University Advanced Quantum Mechanics · unit UAQM-02
Scattering with Spin and Identical Particles
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.
The spin-dependent amplitude g(θ) + i h(θ) σ⋅n: infer an input with uncertainty
For a spin-half particle scattering from a spin-zero target, the amplitude is the 2 × 2 spin matrix M = g(θ) + i h(θ) σ⋅n, where g leaves the spin alone, h acts through the spin component along the normal n to the scattering plane, and an unpolarised beam scatters with dσ/dΩ = |g|² + |h|². Use the supplied topic model Y_out = -2*deg(atan(w*kb^2*sind(th)/4)), where Y_out is SPIN ROTATION ABOUT n (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; unmarked trigonometric functions use radians. th is Marker angle θ, measured in °. Keep kb (Momentum times range kb) = 2.5; w (Spin-orbit depth ratio W/U) = 0.4 fixed. For this exercise use only 94.725 ° ≤ th ≤ 160.875 ° 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 = -47.628653 °, with a bounded reading uncertainty ±2.431 °. 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 0.006615 °, 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.
SPIN ROTATION ABOUT n against th
Predictions from the supplied The spin-dependent amplitude g(θ) + i h(θ) σ⋅n 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): SPIN ROTATION ABOUT n / °
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 / ° | SPIN ROTATION ABOUT n / ° |
|---|---|
| 94.725 | -63.835572394075136 |
| 111.26249999999999 | -60.43779102071531 |
| 127.8 | -52.56463894734074 |
| 144.3375 | -40.04155729129355 |
| 160.875 | -23.144833057771717 |
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(94.725) = -63.8356 ° and Y_out(160.875) = -23.1448 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 94.725 + (-47.6287 − -63.8356)(160.875 − 94.725)/(-23.1448 − -63.8356) = 121.072 °. 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 = 135.077 °; substituting back gives Y_out = -47.628653 °.
- The interpolation error in the input is |th_line − th| = 14.0043 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -50.059653 ° and -45.197653 °. Sort the roots even if the response decreases: 131.66 ° ≤ th ≤ 138.243 °.
- The lower and upper input deviations are 3.41685 ° and 3.16616 °. 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 ≈ 135.077 °; compatible interval [131.66, 138.243] °. Endpoint interpolation gives 121.072 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Partial waves with spin: phase shifts for j = l ± ½: infer an input with uncertainty
For a spin-½ projectile on a spinless target, each orbital angular momentum l ≥ 1 scatters in two independent channels, j = l + ½ and j = l − ½, each with its own phase shift; g(θ) adds them with weights l + 1 and l, and h(θ) adds their differences. Use the supplied topic model Y_out = sind(dp-dm)^2/max(sind(ds)^2 + sind(dp-dm)^2, 1e-9), where Y_out is FLIP SHARE AT 90° (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; sind/cosd/tand use degrees. dp is p₃/₂ phase shift δ₁⁺, measured in °. Keep ds (s₁/₂ phase shift δ₀) = -27 °; dm (p₁/₂ phase shift δ₁⁻) = 8 ° fixed. For this exercise use only 19.125 ° ≤ dp ≤ 94.725 ° 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 = 0.77432949, with a bounded reading uncertainty ±0.0081481. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00756 °, 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.
FLIP SHARE AT 90° against dp
Predictions from the supplied Partial waves with spin: phase shifts for j = l ± ½ 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): FLIP SHARE AT 90°
Horizontal axis (x): dp: p₃/₂ phase shift δ₁⁺ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dp: p₃/₂ phase shift δ₁⁺ / ° | FLIP SHARE AT 90° |
|---|---|
| 19.125 | 0.15299658230910743 |
| 38.025 | 0.5484907092348283 |
| 56.925 | 0.7338480655938994 |
| 75.82499999999999 | 0.806226113749067 |
| 94.725 | 0.8286499146021341 |
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(dp) = 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(19.125) = 0.152997 and Y_out(94.725) = 0.82865. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives dp_line = 19.125 + (0.774329 − 0.152997)(94.725 − 19.125)/(0.82865 − 0.152997) = 88.647 °. 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 dp = 65.241 °; substituting back gives Y_out = 0.77432949.
- The interpolation error in the input is |dp_line − dp| = 23.406 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.76618139 and 0.78247759. Sort the roots even if the response decreases: 63.2667 ° ≤ dp ≤ 67.4352 °.
