University Advanced Quantum Mechanics · topic UAQM-02.06
Identical spin-half fermions: (1/4)|f + f'|^2 + (3/4)|f - f'|^2
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.
Identical spin-half fermions: ¼|f + f′|² + ¾|f − f′|²: predict a finite change
Two identical spin-half fermions under a spin-independent force scatter with f(θ) + f(π − θ) in the singlet and f(θ) − f(π − θ) in the triplet, and an unpolarised pair weights the two cross-sections 1/4 and 3/4. Use the supplied topic model Y_out = 2*pi*(1-pdot)*sind(d0)^2+6*pi*(3+pdot)*sind(d1)^2, where Y_out is k²σ, EVENTS COUNTED ONCE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. d1 is p-wave phase δ₁ (triplet only), measured in °. Keep pdot (Polarisation product P₁⋅P₂) = 0; d0 (s-wave phase δ₀ (singlet only)) = 50 ° fixed. For this exercise use only 1.59375 ° ≤ d1 ≤ 13.4062 ° 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 d1 = 4.1925 ° and 5.84625 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd1 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
k²σ, EVENTS COUNTED ONCE against d1
Predictions from the supplied Identical spin-half fermions: ¼|f + f′|² + ¾|f − f′|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): k²σ, EVENTS COUNTED ONCE
Horizontal axis (x): d1: p-wave phase δ₁ (triplet only) / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d1: p-wave phase δ₁ (triplet only) / ° | k²σ, EVENTS COUNTED ONCE |
|---|---|
| 1.59375 | 3.7308671837031215 |
| 4.546875 | 4.042503089956545 |
| 7.499999999999999 | 4.650549057124126 |
| 10.453124999999998 | 5.548549580681227 |
| 13.406249999999998 | 6.726970765567331 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d1 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d1 = 4.1925 °, substitute into Y_out = 2*pi*(1-pdot)*sind(d0)^2+6*pi*(3+pdot)*sind(d1)^2: Y₀ = 3.98936.
- At d1 = 5.84625 °, the same fixed-parameter model gives Y₁ = 4.27384.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.284474; the input change is Δd1 = 1.65375 °.
- Average sensitivity is ΔY_out/Δd1 = 0.172017 1 per °. Its positive sign means that this response increases 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 = 3.98936, Y₁ = 4.27384; ΔY_out = 0.284474; average sensitivity = 0.172017 1 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.
Identical spin-half fermions: ¼|f + f′|² + ¾|f − f′|²: infer an input with uncertainty
Two identical spin-half fermions under a spin-independent force scatter with f(θ) + f(π − θ) in the singlet and f(θ) − f(π − θ) in the triplet, and an unpolarised pair weights the two cross-sections 1/4 and 3/4. Use the supplied topic model Y_out = 2*pi*(1-pdot)*sind(d0)^2+6*pi*(3+pdot)*sind(d1)^2, where Y_out is k²σ, EVENTS COUNTED ONCE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. d1 is p-wave phase δ₁ (triplet only), measured in °. Keep pdot (Polarisation product P₁⋅P₂) = 0; d0 (s-wave phase δ₀ (singlet only)) = 50 ° fixed. For this exercise use only 1.59375 ° ≤ d1 ≤ 13.4062 ° 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 = 5.0104407, with a bounded reading uncertainty ±0.19194. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00118125 °, 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.
k²σ, EVENTS COUNTED ONCE against d1
Predictions from the supplied Identical spin-half fermions: ¼|f + f′|² + ¾|f − f′|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): k²σ, EVENTS COUNTED ONCE
Horizontal axis (x): d1: p-wave phase δ₁ (triplet only) / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d1: p-wave phase δ₁ (triplet only) / ° | k²σ, EVENTS COUNTED ONCE |
|---|---|
| 1.59375 | 3.7308671837031215 |
| 4.546875 | 4.042503089956545 |
| 7.499999999999999 | 4.650549057124126 |
| 10.453124999999998 | 5.548549580681227 |
| 13.406249999999998 | 6.726970765567331 |
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(d1) = 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(1.59375) = 3.73087 and Y_out(13.4062) = 6.72697. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d1_line = 1.59375 + (5.01044 − 3.73087)(13.4062 − 1.59375)/(6.72697 − 3.73087) = 6.63862 °. 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 d1 = 8.79938 °; substituting back gives Y_out = 5.0104407.
- The interpolation error in the input is |d1_line − d1| = 2.16075 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.8185007 and 5.2023807. Sort the roots even if the response decreases: 8.13156 ° ≤ d1 ≤ 9.42135 °.
- The lower and upper input deviations are 0.667815 ° and 0.62198 °. 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 d1 ≈ 8.79938 °; compatible interval [8.13156, 9.42135] °. Endpoint interpolation gives 6.63862 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.