University Quantum Field Theory I · topic UQFTI-08.06
The Ward identity for external photons: k_mu M^mu = 0 and gauge-independent amplitudes
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.
The Ward identity for external photons: k_μM^μ = 0: predict a finite change
The Ward identity for an external photon says that the amplitude ε_μM^μ obeys k_μM^μ = 0 whenever every charged external particle is on its mass shell, so adding any multiple of k to the polarisation changes nothing. Use the supplied topic model Y_out = -delta/(1+delta), where Y_out is k⋅M OF THE SUM (eg). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. delta is φ⁻ off-shellness δ/(2p₂⋅k), dimensionless. Keep th (φ⁺ angle to the photon θ) = 60 °; c (Gauge shift cω) = 0.5 fixed. For this exercise use only -0.125625 ≤ delta ≤ 0.425625 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 delta = -0.00435 and 0.072825. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdelta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
k⋅M OF THE SUM against delta
Predictions from the supplied The Ward identity for external photons: k_μM^μ = 0 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): k⋅M OF THE SUM / eg
Horizontal axis (x): delta: φ⁻ off-shellness δ/(2p₂⋅k)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| delta: φ⁻ off-shellness δ/(2p₂⋅k) | k⋅M OF THE SUM / eg |
|---|---|
| -0.12562500000000001 | 0.14367405289492496 |
| 0.012187499999999962 | -0.012040753318925557 |
| 0.14999999999999994 | -0.13043478260869562 |
| 0.2878124999999999 | -0.22348944430963355 |
| 0.4256249999999999 | -0.29855326611135463 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of delta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At delta = -0.00435, substitute into Y_out = -delta/(1+delta): Y₀ = 0.00436901 eg.
- At delta = 0.072825, the same fixed-parameter model gives Y₁ = -0.0678815 eg.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0722505 eg; the input change is Δdelta = 0.077175.
- Average sensitivity is ΔY_out/Δdelta = -0.936191 eg per unit input. 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 = 0.00436901 eg, Y₁ = -0.0678815 eg; ΔY_out = -0.0722505 eg; average sensitivity = -0.936191 eg per unit input.
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.
The Ward identity for external photons: k_μM^μ = 0: infer an input with uncertainty
The Ward identity for an external photon says that the amplitude ε_μM^μ obeys k_μM^μ = 0 whenever every charged external particle is on its mass shell, so adding any multiple of k to the polarisation changes nothing. Use the supplied topic model Y_out = -delta/(1+delta), where Y_out is k⋅M OF THE SUM (eg). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. delta is φ⁻ off-shellness δ/(2p₂⋅k), dimensionless. Keep th (φ⁺ angle to the photon θ) = 60 °; c (Gauge shift cω) = 0.5 fixed. For this exercise use only -0.125625 ≤ delta ≤ 0.425625 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 = -0.17398891 eg, with a bounded reading uncertainty ±0.018685 eg. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown delta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000055125, 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.
k⋅M OF THE SUM against delta
Predictions from the supplied The Ward identity for external photons: k_μM^μ = 0 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): k⋅M OF THE SUM / eg
Horizontal axis (x): delta: φ⁻ off-shellness δ/(2p₂⋅k)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| delta: φ⁻ off-shellness δ/(2p₂⋅k) | k⋅M OF THE SUM / eg |
|---|---|
| -0.12562500000000001 | 0.14367405289492496 |
| 0.012187499999999962 | -0.012040753318925557 |
| 0.14999999999999994 | -0.13043478260869562 |
| 0.2878124999999999 | -0.22348944430963355 |
| 0.4256249999999999 | -0.29855326611135463 |
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(delta) = 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(-0.125625) = 0.143674 eg and Y_out(0.425625) = -0.298553 eg. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives delta_line = -0.125625 + (-0.173989 − 0.143674)(0.425625 − -0.125625)/(-0.298553 − 0.143674) = 0.270352. 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 delta = 0.210637; substituting back gives Y_out = -0.17398891 eg.
- The interpolation error in the input is |delta_line − delta| = 0.0597142. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.19267391 eg and -0.15530391 eg. Sort the roots even if the response decreases: 0.183858 ≤ delta ≤ 0.238657.
- The lower and upper input deviations are 0.0267798 and 0.0280194. 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 delta ≈ 0.210637; compatible interval [0.183858, 0.238657]. Endpoint interpolation gives 0.270352.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.