University Quantum Field Theory I · unit UQFTI-13
Vacuum Polarisation, Charge Renormalisation and the Running of alpha
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Why a cutoff fails: a photon mass from the regulator: infer an input with uncertainty
A sharp momentum cutoff regulates the photon self-energy by discarding loop momenta with |l_E| above Λ; because that ball is not invariant under shifts of l, it adds a non-transverse g^μν Λ² term that acts as a photon mass. Use the supplied topic model Y_out = (2/d)*10^(d*nlog)/(10^(2*nlog)+1), where Y_out is LEFTOVER AT Λ (mₑ^(d−2)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nlog is Cutoff log₁₀(Λ/mₑ), dimensionless. Keep d (Dimension d) = 4 fixed. For this exercise use only 0.31875 ≤ nlog ≤ 2.68125 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 = 1654.203 mₑ^(d−2), with a bounded reading uncertainty ±247.87 mₑ^(d−2). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nlog by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, 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.
LEFTOVER AT Λ against nlog
Predictions from the supplied Why a cutoff fails: a photon mass from the regulator 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): LEFTOVER AT Λ / mₑ^(d−2)
Horizontal axis (x): nlog: Cutoff log₁₀(Λ/mₑ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nlog: Cutoff log₁₀(Λ/mₑ) | LEFTOVER AT Λ / mₑ^(d−2) |
|---|---|
| 0.31875 | 1.7636824774054516 |
| 0.9093749999999999 | 32.447203809535196 |
| 1.5 | 499.5004995004995 |
| 2.0906249999999997 | 7589.119549184469 |
| 2.68125 | 115204.14880496229 |
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(nlog) = 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(0.31875) = 1.76368 mₑ^(d−2) and Y_out(2.68125) = 115204 mₑ^(d−2). The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives nlog_line = 0.31875 + (1654.2 − 1.76368)(2.68125 − 0.31875)/(115204 − 1.76368) = 0.352637. 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 nlog = 1.75988; substituting back gives Y_out = 1654.203 mₑ^(d−2).
- The interpolation error in the input is |nlog_line − nlog| = 1.40724. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1406.333 mₑ^(d−2) and 1902.073 mₑ^(d−2). Sort the roots even if the response decreases: 1.72464 ≤ nlog ≤ 1.79019.
- The lower and upper input deviations are 0.0352388 and 0.0303106. 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 nlog ≈ 1.75988; compatible interval [1.72464, 1.79019]. Endpoint interpolation gives 0.352637.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The electron loop: Π₂(q²) in dimensional regularisation: infer an input with uncertainty
Π₂(q²) is the scalar function that multiplies (q²g^μν − q^μq^ν) in the one-loop electron-loop correction to the photon propagator; computed in d = 4 − ε dimensions it is finite except for a single pole in ε. Use the supplied topic model Y_out = 0.51099895*10^lq, where Y_out is Q FOR AN ELECTRON LOOP (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lq is Momentum transfer log₁₀(Q/m), dimensionless. Keep lm (Scale log₁₀(μ/m)) = 0 fixed. For this exercise use only -1.3625 ≤ lq ≤ 3.3625 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 = 16.911028 MeV, with a bounded reading uncertainty ±2.5333 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lq by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725, 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.
Q FOR AN ELECTRON LOOP against lq
Predictions from the supplied The electron loop: Π₂(q²) in dimensional regularisation 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): Q FOR AN ELECTRON LOOP / MeV
Horizontal axis (x): lq: Momentum transfer log₁₀(Q/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Momentum transfer log₁₀(Q/m) | Q FOR AN ELECTRON LOOP / MeV |
|---|---|
| -1.3625 | 0.02217787890116872 |
| -0.18125000000000013 | 0.3366433250745418 |
| 0.9999999999999998 | 5.109989499999997 |
| 2.1812499999999995 | 77.56575207403358 |
| 3.3625 | 1177.3890914669116 |
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(lq) = 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(-1.3625) = 0.0221779 MeV and Y_out(3.3625) = 1177.39 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lq_line = -1.3625 + (16.911 − 0.0221779)(3.3625 − -1.3625)/(1177.39 − 0.0221779) = -1.29472. 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 lq = 1.51975; substituting back gives Y_out = 16.911028 MeV.
- The interpolation error in the input is |lq_line − lq| = 2.81447. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 14.377728 MeV and 19.444328 MeV. Sort the roots even if the response decreases: 1.44927 ≤ lq ≤ 1.58037.
