University Quantum Field Theory I · unit UQFTI-02
Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem
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.
Model interactions and their coupling dimensions: infer an input with uncertainty
A model interaction is a local, Hermitian, Lorentz-invariant polynomial in the fields added to the free Lagrangian; its coupling has mass dimension 4 minus the total dimension of the fields it multiplies, and that dimension sets how the interaction's strength changes with energy. Use the supplied topic model Y_out = 100*gs^2/(96*sqrt(3)*pi*mpsi^2), where Y_out is SCALAR PEAK Γ/m (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. mpsi is ψ mass M, measured in GeV. Keep gs (Scalar Yukawa g, a mass) = 2.5 GeV; gd (Dirac Yukawa g, a pure number) = 0.5 fixed. For this exercise use only 1.2125 GeV ≤ mpsi ≤ 2.7875 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 = 0.2533255 %, with a bounded reading uncertainty ±0.014902 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mpsi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 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.
SCALAR PEAK Γ/m against mpsi
Predictions from the supplied Model interactions and their coupling dimensions 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): SCALAR PEAK Γ/m / %
Horizontal axis (x): mpsi: ψ mass M / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mpsi: ψ mass M / GeV | SCALAR PEAK Γ/m / % |
|---|---|
| 1.2125 | 0.8138320512006709 |
| 1.6062499999999997 | 0.46373727200999953 |
| 1.9999999999999998 | 0.29911506913074665 |
| 2.39375 | 0.2088049075185491 |
| 2.7874999999999996 | 0.15398149509837547 |
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(mpsi) = 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(1.2125) = 0.813832 % and Y_out(2.7875) = 0.153981 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mpsi_line = 1.2125 + (0.253325 − 0.813832)(2.7875 − 1.2125)/(0.153981 − 0.813832) = 2.55038 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 mpsi = 2.17325 GeV; substituting back gives Y_out = 0.2533255 %.
- The interpolation error in the input is |mpsi_line − mpsi| = 0.377125 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.2384235 % and 0.2682275 %. Sort the roots even if the response decreases: 2.11202 GeV ≤ mpsi ≤ 2.24014 GeV.
- The lower and upper input deviations are 0.0612326 GeV and 0.0668872 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 mpsi ≈ 2.17325 GeV; compatible interval [2.11202, 2.24014] GeV. Endpoint interpolation gives 2.55038 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Heisenberg field, |Ω⟩ and correlation functions: infer an input with uncertainty
An n-point correlation function is the vacuum expectation value ⟨Ω|T φ(x₁)…φ(x_n)|Ω⟩ of time-ordered Heisenberg fields, taken in the ground state |Ω⟩ of the full Hamiltonian rather than in the free vacuum |0⟩. Use the supplied topic model Y_out = 1/(1+13*g^2/6144), where Y_out is |⟨0|Ω⟩|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. g is Coupling g = λ/m³, dimensionless. Keep no other adjustable input fixed. For this exercise use only 0.2125 ≤ g ≤ 1.7875 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 = 0.99709591, with a bounded reading uncertainty ±0.00042128. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown g by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
|⟨0|Ω⟩|² against g
Predictions from the supplied The Heisenberg field, |Ω⟩ and correlation functions 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): |⟨0|Ω⟩|²
Horizontal axis (x): g: Coupling g = λ/m³
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| g: Coupling g = λ/m³ | |⟨0|Ω⟩|² |
|---|---|
| 0.2125 | 0.9999044636772146 |
| 0.60625 | 0.9992229337591259 |
| 1 | 0.9978885821016729 |
| 1.3937499999999998 | 0.9959066344368563 |
| 1.7874999999999999 | 0.9932848140686653 |
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(g) = 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(0.2125) = 0.999904 and Y_out(1.7875) = 0.993285. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives g_line = 0.2125 + (0.997096 − 0.999904)(1.7875 − 0.2125)/(0.993285 − 0.999904) = 0.880734. 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 g = 1.17325; substituting back gives Y_out = 0.99709591.
- The interpolation error in the input is |g_line − g| = 0.292516. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99667463 and 0.99751719. Sort the roots even if the response decreases: 1.08459 ≤ g ≤ 1.25573.
