University Quantum Field Theory I · unit UQFTI-14
The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences
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.
The vertex function: form factors F₁(q²) and F₂(q²): infer an input with uncertainty
The form factors F₁(q²) and F₂(q²) are the two real functions of the momentum transfer that, by Lorentz invariance, parity and current conservation, fix the whole coupling of an on-shell spin-½ particle to a photon: ū(p′)[γ^μ F₁ + (iσ^μν q_ν/2m)F₂]u(p). Use the supplied topic model Y_out = (f1-pm^2*f2)^2/(1+pm^2), where Y_out is N AT 180° (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f1 is Dirac form factor F₁, dimensionless. Keep f2 (Pauli form factor F₂) = 1.79; pm (Momentum p/m) = 0.6 fixed. For this exercise use only 0.10625 ≤ f1 ≤ 0.63125 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 = 0.034912066, with a bounded reading uncertainty ±0.0052177. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000525, 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.
N AT 180° against f1
Predictions from the supplied The vertex function: form factors F₁(q²) and F₂(q²) 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): N AT 180°
Horizontal axis (x): f1: Dirac form factor F₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f1: Dirac form factor F₁ | N AT 180° |
|---|---|
| 0.10625 | 0.21294516360294122 |
| 0.2375 | 0.12174088970588234 |
| 0.36875 | 0.05586979595588233 |
| 0.5 | 0.015331882352941173 |
| 0.63125 | 0.00012714889705882343 |
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(f1) = 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(0.10625) = 0.212945 and Y_out(0.63125) = 0.000127149. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f1_line = 0.10625 + (0.0349121 − 0.212945)(0.63125 − 0.10625)/(0.000127149 − 0.212945) = 0.545439. 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 f1 = 0.4265; substituting back gives Y_out = 0.034912066.
- The interpolation error in the input is |f1_line − f1| = 0.118939. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.029694366 and 0.040129766. Sort the roots even if the response decreases: 0.410784 ≤ f1 ≤ 0.443441.
- The lower and upper input deviations are 0.0157161 and 0.0169415. 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 f1 ≈ 0.4265; compatible interval [0.410784, 0.443441]. Endpoint interpolation gives 0.545439.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The one-loop vertex: Feynman parameters and the numerator: infer an input with uncertainty
The one-loop vertex correction is the diagram with a photon exchanged between the incoming and outgoing electron, rewritten with Feynman parameters x + y + z = 1 as one denominator (ℓ² − Δ)³ and a numerator reduced to a γ^μ part and a σ^(μν)q_ν part. Use the supplied topic model Y_out = 4*ln((sqrt(q+4.000001)+sqrt(q+1e-6))/2)/sqrt((q+1e-6)*(q+4.000001)), where Y_out is F₂/(α/2π) AT μ_γ → 0 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm. q is Momentum transfer −q²/m², dimensionless. Keep n (Photon mass μ_γ = m × 10⁻ⁿ, n) = 3 fixed. For this exercise use only 0.31875 ≤ q ≤ 2.68125 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 = 0.78188381, with a bounded reading uncertainty ±0.011356. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, 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.
F₂/(α/2π) AT μ_γ → 0 against q
Predictions from the supplied The one-loop vertex: Feynman parameters and the numerator 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): F₂/(α/2π) AT μ_γ → 0
Horizontal axis (x): q: Momentum transfer −q²/m²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Momentum transfer −q²/m² | F₂/(α/2π) AT μ_γ → 0 |
|---|---|
| 0.31875 | 0.9500455261325605 |
| 0.9093749999999999 | 0.8715312488859378 |
| 1.5 | 0.8068918650697435 |
| 2.0906249999999997 | 0.7525762282959056 |
| 2.68125 | 0.7061766823452367 |
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(q) = 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(0.31875) = 0.950046 and Y_out(2.68125) = 0.706177. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 0.31875 + (0.781884 − 0.950046)(2.68125 − 0.31875)/(0.706177 − 0.950046) = 1.94783. 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 q = 1.75987; substituting back gives Y_out = 0.78188381.
- The interpolation error in the input is |q_line − q| = 0.187956. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.77052781 and 0.79323981. Sort the roots even if the response decreases: 1.63946 ≤ q ≤ 1.8845.
