University Quantum Field Theory I · topic UQFTI-03.05
Momentum-space Feynman rules: conservation at every vertex and one integral per loop
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.
Momentum-space Feynman rules and loop counting: predict a finite change
Momentum-space Feynman rules evaluate a diagram by giving each line a propagator i/(p² − m² + iε) in its own four-momentum, each vertex −iλ with four-momentum conserved, and each of the L = I − V + 1 momenta conservation leaves free an integral ∫d⁴l/(2π)⁴. Use the supplied topic model Y_out = 2*pi/tw, where Y_out is FIRST ZEROS AT ±2π/T (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. tw is Vertex time window T, measured in GeV⁻¹. Keep q (Energy mismatch q⁰ at the marker) = 0.3 GeV fixed. For this exercise use only 7.125 GeV⁻¹ ≤ tw ≤ 22.875 GeV⁻¹ 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 tw = 10.59 GeV⁻¹ and 12.795 GeV⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIRST ZEROS AT ±2π/T against tw
Predictions from the supplied Momentum-space Feynman rules and loop counting 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): FIRST ZEROS AT ±2π/T / GeV
Horizontal axis (x): tw: Vertex time window T / GeV⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tw: Vertex time window T / GeV⁻¹ | FIRST ZEROS AT ±2π/T / GeV |
|---|---|
| 7.125 | 0.8818505694287139 |
| 11.0625 | 0.567971553191375 |
| 15 | 0.41887902047863906 |
| 18.9375 | 0.3317853627553577 |
| 22.875 | 0.27467476752697645 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of tw separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At tw = 10.59 GeV⁻¹, substitute into Y_out = 2*pi/tw: Y₀ = 0.593313 GeV.
- At tw = 12.795 GeV⁻¹, the same fixed-parameter model gives Y₁ = 0.491066 GeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.102247 GeV; the input change is Δtw = 2.205 GeV⁻¹.
- Average sensitivity is ΔY_out/Δtw = -0.0463707 GeV per GeV⁻¹. 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 = 0.593313 GeV, Y₁ = 0.491066 GeV; ΔY_out = -0.102247 GeV; average sensitivity = -0.0463707 GeV per GeV⁻¹.
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.
Momentum-space Feynman rules and loop counting: infer an input with uncertainty
Momentum-space Feynman rules evaluate a diagram by giving each line a propagator i/(p² − m² + iε) in its own four-momentum, each vertex −iλ with four-momentum conserved, and each of the L = I − V + 1 momenta conservation leaves free an integral ∫d⁴l/(2π)⁴. Use the supplied topic model Y_out = 2*pi/tw, where Y_out is FIRST ZEROS AT ±2π/T (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. tw is Vertex time window T, measured in GeV⁻¹. Keep q (Energy mismatch q⁰ at the marker) = 0.3 GeV fixed. For this exercise use only 7.125 GeV⁻¹ ≤ tw ≤ 22.875 GeV⁻¹ 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 = 0.37550786 GeV, with a bounded reading uncertainty ±0.015125 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 GeV⁻¹, 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.
FIRST ZEROS AT ±2π/T against tw
Predictions from the supplied Momentum-space Feynman rules and loop counting 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): FIRST ZEROS AT ±2π/T / GeV
Horizontal axis (x): tw: Vertex time window T / GeV⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tw: Vertex time window T / GeV⁻¹ | FIRST ZEROS AT ±2π/T / GeV |
|---|---|
| 7.125 | 0.8818505694287139 |
| 11.0625 | 0.567971553191375 |
| 15 | 0.41887902047863906 |
| 18.9375 | 0.3317853627553577 |
| 22.875 | 0.27467476752697645 |
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(tw) = 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(7.125) = 0.881851 GeV and Y_out(22.875) = 0.274675 GeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tw_line = 7.125 + (0.375508 − 0.881851)(22.875 − 7.125)/(0.274675 − 0.881851) = 20.2594 GeV⁻¹. 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 tw = 16.7325 GeV⁻¹; substituting back gives Y_out = 0.37550786 GeV.
- The interpolation error in the input is |tw_line − tw| = 3.52691 GeV⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.36038286 GeV and 0.39063286 GeV. Sort the roots even if the response decreases: 16.0846 GeV⁻¹ ≤ tw ≤ 17.4348 GeV⁻¹.
- The lower and upper input deviations are 0.647869 GeV⁻¹ and 0.702251 GeV⁻¹. 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 tw ≈ 16.7325 GeV⁻¹; compatible interval [16.0846, 17.4348] GeV⁻¹. Endpoint interpolation gives 20.2594 GeV⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.