University Quantum Mechanics I · topic UQMI-04.09
Why uncertainty is not measurement error
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.
Why uncertainty is not measurement error: predict a finite change
The uncertainty σₓ of a state is the standard deviation of the positions that state predicts, fixed by ψ alone and read off identically prepared systems each measured once; measurement error is what a particular detector adds on top of it. Use the supplied topic model Y_out = 52.7286/sig, where Y_out is σₚ FLOOR ħ/(2σₓ) (10⁻²⁷ kg m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sig is State spread σₓ, measured in nm. Keep r (Detector resolution r) = 30 nm; n (Systems measured N) = 25 fixed. For this exercise use only 24.25 nm ≤ sig ≤ 55.75 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 sig = 31.18 nm and 35.59 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsig across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
σₚ FLOOR ħ/(2σₓ) against sig
Predictions from the supplied Why uncertainty is not measurement error 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): σₚ FLOOR ħ/(2σₓ) / 10⁻²⁷ kg m s⁻¹
Horizontal axis (x): sig: State spread σₓ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: State spread σₓ / nm | σₚ FLOOR ħ/(2σₓ) / 10⁻²⁷ kg m s⁻¹ |
|---|---|
| 24.25 | 2.174375257731959 |
| 32.125 | 1.6413571984435797 |
| 40 | 1.318215 |
| 47.875 | 1.1013806788511749 |
| 55.75 | 0.9458044843049327 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sig separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sig = 31.18 nm, substitute into Y_out = 52.7286/sig: Y₀ = 1.6911 10⁻²⁷ kg m s⁻¹.
- At sig = 35.59 nm, the same fixed-parameter model gives Y₁ = 1.48156 10⁻²⁷ kg m s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.209547 10⁻²⁷ kg m s⁻¹; the input change is Δsig = 4.41 nm.
- Average sensitivity is ΔY_out/Δsig = -0.0475162 10⁻²⁷ kg m s⁻¹ 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 = 1.6911 10⁻²⁷ kg m s⁻¹, Y₁ = 1.48156 10⁻²⁷ kg m s⁻¹; ΔY_out = -0.209547 10⁻²⁷ kg m s⁻¹; average sensitivity = -0.0475162 10⁻²⁷ kg m s⁻¹ 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.
Why uncertainty is not measurement error: infer an input with uncertainty
The uncertainty σₓ of a state is the standard deviation of the positions that state predicts, fixed by ψ alone and read off identically prepared systems each measured once; measurement error is what a particular detector adds on top of it. Use the supplied topic model Y_out = 52.7286/sig, where Y_out is σₚ FLOOR ħ/(2σₓ) (10⁻²⁷ kg m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sig is State spread σₓ, measured in nm. Keep r (Detector resolution r) = 30 nm; n (Systems measured N) = 25 fixed. For this exercise use only 24.25 nm ≤ sig ≤ 55.75 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 = 1.2131278 10⁻²⁷ kg m s⁻¹, with a bounded reading uncertainty ±0.040098 10⁻²⁷ kg m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sig by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 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.
σₚ FLOOR ħ/(2σₓ) against sig
Predictions from the supplied Why uncertainty is not measurement error 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): σₚ FLOOR ħ/(2σₓ) / 10⁻²⁷ kg m s⁻¹
Horizontal axis (x): sig: State spread σₓ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: State spread σₓ / nm | σₚ FLOOR ħ/(2σₓ) / 10⁻²⁷ kg m s⁻¹ |
|---|---|
| 24.25 | 2.174375257731959 |
| 32.125 | 1.6413571984435797 |
| 40 | 1.318215 |
| 47.875 | 1.1013806788511749 |
| 55.75 | 0.9458044843049327 |
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(sig) = 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(24.25) = 2.17438 10⁻²⁷ kg m s⁻¹ and Y_out(55.75) = 0.945804 10⁻²⁷ kg m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 24.25 + (1.21313 − 2.17438)(55.75 − 24.25)/(0.945804 − 2.17438) = 48.896 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 sig = 43.465 nm; substituting back gives Y_out = 1.2131278 10⁻²⁷ kg m s⁻¹.
- The interpolation error in the input is |sig_line − sig| = 5.43095 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1730298 10⁻²⁷ kg m s⁻¹ and 1.2532258 10⁻²⁷ kg m s⁻¹. Sort the roots even if the response decreases: 42.0743 nm ≤ sig ≤ 44.9508 nm.
- The lower and upper input deviations are 1.3907 nm and 1.48578 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 sig ≈ 43.465 nm; compatible interval [42.0743, 44.9508] nm. Endpoint interpolation gives 48.896 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.