University Advanced Quantum Mechanics · unit UAQM-13
Self-Energy and the Lamb 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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Bound electron's second-order energy in the quantised field: infer an input with uncertainty
The second-order energy of a bound state |n⟩ in the quantised field is the shift δEₙ from virtual emission and reabsorption: the electron emits a photon into any mode, passes through any atomic state m, and takes the photon back, each path weighted by its squared coupling over Eₙ − Eₘ − ħω. Use the supplied topic model Y_out = -ep/(dm+ep), where Y_out is INTEGRAND AT E (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dm is Eₘ − Eₙ, intermediate state, measured in eV. Keep ep (Photon energy E) = 5.25 eV fixed. For this exercise use only -4.41 eV ≤ dm ≤ 9.45 eV 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 = -0.5648441, with a bounded reading uncertainty ±0.031155. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dm by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001386 eV, 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.
INTEGRAND AT E against dm
Predictions from the supplied Bound electron's second-order energy in the quantised field 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): INTEGRAND AT E
Horizontal axis (x): dm: Eₘ − Eₙ, intermediate state / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dm: Eₘ − Eₙ, intermediate state / eV | INTEGRAND AT E |
|---|---|
| -4.41 | -6.250000000000001 |
| -0.9450000000000003 | -1.2195121951219512 |
| 2.5199999999999996 | -0.6756756756756757 |
| 5.984999999999999 | -0.4672897196261682 |
| 9.45 | -0.35714285714285715 |
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(dm) = 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(-4.41) = -6.25 and Y_out(9.45) = -0.357143. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives dm_line = -4.41 + (-0.564844 − -6.25)(9.45 − -4.41)/(-0.357143 − -6.25) = 8.96149 eV. 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 dm = 4.0446 eV; substituting back gives Y_out = -0.5648441.
- The interpolation error in the input is |dm_line − dm| = 4.91689 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.5959991 and -0.5336891. Sort the roots even if the response decreases: 3.55874 eV ≤ dm ≤ 4.58719 eV.
- The lower and upper input deviations are 0.485862 eV and 0.542588 eV. 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 dm ≈ 4.0446 eV; compatible interval [3.55874, 4.58719] eV. Endpoint interpolation gives 8.96149 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
The free electron's self-energy as an electromagnetic mass: infer an input with uncertainty
The electromagnetic mass δm is the part of an electron's inertia supplied by its own field: coupling to transverse photons lowers the energy in proportion to p², so it changes the kinetic energy exactly as a larger mass would. Use the supplied topic model Y_out = 100*4*0.0072973525/(3*pi)*10^lk, where Y_out is DIPOLE MODEL δm/m (%). 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 no other adjustable input fixed. For this exercise use only -0.575 ≤ lk ≤ 2.575 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 = 6.8778736 %, with a bounded reading uncertainty ±1.0193 %. 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.
DIPOLE MODEL δm/m against lk
Predictions from the supplied The free electron's self-energy as an electromagnetic mass 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): DIPOLE MODEL δm/m / %
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) | DIPOLE MODEL δm/m / % |
|---|---|
| -0.575 | 0.08240511871051219 |
| 0.2124999999999999 | 0.5051893532827305 |
| 0.9999999999999998 | 3.097092591623999 |
| 1.7874999999999999 | 18.986905521193922 |
| 2.5749999999999993 | 116.40032404769329 |
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(-0.575) = 0.0824051 % and Y_out(2.575) = 116.4 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lk_line = -0.575 + (6.87787 − 0.0824051)(2.575 − -0.575)/(116.4 − 0.0824051) = -0.390972. 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 = 1.3465; substituting back gives Y_out = 6.8778736 %.
- The interpolation error in the input is |lk_line − lk| = 1.73747. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.8585736 % and 7.8971736 %. Sort the roots even if the response decreases: 1.27684 ≤ lk ≤ 1.40652.
- The lower and upper input deviations are 0.0696623 and 0.0600175. 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 ≈ 1.3465; compatible interval [1.27684, 1.40652]. Endpoint interpolation gives -0.390972.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.