University Quantum Field Theory I · topic UQFTI-03.06
Feynman rules for the scalar Yukawa model: arrows for psi number
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.
Scalar Yukawa Feynman rules: arrows for ψ number: predict a finite change
The scalar Yukawa Feynman rules turn each Wick contraction of −gψ†ψφ theory into a diagram: −ig at every vertex where one arrow enters, one leaves and one φ line attaches, i/(p² − M² + iε) along each arrow and i/(p² − m² + iε) on each φ line. Use the supplied topic model Y_out = p0^2-q^2-mpsi^2, where Y_out is OFF SHELL BY p² − M² (GeV²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p0 is Energy p⁰ carried along the arrow, measured in GeV. Keep q (Spatial momentum |p|) = 0.7 GeV; mpsi (ψ mass M) = 1.6 GeV fixed. For this exercise use only -2.63813 GeV ≤ p0 ≤ -0.175875 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 p0 = -2.09643 GeV and -1.75171 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
OFF SHELL BY p² − M² against p0
Predictions from the supplied Scalar Yukawa Feynman rules: arrows for ψ number 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): OFF SHELL BY p² − M² / GeV²
Horizontal axis (x): p0: Energy p⁰ carried along the arrow / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p0: Energy p⁰ carried along the arrow / GeV | OFF SHELL BY p² − M² / GeV² |
|---|---|
| -2.638125 | 3.909703515625 |
| -2.0225625000000003 | 1.0407590664062512 |
| -1.407 | -1.0703510000000005 |
| -0.7914375 | -2.4236266835937506 |
| -0.175875 | -3.0190679843750003 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p0 = -2.09643 GeV, substitute into Y_out = p0^2-q^2-mpsi^2: Y₀ = 1.34502 GeV².
- At p0 = -1.75171 GeV, the same fixed-parameter model gives Y₁ = 0.0185054 GeV².
- Signed change: ΔY_out = Y₁ − Y₀ = -1.32651 GeV²; the input change is Δp0 = 0.344715 GeV.
- Average sensitivity is ΔY_out/Δp0 = -3.84814 GeV² per GeV. 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 = 1.34502 GeV², Y₁ = 0.0185054 GeV²; ΔY_out = -1.32651 GeV²; average sensitivity = -3.84814 GeV² 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.
Scalar Yukawa Feynman rules: arrows for ψ number: infer an input with uncertainty
The scalar Yukawa Feynman rules turn each Wick contraction of −gψ†ψφ theory into a diagram: −ig at every vertex where one arrow enters, one leaves and one φ line attaches, i/(p² − M² + iε) along each arrow and i/(p² − m² + iε) on each φ line. Use the supplied topic model Y_out = p0^2-q^2-mpsi^2, where Y_out is OFF SHELL BY p² − M² (GeV²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p0 is Energy p⁰ carried along the arrow, measured in GeV. Keep q (Spatial momentum |p|) = 0.7 GeV; mpsi (ψ mass M) = 1.6 GeV fixed. For this exercise use only -2.63813 GeV ≤ p0 ≤ -0.175875 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 = -1.7591575 GeV², with a bounded reading uncertainty ±0.18899 GeV². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000246225 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.
OFF SHELL BY p² − M² against p0
Predictions from the supplied Scalar Yukawa Feynman rules: arrows for ψ number 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): OFF SHELL BY p² − M² / GeV²
Horizontal axis (x): p0: Energy p⁰ carried along the arrow / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p0: Energy p⁰ carried along the arrow / GeV | OFF SHELL BY p² − M² / GeV² |
|---|---|
| -2.638125 | 3.909703515625 |
| -2.0225625000000003 | 1.0407590664062512 |
| -1.407 | -1.0703510000000005 |
| -0.7914375 | -2.4236266835937506 |
| -0.175875 | -3.0190679843750003 |
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(p0) = 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(-2.63813) = 3.9097 GeV² and Y_out(-0.175875) = -3.01907 GeV². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p0_line = -2.63813 + (-1.75916 − 3.9097)(-0.175875 − -2.63813)/(-3.01907 − 3.9097) = -0.623604 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 p0 = -1.13615 GeV; substituting back gives Y_out = -1.7591575 GeV².
- The interpolation error in the input is |p0_line − p0| = 0.512548 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.9481475 GeV² and -1.5701675 GeV². Sort the roots even if the response decreases: -1.21648 GeV ≤ p0 ≤ -1.04969 GeV.
- The lower and upper input deviations are 0.0803312 GeV and 0.0864609 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 p0 ≈ -1.13615 GeV; compatible interval [-1.21648, -1.04969] GeV. Endpoint interpolation gives -0.623604 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.