University Quantum Field Theory I · topic UQFTI-03.07
Reading Feynman rules off a Lagrangian, with derivative couplings and two-point vertices
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.
Reading Feynman rules off a Lagrangian: predict a finite change
Reading Feynman rules off a Lagrangian assigns each term of ℒᵢₙₜ a vertex: i times its coefficient, times the number of ways of attaching identical lines, with −ip_μ for every derivative on a line carrying momentum p into the vertex. Use the supplied topic model Y_out = s*dz-dm, where Y_out is VERTEX ÷ i AT PROBE (m²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Probe momentum p², measured in m². Keep dz (Field counterterm δ_Z) = 0.3; dm (Mass counterterm δₘ) = 0.1 m² fixed. For this exercise use only -0.628125 m² ≤ s ≤ 2.12812 m² 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 s = -0.02175 m² and 0.364125 m². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
VERTEX ÷ i AT PROBE against s
Predictions from the supplied Reading Feynman rules off a Lagrangian 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): VERTEX ÷ i AT PROBE / m²
Horizontal axis (x): s: Probe momentum p² / m²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Probe momentum p² / m² | VERTEX ÷ i AT PROBE / m² |
|---|---|
| -0.628125 | -0.2884375 |
| 0.06093749999999987 | -0.08171875000000005 |
| 0.7499999999999998 | 0.12499999999999992 |
| 1.4390624999999997 | 0.33171874999999984 |
| 2.128125 | 0.5384374999999999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = -0.02175 m², substitute into Y_out = s*dz-dm: Y₀ = -0.106525 m².
- At s = 0.364125 m², the same fixed-parameter model gives Y₁ = 0.0092375 m².
- Signed change: ΔY_out = Y₁ − Y₀ = 0.115762 m²; the input change is Δs = 0.385875 m².
- Average sensitivity is ΔY_out/Δs = 0.3 m² per m². 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 = -0.106525 m², Y₁ = 0.0092375 m²; ΔY_out = 0.115762 m²; average sensitivity = 0.3 m² per m².
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.
Reading Feynman rules off a Lagrangian: infer an input with uncertainty
Reading Feynman rules off a Lagrangian assigns each term of ℒᵢₙₜ a vertex: i times its coefficient, times the number of ways of attaching identical lines, with −ip_μ for every derivative on a line carrying momentum p into the vertex. Use the supplied topic model Y_out = s*dz-dm, where Y_out is VERTEX ÷ i AT PROBE (m²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Probe momentum p², measured in m². Keep dz (Field counterterm δ_Z) = 0.3; dm (Mass counterterm δₘ) = 0.1 m² fixed. For this exercise use only -0.628125 m² ≤ s ≤ 2.12812 m² 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 = 0.21595625 m², with a bounded reading uncertainty ±0.048372 m². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625 m², 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.
VERTEX ÷ i AT PROBE against s
Predictions from the supplied Reading Feynman rules off a Lagrangian 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): VERTEX ÷ i AT PROBE / m²
Horizontal axis (x): s: Probe momentum p² / m²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Probe momentum p² / m² | VERTEX ÷ i AT PROBE / m² |
|---|---|
| -0.628125 | -0.2884375 |
| 0.06093749999999987 | -0.08171875000000005 |
| 0.7499999999999998 | 0.12499999999999992 |
| 1.4390624999999997 | 0.33171874999999984 |
| 2.128125 | 0.5384374999999999 |
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(s) = 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(-0.628125) = -0.288438 m² and Y_out(2.12812) = 0.538437 m². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = -0.628125 + (0.215956 − -0.288438)(2.12812 − -0.628125)/(0.538437 − -0.288438) = 1.05319 m². 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 s = 1.05319 m²; substituting back gives Y_out = 0.21595625 m².
- The interpolation error in the input is |s_line − s| = 2.22045e-16 m². This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.16758425 m² and 0.26432825 m². Sort the roots even if the response decreases: 0.891948 m² ≤ s ≤ 1.21443 m².
- The lower and upper input deviations are 0.16124 m² and 0.16124 m². 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 s ≈ 1.05319 m²; compatible interval [0.891948, 1.21443] m². Endpoint interpolation gives 1.05319 m².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.