- The lower and upper input deviations are 1.9743 ° and 2.1942 °. 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 dp ≈ 65.241 °; compatible interval [63.2667, 67.4352] °. Endpoint interpolation gives 88.647 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Polarisation by scattering and the analysing power: infer an input with uncertainty
Polarisation by scattering is the net spin alignment along the normal n = k × k′/|k × k′| that an unpolarised beam acquires in one scattering; the analysing power is the left-right asymmetry per unit of beam polarisation, and time reversal makes the two equal. Use the supplied topic model Y_out = (1 + 2*sind(d0)*(sind(dp)*sind(d0-dp) - sind(dm)*sind(d0-dm))/max(sind(d0)^2 + sind(dp)^2 + sind(dm)^2 - 2*sind(dp)*sind(dm)*cosd(dp-dm), 0.0001))/2, where Y_out is FRACTION WITH SPIN ALONG n (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; sind/cosd/tand use degrees. dm is p₁/₂ phase shift δ₁/₂, measured in °. Keep d0 (s-wave phase shift δ₀) = -30 °; dp (p₃/₂ phase shift δ₃/₂) = 80 ° fixed. For this exercise use only -17.25 ° ≤ dm ≤ 70.95 ° 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 = 0.76215986, with a bounded reading uncertainty ±0.020895. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dm by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00882 °, 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.
FRACTION WITH SPIN ALONG n against dm
Predictions from the supplied Polarisation by scattering and the analysing power 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): FRACTION WITH SPIN ALONG n
Horizontal axis (x): dm: p₁/₂ phase shift δ₁/₂ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dm: p₁/₂ phase shift δ₁/₂ / ° | FRACTION WITH SPIN ALONG n |
|---|---|
| -17.25 | 0.9014599966235702 |
| 4.799999999999997 | 0.8703998210611346 |
| 26.849999999999994 | 0.807340519879384 |
| 48.89999999999999 | 0.6799031987558322 |
| 70.94999999999999 | 0.4952538179848161 |
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(dm) = 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(-17.25) = 0.90146 and Y_out(70.95) = 0.495254. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dm_line = -17.25 + (0.76216 − 0.90146)(70.95 − -17.25)/(0.495254 − 0.90146) = 12.9964 °. 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 dm = 36.552 °; substituting back gives Y_out = 0.76215986.
- The interpolation error in the input is |dm_line − dm| = 23.5556 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.74126486 and 0.78305486. Sort the roots even if the response decreases: 32.4435 ° ≤ dm ≤ 40.172 °.
- The lower and upper input deviations are 4.10847 ° and 3.61996 °. 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 dm ≈ 36.552 °; compatible interval [32.4435, 40.172] °. Endpoint interpolation gives 12.9964 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Singlet and triplet weights in neutron-proton scattering: infer an input with uncertainty
For two distinguishable spin-halves whose force depends on spin only through S₁⋅S₂, the scattering amplitude is one function on the spin singlet and another on the triplet, and an unpolarised cross-section weights them 1/4 and 3/4, the fractions of spin states each channel holds. Use the supplied topic model Y_out = 100*(1-pp)*as^2/((1-pp)*as^2+(3+pp)*at^2), where Y_out is SINGLET SHARE (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. pp is Polarisation product Pₙ Pₚ, along z, dimensionless. Keep as (Singlet scattering length aₛ) = -23.7 fm; at (Triplet scattering length aₜ) = 5.4 fm fixed. For this exercise use only -0.7875 ≤ pp ≤ 0.7875 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 = 83.384729 %, with a bounded reading uncertainty ±1.5866 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown pp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
SINGLET SHARE against pp
Predictions from the supplied Singlet and triplet weights in neutron-proton scattering 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): SINGLET SHARE / %
Horizontal axis (x): pp: Polarisation product Pₙ Pₚ, along z
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| pp: Polarisation product Pₙ Pₚ, along z | SINGLET SHARE / % |
|---|---|
| -0.7875 | 93.96216721010812 |
| -0.39375000000000004 | 91.15119942941386 |
| -1.1102230246251565e-16 | 86.52433106890336 |
| 0.39374999999999993 | 77.48240452881 |
| 0.7874999999999998 | 51.93984402919683 |
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(pp) = 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(-0.7875) = 93.9622 % and Y_out(0.7875) = 51.9398 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives pp_line = -0.7875 + (83.3847 − 93.9622)(0.7875 − -0.7875)/(51.9398 − 93.9622) = -0.391057. 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 pp = 0.17325; substituting back gives Y_out = 83.384729 %.
- The interpolation error in the input is |pp_line − pp| = 0.564307. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 81.798129 % and 84.971329 %. Sort the roots even if the response decreases: 0.0923292 ≤ pp ≤ 0.243326.
- The lower and upper input deviations are 0.0809208 and 0.0700761. 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 pp ≈ 0.17325; compatible interval [0.0923292, 0.243326]. Endpoint interpolation gives -0.391057.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.