University Physics IV · topic UPIV-05.05
Observables as operators and expectation values
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.
Operators and expectation values: predict a finite change
An observable is represented by a Hermitian linear operator  acting on the wave function, and the mean of many measurements of it on identically prepared systems is the integral ⟨Â⟩ = ∫ψ*(x) Âψ(x) dx. Use the supplied topic model Y_out = 0.5273/max(s,0.01), where Y_out is σ_p = ħ/(2σ_x) (×10⁻²⁵ kg m/s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. s is Packet width σ_x, measured in nm. Keep x0 (Packet centre ⟨x̂⟩) = 1.2 nm; k (Central wavenumber k₀) = 8 nm⁻¹ fixed. For this exercise use only 0.143375 nm ≤ s ≤ 0.316625 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 s = 0.18149 nm and 0.205745 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
σ_p = ħ/(2σ_x) against s
Predictions from the supplied Operators and expectation values 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): σ_p = ħ/(2σ_x) / ×10⁻²⁵ kg m/s
Horizontal axis (x): s: Packet width σ_x / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Packet width σ_x / nm | σ_p = ħ/(2σ_x) / ×10⁻²⁵ kg m/s |
|---|---|
| 0.143375 | 3.677768090671316 |
| 0.1866875 | 2.824506193505189 |
| 0.23000000000000004 | 2.2926086956521736 |
| 0.27331250000000007 | 1.9292933912645776 |
| 0.31662500000000005 | 1.6653770232925382 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 0.18149 nm, substitute into Y_out = 0.5273/max(s,0.01): Y₀ = 2.90539 ×10⁻²⁵ kg m/s.
- At s = 0.205745 nm, the same fixed-parameter model gives Y₁ = 2.56288 ×10⁻²⁵ kg m/s.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.342513 ×10⁻²⁵ kg m/s; the input change is Δs = 0.024255 nm.
- Average sensitivity is ΔY_out/Δs = -14.1213 ×10⁻²⁵ kg m/s 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 = 2.90539 ×10⁻²⁵ kg m/s, Y₁ = 2.56288 ×10⁻²⁵ kg m/s; ΔY_out = -0.342513 ×10⁻²⁵ kg m/s; average sensitivity = -14.1213 ×10⁻²⁵ kg m/s 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.
Operators and expectation values: infer an input with uncertainty
An observable is represented by a Hermitian linear operator  acting on the wave function, and the mean of many measurements of it on identically prepared systems is the integral ⟨Â⟩ = ∫ψ*(x) Âψ(x) dx. Use the supplied topic model Y_out = 0.5273/max(s,0.01), where Y_out is σ_p = ħ/(2σ_x) (×10⁻²⁵ kg m/s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. s is Packet width σ_x, measured in nm. Keep x0 (Packet centre ⟨x̂⟩) = 1.2 nm; k (Central wavenumber k₀) = 8 nm⁻¹ fixed. For this exercise use only 0.143375 nm ≤ s ≤ 0.316625 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 = 2.1171818 ×10⁻²⁵ kg m/s, with a bounded reading uncertainty ±0.067771 ×10⁻²⁵ kg m/s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000017325 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.
σ_p = ħ/(2σ_x) against s
Predictions from the supplied Operators and expectation values 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): σ_p = ħ/(2σ_x) / ×10⁻²⁵ kg m/s
Horizontal axis (x): s: Packet width σ_x / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Packet width σ_x / nm | σ_p = ħ/(2σ_x) / ×10⁻²⁵ kg m/s |
|---|---|
| 0.143375 | 3.677768090671316 |
| 0.1866875 | 2.824506193505189 |
| 0.23000000000000004 | 2.2926086956521736 |
| 0.27331250000000007 | 1.9292933912645776 |
| 0.31662500000000005 | 1.6653770232925382 |
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(s) = 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(0.143375) = 3.67777 ×10⁻²⁵ kg m/s and Y_out(0.316625) = 1.66538 ×10⁻²⁵ kg m/s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.143375 + (2.11718 − 3.67777)(0.316625 − 0.143375)/(1.66538 − 3.67777) = 0.277728 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 s = 0.249057 nm; substituting back gives Y_out = 2.1171818 ×10⁻²⁵ kg m/s.
- The interpolation error in the input is |s_line − s| = 0.0286709 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0494108 ×10⁻²⁵ kg m/s and 2.1849528 ×10⁻²⁵ kg m/s. Sort the roots even if the response decreases: 0.241332 nm ≤ s ≤ 0.257293 nm.
- The lower and upper input deviations are 0.00772505 nm and 0.00823597 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 s ≈ 0.249057 nm; compatible interval [0.241332, 0.257293] nm. Endpoint interpolation gives 0.277728 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.