University Quantum Mechanics II · topic UQMII-15.03
Free radial solutions and the partial-wave expansion of a plane wave
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.
Free radial solutions and the partial-wave expansion: predict a finite change
The partial-wave expansion writes an incident plane wave as a sum of free spherical waves of definite angular momentum, e^(ikz) = Σ iˡ(2l + 1) jₗ(kr) Pₗ(cos θ), each of which a central potential scatters on its own. Use the supplied topic model Y_out = (lmax+0.5)/krange, where Y_out is b / R FOR WAVE L (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. krange is Range of the potential kR, dimensionless. Keep lmax (Highest partial wave kept L) = 2 fixed. For this exercise use only 1.50937 ≤ krange ≤ 8.99062 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 krange = 3.15525 and 4.20262. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δkrange across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
b / R FOR WAVE L against krange
Predictions from the supplied Free radial solutions and the partial-wave expansion 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): b / R FOR WAVE L
Horizontal axis (x): krange: Range of the potential kR
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| krange: Range of the potential kR | b / R FOR WAVE L |
|---|---|
| 1.509375 | 1.6563146997929608 |
| 3.3796874999999997 | 0.7397133610725845 |
| 5.25 | 0.47619047619047616 |
| 7.120312499999999 | 0.35110818520956777 |
| 8.990625 | 0.2780674313521029 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of krange separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At krange = 3.15525, substitute into Y_out = (lmax+0.5)/krange: Y₀ = 0.79233.
- At krange = 4.20262, the same fixed-parameter model gives Y₁ = 0.594866.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.197464; the input change is Δkrange = 1.04737.
- Average sensitivity is ΔY_out/Δkrange = -0.188532 1 per unit input. 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 = 0.79233, Y₁ = 0.594866; ΔY_out = -0.197464; average sensitivity = -0.188532 1 per unit input.
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.
Free radial solutions and the partial-wave expansion: infer an input with uncertainty
The partial-wave expansion writes an incident plane wave as a sum of free spherical waves of definite angular momentum, e^(ikz) = Σ iˡ(2l + 1) jₗ(kr) Pₗ(cos θ), each of which a central potential scatters on its own. Use the supplied topic model Y_out = (lmax+0.5)/krange, where Y_out is b / R FOR WAVE L (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. krange is Range of the potential kR, dimensionless. Keep lmax (Highest partial wave kept L) = 2 fixed. For this exercise use only 1.50937 ≤ krange ≤ 8.99062 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 = 0.4116624, with a bounded reading uncertainty ±0.020039. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown krange by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125, 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.
b / R FOR WAVE L against krange
Predictions from the supplied Free radial solutions and the partial-wave expansion 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): b / R FOR WAVE L
Horizontal axis (x): krange: Range of the potential kR
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| krange: Range of the potential kR | b / R FOR WAVE L |
|---|---|
| 1.509375 | 1.6563146997929608 |
| 3.3796874999999997 | 0.7397133610725845 |
| 5.25 | 0.47619047619047616 |
| 7.120312499999999 | 0.35110818520956777 |
| 8.990625 | 0.2780674313521029 |
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(krange) = 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(1.50937) = 1.65631 and Y_out(8.99062) = 0.278067. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives krange_line = 1.50937 + (0.411662 − 1.65631)(8.99062 − 1.50937)/(0.278067 − 1.65631) = 8.26546. 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 krange = 6.07294; substituting back gives Y_out = 0.4116624.
- The interpolation error in the input is |krange_line − krange| = 2.19252. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.3916234 and 0.4317014. Sort the roots even if the response decreases: 5.79104 ≤ krange ≤ 6.38368.
- The lower and upper input deviations are 0.281898 and 0.310746. 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 krange ≈ 6.07294; compatible interval [5.79104, 6.38368]. Endpoint interpolation gives 8.26546.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.