University Quantum Field Theory I · unit UQFTI-01
Natural Units, Relativistic States and the Feynman Propagator
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.
Natural units, the metric, and back to seconds and barns: infer an input with uncertainty
Natural units set ħ = c = 1, so energy, momentum and mass share the unit GeV while lengths and times are measured in GeV⁻¹; ħc = 197.327 MeV fm and ħ = 6.582 × 10⁻²⁵ GeV s convert back to SI. Use the supplied topic model Y_out = -24.181634-log(a)-k, where Y_out is log₁₀(time/s) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. a is Mantissa a in E = a × 10ᵏ GeV, dimensionless. Keep k (Power of ten k) = -1 fixed. For this exercise use only 1.95519 ≤ a ≤ 9.03481 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 = -23.979162, with a bounded reading uncertainty ±0.023759. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000707962, 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.
log₁₀(time/s) against a
Predictions from the supplied Natural units, the metric, and back to seconds and barns 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): log₁₀(time/s)
Horizontal axis (x): a: Mantissa a in E = a × 10ᵏ GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Mantissa a in E = a × 10ᵏ GeV | log₁₀(time/s) |
|---|---|
| 1.9551875 | -23.47282241201869 |
| 3.72509375 | -23.752771207177016 |
| 5.494999999999999 | -23.92160169675951 |
| 7.264906249999999 | -24.042864014316024 |
| 9.0348125 | -24.137553144051314 |
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(a) = 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(1.95519) = -23.4728 and Y_out(9.03481) = -24.1376. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 1.95519 + (-23.9792 − -23.4728)(9.03481 − 1.95519)/(-24.1376 − -23.4728) = 7.34789. 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 a = 6.27376; substituting back gives Y_out = -23.979162.
- The interpolation error in the input is |a_line − a| = 1.07413. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -24.002921 and -23.955403. Sort the roots even if the response decreases: 5.93976 ≤ a ≤ 6.62654.
- The lower and upper input deviations are 0.334 and 0.352781. 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 a ≈ 6.27376; compatible interval [5.93976, 6.62654]. Endpoint interpolation gives 7.34789.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
What Advanced QM supplies, quoted and restated: infer an input with uncertainty
Advanced QM's free-field toolkit is the set of exact results about non-interacting quantum fields, their modes, vacuum amplitude, commutator, spin sums, statistics and Coulomb-gauge photons, that this course quotes by topic code and builds interactions on without re-deriving. Use the supplied topic model Y_out = exp(-sqrt(max(r*r - t*t, 0))*m/197.327), where Y_out is TAIL e^(−m√(−x²)), SPACELIKE ONLY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value. m is Particle mass m, measured in MeV. Keep t (Time separation t = x⁰ − y⁰) = 0.8 fm; r (Spatial distance r) = 2 fm fixed. For this exercise use only 195.625 MeV ≤ m ≤ 904.375 MeV 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 = 0.0029282981, with a bounded reading uncertainty ±0.00040555. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.070875 MeV, 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.
TAIL e^(−m√(−x²)), SPACELIKE ONLY against m
Predictions from the supplied What Advanced QM supplies, quoted and restated 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): TAIL e^(−m√(−x²)), SPACELIKE ONLY
Horizontal axis (x): m: Particle mass m / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Particle mass m / MeV | TAIL e^(−m√(−x²)), SPACELIKE ONLY |
|---|---|
| 195.625 | 0.16247682798322646 |
| 372.8125 | 0.03133034237161927 |
| 550 | 0.006041417507388916 |
| 727.1874999999999 | 0.0011649641445236124 |
| 904.3749999999999 | 0.00022463957446506368 |
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(m) = 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(195.625) = 0.162477 and Y_out(904.375) = 0.00022464. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 195.625 + (0.0029283 − 0.162477)(904.375 − 195.625)/(0.00022464 − 0.162477) = 892.565 MeV. 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 m = 627.962 MeV; substituting back gives Y_out = 0.0029282981.
- The interpolation error in the input is |m_line − m| = 264.602 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0025227481 and 0.0033338481. Sort the roots even if the response decreases: 614 MeV ≤ m ≤ 644.01 MeV.
