University Physics V · topic UPV-19.07
Angular momentum, spin addition, and selection rules
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.
Angular momentum, spin addition and selection rules: predict a finite change
Angular momentum is conserved in a reaction because the total Ĵ — every spin plus every orbital L of relative motion — generates rotations and commutes with Ĥ, so the labels J and M of the initial state must be reachable by the final state. Use the supplied topic model Y_out = (jt*(jt+1)-j1*(j1+1)-j2*(j2+1))/max(2*sqrt(j1*(j1+1))*sqrt(j2*(j2+1)),0.01), where Y_out is COS θ₁₂ (RAW) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. jt is Trial total J, dimensionless. Keep j1 (First angular momentum j₁) = 1; j2 (Second angular momentum j₂) = 0.5 fixed. For this exercise use only 0.53125 ≤ jt ≤ 4.46875 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 jt = 1.3975 and 1.94875. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δjt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
COS θ₁₂ (RAW) against jt
Predictions from the supplied Angular momentum, spin addition and selection rules 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): COS θ₁₂ (RAW)
Horizontal axis (x): jt: Trial total J
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| jt: Trial total J | COS θ₁₂ (RAW) |
|---|---|
| 0.53125 | -0.7905823828025784 |
| 1.515625 | 0.43386347861064345 |
| 2.5 | 2.449489742783178 |
| 3.484375 | 5.256296409715025 |
| 4.46875 | 8.854283479406185 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of jt separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At jt = 1.3975, substitute into Y_out = (jt*(jt+1)-j1*(j1+1)-j2*(j2+1))/max(2*sqrt(j1*(j1+1))*sqrt(j2*(j2+1)),0.01): Y₀ = 0.245156.
- At jt = 1.94875, the same fixed-parameter model gives Y₁ = 1.22327.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.97811; the input change is Δjt = 0.55125.
- Average sensitivity is ΔY_out/Δjt = 1.77435 1 per unit input. 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 = 0.245156, Y₁ = 1.22327; ΔY_out = 0.97811; average sensitivity = 1.77435 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.
Angular momentum, spin addition and selection rules: infer an input with uncertainty
Angular momentum is conserved in a reaction because the total Ĵ — every spin plus every orbital L of relative motion — generates rotations and commutes with Ĥ, so the labels J and M of the initial state must be reachable by the final state. Use the supplied topic model Y_out = (jt*(jt+1)-j1*(j1+1)-j2*(j2+1))/max(2*sqrt(j1*(j1+1))*sqrt(j2*(j2+1)),0.01), where Y_out is COS θ₁₂ (RAW) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. jt is Trial total J, dimensionless. Keep j1 (First angular momentum j₁) = 1; j2 (Second angular momentum j₂) = 0.5 fixed. For this exercise use only 0.53125 ≤ jt ≤ 4.46875 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 = 3.5870113, with a bounded reading uncertainty ±0.65664. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown jt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
COS θ₁₂ (RAW) against jt
Predictions from the supplied Angular momentum, spin addition and selection rules 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): COS θ₁₂ (RAW)
Horizontal axis (x): jt: Trial total J
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| jt: Trial total J | COS θ₁₂ (RAW) |
|---|---|
| 0.53125 | -0.7905823828025784 |
| 1.515625 | 0.43386347861064345 |
| 2.5 | 2.449489742783178 |
| 3.484375 | 5.256296409715025 |
| 4.46875 | 8.854283479406185 |
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(jt) = 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(0.53125) = -0.790582 and Y_out(4.46875) = 8.85428. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives jt_line = 0.53125 + (3.58701 − -0.790582)(4.46875 − 0.53125)/(8.85428 − -0.790582) = 2.3184. 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 jt = 2.93313; substituting back gives Y_out = 3.5870113.
- The interpolation error in the input is |jt_line − jt| = 0.61473. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.9303713 and 4.2436513. Sort the roots even if the response decreases: 2.69028 ≤ jt ≤ 3.15989.
- The lower and upper input deviations are 0.242841 and 0.226763. 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 jt ≈ 2.93313; compatible interval [2.69028, 3.15989]. Endpoint interpolation gives 2.3184.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.