University Physics V · topic UPV-07.07
Expectation values, the virial split, and uncertainty
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, the virial split and uncertainty: predict a finite change
An expectation value in a Fock state is the diagonal matrix element ⟨n|Â|n⟩; writing  through a and a† reduces it to whichever term raises and lowers equally often, so the calculation becomes counting rather than integration. Use the supplied topic model Y_out = 78.07*sqrt((n+0.5)/max(w,10)), where Y_out is SPREAD Δx (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. w is Level spacing ħω, measured in meV. Keep n (Quantum number n) = 3 fixed. For this exercise use only 96.125 meV ≤ w ≤ 363.875 meV 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 = 155.03 meV and 192.515 meV. 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.
SPREAD Δx against w
Predictions from the supplied Expectation values, the virial split and uncertainty 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): SPREAD Δx / pm
Horizontal axis (x): w: Level spacing ħω / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Level spacing ħω / meV | SPREAD Δx / pm |
|---|---|
| 96.125 | 14.897041414753259 |
| 163.0625 | 11.4377645790819 |
| 230 | 9.630620809372209 |
| 296.9375 | 8.475897140128339 |
| 363.875 | 7.656708130033838 |
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 = 155.03 meV, substitute into Y_out = 78.07*sqrt((n+0.5)/max(w,10)): Y₀ = 11.7303 pm.
- At w = 192.515 meV, the same fixed-parameter model gives Y₁ = 10.5265 pm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.20379 pm; the input change is Δw = 37.485 meV.
- Average sensitivity is ΔY_out/Δw = -0.0321138 pm per meV. 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 = 11.7303 pm, Y₁ = 10.5265 pm; ΔY_out = -1.20379 pm; average sensitivity = -0.0321138 pm per meV.
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, the virial split and uncertainty: infer an input with uncertainty
An expectation value in a Fock state is the diagonal matrix element ⟨n|Â|n⟩; writing  through a and a† reduces it to whichever term raises and lowers equally often, so the calculation becomes counting rather than integration. Use the supplied topic model Y_out = 78.07*sqrt((n+0.5)/max(w,10)), where Y_out is SPREAD Δx (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. w is Level spacing ħω, measured in meV. Keep n (Quantum number n) = 3 fixed. For this exercise use only 96.125 meV ≤ w ≤ 363.875 meV 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 = 9.0675356 pm, with a bounded reading uncertainty ±0.21162 pm. 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 0.026775 meV, 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.
SPREAD Δx against w
Predictions from the supplied Expectation values, the virial split and uncertainty 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): SPREAD Δx / pm
Horizontal axis (x): w: Level spacing ħω / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Level spacing ħω / meV | SPREAD Δx / pm |
|---|---|
| 96.125 | 14.897041414753259 |
| 163.0625 | 11.4377645790819 |
| 230 | 9.630620809372209 |
| 296.9375 | 8.475897140128339 |
| 363.875 | 7.656708130033838 |
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(96.125) = 14.897 pm and Y_out(363.875) = 7.65671 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 96.125 + (9.06754 − 14.897)(363.875 − 96.125)/(7.65671 − 14.897) = 311.702 meV. 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 = 259.453 meV; substituting back gives Y_out = 9.0675356 pm.
- The interpolation error in the input is |w_line − w| = 52.2496 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.8559156 pm and 9.2791556 pm. Sort the roots even if the response decreases: 247.753 meV ≤ w ≤ 272 meV.
- The lower and upper input deviations are 11.6992 meV and 12.5478 meV. 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 ≈ 259.453 meV; compatible interval [247.753, 272] meV. Endpoint interpolation gives 311.702 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.