University Physics V · topic UPV-08.04
Delta-x delta-p at least h-bar over 2, and what it costs
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.
Δx Δp ≥ ħ/2 and the cost of confinement: predict a finite change
The position–momentum uncertainty relation says that for one prepared state, the standard deviation of x̂ and the standard deviation of p̂ — each read from its own ensemble of identical copies — satisfy σ_x σ_p ≥ ħ/2, whatever the state and whatever the apparatus. Use the supplied topic model Y_out = q*q/(2*s*s), where Y_out is KINETIC ⟨T⟩ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Position spread σx / √(ħ/mω), dimensionless. Keep q (Uncertainty product σx σp / ħ) = 0.5 fixed. For this exercise use only 0.480625 ≤ s ≤ 1.81937 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 = 0.77515 and 0.962575. 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.
KINETIC ⟨T⟩ against s
Predictions from the supplied Δx Δp ≥ ħ/2 and the cost of confinement 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): KINETIC ⟨T⟩ / ħω
Horizontal axis (x): s: Position spread σx / √(ħ/mω)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Position spread σx / √(ħ/mω) | KINETIC ⟨T⟩ / ħω |
|---|---|
| 0.48062499999999997 | 0.5411246260744283 |
| 0.8153125 | 0.18804500916058325 |
| 1.15 | 0.09451795841209831 |
| 1.4846875000000002 | 0.056707422616187185 |
| 1.819375 | 0.03776291990449286 |
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 = 0.77515, substitute into Y_out = q*q/(2*s*s): Y₀ = 0.208036 ħω.
- At s = 0.962575, the same fixed-parameter model gives Y₁ = 0.134909 ħω.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.073127 ħω; the input change is Δs = 0.187425.
- Average sensitivity is ΔY_out/Δs = -0.390167 ħω per unit input. 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 = 0.208036 ħω, Y₁ = 0.134909 ħω; ΔY_out = -0.073127 ħω; average sensitivity = -0.390167 ħω per unit input.
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.
Δx Δp ≥ ħ/2 and the cost of confinement: infer an input with uncertainty
The position–momentum uncertainty relation says that for one prepared state, the standard deviation of x̂ and the standard deviation of p̂ — each read from its own ensemble of identical copies — satisfy σ_x σ_p ≥ ħ/2, whatever the state and whatever the apparatus. Use the supplied topic model Y_out = q*q/(2*s*s), where Y_out is KINETIC ⟨T⟩ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Position spread σx / √(ħ/mω), dimensionless. Keep q (Uncertainty product σx σp / ħ) = 0.5 fixed. For this exercise use only 0.480625 ≤ s ≤ 1.81937 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 = 0.074276988 ħω, with a bounded reading uncertainty ±0.0054771 ħω. 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 0.000133875, 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.
KINETIC ⟨T⟩ against s
Predictions from the supplied Δx Δp ≥ ħ/2 and the cost of confinement 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): KINETIC ⟨T⟩ / ħω
Horizontal axis (x): s: Position spread σx / √(ħ/mω)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Position spread σx / √(ħ/mω) | KINETIC ⟨T⟩ / ħω |
|---|---|
| 0.48062499999999997 | 0.5411246260744283 |
| 0.8153125 | 0.18804500916058325 |
| 1.15 | 0.09451795841209831 |
| 1.4846875000000002 | 0.056707422616187185 |
| 1.819375 | 0.03776291990449286 |
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(0.480625) = 0.541125 ħω and Y_out(1.81937) = 0.0377629 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.480625 + (0.074277 − 0.541125)(1.81937 − 0.480625)/(0.0377629 − 0.541125) = 1.72226. 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 = 1.29726; substituting back gives Y_out = 0.074276988 ħω.
- The interpolation error in the input is |s_line − s| = 0.424999. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.068799888 ħω and 0.079754088 ħω. Sort the roots even if the response decreases: 1.25193 ≤ s ≤ 1.34791.
- The lower and upper input deviations are 0.0453369 and 0.0506483. 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 ≈ 1.29726; compatible interval [1.25193, 1.34791]. Endpoint interpolation gives 1.72226.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.