University Quantum Field Theory I · topic UQFTI-13.06
QED's one-loop beta function and its Landau pole
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.
QED's one-loop beta function and its Landau pole: predict a finite change
QED's beta function β(e) = μ de/dμ is the rate at which the renormalised charge must change with the scale μ so that the bare charge e₀ = μ^(ε/2)e/√Z₃ stays fixed; at one loop it is +e³/(12π²) for each charged lepton. Use the supplied topic model Y_out = max(inv-2*nf*p*ln(10)/(3*pi),0), where Y_out is 1/α AT THE PROBE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; max(a,b) selects the larger value. p is Probe scale log₁₀(μ/m), measured in decades. Keep inv (Inverse coupling 1/α at μ = m) = 137; nf (Unit-charge fermions N, each of mass m) = 1 fixed. For this exercise use only 31.875 decades ≤ p ≤ 268.125 decades 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 = 83.85 decades and 116.925 decades. 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.
1/α AT THE PROBE against p
Predictions from the supplied QED's one-loop beta function and its Landau pole 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): 1/α AT THE PROBE
Horizontal axis (x): p: Probe scale log₁₀(μ/m) / decades
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe scale log₁₀(μ/m) / decades | 1/α AT THE PROBE |
|---|---|
| 31.875 | 121.42511852381216 |
| 90.9375 | 92.56577931793468 |
| 150 | 63.70644011205721 |
| 209.0625 | 34.84710090617973 |
| 268.125 | 5.987761700302258 |
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 = 83.85 decades, substitute into Y_out = max(inv-2*nf*p*ln(10)/(3*pi),0): Y₀ = 96.0289.
- At p = 116.925 decades, the same fixed-parameter model gives Y₁ = 79.8677.
- Signed change: ΔY_out = Y₁ − Y₀ = -16.1612; the input change is Δp = 33.075 decades.
- Average sensitivity is ΔY_out/Δp = -0.488624 1 per decades. 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 = 96.0289, Y₁ = 79.8677; ΔY_out = -16.1612; average sensitivity = -0.488624 1 per decades.
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.
QED's one-loop beta function and its Landau pole: infer an input with uncertainty
QED's beta function β(e) = μ de/dμ is the rate at which the renormalised charge must change with the scale μ so that the bare charge e₀ = μ^(ε/2)e/√Z₃ stays fixed; at one loop it is +e³/(12π²) for each charged lepton. Use the supplied topic model Y_out = max(inv-2*nf*p*ln(10)/(3*pi),0), where Y_out is 1/α AT THE PROBE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; max(a,b) selects the larger value. p is Probe scale log₁₀(μ/m), measured in decades. Keep inv (Inverse coupling 1/α at μ = m) = 137; nf (Unit-charge fermions N, each of mass m) = 1 fixed. For this exercise use only 31.875 decades ≤ p ≤ 268.125 decades 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 = 51.008331, with a bounded reading uncertainty ±6.7531. 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 decades, 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.
1/α AT THE PROBE against p
Predictions from the supplied QED's one-loop beta function and its Landau pole 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): 1/α AT THE PROBE
Horizontal axis (x): p: Probe scale log₁₀(μ/m) / decades
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe scale log₁₀(μ/m) / decades | 1/α AT THE PROBE |
|---|---|
| 31.875 | 121.42511852381216 |
| 90.9375 | 92.56577931793468 |
| 150 | 63.70644011205721 |
| 209.0625 | 34.84710090617973 |
| 268.125 | 5.987761700302258 |
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(31.875) = 121.425 and Y_out(268.125) = 5.98776. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 31.875 + (51.0083 − 121.425)(268.125 − 31.875)/(5.98776 − 121.425) = 175.987 decades. 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 = 175.987 decades; substituting back gives Y_out = 51.008331.
- The interpolation error in the input is |p_line − p| = 5.68434e-14 decades. This is within the requested numerical tolerance.
- Invert both limits Y_out = 44.255231 and 57.761431. Sort the roots even if the response decreases: 162.167 decades ≤ p ≤ 189.808 decades.
- The lower and upper input deviations are 13.8207 decades and 13.8207 decades. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer p ≈ 175.987 decades; compatible interval [162.167, 189.808] decades. Endpoint interpolation gives 175.987 decades.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.