- The lower and upper input deviations are 0.0886601 and 0.0824829. 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 g ≈ 1.17325; compatible interval [1.08459, 1.25573]. Endpoint interpolation gives 0.880734.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Interaction-picture fields and the time-ordered U(t, t₀): infer an input with uncertainty
The interaction picture evolves operators with the free Hamiltonian H₀ and puts the rest of the dynamics into U(t, t₀) = e^(iH₀(t−t₀)) e^(−iH(t−t₀)), which equals the time-ordered exponential T exp(−i∫H_I dt′) of the interaction written in free fields. Use the supplied topic model Y_out = exp(-kappa^2*tau^2), where Y_out is P IF U = e^(−iHᵢₙₜ t) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. kappa is Source strength κ, dimensionless. Keep tau (Time since switch-on ωt) = 6.3 rad fixed. For this exercise use only 0.285 ≤ kappa ≤ 0.915 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 = 1.8984723e-8, with a bounded reading uncertainty ±2.8477e-9. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kappa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000063, 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.
P IF U = e^(−iHᵢₙₜ t) against kappa
Predictions from the supplied Interaction-picture fields and the time-ordered U(t, t₀) 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): P IF U = e^(−iHᵢₙₜ t)
Horizontal axis (x): kappa: Source strength κ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kappa: Source strength κ | P IF U = e^(−iHᵢₙₜ t) |
|---|---|
| 0.28500000000000003 | 0.03980271113666721 |
| 0.4425 | 0.0004215593168198012 |
| 0.6000000000000001 | 6.23198966978223e-7 |
| 0.7575000000000001 | 1.2859282967288337e-10 |
| 0.915 | 3.70364009286819e-15 |
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(kappa) = 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(0.285) = 0.0398027 and Y_out(0.915) = 3.70364e-15. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives kappa_line = 0.285 + (1.89847e-8 − 0.0398027)(0.915 − 0.285)/(3.70364e-15 − 0.0398027) = 0.915. 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 kappa = 0.6693; substituting back gives Y_out = 1.8984723e-8.
- The interpolation error in the input is |kappa_line − kappa| = 0.2457. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6137023e-8 and 2.1832423e-8. Sort the roots even if the response decreases: 0.666664 ≤ kappa ≤ 0.672352.
- The lower and upper input deviations are 0.0026358 and 0.00305198. 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 kappa ≈ 0.6693; compatible interval [0.666664, 0.672352]. Endpoint interpolation gives 0.915.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant: infer an input with uncertainty
The S-operator S = T exp(i∫d⁴x ℒ_int) is Lorentz invariant, U(Λ)SU†(Λ) = S, because its one frame-dependent ingredient, the time ordering, changes only for spacelike-separated vertices, where the interaction densities commute. Use the supplied topic model Y_out = dt/dz, where Y_out is β NEEDED TO EQUALISE THE TIMES (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dz is Distance Δz of y along the boost, measured in fm. Keep dt (Time separation cΔt of y) = 1 fm; b (Observer's velocity β = v/c) = 0.25 fixed. For this exercise use only 0.7125 fm ≤ dz ≤ 2.2875 fm 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 = 0.59763932, with a bounded reading uncertainty ±0.024072. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dz by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 fm, 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.
β NEEDED TO EQUALISE THE TIMES against dz
Predictions from the supplied The S-operator T exp(i∫d⁴x ℒ_int) is Lorentz invariant 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): β NEEDED TO EQUALISE THE TIMES
Horizontal axis (x): dz: Distance Δz of y along the boost / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dz: Distance Δz of y along the boost / fm | β NEEDED TO EQUALISE THE TIMES |
|---|---|
| 0.7125 | 1.4035087719298245 |
| 1.10625 | 0.903954802259887 |
| 1.5 | 0.6666666666666666 |
| 1.8937499999999998 | 0.5280528052805281 |
| 2.2874999999999996 | 0.43715846994535523 |
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(dz) = 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(0.7125) = 1.40351 and Y_out(2.2875) = 0.437158. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dz_line = 0.7125 + (0.597639 − 1.40351)(2.2875 − 0.7125)/(0.437158 − 1.40351) = 2.02594 fm. 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 dz = 1.67325 fm; substituting back gives Y_out = 0.59763932.
- The interpolation error in the input is |dz_line − dz| = 0.352691 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.57356732 and 0.62171132. Sort the roots even if the response decreases: 1.60846 fm ≤ dz ≤ 1.74347 fm.
- The lower and upper input deviations are 0.0647865 fm and 0.0702245 fm. 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 dz ≈ 1.67325 fm; compatible interval [1.60846, 1.74347] fm. Endpoint interpolation gives 2.02594 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.