When particles start to interact.

QFT document 4: from free fields to calculable amplitudes. The machinery that connects field operators to physical predictions.

Documents 1-3 quantized free fields. That’s interesting mathematically but physically sterile; particles fly past each other without interacting. Real physics (scattering, decay, binding) requires interactions.

The challenge: the Hamiltonian of an interacting QFT is typically not exactly solvable. No known theory of interacting quantum fields in 4D spacetime admits closed-form solutions for the full Hilbert space. What we can do is perturbation theory; treat the interaction as a small correction to the free theory, expand in powers of the coupling, and compute term by term.

This document builds the three pillars of perturbative QFT:

1. The interaction picture of quantum mechanics (where the simple dynamics comes from the free Hamiltonian)
2. Dyson’s formula for time evolution (a time-ordered exponential of the interaction)
3. Wick’s theorem for reducing operator products to propagators
4. The LSZ reduction formula connecting correlation functions to scattering amplitudes

Together, these convert the abstract “correlation functions of fields” into numbers you can compare to experiment.

Prerequisites and Conventions

• QFT documents 1, 2, 3 (scalar, Dirac, photon quantization)
• Familiarity with the interaction picture from QM
• Conventions as before: $\hbar = c = 1$, $\eta_{\mu\nu} = \text{diag}(+,-,-,-)$, Einstein summation

Table of Contents

1. The Central Problem
2. The Interaction Picture
3. The Time Evolution Operator
4. Dyson’s Formula
5. The S-Matrix
6. Correlation Functions and the Gell-Mann–Low Theorem
7. Wick’s Theorem
8. The LSZ Reduction Formula
9. Cross Sections and Decay Rates
10. Example: First-Order $\phi^4$ Theory
11. Preview: Where Feynman Diagrams Come From
12. Appendix: Formulas and Identities

1. The Central Problem

What We Want to Compute

A typical experimental question: if I collide an electron and a positron at 500 GeV each, what’s the probability of producing a $\mu^+\mu^-$ pair flying in specific directions?

The answer involves computing an amplitude:

$\mathcal{M}(e^+ e^- \to \mu^+ \mu^-) = \langle \mu^+, \mu^- | \hat S | e^+, e^-\rangle$

where $\hat S$ is an operator (the S-matrix) that maps initial states to final states, and the in/out states are the multi-particle states we built in Fock space.

The problem: $\hat S$ is defined by the full interacting time evolution, which we can’t compute exactly. We need to expand it perturbatively.

The Setup

Split the full Hamiltonian into a free part plus an interaction:

$H = H_0 + H_{\text{int}}$

• $H_0$: the free Hamiltonian (known from documents 1-3). Its eigenstates are the Fock states we’ve been working with.
• $H_{\text{int}}$: the interaction term (e.g., $e\bar\psi\gamma^\mu\psi A_\mu$ for QED).

In the free theory, Fock states are energy eigenstates and evolve via simple phases: $|\vec p\rangle \to e^{-iE_pt}|\vec p\rangle$. Multi-particle states are sums of such phases. Boring.

In the interacting theory, Fock states are not eigenstates of the full $H$. They evolve into complicated superpositions involving arbitrary numbers of other particles. That’s scattering.

The question: how do we compute this evolution perturbatively?

Why the Schrödinger Picture Fails (for Us)

In the Schrödinger picture, states evolve and operators are fixed:

$i\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$

For computations, this is awkward: the states we know how to write (multi-particle Fock states) are eigenstates of $H_0$, not $H$. In the Schrödinger picture, time evolution immediately mixes them into unknown superpositions.

The Heisenberg Picture Doesn’t Help Either

In the Heisenberg picture, operators evolve and states are fixed:

$\phi(\vec x, t) = e^{iHt}\phi(\vec x)e^{-iHt}$

This requires using the full Hamiltonian $H$ for time evolution; and we don’t know how to evaluate $e^{-iHt}$ when $H$ is interacting.

The Fix: Interaction Picture

Split the difference: evolve operators with $H_0$ (which we can do), and let the states carry whatever’s left (which is where the interaction enters).

The interaction picture is a computational choice, not a physical one. The physics; measurable quantities; is identical in any picture.

2. The Interaction Picture

Definition

Split the Hamiltonian: $H = H_0 + H_{\text{int}}$. Define operators and states in the interaction picture by:

$\phi_I(t) = e^{iH_0 t}\phi_S e^{-iH_0 t}$

$|\psi(t)\rangle_I = e^{iH_0 t}|\psi(t)\rangle_S$

where subscripts denote picture ($I$ = interaction, $S$ = Schrödinger). This is a “partial unitary transformation”; we’ve absorbed the trivial free evolution into the operators, leaving the states to carry the interaction-induced dynamics.

Equation of Motion for Operators

Differentiate $\phi_I(t)$:

$i\frac{\partial \phi_I}{\partial t} = [\phi_I, H_0]$

Operators in the interaction picture evolve with the free Hamiltonian $H_0$, not $H$. This is crucial: $\phi_I(t)$ satisfies exactly the same equation of motion as the free field in the Heisenberg picture. So we already know how to work with it; it’s the field operator from documents 1-3.

