University Quantum Mechanics II · topic UQMII-11.05
The linearised turning point and the Airy functions
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.
The linearised turning point and the Airy functions: predict a finite change
Linearising at a turning point a, where V(a) = E, replaces V by its tangent line there; in the variable z = (x − a)/l this turns the Schrödinger equation into Airy's equation ψ″ = zψ, whose solution that decays into the forbidden region is Ai(z). Use the supplied topic model Y_out = (5/(48*eta))^(2/3), where Y_out is WKB HOLDS FOR |z| ≥ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. eta is Phase tolerance η, measured in rad. Keep n (Oscillator level n) = 30 fixed. For this exercise use only 0.026375 rad ≤ eta ≤ 0.073625 rad 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 eta = 0.03677 rad and 0.043385 rad. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δeta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WKB HOLDS FOR |z| ≥ against eta
Predictions from the supplied The linearised turning point and the Airy functions 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): WKB HOLDS FOR |z| ≥
Horizontal axis (x): eta: Phase tolerance η / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eta: Phase tolerance η / rad | WKB HOLDS FOR |z| ≥ |
|---|---|
| 0.026375 | 2.4985662294649904 |
| 0.0381875 | 1.9522631624204545 |
| 0.05 | 1.6311948504870264 |
| 0.06181249999999999 | 1.4161251593689625 |
| 0.073625 | 1.2602855963191488 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of eta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At eta = 0.03677 rad, substitute into Y_out = (5/(48*eta))^(2/3): Y₀ = 2.00212.
- At eta = 0.043385 rad, the same fixed-parameter model gives Y₁ = 1.79305.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.209068; the input change is Δeta = 0.006615 rad.
- Average sensitivity is ΔY_out/Δeta = -31.6052 1 per rad. 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 = 2.00212, Y₁ = 1.79305; ΔY_out = -0.209068; average sensitivity = -31.6052 1 per rad.
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.
The linearised turning point and the Airy functions: infer an input with uncertainty
Linearising at a turning point a, where V(a) = E, replaces V by its tangent line there; in the variable z = (x − a)/l this turns the Schrödinger equation into Airy's equation ψ″ = zψ, whose solution that decays into the forbidden region is Ai(z). Use the supplied topic model Y_out = (5/(48*eta))^(2/3), where Y_out is WKB HOLDS FOR |z| ≥ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. eta is Phase tolerance η, measured in rad. Keep n (Oscillator level n) = 30 fixed. For this exercise use only 0.026375 rad ≤ eta ≤ 0.073625 rad 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 = 1.5271191, with a bounded reading uncertainty ±0.040025. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown eta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000004725 rad, 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.
WKB HOLDS FOR |z| ≥ against eta
Predictions from the supplied The linearised turning point and the Airy functions 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): WKB HOLDS FOR |z| ≥
Horizontal axis (x): eta: Phase tolerance η / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eta: Phase tolerance η / rad | WKB HOLDS FOR |z| ≥ |
|---|---|
| 0.026375 | 2.4985662294649904 |
| 0.0381875 | 1.9522631624204545 |
| 0.05 | 1.6311948504870264 |
| 0.06181249999999999 | 1.4161251593689625 |
| 0.073625 | 1.2602855963191488 |
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(eta) = 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(0.026375) = 2.49857 and Y_out(0.073625) = 1.26029. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives eta_line = 0.026375 + (1.52712 − 2.49857)(0.073625 − 0.026375)/(1.26029 − 2.49857) = 0.0634432 rad. 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 eta = 0.0551975 rad; substituting back gives Y_out = 1.5271191.
- The interpolation error in the input is |eta_line − eta| = 0.00824574 rad. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4870941 and 1.5671441. Sort the roots even if the response decreases: 0.0530964 rad ≤ eta ≤ 0.0574409 rad.
- The lower and upper input deviations are 0.00210106 rad and 0.00224338 rad. 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 eta ≈ 0.0551975 rad; compatible interval [0.0530964, 0.0574409] rad. Endpoint interpolation gives 0.0634432 rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.