University Physics V · topic UPV-08.09
Quantum spread against experimental 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.
Quantum spread vs experimental error: predict a finite change
A state's intrinsic spread is the standard deviation of the outcome distribution that ρ and the Hermitian observable A fix between them, σ_A = √[Tr(ρA²) − (Tr ρA)²]; whatever the apparatus and the finite run add sits on top of it. Use the supplied topic model Y_out = hypot(s,r)/max(s,0.01), where Y_out is WIDTH INFLATION (times). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); max(a,b) selects the larger value. s is State spread σ_ψ, measured in nm. Keep r (Detector response σ_R) = 8 nm; n (Shots averaged N) = 25 shots fixed. For this exercise use only 8.55 nm ≤ s ≤ 27.45 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 = 12.708 nm and 15.354 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.
WIDTH INFLATION against s
Predictions from the supplied Quantum spread vs experimental 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): WIDTH INFLATION / times
Horizontal axis (x): s: State spread σ_ψ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: State spread σ_ψ / nm | WIDTH INFLATION / times |
|---|---|
| 8.55 | 1.3694827690099263 |
| 13.275 | 1.1675490538405107 |
| 18 | 1.0943175335329005 |
| 22.725 | 1.060155075692993 |
| 27.45 | 1.0416029373241062 |
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 = 12.708 nm, substitute into Y_out = hypot(s,r)/max(s,0.01): Y₀ = 1.18165 times.
- At s = 15.354 nm, the same fixed-parameter model gives Y₁ = 1.1276 times.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.054053 times; the input change is Δs = 2.646 nm.
- Average sensitivity is ΔY_out/Δs = -0.0204282 times 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 = 1.18165 times, Y₁ = 1.1276 times; ΔY_out = -0.054053 times; average sensitivity = -0.0204282 times 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.
Quantum spread vs experimental error: infer an input with uncertainty
A state's intrinsic spread is the standard deviation of the outcome distribution that ρ and the Hermitian observable A fix between them, σ_A = √[Tr(ρA²) − (Tr ρA)²]; whatever the apparatus and the finite run add sits on top of it. Use the supplied topic model Y_out = hypot(s,r)/max(s,0.01), where Y_out is WIDTH INFLATION (times). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); max(a,b) selects the larger value. s is State spread σ_ψ, measured in nm. Keep r (Detector response σ_R) = 8 nm; n (Shots averaged N) = 25 shots fixed. For this exercise use only 8.55 nm ≤ s ≤ 27.45 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 = 1.0764495 times, with a bounded reading uncertainty ±0.005227 times. 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 0.00189 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.
WIDTH INFLATION against s
Predictions from the supplied Quantum spread vs experimental 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): WIDTH INFLATION / times
Horizontal axis (x): s: State spread σ_ψ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: State spread σ_ψ / nm | WIDTH INFLATION / times |
|---|---|
| 8.55 | 1.3694827690099263 |
| 13.275 | 1.1675490538405107 |
| 18 | 1.0943175335329005 |
| 22.725 | 1.060155075692993 |
| 27.45 | 1.0416029373241062 |
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(8.55) = 1.36948 times and Y_out(27.45) = 1.0416 times. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 8.55 + (1.07645 − 1.36948)(27.45 − 8.55)/(1.0416 − 1.36948) = 25.4413 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 = 20.079 nm; substituting back gives Y_out = 1.0764495 times.
- The interpolation error in the input is |s_line − s| = 5.36234 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0712225 times and 1.0816765 times. Sort the roots even if the response decreases: 19.4015 nm ≤ s ≤ 20.829 nm.
- The lower and upper input deviations are 0.677518 nm and 0.749985 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 ≈ 20.079 nm; compatible interval [19.4015, 20.829] nm. Endpoint interpolation gives 25.4413 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.