University Physics IV · topic UPIV-07.02
The infinite square well: spectrum, normalisation, orthonormality
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 infinite square well: predict a finite change
The infinite square well is the model of a particle that moves freely inside a segment of length L and is forbidden outside it, so its stationary states are the standing sine waves that vanish at both walls. Use the supplied topic model Y_out = 0.376*(2*n+1)/max(w*w,0.0025), where Y_out is GAP TO n+1 (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Well width L, measured in nm. Keep n (Quantum number n) = 2 fixed. For this exercise use only 0.46375 nm ≤ w ≤ 0.93625 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 = 0.5677 nm and 0.63385 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.
GAP TO n+1 against w
Predictions from the supplied The infinite square 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): GAP TO n+1 / 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 | GAP TO n+1 / eV |
|---|---|
| 0.46375 | 8.74158136020517 |
| 0.5818749999999999 | 5.5526263872047785 |
| 0.7 | 3.8367346938775513 |
| 0.818125 | 2.8087851572325575 |
| 0.9362499999999999 | 2.144737709914956 |
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 = 0.5677 nm, substitute into Y_out = 0.376*(2*n+1)/max(w*w,0.0025): Y₀ = 5.83338 eV.
- At w = 0.63385 nm, the same fixed-parameter model gives Y₁ = 4.67934 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.15403 eV; the input change is Δw = 0.06615 nm.
- Average sensitivity is ΔY_out/Δw = -17.4457 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 = 5.83338 eV, Y₁ = 4.67934 eV; ΔY_out = -1.15403 eV; average sensitivity = -17.4457 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.
The infinite square well: infer an input with uncertainty
The infinite square well is the model of a particle that moves freely inside a segment of length L and is forbidden outside it, so its stationary states are the standing sine waves that vanish at both walls. Use the supplied topic model Y_out = 0.376*(2*n+1)/max(w*w,0.0025), where Y_out is GAP TO n+1 (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Well width L, measured in nm. Keep n (Quantum number n) = 2 fixed. For this exercise use only 0.46375 nm ≤ w ≤ 0.93625 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 = 3.3246891 eV, with a bounded reading uncertainty ±0.17699 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 0.00004725 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.
GAP TO n+1 against w
Predictions from the supplied The infinite square 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): GAP TO n+1 / 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 | GAP TO n+1 / eV |
|---|---|
| 0.46375 | 8.74158136020517 |
| 0.5818749999999999 | 5.5526263872047785 |
| 0.7 | 3.8367346938775513 |
| 0.818125 | 2.8087851572325575 |
| 0.9362499999999999 | 2.144737709914956 |
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(0.46375) = 8.74158 eV and Y_out(0.93625) = 2.14474 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.46375 + (3.32469 − 8.74158)(0.93625 − 0.46375)/(2.14474 − 8.74158) = 0.851736 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 = 0.751975 nm; substituting back gives Y_out = 3.3246891 eV.
- The interpolation error in the input is |w_line − w| = 0.0997608 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.1476991 eV and 3.5016791 eV. Sort the roots even if the response decreases: 0.732725 nm ≤ w ≤ 0.772827 nm.
- The lower and upper input deviations are 0.0192504 nm and 0.0208521 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 ≈ 0.751975 nm; compatible interval [0.732725, 0.772827] nm. Endpoint interpolation gives 0.851736 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.