁(∞), because the tails let the well act about 2/κ wider. Use the supplied topic model Y_out = 0.0381*(pi/w)^2, where Y_out is INFINITE-WELL 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 v (Well depth V₀) = 1.2 eV fixed. For this exercise use only 0.659375 nm ≤ w ≤ 1.84062 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 = 0.19747216 eV, with a bounded reading uncertainty ±0.012972 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 0.000118125 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.
INFINITE-WELL E₁ against w
Predictions from the supplied Comparison with the infinite well 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): INFINITE-WELL 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 | INFINITE-WELL E₁ / eV |
|---|---|
| 0.659375 | 0.8648877921561974 |
| 0.9546875 | 0.41257437320253687 |
| 1.25 | 0.2406604337161629 |
| 1.5453125 | 0.15746791815976166 |
| 1.840625 | 0.11099261617078836 |
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(0.659375) = 0.864888 eV and Y_out(1.84062) = 0.110993 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.659375 + (0.197472 − 0.864888)(1.84062 − 0.659375)/(0.110993 − 0.864888) = 1.70512 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 = 1.37994 nm; substituting back gives Y_out = 0.19747216 eV.
- The interpolation error in the input is |w_line − w| = 0.325186 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18450016 eV and 0.21044416 eV. Sort the roots even if the response decreases: 1.33673 nm ≤ w ≤ 1.42762 nm.
- The lower and upper input deviations are 0.0432068 nm and 0.047687 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 ≈ 1.37994 nm; compatible interval [1.33673, 1.42762] nm. Endpoint interpolation gives 1.70512 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.