University Advanced Quantum Mechanics · topic UAQM-13.07
The 2s shift computed: about 1040 MHz
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 2s shift computed: about 1040 MHz: predict a finite change
Bethe's 2s shift is the rise of hydrogen's 2S₁/₂ level obtained by evaluating the mass-subtracted self-energy ΔE = (α⁵mc²/(6π)) ln(K/K₀) with the photon cutoff K set at mc² and the mean excitation energy K₀ taken from hydrogen's spectrum. Use the supplied topic model Y_out = 100*ln(2)/ln(kc*37557.73/k0), where Y_out is SPREAD FROM K ×2 OR ÷2 (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. kc is Photon cutoff K, measured in mc². Keep k0 (Mean excitation energy K₀) = 16.64 Ry fixed. For this exercise use only 0.648438 mc² ≤ kc ≤ 3.60156 mc² 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 kc = 1.29812 mc² and 1.71156 mc². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δkc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPREAD FROM K ×2 OR ÷2 against kc
Predictions from the supplied The 2s shift computed: about 1040 MHz 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 FROM K ×2 OR ÷2 / %
Horizontal axis (x): kc: Photon cutoff K / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kc: Photon cutoff K / mc² | SPREAD FROM K ×2 OR ÷2 / % |
|---|---|
| 0.6484375 | 9.509971924810968 |
| 1.38671875 | 8.611844745475135 |
| 2.125 | 8.178151838254196 |
| 2.8632812499999996 | 7.900199554812575 |
| 3.6015624999999996 | 7.698904077588968 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of kc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At kc = 1.29812 mc², substitute into Y_out = 100*ln(2)/ln(kc*37557.73/k0): Y₀ = 8.68307 %.
- At kc = 1.71156 mc², the same fixed-parameter model gives Y₁ = 8.39239 %.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.290674 %; the input change is Δkc = 0.413438 mc².
- Average sensitivity is ΔY_out/Δkc = -0.703067 % per mc². 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 = 8.68307 %, Y₁ = 8.39239 %; ΔY_out = -0.290674 %; average sensitivity = -0.703067 % per mc².
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 2s shift computed: about 1040 MHz: infer an input with uncertainty
Bethe's 2s shift is the rise of hydrogen's 2S₁/₂ level obtained by evaluating the mass-subtracted self-energy ΔE = (α⁵mc²/(6π)) ln(K/K₀) with the photon cutoff K set at mc² and the mean excitation energy K₀ taken from hydrogen's spectrum. Use the supplied topic model Y_out = 100*ln(2)/ln(kc*37557.73/k0), where Y_out is SPREAD FROM K ×2 OR ÷2 (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. kc is Photon cutoff K, measured in mc². Keep k0 (Mean excitation energy K₀) = 16.64 Ry fixed. For this exercise use only 0.648438 mc² ≤ kc ≤ 3.60156 mc² 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 = 8.0431574 %, with a bounded reading uncertainty ±0.051638 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000295312 mc², 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 FROM K ×2 OR ÷2 against kc
Predictions from the supplied The 2s shift computed: about 1040 MHz 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 FROM K ×2 OR ÷2 / %
Horizontal axis (x): kc: Photon cutoff K / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kc: Photon cutoff K / mc² | SPREAD FROM K ×2 OR ÷2 / % |
|---|---|
| 0.6484375 | 9.509971924810968 |
| 1.38671875 | 8.611844745475135 |
| 2.125 | 8.178151838254196 |
| 2.8632812499999996 | 7.900199554812575 |
| 3.6015624999999996 | 7.698904077588968 |
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(kc) = 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(0.648438) = 9.50997 % and Y_out(3.60156) = 7.6989 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives kc_line = 0.648438 + (8.04316 − 9.50997)(3.60156 − 0.648438)/(7.6989 − 9.50997) = 3.04022 mc². 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 kc = 2.44984 mc²; substituting back gives Y_out = 8.0431574 %.
- The interpolation error in the input is |kc_line − kc| = 0.59038 mc². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7.9915194 % and 8.0947954 %. Sort the roots even if the response decreases: 2.3188 mc² ≤ kc ≤ 2.59013 mc².
- The lower and upper input deviations are 0.131044 mc² and 0.14029 mc². 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 kc ≈ 2.44984 mc²; compatible interval [2.3188, 2.59013] mc². Endpoint interpolation gives 3.04022 mc².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.