University Quantum Field Theory I · unit UQFTI-10
QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation
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.
Møller scattering: t and u channels with opposite signs: infer an input with uncertainty
Møller scattering is elastic e⁻e⁻ → e⁻e⁻; at tree level its amplitude is the t-channel photon exchange minus the u-channel exchange with the two outgoing electrons swapped, the minus sign being Fermi statistics. Use the supplied topic model Y_out = 31.183403*(1-v^2)*(8*(1+v^2)^2*cosd(cut)/(v^4*sind(cut)^2)+4*cosd(cut)+4*ln(tand(cut/2))*(4*v^2+eta*(3*v^2-1)*(1+v^2))/v^4), where Y_out is σ, θ_min TO 90° (mb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; sind/cosd/tand use degrees. v is CM speed of each electron v, measured in c. Keep eta (Relative sign η of the u diagram) = -1; cut (Counting cut θ_min) = 30 ° fixed. For this exercise use only 0.373312 c ≤ v ≤ 0.916687 c 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 = 3583.5677 mb, with a bounded reading uncertainty ±438.8 mb. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000543375 c, 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.
σ, θ_min TO 90° against v
Predictions from the supplied Møller scattering: t and u channels with opposite signs 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): σ, θ_min TO 90° / mb
Horizontal axis (x): v: CM speed of each electron v / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: CM speed of each electron v / c | σ, θ_min TO 90° / mb |
|---|---|
| 0.3733125 | 40920.59470476743 |
| 0.50915625 | 12799.754503273512 |
| 0.645 | 5182.016367267373 |
| 0.78084375 | 2202.5959505857154 |
| 0.9166874999999999 | 658.2220839826692 |
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(v) = 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(0.373312) = 40920.6 mb and Y_out(0.916687) = 658.222 mb. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 0.373312 + (3583.57 − 40920.6)(0.916687 − 0.373312)/(658.222 − 40920.6) = 0.877207 c. 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 v = 0.704771 c; substituting back gives Y_out = 3583.5677 mb.
- The interpolation error in the input is |v_line − v| = 0.172436 c. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3144.7677 mb and 4022.3677 mb. Sort the roots even if the response decreases: 0.686045 c ≤ v ≤ 0.725774 c.
- The lower and upper input deviations are 0.0187265 c and 0.0210029 c. 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 v ≈ 0.704771 c; compatible interval [0.686045, 0.725774] c. Endpoint interpolation gives 0.877207 c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Møller scattering at low energy: Mott's formula recovered: infer an input with uncertainty
Low-energy Møller scattering is tree-level e⁻e⁻ → e⁻e⁻ at α ≪ v ≪ 1: each diagram becomes a Coulomb Born amplitude, the t channel keeping both spins and the u channel swapping them with a minus sign, and the spin average is Mott's formula at first Born order. Use the supplied topic model Y_out = 0.0036487*10^(-lv), where Y_out is η = α/(2v) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lv is log₁₀(v/c), CM speed of each electron, dimensionless. Keep th (Marker angle θ, centre of mass) = 60 ° fixed. For this exercise use only -2.43969 ≤ lv ≤ -0.510313 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 = 0.066820067, with a bounded reading uncertainty ±0.0082507. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lv 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, 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.
η = α/(2v) against lv
Predictions from the supplied Møller scattering at low energy: Mott's formula recovered 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): η = α/(2v)
Horizontal axis (x): lv: log₁₀(v/c), CM speed of each electron
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lv: log₁₀(v/c), CM speed of each electron | η = α/(2v) |
|---|---|
| -2.4396875000000002 | 1.0042125779172937 |
| -1.9573437500000002 | 0.33073633282887455 |
| -1.4750000000000003 | 0.10892765561645967 |
| -0.9926562500000005 | 0.03587520626056285 |
| -0.5103125000000004 | 0.011815460609651168 |
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(lv) = 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(-2.43969) = 1.00421 and Y_out(-0.510313) = 0.0118155. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lv_line = -2.43969 + (0.0668201 − 1.00421)(-0.510313 − -2.43969)/(0.0118155 − 1.00421) = -0.61725. 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 lv = -1.26277; substituting back gives Y_out = 0.066820067.
- The interpolation error in the input is |lv_line − lv| = 0.645519. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.058569367 and 0.075070767. Sort the roots even if the response decreases: -1.31333 ≤ lv ≤ -1.20553.
