University Quantum Field Theory I · topic UQFTI-12.05
Z_2: the electron's field-strength renormalisation and its infrared divergence
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 electron's Z₂ and its infrared divergence: predict a finite change
Z₂ is the electron's field-strength renormalisation, ψ₀ = √Z₂ ψ, chosen so the renormalised propagator has residue 1 at the physical mass; at one loop δ₂ = Z₂ − 1 is the slope dΣ₂/dp̸ of the self-energy at p̸ = m. Use the supplied topic model Y_out = -4*ln(10)*lg, where Y_out is IR LOG 2 ln(m²/μ_γ²) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lg is Photon mass log₁₀(μ_γ/m), dimensionless. Keep uv (Pole term L = 2/ε − γ_E + ln(4πμ²/m²)) = 4 fixed. For this exercise use only -5.575 ≤ lg ≤ -2.425 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 lg = -4.882 and -4.441. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlg across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
IR LOG 2 ln(m²/μ_γ²) against lg
Predictions from the supplied The electron's Z₂ and its infrared divergence 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): IR LOG 2 ln(m²/μ_γ²)
Horizontal axis (x): lg: Photon mass log₁₀(μ_γ/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Photon mass log₁₀(μ_γ/m) | IR LOG 2 ln(m²/μ_γ²) |
|---|---|
| -5.575 | 51.347647573767226 |
| -4.7875000000000005 | 44.094504530835984 |
| -4 | 36.841361487904734 |
| -3.2125000000000004 | 29.588218444973492 |
| -2.4250000000000003 | 22.335075402042246 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lg separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lg = -4.882, substitute into Y_out = -4*ln(10)*lg: Y₀ = 44.9649.
- At lg = -4.441, the same fixed-parameter model gives Y₁ = 40.9031.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.06176; the input change is Δlg = 0.441.
- Average sensitivity is ΔY_out/Δlg = -9.21034 1 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 = 44.9649, Y₁ = 40.9031; ΔY_out = -4.06176; average sensitivity = -9.21034 1 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 electron's Z₂ and its infrared divergence: infer an input with uncertainty
Z₂ is the electron's field-strength renormalisation, ψ₀ = √Z₂ ψ, chosen so the renormalised propagator has residue 1 at the physical mass; at one loop δ₂ = Z₂ − 1 is the slope dΣ₂/dp̸ of the self-energy at p̸ = m. Use the supplied topic model Y_out = -4*ln(10)*lg, where Y_out is IR LOG 2 ln(m²/μ_γ²) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lg is Photon mass log₁₀(μ_γ/m), dimensionless. Keep uv (Pole term L = 2/ε − γ_E + ln(4πμ²/m²)) = 4 fixed. For this exercise use only -5.575 ≤ lg ≤ -2.425 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 = 33.649979, with a bounded reading uncertainty ±1.6972. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lg 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.
IR LOG 2 ln(m²/μ_γ²) against lg
Predictions from the supplied The electron's Z₂ and its infrared divergence 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): IR LOG 2 ln(m²/μ_γ²)
Horizontal axis (x): lg: Photon mass log₁₀(μ_γ/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Photon mass log₁₀(μ_γ/m) | IR LOG 2 ln(m²/μ_γ²) |
|---|---|
| -5.575 | 51.347647573767226 |
| -4.7875000000000005 | 44.094504530835984 |
| -4 | 36.841361487904734 |
| -3.2125000000000004 | 29.588218444973492 |
| -2.4250000000000003 | 22.335075402042246 |
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(lg) = 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(-5.575) = 51.3476 and Y_out(-2.425) = 22.3351. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lg_line = -5.575 + (33.65 − 51.3476)(-2.425 − -5.575)/(22.3351 − 51.3476) = -3.6535. 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 lg = -3.6535; substituting back gives Y_out = 33.649979.
- The interpolation error in the input is |lg_line − lg| = 0. This is within the requested numerical tolerance.
- Invert both limits Y_out = 31.952779 and 35.347179. Sort the roots even if the response decreases: -3.83777 ≤ lg ≤ -3.46923.
- The lower and upper input deviations are 0.184271 and 0.184271. 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 lg ≈ -3.6535; compatible interval [-3.83777, -3.46923]. Endpoint interpolation gives -3.6535.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.