Skip to main content
Course and syllabus guidance

University Quantum Field Theory I · unit UQFTI-05

Cross-Sections, Decay Rates and Unitarity

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.

Extra challenging questions

A selection from this pathway. Open a topic below for its complete practice and challenge set.

Extra challenging · 01

From |M|² to a rate: squared deltas and the flux factor: infer an input with uncertainty

A cross-section is the transition rate per target particle divided by the incident flux; in relativistic normalisation it is dσ = |M|² dΠₙ/(4E₁E₂|v₁ − v₂|), the flux factor 4E₁E₂|v₁ − v₂| undoing the 2E-per-volume normalisation of the two incoming states. Use the supplied topic model Y_out = ((((p+sqrt(p^2+m1^2))/m1)^(2-2*f)-1)/(((p+sqrt(p^2+m1^2))/m1)^(2-2*f)+1)+(((p+sqrt(p^2+m1^2))/m1)^(2*f)-1)/(((p+sqrt(p^2+m1^2))/m1)^(2*f)+1)), where Y_out is |v₁ − v₂| IN THIS FRAME (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Beam momentum in the target frame p, measured in GeV. Keep m1 (Beam particle mass m₁) = 0.14 GeV; f (Observer's boost along the beam, η/Y) = 0 fixed. For this exercise use only 2.57187 GeV ≤ p ≤ 17.9281 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 = 0.99993126, with a bounded reading uncertainty ±0.0000057377. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00153562 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.

Explore the question data

|v₁ − v₂| IN THIS FRAME against p

Predictions from the supplied From |M|² to a rate: squared deltas and the flux factor 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): |v₁ − v₂| IN THIS FRAME

|v₁ − v₂| IN THIS FRAME against pPredictions from the supplied From |M|² to a rate: squared deltas and the flux factor 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. Horizontal axis: p: Beam momentum in the target frame p / GeV. Vertical axis: |v₁ − v₂| IN THIS FRAME. Exact values are available in the data table below.0.9984640.9988550.9992460.9996371.000031.9586.10410.2514.418.54xy

Horizontal axis (x): p: Beam momentum in the target frame p / GeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Beam momentum in the target frame p / GeV: 2.572|v₁ − v₂| IN THIS FRAME: 0.9985
Read the exact data as a table
|v₁ − v₂| IN THIS FRAME against p — supplied values
p: Beam momentum in the target frame p / GeV|v₁ − v₂| IN THIS FRAME
2.5718750.998521700246677
6.4109374999999990.9997616431197397
10.2499999999999980.9999067352382695
14.0890624999999990.9999506337970656
17.9281249999999980.9999695114710327

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(p) = 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
  1. Endpoint predictions are Y_out(2.57187) = 0.998522 and Y_out(17.9281) = 0.99997. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives p_line = 2.57187 + (0.999931 − 0.998522)(17.9281 − 2.57187)/(0.99997 − 0.998522) = 17.5224 GeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives p = 11.9395 GeV; substituting back gives Y_out = 0.99993126.
  4. The interpolation error in the input is |p_line − p| = 5.58292 GeV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.99992552 and 0.999937. Sort the roots even if the response decreases: 11.4703 GeV ≤ p ≤ 12.4714 GeV.
  6. The lower and upper input deviations are 0.46917 GeV and 0.531881 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 p ≈ 11.9395 GeV; compatible interval [11.4703, 12.4714] GeV. Endpoint interpolation gives 17.5224 GeV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Lorentz-invariant phase space and two-body final states: infer an input with uncertainty

Lorentz-invariant phase space dΠₙ is the measure over the momenta of n final particles, d³p/((2π)³ 2E) for each one, constrained by (2π)⁴δ⁴ of total four-momentum, that counts final states with the same weight in every inertial frame. Use the supplied topic model Y_out = sqrt(max(0,(m1+m2+q)^2-(m1+m2)^2)*max(0,(m1+m2+q)^2-(m1-m2)^2))/(2*(m1+m2+q))/(4*pi*(m1+m2+q)), where Y_out is ∫dΠ₂ DISTINCT PAIR (dimensionless). 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 release Q = √s − m₁ − m₂, measured in GeV. Keep m1 (Mass m₁ of product 1) = 0.6 GeV; m2 (Mass m₂ of product 2) = 0.35 GeV fixed. For this exercise use only 0.310312 GeV ≤ q ≤ 2.23969 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 = 0.036448378, with a bounded reading uncertainty ±0.00021267. 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 0.000192937 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.

