University Quantum Field Theory I · topic UQFTI-06.07
A finite prediction: the renormalised one-loop phi phi -> phi phi amplitude
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.
A finite prediction: the one-loop φφ → φφ amplitude: predict a finite change
The renormalised one-loop φφ → φφ amplitude is −λ plus the s-, t- and u-channel bubbles, each subtracted at its own threshold momentum by the counterterm δλ, leaving a finite function of s, t and u in which no regulator appears. Use the supplied topic model Y_out = max(2*sqrt(1+1/max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2)*ln(max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)+sqrt(1+max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2))-2,0), where Y_out is u CHANNEL F(u) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm; max(a,b) selects the larger value; sind/cosd/tand use degrees. th is Scattering angle θ, measured in °. Keep lam (Coupling λ, fixed at threshold) = 2; rs (Marker energy √s / m) = 20 fixed. For this exercise use only 19.125 ° ≤ th ≤ 160.875 ° 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 th = 50.31 ° and 70.155 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
u CHANNEL F(u) against th
Predictions from the supplied A finite prediction: the one-loop φφ → φφ amplitude 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): u CHANNEL F(u)
Horizontal axis (x): th: Scattering angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Scattering angle θ / ° | u CHANNEL F(u) |
|---|---|
| 19.125 | 3.98947107880638 |
| 54.5625 | 3.7885977315331916 |
| 90 | 3.3515428731000076 |
| 125.4375 | 2.5504217614049383 |
| 160.875 | 0.9843198665641486 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of th separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At th = 50.31 °, substitute into Y_out = max(2*sqrt(1+1/max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2)*ln(max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)+sqrt(1+max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2))-2,0): Y₀ = 3.82384.
- At th = 70.155 °, the same fixed-parameter model gives Y₁ = 3.63003.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.19381; the input change is Δth = 19.845 °.
- Average sensitivity is ΔY_out/Δth = -0.0097662 1 per °. 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 = 3.82384, Y₁ = 3.63003; ΔY_out = -0.19381; average sensitivity = -0.0097662 1 per °.
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.
A finite prediction: the one-loop φφ → φφ amplitude: infer an input with uncertainty
The renormalised one-loop φφ → φφ amplitude is −λ plus the s-, t- and u-channel bubbles, each subtracted at its own threshold momentum by the counterterm δλ, leaving a finite function of s, t and u in which no regulator appears. Use the supplied topic model Y_out = max(2*sqrt(1+1/max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2)*ln(max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)+sqrt(1+max(sqrt(rs^2/4-1)*cosd(th/2),1e-6)^2))-2,0), where Y_out is u CHANNEL F(u) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm; max(a,b) selects the larger value; sind/cosd/tand use degrees. th is Scattering angle θ, measured in °. Keep lam (Coupling λ, fixed at threshold) = 2; rs (Marker energy √s / m) = 20 fixed. For this exercise use only 19.125 ° ≤ th ≤ 160.875 ° 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 = 3.0573282, with a bounded reading uncertainty ±0.13982. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.014175 °, 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.
u CHANNEL F(u) against th
Predictions from the supplied A finite prediction: the one-loop φφ → φφ amplitude 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): u CHANNEL F(u)
Horizontal axis (x): th: Scattering angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Scattering angle θ / ° | u CHANNEL F(u) |
|---|---|
| 19.125 | 3.98947107880638 |
| 54.5625 | 3.7885977315331916 |
| 90 | 3.3515428731000076 |
| 125.4375 | 2.5504217614049383 |
| 160.875 | 0.9843198665641486 |
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(th) = 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(19.125) = 3.98947 and Y_out(160.875) = 0.98432. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 19.125 + (3.05733 − 3.98947)(160.875 − 19.125)/(0.98432 − 3.98947) = 63.0933 °. 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 th = 105.593 °; substituting back gives Y_out = 3.0573282.
- The interpolation error in the input is |th_line − th| = 42.4992 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.9175082 and 3.1971482. Sort the roots even if the response decreases: 98.6831 ° ≤ th ≤ 111.782 °.
- The lower and upper input deviations are 6.90938 ° and 6.18994 °. 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 th ≈ 105.593 °; compatible interval [98.6831, 111.782] °. Endpoint interpolation gives 63.0933 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.