University Quantum Field Theory I · unit UQFTI-09
QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes
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.
The QED Lagrangian from local phase invariance: infer an input with uncertainty
The QED Lagrangian, L = ψ̄(iγ^μ D_μ − m)ψ − (1/4)F_μν F^μν with D_μ = ∂_μ + ieA_μ, is the free Dirac Lagrangian made invariant under ψ → e^(iα)ψ, for any function α(x), by a vector field that shifts as A_μ → A_μ − (1/e)∂_μ α. Use the supplied topic model Y_out = 1-(a*n/3)*sin(2*pi*n*xm/3), where Y_out is CANONICAL WAVENUMBER (k). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. n is Phase periods across the window n, dimensionless. Keep a (Phase amplitude a) = 1.5 rad; xm (Marker position x) = 0.8 λ fixed. For this exercise use only 1.2625 ≤ n ≤ 1.7875 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 = 0.62778918 k, with a bounded reading uncertainty ±0.025161 k. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000525, 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.
CANONICAL WAVENUMBER against n
Predictions from the supplied The QED Lagrangian from local phase invariance 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): CANONICAL WAVENUMBER / k
Horizontal axis (x): n: Phase periods across the window n
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Phase periods across the window n | CANONICAL WAVENUMBER / k |
|---|---|
| 1.2625 | 0.4600513107736801 |
| 1.3937499999999998 | 0.4970233410439536 |
| 1.525 | 0.57803894370195 |
| 1.65625 | 0.7032265417827983 |
| 1.7874999999999999 | 0.8694382932223448 |
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(n) = 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(1.2625) = 0.460051 k and Y_out(1.7875) = 0.869438 k. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 1.2625 + (0.627789 − 0.460051)(1.7875 − 1.2625)/(0.869438 − 0.460051) = 1.47761. 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 n = 1.58275; substituting back gives Y_out = 0.62778918 k.
- The interpolation error in the input is |n_line − n| = 0.105142. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.60262818 k and 0.65295018 k. Sort the roots even if the response decreases: 1.55478 ≤ n ≤ 1.60875.
- The lower and upper input deviations are 0.0279714 and 0.0260026. 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 n ≈ 1.58275; compatible interval [1.55478, 1.60875]. Endpoint interpolation gives 1.47761.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
QED Feynman rules: the photon vertex and propagator: infer an input with uncertainty
QED's Feynman rules turn each term of the perturbation series into vertices −ieγ^μ, propagators and external factors, every photon index μ opened at a vertex and closed on a polarisation vector, a photon propagator or the same fermion line. Use the supplied topic model Y_out = sqrt(max(0, 1-(1-b1^2)*(1-b2^2)/(1-b1*b2*cosd(a))^2)), where Y_out is RELATIVE SPEED β_rel (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; sind/cosd/tand use degrees. a is Angle α between the velocities, measured in °. Keep b1 (Speed β₁ of lepton 1) = 0.6; b2 (Speed β₂ of lepton 2) = 0.6 fixed. For this exercise use only 19.125 ° ≤ a ≤ 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 = 0.81208878, with a bounded reading uncertainty ±0.0099761. 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 0.014175 °, 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.
RELATIVE SPEED β_rel against a
Predictions from the supplied QED Feynman rules: the photon vertex and propagator 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): RELATIVE SPEED β_rel
Horizontal axis (x): a: Angle α between the velocities / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Angle α between the velocities / ° | RELATIVE SPEED β_rel |
|---|---|
| 19.125 | 0.24355029786895643 |
| 54.5625 | 0.5880430558493998 |
| 90 | 0.7683749084919419 |
| 125.4375 | 0.8483224592500153 |
| 160.875 | 0.8785961239160369 |
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(19.125) = 0.24355 and Y_out(160.875) = 0.878596. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 19.125 + (0.812089 − 0.24355)(160.875 − 19.125)/(0.878596 − 0.24355) = 146.03 °. 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 = 105.593 °; substituting back gives Y_out = 0.81208878.
- The interpolation error in the input is |a_line − a| = 40.4372 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.80211268 and 0.82206488. Sort the roots even if the response decreases: 101.491 ° ≤ a ≤ 110.148 °.
