University Quantum Mechanics I · topic UQMI-06.03
Normalisation, the ground state, and zero-point energy
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.
Normalisation, the ground state, and zero-point energy: predict a finite change
The zero-point energy of the infinite square well is its lowest allowed energy, E₁ = π²ħ²/(2mL²), held by the normalised ground state ψ₁ = √(2/L) sin(πx/L) and kept by the particle even at absolute zero. Use the supplied topic model Y_out = 2/(d*(1-4*d/3)), where Y_out is TRAPEZOID ⟨T⟩ (ħ²/(2mL²)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Trapezoid ramp length, measured in L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.0978125 L ≤ d ≤ 0.357687 L 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 d = 0.154985 L and 0.191367 L. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TRAPEZOID ⟨T⟩ against d
Predictions from the supplied Normalisation, the ground state, and zero-point energy 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): TRAPEZOID ⟨T⟩ / ħ²/(2mL²)
Horizontal axis (x): d: Trapezoid ramp length / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Trapezoid ramp length / L | TRAPEZOID ⟨T⟩ / ħ²/(2mL²) |
|---|---|
| 0.0978125 | 23.513887124156692 |
| 0.16278125 | 15.692313133692718 |
| 0.22774999999999998 | 12.611142259464794 |
| 0.29271874999999997 | 11.206173002109892 |
| 0.35768749999999994 | 10.689449744150528 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.154985 L, substitute into Y_out = 2/(d*(1-4*d/3)): Y₀ = 16.2657 ħ²/(2mL²).
- At d = 0.191367 L, the same fixed-parameter model gives Y₁ = 14.0313 ħ²/(2mL²).
- Signed change: ΔY_out = Y₁ − Y₀ = -2.23447 ħ²/(2mL²); the input change is Δd = 0.0363825 L.
- Average sensitivity is ΔY_out/Δd = -61.416 ħ²/(2mL²) per L. 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 = 16.2657 ħ²/(2mL²), Y₁ = 14.0313 ħ²/(2mL²); ΔY_out = -2.23447 ħ²/(2mL²); average sensitivity = -61.416 ħ²/(2mL²) per L.
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.
Normalisation, the ground state, and zero-point energy: infer an input with uncertainty
The zero-point energy of the infinite square well is its lowest allowed energy, E₁ = π²ħ²/(2mL²), held by the normalised ground state ψ₁ = √(2/L) sin(πx/L) and kept by the particle even at absolute zero. Use the supplied topic model Y_out = 2/(d*(1-4*d/3)), where Y_out is TRAPEZOID ⟨T⟩ (ħ²/(2mL²)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Trapezoid ramp length, measured in L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.0978125 L ≤ d ≤ 0.357687 L 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 = 11.853593 ħ²/(2mL²), with a bounded reading uncertainty ±0.17462 ħ²/(2mL²). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000259875 L, 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.
TRAPEZOID ⟨T⟩ against d
Predictions from the supplied Normalisation, the ground state, and zero-point energy 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): TRAPEZOID ⟨T⟩ / ħ²/(2mL²)
Horizontal axis (x): d: Trapezoid ramp length / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Trapezoid ramp length / L | TRAPEZOID ⟨T⟩ / ħ²/(2mL²) |
|---|---|
| 0.0978125 | 23.513887124156692 |
| 0.16278125 | 15.692313133692718 |
| 0.22774999999999998 | 12.611142259464794 |
| 0.29271874999999997 | 11.206173002109892 |
| 0.35768749999999994 | 10.689449744150528 |
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(d) = 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(0.0978125) = 23.5139 ħ²/(2mL²) and Y_out(0.357687) = 10.6894 ħ²/(2mL²). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.0978125 + (11.8536 − 23.5139)(0.357687 − 0.0978125)/(10.6894 − 23.5139) = 0.334097 L. 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 d = 0.256336 L; substituting back gives Y_out = 11.853593 ħ²/(2mL²).
- The interpolation error in the input is |d_line − d| = 0.077761 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 11.678973 ħ²/(2mL²) and 12.028213 ħ²/(2mL²). Sort the roots even if the response decreases: 0.248833 L ≤ d ≤ 0.264596 L.
- The lower and upper input deviations are 0.00750356 L and 0.00825974 L. 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 d ≈ 0.256336 L; compatible interval [0.248833, 0.264596] L. Endpoint interpolation gives 0.334097 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.