University Quantum Field Theory I · topic UQFTI-15.06
Bethe's low-energy part with a photon-energy cutoff K, quoted
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's low-energy part with a photon cutoff K: predict a finite change
Bethe's low-energy part is the level shift a bound electron gets from transverse photons below a cutoff K, treated non-relativistically in the dipole approximation; on s states it grows as ln K, at a rate set by the Laplacian of the potential where the electron sits. Use the supplied topic model Y_out = 135.644*(2.302585*lk-2*ln(z)-5.422258), where Y_out is 2s LOW PART ÷ Z⁴ (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lk is Cutoff K, as log₁₀(K/1 eV), dimensionless. Keep z (Nuclear charge Z) = 1 fixed. For this exercise use only 1.53125 ≤ lk ≤ 5.46875 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 lk = 2.3975 and 2.94875. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
2s LOW PART ÷ Z⁴ against lk
Predictions from the supplied Bethe's low-energy part with a photon cutoff K 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): 2s LOW PART ÷ Z⁴ / MHz
Horizontal axis (x): lk: Cutoff K, as log₁₀(K/1 eV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lk: Cutoff K, as log₁₀(K/1 eV) | 2s LOW PART ÷ Z⁴ / MHz |
|---|---|
| 1.53125 | -257.238634550125 |
| 2.515625 | 50.21302019393746 |
| 3.5 | 357.664674938 |
| 4.484375 | 665.1163296820625 |
| 5.46875 | 972.5679844261251 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lk separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lk = 2.3975, substitute into Y_out = 135.644*(2.302585*lk-2*ln(z)-5.422258): Y₀ = 13.3188 MHz.
- At lk = 2.94875, the same fixed-parameter model gives Y₁ = 185.492 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = 172.173 MHz; the input change is Δlk = 0.55125.
- Average sensitivity is ΔY_out/Δlk = 312.332 MHz 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 = 13.3188 MHz, Y₁ = 185.492 MHz; ΔY_out = 172.173 MHz; average sensitivity = 312.332 MHz 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.
Bethe's low-energy part with a photon cutoff K: infer an input with uncertainty
Bethe's low-energy part is the level shift a bound electron gets from transverse photons below a cutoff K, treated non-relativistically in the dipole approximation; on s states it grows as ln K, at a rate set by the Laplacian of the potential where the electron sits. Use the supplied topic model Y_out = 135.644*(2.302585*lk-2*ln(z)-5.422258), where Y_out is 2s LOW PART ÷ Z⁴ (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lk is Cutoff K, as log₁₀(K/1 eV), dimensionless. Keep z (Nuclear charge Z) = 1 fixed. For this exercise use only 1.53125 ≤ lk ≤ 5.46875 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 = 492.9434 MHz, with a bounded reading uncertainty ±71.944 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lk 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.
2s LOW PART ÷ Z⁴ against lk
Predictions from the supplied Bethe's low-energy part with a photon cutoff K 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): 2s LOW PART ÷ Z⁴ / MHz
Horizontal axis (x): lk: Cutoff K, as log₁₀(K/1 eV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lk: Cutoff K, as log₁₀(K/1 eV) | 2s LOW PART ÷ Z⁴ / MHz |
|---|---|
| 1.53125 | -257.238634550125 |
| 2.515625 | 50.21302019393746 |
| 3.5 | 357.664674938 |
| 4.484375 | 665.1163296820625 |
| 5.46875 | 972.5679844261251 |
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(lk) = 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(1.53125) = -257.239 MHz and Y_out(5.46875) = 972.568 MHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lk_line = 1.53125 + (492.943 − -257.239)(5.46875 − 1.53125)/(972.568 − -257.239) = 3.93312. 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 lk = 3.93312; substituting back gives Y_out = 492.9434 MHz.
- The interpolation error in the input is |lk_line − lk| = 4.44089e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 420.9994 MHz and 564.8874 MHz. Sort the roots even if the response decreases: 3.70278 ≤ lk ≤ 4.16347.
- The lower and upper input deviations are 0.230345 and 0.230345. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer lk ≈ 3.93312; compatible interval [3.70278, 4.16347]. Endpoint interpolation gives 3.93312.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.