University Advanced Quantum Mechanics · topic UAQM-13.02
The divergence of the bound self-energy
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 divergence of the bound self-energy: predict a finite change
The linear divergence of the bound self-energy is the part of a level's second-order energy in the vacuum field that grows in proportion to the photon cutoff K: −(4αK/(3πmc²)) times the level's mean kinetic energy, the same for every state with the same ⟨T⟩. Use the supplied topic model Y_out = 510999*10^lk, where Y_out is CUTOFF K (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lk is Cutoff log₁₀(K/mc²), dimensionless. Keep w (Bound-state share of ⟨p²⟩, w) = 0.31; dc (Continuum group's energy) = 39 eV fixed. For this exercise use only -4.41563 ≤ lk ≤ -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 lk = -3.46275 and -2.85638. 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.
CUTOFF K against lk
Predictions from the supplied The divergence of the bound self-energy 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): CUTOFF K / eV
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) | CUTOFF K / eV |
|---|---|
| -4.415625 | 19.624339577198057 |
| -3.3328125 | 237.46934275182676 |
| -2.2500000000000004 | 2873.5585483094287 |
| -1.1671875000000007 | 34772.23053247648 |
| -0.08437500000000053 | 420770.2734698177 |
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 = -3.46275, substitute into Y_out = 510999*10^lk: Y₀ = 176.064 eV.
- At lk = -2.85638, the same fixed-parameter model gives Y₁ = 711.287 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 535.224 eV; the input change is Δlk = 0.606375.
- Average sensitivity is ΔY_out/Δlk = 882.661 eV 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 = 176.064 eV, Y₁ = 711.287 eV; ΔY_out = 535.224 eV; average sensitivity = 882.661 eV 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 divergence of the bound self-energy: infer an input with uncertainty
The linear divergence of the bound self-energy is the part of a level's second-order energy in the vacuum field that grows in proportion to the photon cutoff K: −(4αK/(3πmc²)) times the level's mean kinetic energy, the same for every state with the same ⟨T⟩. Use the supplied topic model Y_out = 510999*10^lk, where Y_out is CUTOFF K (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lk is Cutoff log₁₀(K/mc²), dimensionless. Keep w (Bound-state share of ⟨p²⟩, w) = 0.31; dc (Continuum group's energy) = 39 eV fixed. For this exercise use only -4.41563 ≤ lk ≤ -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 = 8607.1139 eV, with a bounded reading uncertainty ±1288.1 eV. 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 0.000433125, 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.
CUTOFF K against lk
Predictions from the supplied The divergence of the bound self-energy 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): CUTOFF K / eV
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) | CUTOFF K / eV |
|---|---|
| -4.415625 | 19.624339577198057 |
| -3.3328125 | 237.46934275182676 |
| -2.2500000000000004 | 2873.5585483094287 |
| -1.1671875000000007 | 34772.23053247648 |
| -0.08437500000000053 | 420770.2734698177 |
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(-4.41563) = 19.6243 eV and Y_out(-0.084375) = 420770 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lk_line = -4.41563 + (8607.11 − 19.6243)(-0.084375 − -4.41563)/(420770 − 19.6243) = -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 lk = -1.77356; substituting back gives Y_out = 8607.1139 eV.
- The interpolation error in the input is |lk_line − lk| = 2.55366. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7319.0139 eV and 9895.2139 eV. Sort the roots even if the response decreases: -1.84397 ≤ lk ≤ -1.71299.
- The lower and upper input deviations are 0.070405 and 0.0605676. 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 ≈ -1.77356; compatible interval [-1.84397, -1.71299]. Endpoint interpolation gives -4.32722.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.