University Advanced Quantum Mechanics · topic UAQM-07.01
Conserved operators in a central field: J and parity, but not L
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.
Conserved operators in a central field: J and parity, not L: predict a finite change
In a central electrostatic potential the Dirac Hamiltonian commutes with the total angular momentum J = L + (ħ/2)Σ and with the parity operator γ⁰P₀, but not with L² or S⋅L, so states are labelled by j, m_j and parity η instead of by l. Use the supplied topic model Y_out = (2*j+1)*sqrt((wl/100)*(1-wl/100)), where Y_out is SPREAD ΔL² (ħ²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. wl is Lower-component weight w, measured in %. Keep j (Total angular momentum j) = 1.5; par (Parity η) = -1 fixed. For this exercise use only 3.1875 % ≤ wl ≤ 26.8125 % 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 wl = 8.385 % and 11.6925 %. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δwl across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPREAD ΔL² against wl
Predictions from the supplied Conserved operators in a central field: J and parity, not L 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): SPREAD ΔL² / ħ²
Horizontal axis (x): wl: Lower-component weight w / %
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| wl: Lower-component weight w / % | SPREAD ΔL² / ħ² |
|---|---|
| 3.1875 | 0.7026690188132675 |
| 9.09375 | 1.1500808395499857 |
| 14.999999999999998 | 1.42828568570857 |
| 20.906249999999996 | 1.6265564661271368 |
| 26.812499999999996 | 1.7719322080711777 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of wl separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At wl = 8.385 %, substitute into Y_out = (2*j+1)*sqrt((wl/100)*(1-wl/100)): Y₀ = 1.10865 ħ².
- At wl = 11.6925 %, the same fixed-parameter model gives Y₁ = 1.28532 ħ².
- Signed change: ΔY_out = Y₁ − Y₀ = 0.176673 ħ²; the input change is Δwl = 3.3075 %.
- Average sensitivity is ΔY_out/Δwl = 0.0534158 ħ² 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 = 1.10865 ħ², Y₁ = 1.28532 ħ²; ΔY_out = 0.176673 ħ²; average sensitivity = 0.0534158 ħ² 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.
Conserved operators in a central field: J and parity, not L: infer an input with uncertainty
In a central electrostatic potential the Dirac Hamiltonian commutes with the total angular momentum J = L + (ħ/2)Σ and with the parity operator γ⁰P₀, but not with L² or S⋅L, so states are labelled by j, m_j and parity η instead of by l. Use the supplied topic model Y_out = (2*j+1)*sqrt((wl/100)*(1-wl/100)), where Y_out is SPREAD ΔL² (ħ²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. wl is Lower-component weight w, measured in %. Keep j (Total angular momentum j) = 1.5; par (Parity η) = -1 fixed. For this exercise use only 3.1875 % ≤ wl ≤ 26.8125 % 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 = 1.5232381 ħ², with a bounded reading uncertainty ±0.037304 ħ². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown wl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 %, 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.
SPREAD ΔL² against wl
Predictions from the supplied Conserved operators in a central field: J and parity, not L 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): SPREAD ΔL² / ħ²
Horizontal axis (x): wl: Lower-component weight w / %
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| wl: Lower-component weight w / % | SPREAD ΔL² / ħ² |
|---|---|
| 3.1875 | 0.7026690188132675 |
| 9.09375 | 1.1500808395499857 |
| 14.999999999999998 | 1.42828568570857 |
| 20.906249999999996 | 1.6265564661271368 |
| 26.812499999999996 | 1.7719322080711777 |
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(wl) = 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(3.1875) = 0.702669 ħ² and Y_out(26.8125) = 1.77193 ħ². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives wl_line = 3.1875 + (1.52324 − 0.702669)(26.8125 − 3.1875)/(1.77193 − 0.702669) = 21.3177 %. 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 wl = 17.5987 %; substituting back gives Y_out = 1.5232381 ħ².
- The interpolation error in the input is |wl_line − wl| = 3.71894 %. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4859341 ħ² and 1.5605421 ħ². Sort the roots even if the response decreases: 16.5336 % ≤ wl ≤ 18.7279 %.
- The lower and upper input deviations are 1.06515 % and 1.12918 %. 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 wl ≈ 17.5987 %; compatible interval [16.5336, 18.7279] %. Endpoint interpolation gives 21.3177 %.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.