University Quantum Field Theory I · topic UQFTI-05.08
The optical theorem at one loop: the imaginary part of the phi^4 bubble
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.
Optical theorem at one loop: the φ⁴ bubble: predict a finite change
The imaginary part of the φ⁴ bubble is the piece of the one-loop amplitude in which both internal particles are on shell; it is finite, vanishes below s = 4m², and unitarity fixes it at half of λ² integrated over two-body phase space. Use the supplied topic model Y_out = 1000*sqrt(max(0,1-4*m^2/rs^2))/(32*pi), where Y_out is Im Mₛ ÷ λ² (× 10⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. rs is Collision energy √s, measured in GeV. Keep m (Scalar mass m) = 0.5 GeV; eps (Regulator ε (→ 0 at the end)) = 0.08 GeV fixed. For this exercise use only 1.044 GeV ≤ rs ≤ 2.745 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 rs = 1.41822 GeV and 1.65636 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
Im Mₛ ÷ λ² against rs
Predictions from the supplied Optical theorem at one loop: the φ⁴ bubble 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): Im Mₛ ÷ λ² / × 10⁻³
Horizontal axis (x): rs: Collision energy √s / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Collision energy √s / GeV | Im Mₛ ÷ λ² / × 10⁻³ |
|---|---|
| 1.044 | 2.8573696947547074 |
| 1.46925 | 7.287677094166657 |
| 1.8944999999999999 | 8.448556033085378 |
| 2.31975 | 8.975476015322316 |
| 2.7449999999999997 | 9.263634828216503 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rs separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rs = 1.41822 GeV, substitute into Y_out = 1000*sqrt(max(0,1-4*m^2/rs^2))/(32*pi): Y₀ = 7.05354 × 10⁻³.
- At rs = 1.65636 GeV, the same fixed-parameter model gives Y₁ = 7.92976 × 10⁻³.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.876223 × 10⁻³; the input change is Δrs = 0.23814 GeV.
- Average sensitivity is ΔY_out/Δrs = 3.67944 × 10⁻³ 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 = 7.05354 × 10⁻³, Y₁ = 7.92976 × 10⁻³; ΔY_out = 0.876223 × 10⁻³; average sensitivity = 3.67944 × 10⁻³ 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.
Optical theorem at one loop: the φ⁴ bubble: infer an input with uncertainty
The imaginary part of the φ⁴ bubble is the piece of the one-loop amplitude in which both internal particles are on shell; it is finite, vanishes below s = 4m², and unitarity fixes it at half of λ² integrated over two-body phase space. Use the supplied topic model Y_out = 1000*sqrt(max(0,1-4*m^2/rs^2))/(32*pi), where Y_out is Im Mₛ ÷ λ² (× 10⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. rs is Collision energy √s, measured in GeV. Keep m (Scalar mass m) = 0.5 GeV; eps (Regulator ε (→ 0 at the end)) = 0.08 GeV fixed. For this exercise use only 1.044 GeV ≤ rs ≤ 2.745 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 = 8.724187 × 10⁻³, with a bounded reading uncertainty ±0.080917 × 10⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rs by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001701 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.
Im Mₛ ÷ λ² against rs
Predictions from the supplied Optical theorem at one loop: the φ⁴ bubble 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): Im Mₛ ÷ λ² / × 10⁻³
Horizontal axis (x): rs: Collision energy √s / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Collision energy √s / GeV | Im Mₛ ÷ λ² / × 10⁻³ |
|---|---|
| 1.044 | 2.8573696947547074 |
| 1.46925 | 7.287677094166657 |
| 1.8944999999999999 | 8.448556033085378 |
| 2.31975 | 8.975476015322316 |
| 2.7449999999999997 | 9.263634828216503 |
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(rs) = 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(1.044) = 2.85737 × 10⁻³ and Y_out(2.745) = 9.26363 × 10⁻³. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rs_line = 1.044 + (8.72419 − 2.85737)(2.745 − 1.044)/(9.26363 − 2.85737) = 2.60177 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 rs = 2.08161 GeV; substituting back gives Y_out = 8.724187 × 10⁻³.
- The interpolation error in the input is |rs_line − rs| = 0.520155 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.64327 × 10⁻³ and 8.805104 × 10⁻³. Sort the roots even if the response decreases: 2.02037 GeV ≤ rs ≤ 2.14944 GeV.
- The lower and upper input deviations are 0.061241 GeV and 0.0678272 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 rs ≈ 2.08161 GeV; compatible interval [2.02037, 2.14944] GeV. Endpoint interpolation gives 2.60177 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.