University Advanced Quantum Mechanics · topic UAQM-13.05
The sum rule that confines the shift to s states
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.
The sum rule that confines the shift to s states: predict a finite change
The sum rule says that Σₘ |⟨m|p|n⟩|²(Eₘ − Eₙ), the weight multiplying ln K in Bethe's shift, equals (ħ²/2)⟨n|∇²V|n⟩, which for a Coulomb potential is fixed by the electron's probability density at the nucleus. Use the supplied topic model Y_out = 52917.7*10^lr, where Y_out is SOURCE RADIUS R (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lr is Source radius, log₁₀(R/a₀), dimensionless. Keep no other adjustable input fixed. For this exercise use only -4.41563 ≤ lr ≤ -0.084375 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 lr = -3.46275 and -2.85638. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SOURCE RADIUS R against lr
Predictions from the supplied The sum rule that confines the shift to s states 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): SOURCE RADIUS R / fm
Horizontal axis (x): lr: Source radius, log₁₀(R/a₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lr: Source radius, log₁₀(R/a₀) | SOURCE RADIUS R / fm |
|---|---|
| -4.415625 | 2.032244514068117 |
| -3.3328125 | 24.591694776190057 |
| -2.2500000000000004 | 297.57809544025304 |
| -1.1671875000000007 | 3600.9198915231354 |
| -0.08437500000000053 | 43573.85259148016 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lr separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lr = -3.46275, substitute into Y_out = 52917.7*10^lr: Y₀ = 18.2327 fm.
- At lr = -2.85638, the same fixed-parameter model gives Y₁ = 73.659 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = 55.4263 fm; the input change is Δlr = 0.606375.
- Average sensitivity is ΔY_out/Δlr = 91.406 fm 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 = 18.2327 fm, Y₁ = 73.659 fm; ΔY_out = 55.4263 fm; average sensitivity = 91.406 fm 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.
The sum rule that confines the shift to s states: infer an input with uncertainty
The sum rule says that Σₘ |⟨m|p|n⟩|²(Eₘ − Eₙ), the weight multiplying ln K in Bethe's shift, equals (ħ²/2)⟨n|∇²V|n⟩, which for a Coulomb potential is fixed by the electron's probability density at the nucleus. Use the supplied topic model Y_out = 52917.7*10^lr, where Y_out is SOURCE RADIUS R (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lr is Source radius, log₁₀(R/a₀), dimensionless. Keep no other adjustable input fixed. For this exercise use only -4.41563 ≤ lr ≤ -0.084375 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 = 891.32987 fm, with a bounded reading uncertainty ±133.39 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lr by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000433125, 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.
SOURCE RADIUS R against lr
Predictions from the supplied The sum rule that confines the shift to s states 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): SOURCE RADIUS R / fm
Horizontal axis (x): lr: Source radius, log₁₀(R/a₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lr: Source radius, log₁₀(R/a₀) | SOURCE RADIUS R / fm |
|---|---|
| -4.415625 | 2.032244514068117 |
| -3.3328125 | 24.591694776190057 |
| -2.2500000000000004 | 297.57809544025304 |
| -1.1671875000000007 | 3600.9198915231354 |
| -0.08437500000000053 | 43573.85259148016 |
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(lr) = 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(-4.41563) = 2.03224 fm and Y_out(-0.084375) = 43573.9 fm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lr_line = -4.41563 + (891.33 − 2.03224)(-0.084375 − -4.41563)/(43573.9 − 2.03224) = -4.32722. 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 lr = -1.77356; substituting back gives Y_out = 891.32987 fm.
- The interpolation error in the input is |lr_line − lr| = 2.55366. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 757.93987 fm and 1024.7199 fm. Sort the roots even if the response decreases: -1.84397 ≤ lr ≤ -1.713.
- The lower and upper input deviations are 0.0704037 and 0.0605667. 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 lr ≈ -1.77356; compatible interval [-1.84397, -1.713]. Endpoint interpolation gives -4.32722.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.