University Quantum Field Theory I · topic UQFTI-03.02
Symmetry factors: where the 1/4! fails to cancel
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.
Symmetry factors: where the 1/4! fails to cancel: predict a finite change
A diagram's symmetry factor S is the number of ways to permute its vertices, and the line ends at each vertex, that give back exactly the same contraction, every line end still paired with the same partner and external points held fixed; the diagram enters the series with weight 1/S. Use the supplied topic model Y_out = 0.5+1/ss, where Y_out is λ² COEFFICIENT (EXACT 0.667) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. ss is Sunset symmetry factor S, dimensionless. Keep st (Tadpole symmetry factor S) = 1 fixed. For this exercise use only 1 ≤ ss ≤ 12, at the specified test settings separated by 1. 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 ss = 3 and 4. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δss across this interval. State whether the response increases or decreases; explain why a finite difference between these settings does not require interpolation between them.
λ² COEFFICIENT (EXACT 0.667) against ss
Predictions from the supplied Symmetry factors: where the 1/4! fails to cancel 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): λ² COEFFICIENT (EXACT 0.667)
Horizontal axis (x): ss: Sunset symmetry factor S
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ss: Sunset symmetry factor S | λ² COEFFICIENT (EXACT 0.667) |
|---|---|
| 1 | 1.5 |
| 4 | 0.75 |
| 7 | 0.6428571428571428 |
| 9 | 0.6111111111111112 |
| 12 | 0.5833333333333334 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ss separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ss = 3, substitute into Y_out = 0.5+1/ss: Y₀ = 0.833333.
- At ss = 4, the same fixed-parameter model gives Y₁ = 0.75.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0833333; the input change is Δss = 1.
- Average sensitivity is ΔY_out/Δss = -0.0833333 1 per unit input. Its negative sign means that this response decreases over the specified interval.
- This finite difference compares two specified test settings without interpolating between them. A point derivative is not required. Any quantum numbers or particle counts must also obey their physical admissibility conditions.
Answer Y₀ = 0.833333, Y₁ = 0.75; ΔY_out = -0.0833333; average sensitivity = -0.0833333 1 per unit input.
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.
Symmetry factors: where the 1/4! fails to cancel: infer an input with uncertainty
A diagram's symmetry factor S is the number of ways to permute its vertices, and the line ends at each vertex, that give back exactly the same contraction, every line end still paired with the same partner and external points held fixed; the diagram enters the series with weight 1/S. Use the supplied topic model Y_out = 0.5+1/ss, where Y_out is λ² COEFFICIENT (EXACT 0.667) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. ss is Sunset symmetry factor S, dimensionless. Keep st (Tadpole symmetry factor S) = 1 fixed. For this exercise use only 1 ≤ ss ≤ 12, at the specified test settings separated by 1. 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 = 0.625, with a bounded reading uncertainty ±0.073333. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of ss consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
λ² COEFFICIENT (EXACT 0.667) against ss
Predictions from the supplied Symmetry factors: where the 1/4! fails to cancel 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): λ² COEFFICIENT (EXACT 0.667)
Horizontal axis (x): ss: Sunset symmetry factor S
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ss: Sunset symmetry factor S | λ² COEFFICIENT (EXACT 0.667) |
|---|---|
| 1 | 1.5 |
| 4 | 0.75 |
| 7 | 0.6428571428571428 |
| 9 | 0.6111111111111112 |
| 12 | 0.5833333333333334 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(1) = 1.5 and Y_out(12) = 0.583333. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives ss_line = 1 + (0.625 − 1.5)(12 − 1)/(0.583333 − 1.5) = 11.5. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to ss = 8; only the allowed settings 6, 7, 8, 9, 10, 11, 12 lie inside the stated reading bound.
- The interpolation error in the input is |ss_line − ss| = 3.5.
- The compatible allowed values span 6 to 12. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible ss: 6, 7, 8, 9, 10, 11, 12. Endpoint interpolation predicts 11.5; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.