University Quantum Field Theory I · unit UQFTI-08
Covariant Quantisation of the Electromagnetic Field
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 Maxwell Lagrangian and the missing momentum of A⁰: infer an input with uncertainty
The momentum conjugate to A₀ in the Maxwell Lagrangian is identically zero, because ∂₀A₀ appears only in F₀₀ = 0; A⁰ is therefore not a dynamical coordinate but a Lagrange multiplier enforcing Gauss's law, a constraint on the fields at one instant. Use the supplied topic model Y_out = lam*pi/(2*w), where Y_out is PEAK δA⁰ = ∂ₜΛ (ρ₀/k²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Ramp duration kΔt, dimensionless. Keep lam (Final gauge amplitude ℓ) = 0.6 ρ₀/k³; t0 (Switch-on time kt₀) = 2 fixed. For this exercise use only 1.31875 ≤ w ≤ 3.68125 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 = 0.34149293 ρ₀/k², with a bounded reading uncertainty ±0.012821 ρ₀/k². 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 0.00023625, 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.
PEAK δA⁰ = ∂ₜΛ against w
Predictions from the supplied The Maxwell Lagrangian and the missing momentum of A⁰ 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): PEAK δA⁰ = ∂ₜΛ / ρ₀/k²
Horizontal axis (x): w: Ramp duration kΔt
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Ramp duration kΔt | PEAK δA⁰ = ∂ₜΛ / ρ₀/k² |
|---|---|
| 1.31875 | 0.7146751060299055 |
| 1.909375 | 0.4936053923807203 |
| 2.5 | 0.37699111843077515 |
| 3.090625 | 0.30494731521195156 |
| 3.68125 | 0.25602113305994917 |
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(1.31875) = 0.714675 ρ₀/k² and Y_out(3.68125) = 0.256021 ρ₀/k². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 1.31875 + (0.341493 − 0.714675)(3.68125 − 1.31875)/(0.256021 − 0.714675) = 3.24099. 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 = 2.75987; substituting back gives Y_out = 0.34149293 ρ₀/k².
- The interpolation error in the input is |w_line − w| = 0.481115. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.32867193 ρ₀/k² and 0.35431393 ρ₀/k². Sort the roots even if the response decreases: 2.66001 ≤ w ≤ 2.86753.
- The lower and upper input deviations are 0.0998672 and 0.107659. 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 ≈ 2.75987; compatible interval [2.66001, 2.86753]. Endpoint interpolation gives 3.24099.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Gauge fixing, Feynman gauge and photons of negative norm: infer an input with uncertainty
Feynman-gauge quantisation adds −½(∂⋅A)² to Maxwell's Lagrangian so that all four components of A_μ have conjugate momenta, then imposes covariant commutators [a_μ(k), a_ν†(k′)] = −g_μν(2π)³2|k|δ³(k − k′), which give the timelike photon a negative norm. Use the supplied topic model Y_out = (1-max(0,1-mu)*(1-v))*n, where Y_out is ENERGY E_n (|k|). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. n is Quanta n in the mode, dimensionless. Keep mu (Component μ of a_μ(k)) = 0; v (Vacuum reading v (+1: a₀|0⟩ = 0)) = 1 fixed. For this exercise use only 0 ≤ n ≤ 6, at the specified test settings separated by 1. 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 = 4 |k|, with a bounded reading uncertainty ±0.48 |k|. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
ENERGY E_n against n
Predictions from the supplied Gauge fixing, Feynman gauge and photons of negative norm 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 E_n / |k|
Horizontal axis (x): n: Quanta n in the mode
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Quanta n in the mode | ENERGY E_n / |k| |
|---|---|
| 0 | 0 |
| 2 | 2 |
| 3 | 3 |
| 5 | 5 |
| 6 | 6 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(0) = 0 |k| and Y_out(6) = 6 |k|. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 0 + (4 − 0)(6 − 0)/(6 − 0) = 4. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to n = 4; only the allowed settings 4 lie inside the stated reading bound.
- The interpolation error in the input is |n_line − n| = 0.
