University Advanced Quantum Mechanics · topic UAQM-13.08
Bethe against Welton, and the ground-state shift
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.
Bethe against Welton, and the ground-state shift: predict a finite change
Bethe's mean excitation energy K₀ is the lower cutoff of the self-energy logarithm: a geometric mean of a level's transition energies weighted by |pₘₙ|²(Eₘ − Eₙ), replacing the atomic frequency Welton had to choose by hand. Use the supplied topic model Y_out = 1.08515*ln(37557.7*k/(w*z*z)), where Y_out is WELTON 1s SHIFT ÷ Z⁴ (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. w is Welton's lower cutoff ħωₘᵢₙ, measured in Z² Ry. Keep z (Nuclear charge Z) = 1; k (Upper cutoff K) = 1 mc² fixed. For this exercise use only 4.69687 Z² Ry ≤ w ≤ 35.8031 Z² Ry 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 w = 11.5402 Z² Ry and 15.8951 Z² Ry. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WELTON 1s SHIFT ÷ Z⁴ against w
Predictions from the supplied Bethe against Welton, and the ground-state shift 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): WELTON 1s SHIFT ÷ Z⁴ / GHz
Horizontal axis (x): w: Welton's lower cutoff ħωₘᵢₙ / Z² Ry
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Welton's lower cutoff ħωₘᵢₙ / Z² Ry | WELTON 1s SHIFT ÷ Z⁴ / GHz |
|---|---|
| 4.6968749999999995 | 9.751956898257752 |
| 12.4734375 | 8.692086564233177 |
| 20.249999999999996 | 8.166273431089033 |
| 28.026562499999997 | 7.813601929907504 |
| 35.803124999999994 | 7.547867729627224 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 11.5402 Z² Ry, substitute into Y_out = 1.08515*ln(37557.7*k/(w*z*z)): Y₀ = 8.77647 GHz.
- At w = 15.8951 Z² Ry, the same fixed-parameter model gives Y₁ = 8.42903 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.347434 GHz; the input change is Δw = 4.35487 Z² Ry.
- Average sensitivity is ΔY_out/Δw = -0.0797805 GHz per Z² Ry. Its negative sign means that this response decreases 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 = 8.77647 GHz, Y₁ = 8.42903 GHz; ΔY_out = -0.347434 GHz; average sensitivity = -0.0797805 GHz per Z² Ry.
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.
Bethe against Welton, and the ground-state shift: infer an input with uncertainty
Bethe's mean excitation energy K₀ is the lower cutoff of the self-energy logarithm: a geometric mean of a level's transition energies weighted by |pₘₙ|²(Eₘ − Eₙ), replacing the atomic frequency Welton had to choose by hand. Use the supplied topic model Y_out = 1.08515*ln(37557.7*k/(w*z*z)), where Y_out is WELTON 1s SHIFT ÷ Z⁴ (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. w is Welton's lower cutoff ħωₘᵢₙ, measured in Z² Ry. Keep z (Nuclear charge Z) = 1; k (Upper cutoff K) = 1 mc² fixed. For this exercise use only 4.69687 Z² Ry ≤ w ≤ 35.8031 Z² Ry 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 = 7.9968545 GHz, with a bounded reading uncertainty ±0.067348 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00311062 Z² Ry, 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.
WELTON 1s SHIFT ÷ Z⁴ against w
Predictions from the supplied Bethe against Welton, and the ground-state shift 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): WELTON 1s SHIFT ÷ Z⁴ / GHz
Horizontal axis (x): w: Welton's lower cutoff ħωₘᵢₙ / Z² Ry
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Welton's lower cutoff ħωₘᵢₙ / Z² Ry | WELTON 1s SHIFT ÷ Z⁴ / GHz |
|---|---|
| 4.6968749999999995 | 9.751956898257752 |
| 12.4734375 | 8.692086564233177 |
| 20.249999999999996 | 8.166273431089033 |
| 28.026562499999997 | 7.813601929907504 |
| 35.803124999999994 | 7.547867729627224 |
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(w) = 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(4.69687) = 9.75196 GHz and Y_out(35.8031) = 7.54787 GHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 4.69687 + (7.99685 − 9.75196)(35.8031 − 4.69687)/(7.54787 − 9.75196) = 29.4666 Z² Ry. 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 w = 23.6717 Z² Ry; substituting back gives Y_out = 7.9968545 GHz.
- The interpolation error in the input is |w_line − w| = 5.7949 Z² Ry. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7.9295065 GHz and 8.0642025 GHz. Sort the roots even if the response decreases: 22.2472 Z² Ry ≤ w ≤ 25.1874 Z² Ry.
- The lower and upper input deviations are 1.42448 Z² Ry and 1.51569 Z² Ry. 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 w ≈ 23.6717 Z² Ry; compatible interval [22.2472, 25.1874] Z² Ry. Endpoint interpolation gives 29.4666 Z² Ry.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.