University Quantum Field Theory I · topic UQFTI-01.07
The retarded propagator, the Pauli-Jordan function and where causality lives
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 retarded propagator and where causality lives: predict a finite change
The retarded propagator D_R(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩ is the field's commutator kept only when x is later than y: a Green's function of the Klein-Gordon operator that gives the response at x to a disturbance at y and vanishes outside the forward light cone. Use the supplied topic model Y_out = 197.327*max(sign(t),0)*sin(sqrt(m*m+p*p)*t/197.327)/sqrt(m*m+p*p), where Y_out is i D_R AT t (fm). 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; unmarked trigonometric functions use radians. m is Field mass m, measured in MeV. Keep p (Mode momentum |p|) = 0 MeV; t (Probe time t) = 3 fm/c fixed. For this exercise use only 142.5 MeV ≤ m ≤ 289.5 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. Calculate Y_out at m = 174.84 MeV and 195.42 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
i D_R AT t against m
Predictions from the supplied The retarded propagator and where causality lives 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): i D_R AT t / fm
Horizontal axis (x): m: Field mass m / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Field mass m / MeV | i D_R AT t / fm |
|---|---|
| 142.5 | 1.1462681314145868 |
| 179.25 | 0.445281620918811 |
| 216 | -0.12955686962980148 |
| 252.75 | -0.5035587821135885 |
| 289.5 | -0.6489011856625237 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 174.84 MeV, substitute into Y_out = 197.327*max(sign(t),0)*sin(sqrt(m*m+p*p)*t/197.327)/sqrt(m*m+p*p): Y₀ = 0.524638 fm.
- At m = 195.42 MeV, the same fixed-parameter model gives Y₁ = 0.171416 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.353223 fm; the input change is Δm = 20.58 MeV.
- Average sensitivity is ΔY_out/Δm = -0.0171634 fm per MeV. 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 = 0.524638 fm, Y₁ = 0.171416 fm; ΔY_out = -0.353223 fm; average sensitivity = -0.0171634 fm per MeV.
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 retarded propagator and where causality lives: infer an input with uncertainty
The retarded propagator D_R(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩ is the field's commutator kept only when x is later than y: a Green's function of the Klein-Gordon operator that gives the response at x to a disturbance at y and vanishes outside the forward light cone. Use the supplied topic model Y_out = 197.327*max(sign(t),0)*sin(sqrt(m*m+p*p)*t/197.327)/sqrt(m*m+p*p), where Y_out is i D_R AT t (fm). 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; unmarked trigonometric functions use radians. m is Field mass m, measured in MeV. Keep p (Mode momentum |p|) = 0 MeV; t (Probe time t) = 3 fm/c fixed. For this exercise use only 142.5 MeV ≤ m ≤ 289.5 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 = -0.32166259 fm, with a bounded reading uncertainty ±0.049086 fm. 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 0.0147 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.
i D_R AT t against m
Predictions from the supplied The retarded propagator and where causality lives 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): i D_R AT t / fm
Horizontal axis (x): m: Field mass m / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Field mass m / MeV | i D_R AT t / fm |
|---|---|
| 142.5 | 1.1462681314145868 |
| 179.25 | 0.445281620918811 |
| 216 | -0.12955686962980148 |
| 252.75 | -0.5035587821135885 |
| 289.5 | -0.6489011856625237 |
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(142.5) = 1.14627 fm and Y_out(289.5) = -0.648901 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 142.5 + (-0.321663 − 1.14627)(289.5 − 142.5)/(-0.648901 − 1.14627) = 262.704 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 = 232.17 MeV; substituting back gives Y_out = -0.32166259 fm.
- The interpolation error in the input is |m_line − m| = 30.5336 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.37074859 fm and -0.27257659 fm. Sort the roots even if the response decreases: 227.685 MeV ≤ m ≤ 236.995 MeV.
- The lower and upper input deviations are 4.48528 MeV and 4.82487 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 ≈ 232.17 MeV; compatible interval [227.685, 236.995] MeV. Endpoint interpolation gives 262.704 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.