University Quantum Field Theory I · topic UQFTI-05.03
Decay rates, and the scalar's decay to psi psi-bar in seconds
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.
Decay rates, and the scalar's decay to ψψ̄ in seconds: predict a finite change
A particle's decay rate Γ is its probability per unit time of turning into the final states open to it, Γ = (1/(2E))∫|ℳ|² dΠₙ summed over channels, and at rest its mean life is τ = ħ/Γ. Use the supplied topic model Y_out = 0.197327*16*pi*p/max(sqrt(1-4*mpsi^2/(2*mpsi+q)^2),0.01), where Y_out is LAB DECAY LENGTH (fm). 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. q is Energy released Q = m − 2M, measured in GeV. Keep mpsi (ψ mass M) = 0.3 GeV; p (Lab momentum of φ, p) = 2 GeV fixed. For this exercise use only 0.2835 GeV ≤ q ≤ 2.2365 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 q = 0.71316 GeV and 0.98658 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δq across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LAB DECAY LENGTH against q
Predictions from the supplied Decay rates, and the scalar's decay to ψψ̄ in seconds 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): LAB DECAY LENGTH / fm
Horizontal axis (x): q: Energy released Q = m − 2M / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Energy released Q = m − 2M / GeV | LAB DECAY LENGTH / fm |
|---|---|
| 0.28350000000000003 | 27.025429905184744 |
| 0.7717499999999999 | 22.059572372297623 |
| 1.26 | 20.95783816277657 |
| 1.74825 | 20.518555023619562 |
| 2.2365 | 20.296751138934933 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of q separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At q = 0.71316 GeV, substitute into Y_out = 0.197327*16*pi*p/max(sqrt(1-4*mpsi^2/(2*mpsi+q)^2),0.01): Y₀ = 22.3015 fm.
- At q = 0.98658 GeV, the same fixed-parameter model gives Y₁ = 21.4289 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.87266 fm; the input change is Δq = 0.27342 GeV.
- Average sensitivity is ΔY_out/Δq = -3.19165 fm 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 = 22.3015 fm, Y₁ = 21.4289 fm; ΔY_out = -0.87266 fm; average sensitivity = -3.19165 fm 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.
Decay rates, and the scalar's decay to ψψ̄ in seconds: infer an input with uncertainty
A particle's decay rate Γ is its probability per unit time of turning into the final states open to it, Γ = (1/(2E))∫|ℳ|² dΠₙ summed over channels, and at rest its mean life is τ = ħ/Γ. Use the supplied topic model Y_out = 0.197327*16*pi*p/max(sqrt(1-4*mpsi^2/(2*mpsi+q)^2),0.01), where Y_out is LAB DECAY LENGTH (fm). 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. q is Energy released Q = m − 2M, measured in GeV. Keep mpsi (ψ mass M) = 0.3 GeV; p (Lab momentum of φ, p) = 2 GeV fixed. For this exercise use only 0.2835 GeV ≤ q ≤ 2.2365 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 = 20.722866 fm, with a bounded reading uncertainty ±0.063917 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001953 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.
LAB DECAY LENGTH against q
Predictions from the supplied Decay rates, and the scalar's decay to ψψ̄ in seconds 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): LAB DECAY LENGTH / fm
Horizontal axis (x): q: Energy released Q = m − 2M / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Energy released Q = m − 2M / GeV | LAB DECAY LENGTH / fm |
|---|---|
| 0.28350000000000003 | 27.025429905184744 |
| 0.7717499999999999 | 22.059572372297623 |
| 1.26 | 20.95783816277657 |
| 1.74825 | 20.518555023619562 |
| 2.2365 | 20.296751138934933 |
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(q) = 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(0.2835) = 27.0254 fm and Y_out(2.2365) = 20.2968 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 0.2835 + (20.7229 − 27.0254)(2.2365 − 0.2835)/(20.2968 − 27.0254) = 2.11282 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 q = 1.47483 GeV; substituting back gives Y_out = 20.722866 fm.
- The interpolation error in the input is |q_line − q| = 0.637991 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.658949 fm and 20.786783 fm. Sort the roots even if the response decreases: 1.40836 GeV ≤ q ≤ 1.54909 GeV.
- The lower and upper input deviations are 0.0664677 GeV and 0.074257 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 q ≈ 1.47483 GeV; compatible interval [1.40836, 1.54909] GeV. Endpoint interpolation gives 2.11282 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.