University Quantum Mechanics I · topic UQMI-04.08
Delta-x Delta-p >= hbar/2 and the price of confinement
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 price of confinement: predict a finite change
For every normalisable state, the standard deviations of the position outcomes and of the momentum outcomes, each measured on separate copies of that state, obey σₓσₚ ≥ ħ/2, so narrowing one histogram forces the other to widen. Use the supplied topic model Y_out = 38.10*((1+c^2)/(4*sx^2)+p0^2), where Y_out is ELECTRON ⟨T⟩ (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sx is Position spread σₓ, measured in nm. Keep c (Phase curvature c) = 0; p0 (Mean momentum ⟨p⟩) = 0 ħ nm⁻¹ fixed. For this exercise use only 0.303125 nm ≤ sx ≤ 0.696875 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 sx = 0.38975 nm and 0.444875 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsx across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ELECTRON ⟨T⟩ against sx
Predictions from the supplied Δx Δp ≥ ħ/2 and the price 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): ELECTRON ⟨T⟩ / meV
Horizontal axis (x): sx: Position spread σₓ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sx: Position spread σₓ / nm | ELECTRON ⟨T⟩ / meV |
|---|---|
| 0.303125 | 103.66245084493572 |
| 0.40156249999999993 | 59.06887310935823 |
| 0.49999999999999994 | 38.10000000000001 |
| 0.5984375 | 26.59667732413474 |
| 0.6968749999999999 | 19.61350519817411 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sx separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sx = 0.38975 nm, substitute into Y_out = 38.10*((1+c^2)/(4*sx^2)+p0^2): Y₀ = 62.7036 meV.
- At sx = 0.444875 nm, the same fixed-parameter model gives Y₁ = 48.127 meV.
- Signed change: ΔY_out = Y₁ − Y₀ = -14.5766 meV; the input change is Δsx = 0.055125 nm.
- Average sensitivity is ΔY_out/Δsx = -264.428 meV 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 = 62.7036 meV, Y₁ = 48.127 meV; ΔY_out = -14.5766 meV; average sensitivity = -264.428 meV 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.
Δx Δp ≥ ħ/2 and the price of confinement: infer an input with uncertainty
For every normalisable state, the standard deviations of the position outcomes and of the momentum outcomes, each measured on separate copies of that state, obey σₓσₚ ≥ ħ/2, so narrowing one histogram forces the other to widen. Use the supplied topic model Y_out = 38.10*((1+c^2)/(4*sx^2)+p0^2), where Y_out is ELECTRON ⟨T⟩ (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sx is Position spread σₓ, measured in nm. Keep c (Phase curvature c) = 0; p0 (Mean momentum ⟨p⟩) = 0 ħ nm⁻¹ fixed. For this exercise use only 0.303125 nm ≤ sx ≤ 0.696875 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 = 32.26752 meV, with a bounded reading uncertainty ±1.8981 meV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sx by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000039375 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.
ELECTRON ⟨T⟩ against sx
Predictions from the supplied Δx Δp ≥ ħ/2 and the price 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): ELECTRON ⟨T⟩ / meV
Horizontal axis (x): sx: Position spread σₓ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sx: Position spread σₓ / nm | ELECTRON ⟨T⟩ / meV |
|---|---|
| 0.303125 | 103.66245084493572 |
| 0.40156249999999993 | 59.06887310935823 |
| 0.49999999999999994 | 38.10000000000001 |
| 0.5984375 | 26.59667732413474 |
| 0.6968749999999999 | 19.61350519817411 |
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(sx) = 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(0.303125) = 103.662 meV and Y_out(0.696875) = 19.6135 meV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sx_line = 0.303125 + (32.2675 − 103.662)(0.696875 − 0.303125)/(19.6135 − 103.662) = 0.637594 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 sx = 0.543313 nm; substituting back gives Y_out = 32.26752 meV.
- The interpolation error in the input is |sx_line − sx| = 0.0942813 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 30.36942 meV and 34.16562 meV. Sort the roots even if the response decreases: 0.528005 nm ≤ sx ≤ 0.560034 nm.
- The lower and upper input deviations are 0.0153077 nm and 0.0167213 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 sx ≈ 0.543313 nm; compatible interval [0.528005, 0.560034] nm. Endpoint interpolation gives 0.637594 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.