University Quantum Mechanics II · unit UQMII-15
Quantum Scattering Theory
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Cross-sections and the classical benchmarks: infer an input with uncertainty
The differential cross-section dσ/dΩ is the rate at which particles scatter into unit solid angle about a direction, divided by the incident flux; it is an area per steradian that belongs to the interaction, and its integral σ is the beam area the target removes. Use the supplied topic model Y_out = 32.35/(en*en*sind(th/2)^4), where Y_out is RUTHERFORD dσ/dΩ(θ₀) (b sr⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. en is CM energy E, α on gold, measured in MeV. Keep th (Marker angle θ₀) = 60 °; rs (Hard-sphere radius R) = 10 fm fixed. For this exercise use only 2.4875 MeV ≤ en ≤ 13.5125 MeV 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 = 6.098397 b sr⁻¹, with a bounded reading uncertainty ±0.48954 b sr⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0011025 MeV, 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.
RUTHERFORD dσ/dΩ(θ₀) against en
Predictions from the supplied Cross-sections and the classical benchmarks 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): RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹
Horizontal axis (x): en: CM energy E, α on gold / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: CM energy E, α on gold / MeV | RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹ |
|---|---|
| 2.4875 | 83.65041286836197 |
| 5.2437499999999995 | 18.823930526294866 |
| 7.999999999999999 | 8.087500000000006 |
| 10.756249999999998 | 4.473758044405494 |
| 13.5125 | 2.8348028044749074 |
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(en) = 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(2.4875) = 83.6504 b sr⁻¹ and Y_out(13.5125) = 2.8348 b sr⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 2.4875 + (6.0984 − 83.6504)(13.5125 − 2.4875)/(2.8348 − 83.6504) = 13.0673 MeV. 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 en = 9.21275 MeV; substituting back gives Y_out = 6.098397 b sr⁻¹.
- The interpolation error in the input is |en_line − en| = 3.85453 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.608857 b sr⁻¹ and 6.587937 b sr⁻¹. Sort the roots even if the response decreases: 8.86385 MeV ≤ en ≤ 9.60638 MeV.
- The lower and upper input deviations are 0.3489 MeV and 0.393634 MeV. 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 en ≈ 9.21275 MeV; compatible interval [8.86385, 9.60638] MeV. Endpoint interpolation gives 13.0673 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The scattering amplitude and dσ/dΩ = |f(θ)|²: infer an input with uncertainty
The scattering amplitude f(θ) is the coefficient, with the dimension of a length, of the outgoing wave e^(ikr)/r that a stationary scattering state carries far from the target, and its squared modulus is the differential cross-section. Use the supplied topic model Y_out = 4*exp(-0.5*(sind(30)/b)^2), where Y_out is CROSS-TERM SIZE AT 30° (× |f|²/r²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; sind/cosd/tand use degrees. b is Beam half-width a/r, dimensionless. Keep kr (Detector distance kr) = 60; ph (Phase of f) = 90 ° fixed. For this exercise use only 0.126562 ≤ b ≤ 0.323437 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 = 0.51258002 × |f|²/r², with a bounded reading uncertainty ±0.076642 × |f|²/r². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000196875, 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.
CROSS-TERM SIZE AT 30° against b
Predictions from the supplied The scattering amplitude and dσ/dΩ = |f(θ)|² 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): CROSS-TERM SIZE AT 30° / × |f|²/r²
Horizontal axis (x): b: Beam half-width a/r
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Beam half-width a/r | CROSS-TERM SIZE AT 30° / × |f|²/r² |
|---|---|
| 0.1265625 | 0.0016329058831063635 |
| 0.17578124999999997 | 0.07000856147362694 |
| 0.22499999999999998 | 0.33863195449011974 |
| 0.27421874999999996 | 0.758790312154545 |
| 0.32343749999999993 | 1.210944537566904 |
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(b) = 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(0.126562) = 0.00163291 × |f|²/r² and Y_out(0.323437) = 1.21094 × |f|²/r². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.126562 + (0.51258 − 0.00163291)(0.323437 − 0.126562)/(1.21094 − 0.00163291) = 0.209744. 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 b = 0.246656; substituting back gives Y_out = 0.51258002 × |f|²/r².
- The interpolation error in the input is |b_line − b| = 0.036912. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.43593802 × |f|²/r² and 0.58922202 × |f|²/r². Sort the roots even if the response decreases: 0.237474 ≤ b ≤ 0.255472.
- The lower and upper input deviations are 0.00918213 and 0.00881537. 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 b ≈ 0.246656; compatible interval [0.237474, 0.255472]. Endpoint interpolation gives 0.209744.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Phase shifts, partial cross-sections and the unitarity limit: infer an input with uncertainty
The phase shift δₗ is the angle by which a central potential advances or delays the lth partial wave outside its range; that wave then adds 4π(2l + 1) sin²δₗ/k² to the cross-section, and never more than 4π(2l + 1)/k². Use the supplied topic model Y_out = 4*sin(atan(k*tan(sqrt(max(k^2+s,1e-9)))/sqrt(max(k^2+s,1e-9))*(1-2/(exp(2*sqrt(max(-k^2-s,1e-9)))+1))/sqrt(max(-k^2-s,1e-9)))-k)^2/k^2, where Y_out is σ₀ ÷ πa² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value; unmarked trigonometric functions use radians. k is Energy as ka (wave number × range), dimensionless. Keep s (Strength 2mV₀a²/ħ² (negative: barrier)) = 2 fixed. For this exercise use only 0.4975 ≤ k ≤ 2.7025 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 = 0.3707703, with a bounded reading uncertainty ±0.045527. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002205, 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.
σ₀ ÷ πa² against k
Predictions from the supplied Phase shifts, partial cross-sections and the unitarity limit 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): σ₀ ÷ πa²
Horizontal axis (x): k: Energy as ka (wave number × range)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Energy as ka (wave number × range) | σ₀ ÷ πa² |
|---|---|
| 0.4975 | 9.287554983431015 |
| 1.0487499999999998 | 1.9952676228337296 |
| 1.5999999999999999 | 0.6064813491300477 |
| 2.1512499999999997 | 0.19957730249931588 |
| 2.7024999999999997 | 0.06725868733693541 |
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(k) = 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(0.4975) = 9.28755 and Y_out(2.7025) = 0.0672587. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 0.4975 + (0.37077 − 9.28755)(2.7025 − 0.4975)/(0.0672587 − 9.28755) = 2.62992. 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 k = 1.84255; substituting back gives Y_out = 0.3707703.
- The interpolation error in the input is |k_line − k| = 0.787366. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.3252433 and 0.4162973. Sort the roots even if the response decreases: 1.78513 ≤ k ≤ 1.90765.
- The lower and upper input deviations are 0.0574154 and 0.0651032. 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 k ≈ 1.84255; compatible interval [1.78513, 1.90765]. Endpoint interpolation gives 2.62992.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.