advanced26 min studyLesson 45 in Physics

Interactions and perturbation theory

Build the Dyson expansion, test what a small coupling really controls, and separate amplitudes from probabilities.

On this page

What you’ll understand

  • Explain the interaction picture and time-ordered Dyson series.
  • Track the first orders of an amplitude and its probability.
  • Recognize ultraviolet, secular, and strong-coupling limitations of a naive expansion.

Free fields tell us what independent excitations are. Interactions tell us how preparation and detection can be connected by a nontrivial amplitude. Exact solutions are rare, so we expand around a solvable theory. This is useful only when we understand the expansion parameter, the observable being approximated, and the regimes where additional contributions are not actually small.

Split the Hamiltonian, not reality

Use ℏ=c=1\hbar=c=1, metric (+−−−)(+---), and a real scalar with Lint=−λϕ4/4!\mathcal L_{\mathrm{int}}=-\lambda\phi^4/4!. The free Hamiltonian evolves interaction-picture operators, while the interaction evolves the states. For this interaction without derivative couplings, HI(t)=∫d3x λϕI4/4!H_I(t)=\int d^3x\,\lambda\phi_I^4/4!. The sign follows from the Legendre transform. For derivative interactions, simply writing HI=−LIH_I=-L_I can fail and must be checked.

The interaction-picture evolution operator obeys a first-order operator differential equation. Operators at different times generally do not commute, so the solution is not an ordinary exponential with its ordering ignored. Time ordering makes the later Hamiltonian act on the left. Expanding the ordered exponential inserts increasing numbers of interaction vertices, with integrals over every possible insertion time and position.

UI(tf,ti)=Texp⁡ ⁣[−i∫titfdt HI(t)]=1−i∫dt HI(t)+(−i)22!∫dt dt′ T[HI(t)HI(t′)]+⋯U_I(t_f,t_i)=T\exp\!\left[-i\int_{t_i}^{t_f}dt\,H_I(t)\right]=1-i\int dt\,H_I(t)+\frac{(-i)^2}{2!}\int dt\,dt'\,T[H_I(t)H_I(t')]+\cdots
Each time integral on the right runs over the stated initial-to-final interval. This is a formal perturbative series; regulated products and a state prescription are needed for QFT calculations.↔ Scroll wide equations horizontally.

Expand the quantity you actually observe

An amplitude is complex, and a probability is its modulus squared. If a transition has no free contribution and its first amplitude is proportional to λ\lambda, the leading probability is proportional to λ2\lambda^2. If a free amplitude is already present, its interference with a first-order correction can instead change the probability at order λ\lambda. Counting vertices without identifying the observable can therefore give the wrong leading power.

For scalar quartic interactions in four spacetime dimensions, λ\lambda has zero mass dimension. A loop often introduces a factor of order 1/(16π2)1/(16\pi^2), but that is a useful estimate rather than a universal convergence theorem. Large logarithms of scale ratios, many available states, thresholds, or long evolution times can enhance nominally higher-order terms. The meaningful question is whether the neglected contribution is small relative to the retained prediction in the regime being studied.

A=A0+λA1+λ2A2+⋯ ,∣A∣2=∣A0∣2+2λRe⁡(A0∗A1)+O(λ2)\mathcal A=\mathcal A_0+\lambda\mathcal A_1+\lambda^2\mathcal A_2+\cdots,\qquad |\mathcal A|^2=|\mathcal A_0|^2+2\lambda\operatorname{Re}(\mathcal A_0^*\mathcal A_1)+O(\lambda^2)
The coupling is taken real here. When A₀ is zero, the first displayed probability terms vanish and the leading nonzero term is λ²|A₁|².↔ Scroll wide equations horizontally.

A controlled approximation has an error story

At coincident field insertions, perturbative expressions develop ultraviolet divergences. A regulator makes intermediate expressions defined; renormalization relates parameters to measurements and organizes the allowed counterterms. A divergent bare integral is not itself an infinite prediction for a detector. Conversely, discarding a divergence without specifying a renormalization prescription does not produce a uniquely meaningful finite answer. The next module develops this distinction systematically.