Explore the question data

∫dΠ₂ DISTINCT PAIR against q

Predictions from the supplied Lorentz-invariant phase space and two-body final states 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): ∫dΠ₂ DISTINCT PAIR

∫dΠ₂ DISTINCT PAIR against qPredictions from the supplied Lorentz-invariant phase space and two-body final states 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. Horizontal axis: q: Energy release Q = √s − m₁ − m₂ / GeV. Vertical axis: ∫dΠ₂ DISTINCT PAIR. Exact values are available in the data table below.0.025140.028440.031750.035050.038360.23310.75411.2751.7962.317xy

Horizontal axis (x): q: Energy release Q = √s − m₁ − m₂ / GeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1q: Energy release Q = √s − m₁ − m₂ / GeV: 0.3103∫dΠ₂ DISTINCT PAIR: 0.02563
Read the exact data as a table
∫dΠ₂ DISTINCT PAIR against q — supplied values
q: Energy release Q = √s − m₁ − m₂ / GeV∫dΠ₂ DISTINCT PAIR
0.31031250.02562662415303772
0.79265624999999980.03301153201880001
1.2750.03575182521782777
1.75734374999999980.037099533534971786
2.23968749999999960.03786617489029977

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
  1. Endpoint predictions are Y_out(0.310312) = 0.0256266 and Y_out(2.23969) = 0.0378662. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives q_line = 0.310312 + (0.0364484 − 0.0256266)(2.23969 − 0.310312)/(0.0378662 − 0.0256266) = 2.01619 GeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives q = 1.48723 GeV; substituting back gives Y_out = 0.036448378.
  4. The interpolation error in the input is |q_line − q| = 0.528963 GeV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.036235708 and 0.036661048. Sort the roots even if the response decreases: 1.41575 GeV ≤ q ≤ 1.56599 GeV.
  6. The lower and upper input deviations are 0.0714853 GeV and 0.0787626 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 ≈ 1.48723 GeV; compatible interval [1.41575, 1.56599] GeV. Endpoint interpolation gives 2.01619 GeV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

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.

Explore the question data

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

LAB DECAY LENGTH against qPredictions 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. Horizontal axis: q: Energy released Q = m − 2M / GeV. Vertical axis: LAB DECAY LENGTH / fm. Exact values are available in the data table below.20.0321.8423.6625.4827.290.20540.73271.261.7872.315xy

Horizontal axis (x): q: Energy released Q = m − 2M / GeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1q: Energy released Q = m − 2M / GeV: 0.2835LAB DECAY LENGTH / fm: 27.03
Read the exact data as a table
LAB DECAY LENGTH against q — supplied values
q: Energy released Q = m − 2M / GeVLAB DECAY LENGTH / fm
0.2835000000000000327.025429905184744
0.771749999999999922.059572372297623
1.2620.95783816277657
1.7482520.518555023619562
2.236520.296751138934933

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |q_line − q| = 0.637991 GeV. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

2 → 2 cross-sections and φ⁴ in microbarns: infer an input with uncertainty

A 2 → 2 differential cross-section is the rate of scatters into a solid angle per target particle per unit incident flux; in the centre-of-mass frame it equals |M|² times |p_f|/|pᵢ|, divided by 64π² s. Use the supplied topic model Y_out = lam^2*389.4*sqrt(max(rs^2/4-mf^2,0))/(32*pi*rs^2*sqrt(rs^2/4-0.0625)), where Y_out is CROSS-SECTION AT MARKER (μb). 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 Centre-of-mass energy, measured in GeV. Keep lam (Coupling strength) = 1; mf (Final-pair mass) = 0.25 GeV fixed. For this exercise use only 1.55406 GeV ≤ rs ≤ 8.99594 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 = 0.10431513 μb, with a bounded reading uncertainty ±0.0084678 μb. 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 0.000744188 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.

Explore the question data

CROSS-SECTION AT MARKER against rs

Predictions from the supplied 2 → 2 cross-sections and φ⁴ in microbarns 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): CROSS-SECTION AT MARKER / μb

CROSS-SECTION AT MARKER against rsPredictions from the supplied 2 → 2 cross-sections and φ⁴ in microbarns 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. Horizontal axis: rs: Centre-of-mass energy / GeV. Vertical axis: CROSS-SECTION AT MARKER / μb. Exact values are available in the data table below.-0.014380.40570.82581.2461.6661.2563.2665.2757.2849.294xy

Horizontal axis (x): rs: Centre-of-mass energy / GeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1rs: Centre-of-mass energy / GeV: 1.554CROSS-SECTION AT MARKER / μb: 1.604
Read the exact data as a table
CROSS-SECTION AT MARKER against rs — supplied values
rs: Centre-of-mass energy / GeVCROSS-SECTION AT MARKER / μb
1.55406251.603832960103594
3.414531250.33222622591542206
5.2750.1392038248017431
7.135468750.07607659724276791
8.99593750.04786336600253621

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
  1. Endpoint predictions are Y_out(1.55406) = 1.60383 μb and Y_out(8.99594) = 0.0478634 μb. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives rs_line = 1.55406 + (0.104315 − 1.60383)(8.99594 − 1.55406)/(0.0478634 − 1.60383) = 8.72594 GeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives rs = 6.09361 GeV; substituting back gives Y_out = 0.10431513 μb.
  4. The interpolation error in the input is |rs_line − rs| = 2.63233 GeV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.09584733 μb and 0.11278293 μb. Sort the roots even if the response decreases: 5.86039 GeV ≤ rs ≤ 6.35709 GeV.
  6. The lower and upper input deviations are 0.233218 GeV and 0.263479 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 ≈ 6.09361 GeV; compatible interval [5.86039, 6.35709] GeV. Endpoint interpolation gives 8.72594 GeV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Choose a topic