University Physics IV · topic UPIV-07.04
Expectation values and the uncertainty product for well states
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.
Expectation values in a box: predict a finite change
For a well eigenstate, the expectation value of an observable is the probability-weighted mean of the values a measurement could return, and its uncertainty is the standard deviation of that same distribution, σ_A = √(⟨A²⟩ − ⟨A⟩²). Use the supplied topic model Y_out = 0.3760*n*n/(w*w), where Y_out is ELECTRON ENERGY E_n (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 n (Quantum number n) = 1 fixed. For this exercise use only 0.30625 nm ≤ w ≤ 1.09375 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 = 0.4795 nm and 0.58975 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.
ELECTRON ENERGY E_n against w
Predictions from the supplied Expectation values in a box 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): ELECTRON ENERGY E_n / 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 | ELECTRON ENERGY E_n / eV |
|---|---|
| 0.30625 | 4.0089962515618485 |
| 0.503125 | 1.4853747926391727 |
| 0.7 | 0.7673469387755103 |
| 0.896875 | 0.4674379924486154 |
| 1.09375 | 0.314305306122449 |
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 = 0.4795 nm, substitute into Y_out = 0.3760*n*n/(w*w): Y₀ = 1.63535 eV.
- At w = 0.58975 nm, the same fixed-parameter model gives Y₁ = 1.08107 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.554284 eV; the input change is Δw = 0.11025 nm.
- Average sensitivity is ΔY_out/Δw = -5.02752 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 = 1.63535 eV, Y₁ = 1.08107 eV; ΔY_out = -0.554284 eV; average sensitivity = -5.02752 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.
Expectation values in a box: infer an input with uncertainty
For a well eigenstate, the expectation value of an observable is the probability-weighted mean of the values a measurement could return, and its uncertainty is the standard deviation of that same distribution, σ_A = √(⟨A²⟩ − ⟨A⟩²). Use the supplied topic model Y_out = 0.3760*n*n/(w*w), where Y_out is ELECTRON ENERGY E_n (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 n (Quantum number n) = 1 fixed. For this exercise use only 0.30625 nm ≤ w ≤ 1.09375 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 = 0.6076484 eV, with a bounded reading uncertainty ±0.044001 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.
ELECTRON ENERGY E_n against w
Predictions from the supplied Expectation values in a box 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): ELECTRON ENERGY E_n / 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 | ELECTRON ENERGY E_n / eV |
|---|---|
| 0.30625 | 4.0089962515618485 |
| 0.503125 | 1.4853747926391727 |
| 0.7 | 0.7673469387755103 |
| 0.896875 | 0.4674379924486154 |
| 1.09375 | 0.314305306122449 |
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(0.30625) = 4.009 eV and Y_out(1.09375) = 0.314305 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.30625 + (0.607648 − 4.009)(1.09375 − 0.30625)/(0.314305 − 4.009) = 1.03123 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 = 0.786625 nm; substituting back gives Y_out = 0.6076484 eV.
- The interpolation error in the input is |w_line − w| = 0.244601 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5636474 eV and 0.6516494 eV. Sort the roots even if the response decreases: 0.759603 nm ≤ w ≤ 0.816752 nm.
- The lower and upper input deviations are 0.0270216 nm and 0.0301269 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 ≈ 0.786625 nm; compatible interval [0.759603, 0.816752] nm. Endpoint interpolation gives 1.03123 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.