University Advanced Quantum Mechanics · topic UAQM-02.07
The exact Coulomb amplitude in parabolic coordinates
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 exact Coulomb amplitude in parabolic coordinates: predict a finite change
The exact Coulomb amplitude f_C(θ) is the coefficient of the outgoing wave in the exact Schrödinger solution for two point charges: its modulus is Rutherford's at every energy, and its phase, −η ln sin²(θ/2) + 2σ₀, is what perturbation theory misses. Use the supplied topic model Y_out = -eta*ln(sind(th/2)^2), where Y_out is LOG PHASE AT θ (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; sind/cosd/tand use degrees. th is Marker angle θ, measured in °. Keep eta (Sommerfeld parameter η) = 1.5; d0 (Nuclear s-wave shift δ₀) = 20 ° fixed. For this exercise use only 28.0625 ° ≤ th ≤ 161.938 ° 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 th = 57.515 ° and 76.2575 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LOG PHASE AT θ against th
Predictions from the supplied The exact Coulomb amplitude in parabolic coordinates 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): LOG PHASE AT θ / rad
Horizontal axis (x): th: Marker angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Marker angle θ / ° | LOG PHASE AT θ / rad |
|---|---|
| 28.0625 | 4.250866039292809 |
| 61.53125 | 2.0110621830102895 |
| 95 | 0.9143734576310785 |
| 128.46875 | 0.31414973819852343 |
| 161.9375 | 0.037423840588956564 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of th separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At th = 57.515 °, substitute into Y_out = -eta*ln(sind(th/2)^2): Y₀ = 2.19502 rad.
- At th = 76.2575 °, the same fixed-parameter model gives Y₁ = 1.44657 rad.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.748453 rad; the input change is Δth = 18.7425 °.
- Average sensitivity is ΔY_out/Δth = -0.0399335 rad per °. 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 = 2.19502 rad, Y₁ = 1.44657 rad; ΔY_out = -0.748453 rad; average sensitivity = -0.0399335 rad per °.
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 exact Coulomb amplitude in parabolic coordinates: infer an input with uncertainty
The exact Coulomb amplitude f_C(θ) is the coefficient of the outgoing wave in the exact Schrödinger solution for two point charges: its modulus is Rutherford's at every energy, and its phase, −η ln sin²(θ/2) + 2σ₀, is what perturbation theory misses. Use the supplied topic model Y_out = -eta*ln(sind(th/2)^2), where Y_out is LOG PHASE AT θ (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; sind/cosd/tand use degrees. th is Marker angle θ, measured in °. Keep eta (Sommerfeld parameter η) = 1.5; d0 (Nuclear s-wave shift δ₀) = 20 ° fixed. For this exercise use only 28.0625 ° ≤ th ≤ 161.938 ° 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 = 0.60348769 rad, with a bounded reading uncertainty ±0.08491 rad. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0133875 °, 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.
LOG PHASE AT θ against th
Predictions from the supplied The exact Coulomb amplitude in parabolic coordinates 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): LOG PHASE AT θ / rad
Horizontal axis (x): th: Marker angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Marker angle θ / ° | LOG PHASE AT θ / rad |
|---|---|
| 28.0625 | 4.250866039292809 |
| 61.53125 | 2.0110621830102895 |
| 95 | 0.9143734576310785 |
| 128.46875 | 0.31414973819852343 |
| 161.9375 | 0.037423840588956564 |
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(th) = 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(28.0625) = 4.25087 rad and Y_out(161.938) = 0.0374238 rad. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 28.0625 + (0.603488 − 4.25087)(161.938 − 28.0625)/(0.0374238 − 4.25087) = 143.952 °. 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 th = 109.726 °; substituting back gives Y_out = 0.60348769 rad.
- The interpolation error in the input is |th_line − th| = 34.2255 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.51857769 rad and 0.68839769 rad. Sort the roots even if the response decreases: 105.303 ° ≤ th ≤ 114.546 °.
- The lower and upper input deviations are 4.42366 ° and 4.81973 °. 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 th ≈ 109.726 °; compatible interval [105.303, 114.546] °. Endpoint interpolation gives 143.952 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.