University Quantum Field Theory I · topic UQFTI-01.05
The time-ordered product and the Feynman propagator D_F(x - y)
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.
Time ordering and the Feynman propagator D_F(x − y): predict a finite change
The Feynman propagator D_F(x − y) is the vacuum expectation of the time-ordered product of two free fields: the amplitude for a quantum created at whichever of x and y is earlier to be absorbed at the later one. Use the supplied topic model Y_out = dt/dx, where Y_out is β TO REVERSE (NONE IF |β| ≥ 1) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dx is Spatial separation x¹ − y¹, measured in fm. Keep dt (Time separation x⁰ − y⁰, as ct) = 0.75 fm; beta (Observer's boost β along x¹) = 0.3 fixed. For this exercise use only 0.489062 fm ≤ dx ≤ 2.26094 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 dx = 0.878875 fm and 1.12694 fm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdx across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
β TO REVERSE (NONE IF |β| ≥ 1) against dx
Predictions from the supplied Time ordering and the Feynman propagator D_F(x − y) 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): β TO REVERSE (NONE IF |β| ≥ 1)
Horizontal axis (x): dx: Spatial separation x¹ − y¹ / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dx: Spatial separation x¹ − y¹ / fm | β TO REVERSE (NONE IF |β| ≥ 1) |
|---|---|
| 0.48906249999999996 | 1.5335463258785944 |
| 0.9320312499999999 | 0.8046940486169322 |
| 1.375 | 0.5454545454545454 |
| 1.81796875 | 0.4125483455092394 |
| 2.2609375 | 0.331720801658604 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dx separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dx = 0.878875 fm, substitute into Y_out = dt/dx: Y₀ = 0.853364.
- At dx = 1.12694 fm, the same fixed-parameter model gives Y₁ = 0.66552.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.187843; the input change is Δdx = 0.248063 fm.
- Average sensitivity is ΔY_out/Δdx = -0.757241 1 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 = 0.853364, Y₁ = 0.66552; ΔY_out = -0.187843; average sensitivity = -0.757241 1 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.
Time ordering and the Feynman propagator D_F(x − y): infer an input with uncertainty
The Feynman propagator D_F(x − y) is the vacuum expectation of the time-ordered product of two free fields: the amplitude for a quantum created at whichever of x and y is earlier to be absorbed at the later one. Use the supplied topic model Y_out = dt/dx, where Y_out is β TO REVERSE (NONE IF |β| ≥ 1) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dx is Spatial separation x¹ − y¹, measured in fm. Keep dt (Time separation x⁰ − y⁰, as ct) = 0.75 fm; beta (Observer's boost β along x¹) = 0.3 fixed. For this exercise use only 0.489062 fm ≤ dx ≤ 2.26094 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 = 0.47773553, with a bounded reading uncertainty ±0.021902. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dx by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000177187 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.
β TO REVERSE (NONE IF |β| ≥ 1) against dx
Predictions from the supplied Time ordering and the Feynman propagator D_F(x − y) 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): β TO REVERSE (NONE IF |β| ≥ 1)
Horizontal axis (x): dx: Spatial separation x¹ − y¹ / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dx: Spatial separation x¹ − y¹ / fm | β TO REVERSE (NONE IF |β| ≥ 1) |
|---|---|
| 0.48906249999999996 | 1.5335463258785944 |
| 0.9320312499999999 | 0.8046940486169322 |
| 1.375 | 0.5454545454545454 |
| 1.81796875 | 0.4125483455092394 |
| 2.2609375 | 0.331720801658604 |
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(dx) = 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(0.489062) = 1.53355 and Y_out(2.26094) = 0.331721. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dx_line = 0.489062 + (0.477736 − 1.53355)(2.26094 − 0.489062)/(0.331721 − 1.53355) = 2.04567 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 dx = 1.56991 fm; substituting back gives Y_out = 0.47773553.
- The interpolation error in the input is |dx_line − dx| = 0.475759 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.45583353 and 0.49963753. Sort the roots even if the response decreases: 1.50109 fm ≤ dx ≤ 1.64534 fm.
- The lower and upper input deviations are 0.0688181 fm and 0.0754312 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 dx ≈ 1.56991 fm; compatible interval [1.50109, 1.64534] fm. Endpoint interpolation gives 2.04567 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.