University Quantum Mechanics I · topic UQMI-09.01
Why every stable minimum is an oscillator
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 every stable minimum is an oscillator: predict a finite change
The harmonic approximation replaces a potential near a stable minimum x₀ by the parabola with the same curvature, V ≈ V(x₀) + ½V″(x₀)(x − x₀)², so the motion there is an oscillator with ω = √(V″(x₀)/m). Use the supplied topic model Y_out = (depth-(min(n+0.5,2*depth)-2*depth)^2/(4*depth)), where Y_out is MORSE Eₙ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. depth is Well depth Dₑ, measured in ħω. Keep n (Level n) = 2 fixed. For this exercise use only 2.22187 ħω ≤ depth ≤ 11.2781 ħω 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 depth = 4.21425 ħω and 5.48212 ħω. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdepth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MORSE Eₙ against depth
Predictions from the supplied Why every stable minimum is an oscillator 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): MORSE Eₙ / ħω
Horizontal axis (x): depth: Well depth Dₑ / ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| depth: Well depth Dₑ / ħω | MORSE Eₙ / ħω |
|---|---|
| 2.221875 | 1.7967651195499297 |
| 4.4859374999999995 | 2.1516893068617207 |
| 6.749999999999999 | 2.268518518518519 |
| 9.014062499999998 | 2.3266597330559886 |
| 11.278125 | 2.3614574674425057 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of depth separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At depth = 4.21425 ħω, substitute into Y_out = (depth-(min(n+0.5,2*depth)-2*depth)^2/(4*depth)): Y₀ = 2.12923 ħω.
- At depth = 5.48212 ħω, the same fixed-parameter model gives Y₁ = 2.21498 ħω.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0857486 ħω; the input change is Δdepth = 1.26787 ħω.
- Average sensitivity is ΔY_out/Δdepth = 0.0676318 ħω per ħω. Its positive sign means that this response increases 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 = 2.12923 ħω, Y₁ = 2.21498 ħω; ΔY_out = 0.0857486 ħω; average sensitivity = 0.0676318 ħω per ħω.
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 every stable minimum is an oscillator: infer an input with uncertainty
The harmonic approximation replaces a potential near a stable minimum x₀ by the parabola with the same curvature, V ≈ V(x₀) + ½V″(x₀)(x − x₀)², so the motion there is an oscillator with ω = √(V″(x₀)/m). Use the supplied topic model Y_out = (depth-(min(n+0.5,2*depth)-2*depth)^2/(4*depth)), where Y_out is MORSE Eₙ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. depth is Well depth Dₑ, measured in ħω. Keep n (Level n) = 2 fixed. For this exercise use only 2.22187 ħω ≤ depth ≤ 11.2781 ħω 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 = 2.2982879 ħω, with a bounded reading uncertainty ±0.0094754 ħω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown depth by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000905625 ħω, 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.
MORSE Eₙ against depth
Predictions from the supplied Why every stable minimum is an oscillator 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): MORSE Eₙ / ħω
Horizontal axis (x): depth: Well depth Dₑ / ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| depth: Well depth Dₑ / ħω | MORSE Eₙ / ħω |
|---|---|
| 2.221875 | 1.7967651195499297 |
| 4.4859374999999995 | 2.1516893068617207 |
| 6.749999999999999 | 2.268518518518519 |
| 9.014062499999998 | 2.3266597330559886 |
| 11.278125 | 2.3614574674425057 |
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(depth) = 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(2.22187) = 1.79677 ħω and Y_out(11.2781) = 2.36146 ħω. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives depth_line = 2.22187 + (2.29829 − 1.79677)(11.2781 − 2.22187)/(2.36146 − 1.79677) = 10.265 ħω. 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 depth = 7.74619 ħω; substituting back gives Y_out = 2.2982879 ħω.
- The interpolation error in the input is |depth_line − depth| = 2.51885 ħω. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.2888125 ħω and 2.3077633 ħω. Sort the roots even if the response decreases: 7.39864 ħω ≤ depth ≤ 8.128 ħω.
- The lower and upper input deviations are 0.34755 ħω and 0.381812 ħω. 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 depth ≈ 7.74619 ħω; compatible interval [7.39864, 8.128] ħω. Endpoint interpolation gives 10.265 ħω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.