University Quantum Field Theory I · topic UQFTI-01.08
The complex field's propagator: particles forward and antiparticles backward in time
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.
Complex propagator: particles forward, antiparticles back: predict a finite change
For a complex scalar field, D_F(x − y) = ⟨0|Tφ(x)φ†(y)|0⟩ is the amplitude for one unit of charge q to pass from y to x: carried by a particle when y is earlier, by its antiparticle when x is earlier, with one invariant function for both. Use the supplied topic model Y_out = -1/(4*pi^2*(t*t-r*r)), where Y_out is Re D_F IF MASSLESS (fm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Distance |x − y|, measured in fm. Keep t (Time of x after y, x⁰ − y⁰) = -1.9 fm fixed. For this exercise use only 0.31875 fm ≤ r ≤ 1.815 fm 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 r = 0.647925 fm and 0.8574 fm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
Re D_F IF MASSLESS against r
Predictions from the supplied Complex propagator: particles forward, antiparticles back 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): Re D_F IF MASSLESS / fm⁻²
Horizontal axis (x): r: Distance |x − y| / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Distance |x − y| / fm | Re D_F IF MASSLESS / fm⁻² |
|---|---|
| 0.31875 | -0.007219902859332649 |
| 0.6928124999999999 | -0.00809271827053766 |
| 1.066875 | -0.010247804872710096 |
| 1.4409375 | -0.01651581823462236 |
| 1.815 | -0.08021628029636435 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 0.647925 fm, substitute into Y_out = -1/(4*pi^2*(t*t-r*r)): Y₀ = -0.00794005 fm⁻².
- At r = 0.8574 fm, the same fixed-parameter model gives Y₁ = -0.00881095 fm⁻².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0008709 fm⁻²; the input change is Δr = 0.209475 fm.
- Average sensitivity is ΔY_out/Δr = -0.00415754 fm⁻² per fm. 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 = -0.00794005 fm⁻², Y₁ = -0.00881095 fm⁻²; ΔY_out = -0.0008709 fm⁻²; average sensitivity = -0.00415754 fm⁻² per fm.
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.
Complex propagator: particles forward, antiparticles back: infer an input with uncertainty
For a complex scalar field, D_F(x − y) = ⟨0|Tφ(x)φ†(y)|0⟩ is the amplitude for one unit of charge q to pass from y to x: carried by a particle when y is earlier, by its antiparticle when x is earlier, with one invariant function for both. Use the supplied topic model Y_out = -1/(4*pi^2*(t*t-r*r)), where Y_out is Re D_F IF MASSLESS (fm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Distance |x − y|, measured in fm. Keep t (Time of x after y, x⁰ − y⁰) = -1.9 fm fixed. For this exercise use only 0.31875 fm ≤ r ≤ 1.815 fm 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 = -0.012099496 fm⁻², with a bounded reading uncertainty ±0.00073194 fm⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000149625 fm, 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.
Re D_F IF MASSLESS against r
Predictions from the supplied Complex propagator: particles forward, antiparticles back 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): Re D_F IF MASSLESS / fm⁻²
Horizontal axis (x): r: Distance |x − y| / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Distance |x − y| / fm | Re D_F IF MASSLESS / fm⁻² |
|---|---|
| 0.31875 | -0.007219902859332649 |
| 0.6928124999999999 | -0.00809271827053766 |
| 1.066875 | -0.010247804872710096 |
| 1.4409375 | -0.01651581823462236 |
| 1.815 | -0.08021628029636435 |
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(r) = 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(0.31875) = -0.0072199 fm⁻² and Y_out(1.815) = -0.0802163 fm⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.31875 + (-0.0120995 − -0.0072199)(1.815 − 0.31875)/(-0.0802163 − -0.0072199) = 0.41877 fm. 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 r = 1.23146 fm; substituting back gives Y_out = -0.012099496 fm⁻².
- The interpolation error in the input is |r_line − r| = 0.812693 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.012831436 fm⁻² and -0.011367556 fm⁻². Sort the roots even if the response decreases: 1.17546 fm ≤ r ≤ 1.27903 fm.
- The lower and upper input deviations are 0.0560041 fm and 0.0475679 fm. 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 r ≈ 1.23146 fm; compatible interval [1.17546, 1.27903] fm. Endpoint interpolation gives 0.41877 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.