- The compatible allowed values span 4 to 4. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible n: 4. Endpoint interpolation predicts 4; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Gupta-Bleuler condition and the physical subspace: infer an input with uncertainty
The Gupta-Bleuler condition admits a state of the covariantly quantised photon field only if the positive-frequency part of ∂⋅A annihilates it, which allows timelike and longitudinal photons only in the zero-norm combination a(0)† − a(3)†. Use the supplied topic model Y_out = (100*(t*t-cosd(2*theta)))/100, where Y_out is NORM ⟨ψ|ψ⟩ = ⟨H⟩/|k| (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. t is Transverse amplitude t, dimensionless. Keep theta (Mixing angle θ of a(0)† and a(3)†) = 0 ° fixed. For this exercise use only 0.10625 ≤ t ≤ 0.89375 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 = -0.65587111, with a bounded reading uncertainty ±0.049926. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
NORM ⟨ψ|ψ⟩ = ⟨H⟩/|k| (before display rounding) against t
Predictions from the supplied The Gupta-Bleuler condition and the physical subspace 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): NORM ⟨ψ|ψ⟩ = ⟨H⟩/|k| (before display rounding)
Horizontal axis (x): t: Transverse amplitude t
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Transverse amplitude t | NORM ⟨ψ|ψ⟩ = ⟨H⟩/|k| (before display rounding) |
|---|---|
| 0.10625 | -0.9887109375 |
| 0.303125 | -0.908115234375 |
| 0.5 | -0.75 |
| 0.6968749999999999 | -0.5143652343750001 |
| 0.8937499999999999 | -0.20121093750000008 |
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(t) = 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(0.10625) = -0.988711 and Y_out(0.89375) = -0.201211. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.10625 + (-0.655871 − -0.988711)(0.89375 − 0.10625)/(-0.201211 − -0.988711) = 0.43909. 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 t = 0.586625; substituting back gives Y_out = -0.65587111.
- The interpolation error in the input is |t_line − t| = 0.147535. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.70579711 and -0.60594511. Sort the roots even if the response decreases: 0.542405 ≤ t ≤ 0.627738.
- The lower and upper input deviations are 0.0442203 and 0.0411129. 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 t ≈ 0.586625; compatible interval [0.542405, 0.627738]. Endpoint interpolation gives 0.43909.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The photon propagator in covariant gauges and Coulomb gauge: infer an input with uncertainty
The photon propagator is the Fourier transform of ⟨0|T A_μ(x) A_ν(y)|0⟩, −i[g_μν − (1 − ξ)k_μ k_ν/k²]/(k² + iε) in the covariant gauge labelled ξ; ξ changes only the part along k, which a conserved current never sees. Use the supplied topic model Y_out = 1-t/(1-u^2), where Y_out is TOTAL WITH TRANSVERSE (J⁰J′⁰/|k|²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Marker at ω/|k|, dimensionless. Keep xi (Gauge parameter ξ) = 1; t (Transverse overlap J_T⋅J′_T ÷ J⁰J′⁰) = 0.3 fixed. For this exercise use only 0.074375 ≤ u ≤ 0.625625 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 = 0.63915281 J⁰J′⁰/|k|², with a bounded reading uncertainty ±0.0088768 J⁰J′⁰/|k|². 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 0.000055125, 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.
TOTAL WITH TRANSVERSE against u
Predictions from the supplied The photon propagator in covariant gauges and Coulomb gauge 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): TOTAL WITH TRANSVERSE / J⁰J′⁰/|k|²
Horizontal axis (x): u: Marker at ω/|k|
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Marker at ω/|k| | TOTAL WITH TRANSVERSE / J⁰J′⁰/|k|² |
|---|---|
| 0.074375 | 0.6983312770367647 |
| 0.21218749999999997 | 0.6858561325392611 |
| 0.35 | 0.6581196581196582 |
| 0.48781249999999987 | 0.6063193434452363 |
| 0.6256249999999999 | 0.5070600173684323 |
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(0.074375) = 0.698331 J⁰J′⁰/|k|² and Y_out(0.625625) = 0.50706 J⁰J′⁰/|k|². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 0.074375 + (0.639153 − 0.698331)(0.625625 − 0.074375)/(0.50706 − 0.698331) = 0.244929. 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 = 0.410637; substituting back gives Y_out = 0.63915281 J⁰J′⁰/|k|².
- The interpolation error in the input is |u_line − u| = 0.165708. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.63027601 J⁰J′⁰/|k|² and 0.64802961 J⁰J′⁰/|k|². Sort the roots even if the response decreases: 0.38426 ≤ u ≤ 0.434262.
- The lower and upper input deviations are 0.0263777 and 0.023625. 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 ≈ 0.410637; compatible interval [0.38426, 0.434262]. Endpoint interpolation gives 0.244929.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.