- The lower and upper input deviations are 0.0704797 and 0.0606229. 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 lq ≈ 1.51975; compatible interval [1.44927, 1.58037]. Endpoint interpolation gives -1.29472.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Charge renormalisation: α = 1/137.036 defined at q² = 0: infer an input with uncertainty
Charge renormalisation defines the physical charge by photon exchange at zero momentum transfer, e² = Z₃e₀² with Z₃ = 1/(1 − Π(0)), so α = e²/4π = 1/137.036 is the constant of long-range Coulomb forces and of Thomson scattering, not the coefficient in the Lagrangian. Use the supplied topic model Y_out = -1.548546*(2.302585*lam+7.579143)+0.774273*(4*0.01^lq-5/3+(1-4*0.01^lq/2)*sqrt(1+4*0.01^lq)*(2*ln(1+sqrt(1+4*0.01^lq))-ln(4*0.01^lq))), where Y_out is REGULATED Π₂(−Q²) (× 10⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm. lq is Momentum transfer log₁₀(Q/m), dimensionless. Keep lam (Cutoff log₁₀(Λ/GeV)) = 3 fixed. For this exercise use only -0.46875 ≤ lq ≤ 3.46875 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 = -16.830587 × 10⁻³, with a bounded reading uncertainty ±0.82123 × 10⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lq by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
REGULATED Π₂(−Q²) against lq
Predictions from the supplied Charge renormalisation: α = 1/137.036 defined at q² = 0 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): REGULATED Π₂(−Q²) / × 10⁻³
Horizontal axis (x): lq: Momentum transfer log₁₀(Q/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Momentum transfer log₁₀(Q/m) | REGULATED Π₂(−Q²) / × 10⁻³ |
|---|---|
| -0.46875 | -22.415963159939576 |
| 0.515625 | -21.561592724422834 |
| 1.5 | -18.370983284358914 |
| 2.484375 | -14.86559911665124 |
| 3.46875 | -11.3557029816164 |
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(lq) = 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(-0.46875) = -22.416 × 10⁻³ and Y_out(3.46875) = -11.3557 × 10⁻³. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lq_line = -0.46875 + (-16.8306 − -22.416)(3.46875 − -0.46875)/(-11.3557 − -22.416) = 1.51967. 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 lq = 1.93313; substituting back gives Y_out = -16.830587 × 10⁻³.
- The interpolation error in the input is |lq_line − lq| = 0.413457. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -17.651817 × 10⁻³ and -16.009357 × 10⁻³. Sort the roots even if the response decreases: 1.70247 ≤ lq ≤ 2.16356.
- The lower and upper input deviations are 0.230651 and 0.230432. 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 lq ≈ 1.93313; compatible interval [1.70247, 2.16356]. Endpoint interpolation gives 1.51967.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The imaginary part of vacuum polarisation is pair creation: infer an input with uncertainty
The imaginary part of the renormalised photon self-energy Π̂₂(q²) is the piece that exists only above q² = 4m², where a virtual photon is heavy enough to become a real lepton pair; the optical theorem makes it the rate for producing that pair. Use the supplied topic model Y_out = 2.43245*sqrt(max(0,1-4*0.01^ls))*(1+2*0.01^ls), where Y_out is −Im Π̂₂ AT THE DOT (× 10⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. ls is Dot position: log₁₀(√s/m), dimensionless. Keep lq (Spacelike Q: log₁₀(Q/m)) = 2 fixed. For this exercise use only 0.444375 ≤ ls ≤ 2.25562 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 = 2.4324407 × 10⁻³, with a bounded reading uncertainty ±0.0000013929 × 10⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ls by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000181125, 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.
−Im Π̂₂ AT THE DOT against ls
Predictions from the supplied The imaginary part of vacuum polarisation is pair creation 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): −Im Π̂₂ AT THE DOT / × 10⁻³
Horizontal axis (x): ls: Dot position: log₁₀(√s/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ls: Dot position: log₁₀(√s/m) | −Im Π̂₂ AT THE DOT / × 10⁻³ |
|---|---|
| 0.44437499999999996 | 2.1277971190358143 |
| 0.8971874999999999 | 2.4286041917809063 |
| 1.3499999999999999 | 2.4323917421820367 |
| 1.8028125 | 2.4324491023876686 |
| 2.2556249999999998 | 2.4324499861416022 |
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(ls) = 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(0.444375) = 2.1278 × 10⁻³ and Y_out(2.25562) = 2.43245 × 10⁻³. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ls_line = 0.444375 + (2.43244 − 2.1278)(2.25562 − 0.444375)/(2.43245 − 2.1278) = 2.25557. 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 ls = 1.54904; substituting back gives Y_out = 2.4324407 × 10⁻³.
- The interpolation error in the input is |ls_line − ls| = 0.706526. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.4324393 × 10⁻³ and 2.4324421 × 10⁻³. Sort the roots even if the response decreases: 1.5339 ≤ ls ≤ 1.56665.
- The lower and upper input deviations are 0.0151448 and 0.0176074. 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 ls ≈ 1.54904; compatible interval [1.5339, 1.56665]. Endpoint interpolation gives 2.25557.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.