University Quantum Field Theory I · topic UQFTI-02.04
The S-operator T exp(i integral of L_int d^4x) is Lorentz invariant
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 S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant: predict a finite change
The S-operator S = T exp(i∫d⁴x ℒ_int) is Lorentz invariant, U(Λ)SU†(Λ) = S, because its one frame-dependent ingredient, the time ordering, changes only for spacelike-separated vertices, where the interaction densities commute. Use the supplied topic model Y_out = dt/dz, where Y_out is β NEEDED TO EQUALISE THE TIMES (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dz is Distance Δz of y along the boost, measured in fm. Keep dt (Time separation cΔt of y) = 1 fm; b (Observer's velocity β = v/c) = 0.25 fixed. For this exercise use only 0.7125 fm ≤ dz ≤ 2.2875 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 dz = 1.059 fm and 1.2795 fm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
β NEEDED TO EQUALISE THE TIMES against dz
Predictions from the supplied The S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant 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 NEEDED TO EQUALISE THE TIMES
Horizontal axis (x): dz: Distance Δz of y along the boost / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dz: Distance Δz of y along the boost / fm | β NEEDED TO EQUALISE THE TIMES |
|---|---|
| 0.7125 | 1.4035087719298245 |
| 1.10625 | 0.903954802259887 |
| 1.5 | 0.6666666666666666 |
| 1.8937499999999998 | 0.5280528052805281 |
| 2.2874999999999996 | 0.43715846994535523 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dz separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dz = 1.059 fm, substitute into Y_out = dt/dz: Y₀ = 0.944287.
- At dz = 1.2795 fm, the same fixed-parameter model gives Y₁ = 0.781555.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.162732; the input change is Δdz = 0.2205 fm.
- Average sensitivity is ΔY_out/Δdz = -0.738013 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 = 0.944287, Y₁ = 0.781555; ΔY_out = -0.162732; average sensitivity = -0.738013 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.
The S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant: infer an input with uncertainty
The S-operator S = T exp(i∫d⁴x ℒ_int) is Lorentz invariant, U(Λ)SU†(Λ) = S, because its one frame-dependent ingredient, the time ordering, changes only for spacelike-separated vertices, where the interaction densities commute. Use the supplied topic model Y_out = dt/dz, where Y_out is β NEEDED TO EQUALISE THE TIMES (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dz is Distance Δz of y along the boost, measured in fm. Keep dt (Time separation cΔt of y) = 1 fm; b (Observer's velocity β = v/c) = 0.25 fixed. For this exercise use only 0.7125 fm ≤ dz ≤ 2.2875 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 = 0.59763932, with a bounded reading uncertainty ±0.024072. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dz by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 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 NEEDED TO EQUALISE THE TIMES against dz
Predictions from the supplied The S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant 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 NEEDED TO EQUALISE THE TIMES
Horizontal axis (x): dz: Distance Δz of y along the boost / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dz: Distance Δz of y along the boost / fm | β NEEDED TO EQUALISE THE TIMES |
|---|---|
| 0.7125 | 1.4035087719298245 |
| 1.10625 | 0.903954802259887 |
| 1.5 | 0.6666666666666666 |
| 1.8937499999999998 | 0.5280528052805281 |
| 2.2874999999999996 | 0.43715846994535523 |
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(dz) = 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(0.7125) = 1.40351 and Y_out(2.2875) = 0.437158. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dz_line = 0.7125 + (0.597639 − 1.40351)(2.2875 − 0.7125)/(0.437158 − 1.40351) = 2.02594 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 dz = 1.67325 fm; substituting back gives Y_out = 0.59763932.
- The interpolation error in the input is |dz_line − dz| = 0.352691 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.57356732 and 0.62171132. Sort the roots even if the response decreases: 1.60846 fm ≤ dz ≤ 1.74347 fm.
- The lower and upper input deviations are 0.0647865 fm and 0.0702245 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 dz ≈ 1.67325 fm; compatible interval [1.60846, 1.74347] fm. Endpoint interpolation gives 2.02594 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.