Even after renormalization, many QFT series are asymptotic rather than convergent. More terms can improve an estimate up to an appropriate order without promising convergence at arbitrarily high order. Near strong coupling or nonperturbative effects, expansion around the free vacuum may miss the important physics. We will state tree-level or leading-order assumptions explicitly and avoid interpreting a truncated expansion as a complete microscopic history.

When a first correction changes a probability

Consider a finite-dimensional transition amplitude A(λ)=0.6+0.2iλ+O(λ2)\mathcal A(\lambda)=0.6+0.2i\lambda+O(\lambda^2). What part of its probability can be determined from this information? Compare a channel with A0=0\mathcal A_0=0.

  1. The zeroth-order probability is 0.62=0.360.6^2=0.36. The linear correction is 2λRe⁡(0.6×0.2i)=02\lambda\operatorname{Re}(0.6\times0.2i)=0, because this correction is ninety degrees out of phase.

  2. Squaring only the known terms produces 0.04λ20.04\lambda^2, but that is not the complete second-order coefficient. The unknown amplitude λ2A2\lambda^2\mathcal A_2 can interfere with 0.60.6 at the same order.

  3. If instead A0=0\mathcal A_0=0 and A1=0.2i\mathcal A_1=0.2i, the leading nonzero probability is 0.04λ20.04\lambda^2. There is no zeroth-order amplitude for the unknown second-order term to interfere with.

The result

The first channel is known only through a vanishing linear correction; its full quadratic correction is undetermined. The second channel’s leading quadratic probability is determined.

Explore the model

Interactive lab

Interactive lab · scattering

Read a coupling power without overinterpreting it

Change the toy coupling g while holding the angle fixed. The displayed dimensionless weight is g⁴(1 + cos²θ), so halving g divides this illustrative weight by sixteen. Then vary the angle to separate coupling dependence from angular dependence.

Read a coupling power without overinterpreting itA schematic term diagram with dimensionless weight g⁴(1 + cos²θ). Lines denote terms, not particle paths; this is not a process cross-section or loop calculation. Current readings: Coupling factor g⁴: 0.24; Angular factor 1 + cos²θ: 1.329; Illustrative weight: 0.319. All parameters are adjustable with the labeled controls below.Incoming termsOutgoing termsθSchematic interaction term · no particle trajectories
Incoming termsOutgoing termsAngle θ
Coupling factor g⁴
0.24
Angular factor 1 + cos²θ
1.329
Illustrative weight
0.319
t = 0 s

A schematic term diagram with dimensionless weight g⁴(1 + cos²θ). Lines denote terms, not particle paths; this is not a process cross-section or loop calculation. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis toy weight is neither a φ⁴ cross section nor a Dyson-series solver. It makes power counting visible; the worked amplitude example explains why the physical power depends on the channel and interference.

Check your understanding

You know A₀ and A₁ for a channel with A₀ ≠ 0. Can |A₁|² alone give the complete second-order probability correction?

Try it yourself

Suppose a correction behaves as ϵlog⁡(Q/μ)\epsilon\log(Q/\mu) with ϵ=0.02\epsilon=0.02. Compare Q/μ=2Q/\mu=2 with Q/μ=e100Q/\mu=e^{100}. What assumption fails in the second case?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Evaluate the full dimensionless product rather than looking at epsilon alone.

Compare with the solution

For a ratio of two, the product is about 0.01390.0139, plausibly a small correction. For e100e^{100}, it is 22, so the logarithm cancels the smallness of the coefficient. A fixed-order expansion in this combination is not controlled; choosing scales appropriately or resumming logarithms may be required. This estimate diagnoses a problem but does not itself perform a resummation.

Take it with you

  • Time ordering is required because interaction Hamiltonians need not commute.
  • Amplitude order and observable order are related by interference, not a universal squaring rule.
  • Small couplings require checks for enhanced terms, ultraviolet structure, and nonperturbative limits.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.