University Quantum Field Theory I · topic UQFTI-03.03
Vacuum bubbles exponentiate and cancel against the denominator
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.
Vacuum bubbles exponentiate and cancel: predict a finite change
A vacuum bubble is a piece of a diagram joined to none of the external points; summed over every number of copies, the bubbles multiply each diagram by the same factor exp(Σᵢ Vᵢ), which is exactly the Gell-Mann-Low denominator, so they cancel. Use the supplied topic model Y_out = 20*2.60958*lc^4*box^3/197.327, where Y_out is PHASE E₀t AT t = 20 fm/c (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lc is Cutoff Λ on ⟨φ²⟩, measured in GeV. Keep box (Box side L) = 3 fm; nb (Bubbles kept N) = 2 fixed. For this exercise use only 0.574375 GeV ≤ lc ≤ 1.12562 GeV 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 lc = 0.69565 GeV and 0.772825 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PHASE E₀t AT t = 20 fm/c against lc
Predictions from the supplied Vacuum bubbles exponentiate and 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): PHASE E₀t AT t = 20 fm/c / rad
Horizontal axis (x): lc: Cutoff Λ on ⟨φ²⟩ / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lc: Cutoff Λ on ⟨φ²⟩ / GeV | PHASE E₀t AT t = 20 fm/c / rad |
|---|---|
| 0.574375 | 0.777248650580159 |
| 0.7121875 | 1.8371949764764257 |
| 0.8499999999999999 | 3.72780824586853 |
| 0.9878124999999999 | 6.7994836196286155 |
| 1.1256249999999999 | 11.464438329420824 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lc = 0.69565 GeV, substitute into Y_out = 20*2.60958*lc^4*box^3/197.327: Y₀ = 1.6724 rad.
- At lc = 0.772825 GeV, the same fixed-parameter model gives Y₁ = 2.54743 rad.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.875028 rad; the input change is Δlc = 0.077175 GeV.
- Average sensitivity is ΔY_out/Δlc = 11.3382 rad per GeV. Its positive sign means that this response increases 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 = 1.6724 rad, Y₁ = 2.54743 rad; ΔY_out = 0.875028 rad; average sensitivity = 11.3382 rad per GeV.
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.
Vacuum bubbles exponentiate and cancel: infer an input with uncertainty
A vacuum bubble is a piece of a diagram joined to none of the external points; summed over every number of copies, the bubbles multiply each diagram by the same factor exp(Σᵢ Vᵢ), which is exactly the Gell-Mann-Low denominator, so they cancel. Use the supplied topic model Y_out = 20*2.60958*lc^4*box^3/197.327, where Y_out is PHASE E₀t AT t = 20 fm/c (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lc is Cutoff Λ on ⟨φ²⟩, measured in GeV. Keep box (Box side L) = 3 fm; nb (Bubbles kept N) = 2 fixed. For this exercise use only 0.574375 GeV ≤ lc ≤ 1.12562 GeV 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 = 4.9108875 rad, with a bounded reading uncertainty ±0.62005 rad. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000055125 GeV, 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.
PHASE E₀t AT t = 20 fm/c against lc
Predictions from the supplied Vacuum bubbles exponentiate and 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): PHASE E₀t AT t = 20 fm/c / rad
Horizontal axis (x): lc: Cutoff Λ on ⟨φ²⟩ / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lc: Cutoff Λ on ⟨φ²⟩ / GeV | PHASE E₀t AT t = 20 fm/c / rad |
|---|---|
| 0.574375 | 0.777248650580159 |
| 0.7121875 | 1.8371949764764257 |
| 0.8499999999999999 | 3.72780824586853 |
| 0.9878124999999999 | 6.7994836196286155 |
| 1.1256249999999999 | 11.464438329420824 |
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(lc) = 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(0.574375) = 0.777249 rad and Y_out(1.12562) = 11.4644 rad. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lc_line = 0.574375 + (4.91089 − 0.777249)(1.12562 − 0.574375)/(11.4644 − 0.777249) = 0.78759 GeV. 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 lc = 0.910638 GeV; substituting back gives Y_out = 4.9108875 rad.
- The interpolation error in the input is |lc_line − lc| = 0.123048 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.2908375 rad and 5.5309375 rad. Sort the roots even if the response decreases: 0.880422 GeV ≤ lc ≤ 0.938113 GeV.
- The lower and upper input deviations are 0.0302152 GeV and 0.0274756 GeV. 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 lc ≈ 0.910638 GeV; compatible interval [0.880422, 0.938113] GeV. Endpoint interpolation gives 0.78759 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.