- The lower and upper input deviations are 0.120414 and 0.124628. 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 q ≈ 1.75987; compatible interval [1.63946, 1.8845]. Endpoint interpolation gives 1.94783.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
F₂(0) = α/2π: Schwinger's anomalous moment derived: infer an input with uncertainty
Schwinger's anomalous moment is the one-loop value of the vertex's magnetic form factor at zero momentum transfer, F₂(0) = α/2π, which raises a charged lepton's g from Dirac's 2 to 2 + α/π. Use the supplied topic model Y_out = 10^(u-(-3.6967986+3.9500859*n-0.5448673*n*n)), where Y_out is Q / m (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Momentum transfer log₁₀(Q/MeV), dimensionless. Keep n (Lepton: 1 = e, 2 = μ, 3 = τ) = 1 fixed. For this exercise use only -0.46875 ≤ u ≤ 3.46875 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 = 167.76641, with a bounded reading uncertainty ±25.065. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u 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.
Q / m against u
Predictions from the supplied F₂(0) = α/2π: Schwinger's anomalous moment derived 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): Q / m
Horizontal axis (x): u: Momentum transfer log₁₀(Q/MeV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Momentum transfer log₁₀(Q/MeV) | Q / m |
|---|---|
| -0.46875 | 0.6650127929944031 |
| 0.515625 | 6.415123181516519 |
| 1.5 | 61.88423120211609 |
| 2.484375 | 596.9734271839249 |
| 3.46875 | 5758.773533111853 |
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(u) = 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(-0.46875) = 0.665013 and Y_out(3.46875) = 5758.77. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives u_line = -0.46875 + (167.766 − 0.665013)(3.46875 − -0.46875)/(5758.77 − 0.665013) = -0.354483. 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 u = 1.93313; substituting back gives Y_out = 167.76641.
- The interpolation error in the input is |u_line − u| = 2.28761. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 142.70141 and 192.83141. Sort the roots even if the response decreases: 1.86285 ≤ u ≤ 1.9936.
- The lower and upper input deviations are 0.0702767 and 0.0604728. 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 u ≈ 1.93313; compatible interval [1.86285, 1.9936]. Endpoint interpolation gives -0.354483.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
g − 2 against experiment: a_e and the series in α/π: infer an input with uncertainty
The electron's anomaly a_e = (g − 2)/2 = F₂(0) is predicted as a series in α/π with computed coefficients, plus small terms from loops of muons, tau leptons and hadrons and from weak bosons, and is measured in a Penning trap as the ratio ν_a/ν_c of two frequencies. Use the supplied topic model Y_out = (0.0084498*(206-166.25)+0.06762*(6.737-c5)+1.6927*(1-kh))/hypot(u,0.0929), where Y_out is RUBIDIUM RESIDUAL (σ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). u is Trap uncertainty on a_e, measured in × 10⁻¹². Keep c5 (Five-loop coefficient C₅) = 6.74; kh (Hadronic term ÷ 1.693 × 10⁻¹²) = 1 fixed. For this exercise use only 0.0978125 × 10⁻¹² ≤ u ≤ 0.452187 × 10⁻¹² 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 = 1.0251659 σ, with a bounded reading uncertainty ±0.044702 σ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000354375 × 10⁻¹², 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.
RUBIDIUM RESIDUAL against u
Predictions from the supplied g − 2 against experiment: a_e and the series in α/π 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): RUBIDIUM RESIDUAL / σ
Horizontal axis (x): u: Trap uncertainty on a_e / × 10⁻¹²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Trap uncertainty on a_e / × 10⁻¹² | RUBIDIUM RESIDUAL / σ |
|---|---|
| 0.0978125 | 2.4883587383891195 |
| 0.18640625 | 1.6117137962203707 |
| 0.275 | 1.1564378189646878 |
| 0.36359374999999994 | 0.8944834763417981 |
| 0.45218749999999996 | 0.7271523955226936 |
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(u) = 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(0.0978125) = 2.48836 σ and Y_out(0.452187) = 0.727152 σ. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 0.0978125 + (1.02517 − 2.48836)(0.452187 − 0.0978125)/(0.727152 − 2.48836) = 0.392224 × 10⁻¹². 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 u = 0.313981 × 10⁻¹²; substituting back gives Y_out = 1.0251659 σ.
- The interpolation error in the input is |u_line − u| = 0.0782425 × 10⁻¹². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.9804639 σ and 1.0698679 σ. Sort the roots even if the response decreases: 0.299686 × 10⁻¹² ≤ u ≤ 0.32952 × 10⁻¹².
- The lower and upper input deviations are 0.0142948 × 10⁻¹² and 0.0155389 × 10⁻¹². 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 u ≈ 0.313981 × 10⁻¹²; compatible interval [0.299686, 0.32952] × 10⁻¹². Endpoint interpolation gives 0.392224 × 10⁻¹².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.