- The lower and upper input deviations are 4.10155 ° and 4.55578 °. 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 ≈ 105.593 °; compatible interval [101.491, 110.148] °. Endpoint interpolation gives 146.03 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Photon exchange between static charges: like charges repel: infer an input with uncertainty
One-photon exchange between two slow charged fermions acts, at first Born order, as the potential V(r) = +α/r between like charges and −α/r between a particle and an antiparticle, the sign fixed by the propagator's g₀₀ and by the ordering of the fermion operators. Use the supplied topic model Y_out = (-pair*(spin+(1-spin)*pair)*(1-2*spin))*14.40/d, where Y_out is V(r) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Separation r, measured in Å. Keep pair (Muon charge ÷ electron's: +1 μ⁻, −1 μ⁺) = 1; spin (Boson spin: 0 scalar, 1 photon) = 1 fixed. For this exercise use only 0.855 Å ≤ d ≤ 2.745 Å 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 = 7.1716719 eV, with a bounded reading uncertainty ±0.28887 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000189 Å, 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.
V(r) against d
Predictions from the supplied Photon exchange between static charges: like charges repel 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): V(r) / eV
Horizontal axis (x): d: Separation r / Å
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation r / Å | V(r) / eV |
|---|---|
| 0.855 | 16.842105263157894 |
| 1.3275 | 10.847457627118645 |
| 1.7999999999999998 | 8.000000000000002 |
| 2.2725 | 6.336633663366337 |
| 2.7449999999999997 | 5.245901639344263 |
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(d) = 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(0.855) = 16.8421 eV and Y_out(2.745) = 5.2459 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.855 + (7.17167 − 16.8421)(2.745 − 0.855)/(5.2459 − 16.8421) = 2.43113 Å. 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 d = 2.0079 Å; substituting back gives Y_out = 7.1716719 eV.
- The interpolation error in the input is |d_line − d| = 0.42323 Å. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.8828019 eV and 7.4605419 eV. Sort the roots even if the response decreases: 1.93015 Å ≤ d ≤ 2.09217 Å.
- The lower and upper input deviations are 0.0777453 Å and 0.0842712 Å. 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 d ≈ 2.0079 Å; compatible interval [1.93015, 2.09217] Å. Endpoint interpolation gives 2.43113 Å.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Tree-level Ward identity: k_μ M^μ = 0 on a fermion line: infer an input with uncertainty
At tree level, replacing a photon's polarisation ε_μ by its momentum k_μ and summing over every place it can attach along one open fermion line with on-shell ends gives exactly zero, whatever else attaches to the line. Use the supplied topic model Y_out = 1-sqrt(1-0.261121*(1-off))-0.15-(1-w-sqrt((1-w)^2-0.261121)), where Y_out is SECOND VERTEX SUPPLIES Q₂ (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. w is Photon energy ω, measured in MeV. Keep phi (Polarisation angle φ of Γ₂) = 0 °; off (Incoming leg off shell, 1 − p²/m²) = 0 fixed. For this exercise use only 0.0925 MeV ≤ w ≤ 0.4075 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 = -0.22432645 MeV, with a bounded reading uncertainty ±0.0085804 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 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.
SECOND VERTEX SUPPLIES Q₂ against w
Predictions from the supplied Tree-level Ward identity: k_μ M^μ = 0 on a fermion line 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): SECOND VERTEX SUPPLIES Q₂ / MeV
Horizontal axis (x): w: Photon energy ω / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Photon energy ω / MeV | SECOND VERTEX SUPPLIES Q₂ / MeV |
|---|---|
| 0.0925 | -0.16712387965162842 |
| 0.17124999999999999 | -0.18586955389491996 |
| 0.24999999999999997 | -0.21060074853816255 |
| 0.32875 | -0.2455657780528335 |
| 0.4075 | -0.30218864782631316 |
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(w) = 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(0.0925) = -0.167124 MeV and Y_out(0.4075) = -0.302189 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.0925 + (-0.224326 − -0.167124)(0.4075 − 0.0925)/(-0.302189 − -0.167124) = 0.225909 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 w = 0.28465 MeV; substituting back gives Y_out = -0.22432645 MeV.
- The interpolation error in the input is |w_line − w| = 0.0587413 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.23290685 MeV and -0.21574605 MeV. Sort the roots even if the response decreases: 0.263637 MeV ≤ w ≤ 0.303719 MeV.
- The lower and upper input deviations are 0.0210132 MeV and 0.0190689 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 w ≈ 0.28465 MeV; compatible interval [0.263637, 0.303719] MeV. Endpoint interpolation gives 0.225909 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.