- The lower and upper input deviations are 0.0505639 and 0.0572364. 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 lv ≈ -1.26277; compatible interval [-1.31333, -1.20553]. Endpoint interpolation gives -0.61725.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Bhabha scattering: s and t channels interfering: infer an input with uncertainty
Bhabha scattering is elastic electron-positron scattering, e⁺e⁻ → e⁺e⁻, whose tree-level amplitude is the t-channel photon-exchange diagram minus the s-channel annihilation diagram. Use the supplied topic model Y_out = (4+(1+cosd(theta))^2)/(1-cosd(theta))^2, where Y_out is EXCHANGE (s²+u²)/t² (α²/(2s)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Scattering angle θ, measured in °. Keep eta (Relative sign η of the s diagram) = -1 fixed. For this exercise use only 45.9375 ° ≤ theta ≤ 164.063 ° 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 = 1.9830646 α²/(2s), with a bounded reading uncertainty ±0.14147 α²/(2s). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0118125 °, 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.
EXCHANGE (s²+u²)/t² against theta
Predictions from the supplied Bhabha scattering: s and t channels interfering 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): EXCHANGE (s²+u²)/t² / α²/(2s)
Horizontal axis (x): theta: Scattering angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Scattering angle θ / ° | EXCHANGE (s²+u²)/t² / α²/(2s) |
|---|---|
| 45.9375 | 74.11472969437779 |
| 75.46875 | 9.916923843485383 |
| 105 | 2.870930405038196 |
| 134.53125 | 1.412797434581187 |
| 164.0625 | 1.0399594102515757 |
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(theta) = 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(45.9375) = 74.1147 α²/(2s) and Y_out(164.063) = 1.03996 α²/(2s). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 45.9375 + (1.98306 − 74.1147)(164.063 − 45.9375)/(1.03996 − 74.1147) = 162.538 °. 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 theta = 117.994 °; substituting back gives Y_out = 1.9830646 α²/(2s).
- The interpolation error in the input is |theta_line − theta| = 44.5442 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8415946 α²/(2s) and 2.1245346 α²/(2s). Sort the roots even if the response decreases: 115.296 ° ≤ theta ≤ 121.083 °.
- The lower and upper input deviations are 2.69817 ° and 3.08895 °. 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 theta ≈ 117.994 °; compatible interval [115.296, 121.083] °. Endpoint interpolation gives 162.538 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Compton amplitude: two diagrams, one Ward identity: infer an input with uncertainty
The Compton amplitude is the sum of QED's two order-e² diagrams for γe⁻ → γe⁻, written as a tensor M^μν in the two photon indices; gauge invariance demands k_μ M^μν = 0 and k′_ν M^μν = 0, which the sum obeys and each diagram alone violates. Use the supplied topic model Y_out = (1000*(-((2-(w/(1+w*(1-cosd(th)))))*cosd(th)+(w/(1+w*(1-cosd(th))))+2*lam*sind(th))/sqrt(2*(2+w-(w/(1+w*(1-cosd(th))))))))/1000, where Y_out is s-CHANNEL M_s ÷ e² (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees. w is Photon energy ω ÷ m (m = 511 keV), dimensionless. Keep lam (Gauge shift λ in ε → ε + λk/ω) = 1; th (Marker angle θ) = 90 ° fixed. For this exercise use only 0.301875 ≤ w ≤ 0.651 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 = -1.1218854, with a bounded reading uncertainty ±0.00076896. 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 0.0000349125, 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.
s-CHANNEL M_s ÷ e² (before display rounding) against w
Predictions from the supplied The Compton amplitude: two diagrams, one Ward identity 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): s-CHANNEL M_s ÷ e² (before display rounding)
Horizontal axis (x): w: Photon energy ω ÷ m (m = 511 keV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Photon energy ω ÷ m (m = 511 keV) | s-CHANNEL M_s ÷ e² (before display rounding) |
|---|---|
| 0.301875 | -1.0969083083948363 |
| 0.38915625 | -1.1102125540316012 |
| 0.47643749999999996 | -1.1191286293914904 |
| 0.5637187499999999 | -1.1245010402395805 |
| 0.6509999999999999 | -1.1270117719955641 |
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(0.301875) = -1.09691 and Y_out(0.651) = -1.12701. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.301875 + (-1.12189 − -1.09691)(0.651 − 0.301875)/(-1.12701 − -1.09691) = 0.591547. 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 = 0.514841; substituting back gives Y_out = -1.1218854.
- The interpolation error in the input is |w_line − w| = 0.0767055. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.1226544 and -1.1211164. Sort the roots even if the response decreases: 0.503174 ≤ w ≤ 0.527492.
- The lower and upper input deviations are 0.0116671 and 0.0126509. 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 ≈ 0.514841; compatible interval [0.503174, 0.527492]. Endpoint interpolation gives 0.591547.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.