University Advanced Quantum Mechanics · topic UAQM-13.04
Mass renormalisation: Bethe's subtraction
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.
Mass renormalisation: Bethe's subtraction: predict a finite change
Mass renormalisation rewrites the Hamiltonian in the measured electron mass, so the bare kinetic energy p²/2m₀ becomes p²/2m plus a counterterm (δm/m)p²/2m, which removes from every level the self-energy that the free electron already carries as mass. Use the supplied topic model Y_out = (-510998.95*10^lk + d*(ln(abs(d+510998.95*10^lk)) - ln(max(abs(d),1e-6))))/1000, where Y_out is UNSUBTRACTED ∫₀^K (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; abs(x) means |x|; max(a,b) selects the larger value. lk is Cutoff, log₁₀(K/mc²), dimensionless. Keep d (Interval Δ = Eₘ − Eₙ) = 10.2 eV fixed. For this exercise use only -3.275 ≤ lk ≤ -0.125 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 = -2.582 and -2.141. 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.
UNSUBTRACTED ∫₀^K against lk
Predictions from the supplied Mass renormalisation: Bethe's subtraction 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): UNSUBTRACTED ∫₀^K / keV
Horizontal axis (x): lk: Cutoff, log₁₀(K/mc²)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lk: Cutoff, log₁₀(K/mc²) | UNSUBTRACTED ∫₀^K / keV |
|---|---|
| -3.2750000000000004 | -0.23744107157108665 |
| -2.4875000000000003 | -1.6110845720568523 |
| -1.7000000000000004 | -10.125304407657922 |
| -0.9125000000000005 | -62.41680329514624 |
| -0.12500000000000044 | -383.08770740649555 |
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 = -2.582, substitute into Y_out = (-510998.95*10^lk + d*(ln(abs(d+510998.95*10^lk)) - ln(max(abs(d),1e-6))))/1000: Y₀ = -1.28807 keV.
- At lk = -2.141, the same fixed-parameter model gives Y₁ = -3.63322 keV.
- Signed change: ΔY_out = Y₁ − Y₀ = -2.34515 keV; the input change is Δlk = 0.441.
- Average sensitivity is ΔY_out/Δlk = -5.3178 keV per unit input. 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 = -1.28807 keV, Y₁ = -3.63322 keV; ΔY_out = -2.34515 keV; average sensitivity = -5.3178 keV 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.
Mass renormalisation: Bethe's subtraction: infer an input with uncertainty
Mass renormalisation rewrites the Hamiltonian in the measured electron mass, so the bare kinetic energy p²/2m₀ becomes p²/2m plus a counterterm (δm/m)p²/2m, which removes from every level the self-energy that the free electron already carries as mass. Use the supplied topic model Y_out = (-510998.95*10^lk + d*(ln(abs(d+510998.95*10^lk)) - ln(max(abs(d),1e-6))))/1000, where Y_out is UNSUBTRACTED ∫₀^K (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; abs(x) means |x|; max(a,b) selects the larger value. lk is Cutoff, log₁₀(K/mc²), dimensionless. Keep d (Interval Δ = Eₘ − Eₙ) = 10.2 eV fixed. For this exercise use only -3.275 ≤ lk ≤ -0.125 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 = -22.563675 keV, with a bounded reading uncertainty ±3.3489 keV. 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 0.000315, 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.
UNSUBTRACTED ∫₀^K against lk
Predictions from the supplied Mass renormalisation: Bethe's subtraction 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): UNSUBTRACTED ∫₀^K / keV
Horizontal axis (x): lk: Cutoff, log₁₀(K/mc²)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lk: Cutoff, log₁₀(K/mc²) | UNSUBTRACTED ∫₀^K / keV |
|---|---|
| -3.2750000000000004 | -0.23744107157108665 |
| -2.4875000000000003 | -1.6110845720568523 |
| -1.7000000000000004 | -10.125304407657922 |
| -0.9125000000000005 | -62.41680329514624 |
| -0.12500000000000044 | -383.08770740649555 |
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(-3.275) = -0.237441 keV and Y_out(-0.125) = -383.088 keV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lk_line = -3.275 + (-22.5637 − -0.237441)(-0.125 − -3.275)/(-383.088 − -0.237441) = -3.09131. 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 = -1.3535; substituting back gives Y_out = -22.563675 keV.
- The interpolation error in the input is |lk_line − lk| = 1.73781. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -25.912575 keV and -19.214775 keV. Sort the roots even if the response decreases: -1.42305 ≤ lk ≤ -1.29357.
- The lower and upper input deviations are 0.0695486 and 0.0599294. 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 lk ≈ -1.3535; compatible interval [-1.42305, -1.29357]. Endpoint interpolation gives -3.09131.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.