- The lower and upper input deviations are 13.9629 MeV and 16.0478 MeV. 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 m ≈ 627.962 MeV; compatible interval [614, 644.01] MeV. Endpoint interpolation gives 892.565 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Invariant normalisation and the Poincaré group on Fock space: infer an input with uncertainty
Invariant normalisation is the choice ⟨p|q⟩ = (2π)³ 2E_p δ³(p − q) for one-particle states, the one normalisation every inertial frame agrees on, so that a Lorentz transformation acts on states as a unitary operator that simply relabels p as Λp. Use the supplied topic model Y_out = (exp(y0+eta)+exp(-(y0+eta)))/(exp(y0)+exp(-(y0))), where Y_out is ENERGY RATIO E′/E (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. eta is Boost rapidity η, dimensionless. Keep y0 (Packet rapidity y₀) = 0.3; m (Mass m) = 0.5 GeV fixed. For this exercise use only -0.2625 ≤ eta ≤ 0.7875 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 = 1.185054, with a bounded reading uncertainty ±0.034163. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown eta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000105, 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.
ENERGY RATIO E′/E against eta
Predictions from the supplied Invariant normalisation and the Poincaré group on Fock space 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): ENERGY RATIO E′/E
Horizontal axis (x): eta: Boost rapidity η
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eta: Boost rapidity η | ENERGY RATIO E′/E |
|---|---|
| -0.26249999999999996 | 0.957300619728209 |
| 0 | 1 |
| 0.26249999999999996 | 1.1120022128088525 |
| 0.5249999999999999 | 1.301069328752839 |
| 0.7874999999999999 | 1.5803042345064902 |
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(eta) = 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(-0.2625) = 0.957301 and Y_out(0.7875) = 1.5803. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives eta_line = -0.2625 + (1.18505 − 0.957301)(0.7875 − -0.2625)/(1.5803 − 0.957301) = 0.121352. 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 eta = 0.378; substituting back gives Y_out = 1.185054.
- The interpolation error in the input is |eta_line − eta| = 0.256648. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.150891 and 1.219217. Sort the roots even if the response decreases: 0.326973 ≤ eta ≤ 0.424958.
- The lower and upper input deviations are 0.0510275 and 0.0469578. 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 eta ≈ 0.378; compatible interval [0.326973, 0.424958]. Endpoint interpolation gives 0.121352.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Wigner's little group: mass, spin and helicity: infer an input with uncertainty
The little group of a four-momentum k is the set of Lorentz transformations that leave k unchanged; its unitary representations supply the spin label of the states |p, σ⟩: 2j + 1 rotation states when m > 0, one Lorentz-invariant helicity when m = 0. Use the supplied topic model Y_out = deg(atan2(beta/sqrt(1-beta^2)*sqrt(1+u^2), u)), where Y_out is MOMENTUM TURN α (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; deg converts radians to degrees; atan2(y,x) is the angle of the vector (x,y), in radians. u is Particle momentum |p|/m, dimensionless. Keep beta (Observer speed β along −y) = 0.8; j (Spin j) = 0.5 fixed. For this exercise use only 1.15188 ≤ u ≤ 8.94812 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 = 53.5178 °, with a bounded reading uncertainty ±0.032576 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000779625, 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.
MOMENTUM TURN α against u
Predictions from the supplied Wigner's little group: mass, spin and helicity 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): MOMENTUM TURN α / °
Horizontal axis (x): u: Particle momentum |p|/m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Particle momentum |p|/m | MOMENTUM TURN α / ° |
|---|---|
| 1.1518750000000002 | 60.47491427260283 |
| 3.1009374999999997 | 54.480685679758245 |
| 5.05 | 53.65754953171523 |
| 6.999062499999999 | 53.40758615011419 |
| 8.948125 | 53.30062888379445 |
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(u) = 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(1.15188) = 60.4749 ° and Y_out(8.94812) = 53.3006 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 1.15188 + (53.5178 − 60.4749)(8.94812 − 1.15188)/(53.3006 − 60.4749) = 8.71213. 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 u = 5.90758; substituting back gives Y_out = 53.5178 °.
- The interpolation error in the input is |u_line − u| = 2.80454. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 53.485224 ° and 53.550376 °. Sort the roots even if the response decreases: 5.67014 ≤ u ≤ 6.17681.
- The lower and upper input deviations are 0.237447 and 0.269229. 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 u ≈ 5.90758; compatible interval [5.67014, 6.17681]. Endpoint interpolation gives 8.71213.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.