University Quantum Field Theory I · topic UQFTI-11.03
Time reversal on quantised fields: an antiunitary T
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 reversal on quantised fields: an antiunitary T: predict a finite change
Time reversal on quantum fields is the antiunitary operator T that leaves the vacuum and every energy unchanged, reverses every momentum and spin, and complex-conjugates every c-number it passes, so that T φ(t, x) T⁻¹ = φ(−t, x). Use the supplied topic model Y_out = 2*pi*197.327/p, where Y_out is WAVELENGTH 2πħc/p (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Momentum p, measured in MeV. Keep t (Time t) = 3 fm/c; delta (Phase δ of the coherent state) = 60 ° fixed. For this exercise use only 131.875 MeV ≤ p ≤ 368.125 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 p = 183.85 MeV and 216.925 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WAVELENGTH 2πħc/p against p
Predictions from the supplied Time reversal on quantised fields: an antiunitary T 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): WAVELENGTH 2πħc/p / fm
Horizontal axis (x): p: Momentum p / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Momentum p / MeV | WAVELENGTH 2πħc/p / fm |
|---|---|
| 131.875 | 9.401646309837545 |
| 190.9375 | 6.493444750820694 |
| 250 | 4.959368428439305 |
| 309.0625 | 4.011622591255252 |
| 368.125 | 3.3679921415547063 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 183.85 MeV, substitute into Y_out = 2*pi*197.327/p: Y₀ = 6.74377 fm.
- At p = 216.925 MeV, the same fixed-parameter model gives Y₁ = 5.71553 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.02824 fm; the input change is Δp = 33.075 MeV.
- Average sensitivity is ΔY_out/Δp = -0.031088 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 = 6.74377 fm, Y₁ = 5.71553 fm; ΔY_out = -1.02824 fm; average sensitivity = -0.031088 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.
Time reversal on quantised fields: an antiunitary T: infer an input with uncertainty
Time reversal on quantum fields is the antiunitary operator T that leaves the vacuum and every energy unchanged, reverses every momentum and spin, and complex-conjugates every c-number it passes, so that T φ(t, x) T⁻¹ = φ(−t, x). Use the supplied topic model Y_out = 2*pi*197.327/p, where Y_out is WAVELENGTH 2πħc/p (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Momentum p, measured in MeV. Keep t (Time t) = 3 fm/c; delta (Phase δ of the coherent state) = 60 ° fixed. For this exercise use only 131.875 MeV ≤ p ≤ 368.125 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 = 4.492385 fm, with a bounded reading uncertainty ±0.16866 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.023625 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.
WAVELENGTH 2πħc/p against p
Predictions from the supplied Time reversal on quantised fields: an antiunitary T 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): WAVELENGTH 2πħc/p / fm
Horizontal axis (x): p: Momentum p / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Momentum p / MeV | WAVELENGTH 2πħc/p / fm |
|---|---|
| 131.875 | 9.401646309837545 |
| 190.9375 | 6.493444750820694 |
| 250 | 4.959368428439305 |
| 309.0625 | 4.011622591255252 |
| 368.125 | 3.3679921415547063 |
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(p) = 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(131.875) = 9.40165 fm and Y_out(368.125) = 3.36799 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 131.875 + (4.49238 − 9.40165)(368.125 − 131.875)/(3.36799 − 9.40165) = 324.099 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 p = 275.988 MeV; substituting back gives Y_out = 4.492385 fm.
- The interpolation error in the input is |p_line − p| = 48.1115 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.323725 fm and 4.661045 fm. Sort the roots even if the response decreases: 266.001 MeV ≤ p ≤ 286.753 MeV.
- The lower and upper input deviations are 9.98661 MeV and 10.7657 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 p ≈ 275.988 MeV; compatible interval [266.001, 286.753] MeV. Endpoint interpolation gives 324.099 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.