University Quantum Field Theory I · topic UQFTI-03.04
Connected diagrams and the four-point function at order lambda
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.
Connected diagrams and the four-point function at order λ: predict a finite change
The connected four-point function is the four-point correlation function minus every product of lower correlation functions; in φ⁴ that means minus the three products of two-point functions, which leaves exactly the diagrams joining all four external points into one piece. Use the supplied topic model Y_out = gap/(0.197327/m*sqrt(1.25-cos(10.13546*m))), where Y_out is 2 → 2 STRETCH ÷ REST (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. m is Mass m, measured in GeV. Keep z0 (Vertex time cz⁰) = 0.25 fm; gap (Gap between the pairs c(t₃ − t₂)) = 4 fm fixed. For this exercise use only 0.174375 GeV ≤ m ≤ 0.615375 GeV 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 m = 0.271395 GeV and 0.333135 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
2 → 2 STRETCH ÷ REST against m
Predictions from the supplied Connected diagrams and the four-point function at order λ 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): 2 → 2 STRETCH ÷ REST
Horizontal axis (x): m: Mass m / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Mass m / GeV | 2 → 2 STRETCH ÷ REST |
|---|---|
| 0.174375 | 2.9402011968559982 |
| 0.284625 | 3.874744043197866 |
| 0.394875 | 5.804087093640198 |
| 0.505125 | 11.080859295633534 |
| 0.615375 | 24.89564636316062 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 0.271395 GeV, substitute into Y_out = gap/(0.197327/m*sqrt(1.25-cos(10.13546*m))): Y₀ = 3.73068.
- At m = 0.333135 GeV, the same fixed-parameter model gives Y₁ = 4.52969.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.799012; the input change is Δm = 0.06174 GeV.
- Average sensitivity is ΔY_out/Δm = 12.9416 1 per GeV. 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 = 3.73068, Y₁ = 4.52969; ΔY_out = 0.799012; average sensitivity = 12.9416 1 per GeV.
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.
Connected diagrams and the four-point function at order λ: infer an input with uncertainty
The connected four-point function is the four-point correlation function minus every product of lower correlation functions; in φ⁴ that means minus the three products of two-point functions, which leaves exactly the diagrams joining all four external points into one piece. Use the supplied topic model Y_out = gap/(0.197327/m*sqrt(1.25-cos(10.13546*m))), where Y_out is 2 → 2 STRETCH ÷ REST (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. m is Mass m, measured in GeV. Keep z0 (Vertex time cz⁰) = 0.25 fm; gap (Gap between the pairs c(t₃ − t₂)) = 4 fm fixed. For this exercise use only 0.174375 GeV ≤ m ≤ 0.615375 GeV 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 = 7.4212541, with a bounded reading uncertainty ±0.67216. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000441 GeV, 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.
2 → 2 STRETCH ÷ REST against m
Predictions from the supplied Connected diagrams and the four-point function at order λ 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): 2 → 2 STRETCH ÷ REST
Horizontal axis (x): m: Mass m / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Mass m / GeV | 2 → 2 STRETCH ÷ REST |
|---|---|
| 0.174375 | 2.9402011968559982 |
| 0.284625 | 3.874744043197866 |
| 0.394875 | 5.804087093640198 |
| 0.505125 | 11.080859295633534 |
| 0.615375 | 24.89564636316062 |
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(m) = 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(0.174375) = 2.9402 and Y_out(0.615375) = 24.8956. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 0.174375 + (7.42125 − 2.9402)(0.615375 − 0.174375)/(24.8956 − 2.9402) = 0.264382 GeV. 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 m = 0.443385 GeV; substituting back gives Y_out = 7.4212541.
- The interpolation error in the input is |m_line − m| = 0.179003 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.7490941 and 8.0934141. Sort the roots even if the response decreases: 0.425848 GeV ≤ m ≤ 0.458278 GeV.
- The lower and upper input deviations are 0.0175367 GeV and 0.0148934 GeV. 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 m ≈ 0.443385 GeV; compatible interval [0.425848, 0.458278] GeV. Endpoint interpolation gives 0.264382 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.