University Physics V · topic UPV-05.07
Expectation values, zero-point energy, and the dx dp product
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 and the σ_x σ_p product: predict a finite change
An expectation value is the sandwich ⟨A⟩ = ⟨ψ|Â|ψ⟩ — the mean of many measurements of A on identically prepared copies — and its uncertainty is the standard deviation σ_A = √(⟨²⟩ − ⟨A⟩²) of that same distribution. Use the supplied topic model Y_out = 0.376*n^2/(max(l,0.05)*max(l,0.05)), where Y_out is E_n, electron (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. l is Box width L, measured in nm. Keep n (Quantum number n) = 1 fixed. For this exercise use only 0.695625 nm ≤ l ≤ 1.40437 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 l = 0.85155 nm and 0.950775 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δl across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E_n, electron against l
Predictions from the supplied Expectation values and the σ_x σ_p product 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): E_n, electron / eV
Horizontal axis (x): l: Box width L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| l: Box width L / nm | E_n, electron / eV |
|---|---|
| 0.6956249999999999 | 0.7770294542404598 |
| 0.8728125 | 0.493566789973758 |
| 1.0499999999999998 | 0.34104308390022686 |
| 1.2271874999999999 | 0.2496697917540052 |
| 1.404375 | 0.19064335199244054 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of l separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At l = 0.85155 nm, substitute into Y_out = 0.376*n^2/(max(l,0.05)*max(l,0.05)): Y₀ = 0.518522 eV.
- At l = 0.950775 nm, the same fixed-parameter model gives Y₁ = 0.415942 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.102581 eV; the input change is Δl = 0.099225 nm.
- Average sensitivity is ΔY_out/Δl = -1.03382 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 = 0.518522 eV, Y₁ = 0.415942 eV; ΔY_out = -0.102581 eV; average sensitivity = -1.03382 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 and the σ_x σ_p product: infer an input with uncertainty
An expectation value is the sandwich ⟨A⟩ = ⟨ψ|Â|ψ⟩ — the mean of many measurements of A on identically prepared copies — and its uncertainty is the standard deviation σ_A = √(⟨²⟩ − ⟨A⟩²) of that same distribution. Use the supplied topic model Y_out = 0.376*n^2/(max(l,0.05)*max(l,0.05)), where Y_out is E_n, electron (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. l is Box width L, measured in nm. Keep n (Quantum number n) = 1 fixed. For this exercise use only 0.695625 nm ≤ l ≤ 1.40437 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 = 0.29552792 eV, with a bounded reading uncertainty ±0.015733 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown l by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000070875 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.
E_n, electron against l
Predictions from the supplied Expectation values and the σ_x σ_p product 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): E_n, electron / eV
Horizontal axis (x): l: Box width L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| l: Box width L / nm | E_n, electron / eV |
|---|---|
| 0.6956249999999999 | 0.7770294542404598 |
| 0.8728125 | 0.493566789973758 |
| 1.0499999999999998 | 0.34104308390022686 |
| 1.2271874999999999 | 0.2496697917540052 |
| 1.404375 | 0.19064335199244054 |
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(l) = 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(0.695625) = 0.777029 eV and Y_out(1.40437) = 0.190643 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives l_line = 0.695625 + (0.295528 − 0.777029)(1.40437 − 0.695625)/(0.190643 − 0.777029) = 1.2776 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 l = 1.12796 nm; substituting back gives Y_out = 0.29552792 eV.
- The interpolation error in the input is |l_line − l| = 0.149641 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.27979492 eV and 0.31126092 eV. Sort the roots even if the response decreases: 1.09909 nm ≤ l ≤ 1.15924 nm.
- The lower and upper input deviations are 0.0288766 nm and 0.0312792 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 l ≈ 1.12796 nm; compatible interval [1.09909, 1.15924] nm. Endpoint interpolation gives 1.2776 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.