Equation of Motion for States

Differentiate $|\psi(t)\rangle_I$:

$i\frac{\partial}{\partial t}|\psi(t)\rangle_I = -H_0 e^{iH_0 t}|\psi(t)\rangle_S + e^{iH_0 t}H|\psi(t)\rangle_S$

$= e^{iH_0 t}(H - H_0)|\psi(t)\rangle_S = e^{iH_0 t}H_{\text{int}}e^{-iH_0 t}|\psi(t)\rangle_I$

Define the interaction Hamiltonian in the interaction picture:

$H_I(t) \equiv e^{iH_0 t}H_{\text{int}}e^{-iH_0 t}$

Then:

$\boxed{i\frac{\partial}{\partial t}|\psi(t)\rangle_I = H_I(t)|\psi(t)\rangle_I}$

States in the interaction picture evolve with $H_I(t)$; the interaction Hamiltonian in which the fundamental fields have been replaced by their free-theory (interaction picture) versions.

Summary

The interaction picture separates the “boring” part (free evolution) from the “interesting” part (interaction-induced scattering). We’ve handed the easy bit to the operators and left the hard bit for the states.

Why This Helps

In the interaction picture:

1. Field operators $\phi_I(x)$, $\psi_I(x)$, $A_I^\mu(x)$ are free-field operators; exactly the ones from documents 1-3. We know their commutators, their vacuum expectation values, their propagators.

2. The interaction Hamiltonian $H_I(t)$ is a product of these free-field operators (with a coupling constant). For example, QED:

$H_I(t) = e\int d^3x\, \bar\psi_I(x)\gamma^\mu\psi_I(x)A_{I\mu}(x)$

where every field is free.

3. Time evolution of states is governed by a differential equation whose right side involves only free-field operators. We can hope to solve this order by order in the coupling $e$.

The price: everything is now frame-dependent. But measurable quantities come out the same as Lorentz-covariant calculations would give.

3. The Time Evolution Operator

Defining $U$

Define the interaction-picture time evolution operator $U(t, t_0)$ by:

$|\psi(t)\rangle_I = U(t, t_0)|\psi(t_0)\rangle_I$

From the state’s equation of motion:

$i\frac{\partial U(t, t_0)}{\partial t} = H_I(t)U(t, t_0), \quad U(t_0, t_0) = 1$

Why Not Just Exponentiate?

In ordinary QM (time-independent Hamiltonian), we’d write $U(t, t_0) = e^{-iH(t - t_0)}$. Why not here?

The problem: $H_I(t)$ depends on $t$, and operators at different times generally don’t commute:

$[H_I(t_1), H_I(t_2)] \neq 0 \text{ in general}$

For an ordinary function $f(t)$, $\exp\int f(t)dt$ makes sense. For a non-commuting operator-valued function $H_I(t)$, the naive exponential doesn’t solve the differential equation; $\frac{d}{dt}e^{-i\int_{t_0}^t H_I(t')dt'}$ is a mess.

The Dyson Series

Formally integrate the differential equation:

$U(t, t_0) = 1 - i\int_{t_0}^t dt_1\, H_I(t_1) U(t_1, t_0)$

This is an implicit equation; $U$ appears on both sides. Iterate by substituting the right side into itself:

$U(t, t_0) = 1 - i\int_{t_0}^t dt_1 H_I(t_1) + (-i)^2\int_{t_0}^t dt_1\int_{t_0}^{t_1} dt_2\, H_I(t_1)H_I(t_2) + \cdots$

This is an expansion in powers of $H_I$. If the coupling is small, higher terms are suppressed. Each $H_I$ brings a factor of the coupling constant (e.g., $e$ in QED).

The Nested Integration Domain

The $n$-th term has nested integrations:

$(-i)^n\int_{t_0}^t dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_n\, H_I(t_1)H_I(t_2)\cdots H_I(t_n)$

with $t_1 > t_2 > \cdots > t_n$ (later times on the left). This ordering matters because the operators don’t commute.

The nested-integral form is awkward. We’d prefer a symmetrical expression where every $t_i$ is integrated from $t_0$ to $t$. That’s what Dyson’s formula provides.

4. Dyson’s Formula

The Time-Ordering Operator

Define the time-ordering operator $T$ acting on a product of operators:

$T\{A(t_1)A(t_2)\cdots A(t_n)\} = A(t_{\sigma(1)})A(t_{\sigma(2)})\cdots A(t_{\sigma(n)})$

where $\sigma$ is the permutation that arranges the arguments in decreasing order: $t_{\sigma(1)} > t_{\sigma(2)} > \cdots$.

For two operators:

$T\{A(t_1)A(t_2)\} = \theta(t_1 - t_2)A(t_1)A(t_2) + \theta(t_2 - t_1)A(t_2)A(t_1)$

For fermionic operators, we include a minus sign for each swap; but $H_I$ is bosonic (contains even products of fermions), so this subtlety doesn’t affect Dyson’s formula as stated below.

The Dyson Formula

Using the time-ordering operator:

$\boxed{U(t, t_0) = T\exp\left[-i\int_{t_0}^t dt'\, H_I(t')\right]}$

Expanded:

$U(t, t_0) = \sum_{n=0}^\infty \frac{(-i)^n}{n!}\int_{t_0}^t dt_1\cdots\int_{t_0}^t dt_n\, T\{H_I(t_1)\cdots H_I(t_n)\}$

Why the $1/n!$?

Consider the $n$-th term. We want to integrate over the unordered region $t_1, t_2, \ldots, t_n \in [t_0, t]$. This region has $n!$ different orderings, but time ordering treats each by arranging into a specific ordering. By symmetry of the integrand under time ordering (which gives the same operator product regardless of which ordering of integration variables we start with), each of the $n!$ orderings contributes equally. So integrating over the full unordered region gives $n!$ times the original ordered result; we divide by $n!$ to compensate.

Formally:

$\int_{t_0}^t dt_1\int_{t_0}^{t_1}dt_2\cdots H_I(t_1)H_I(t_2)\cdots = \frac{1}{n!}\int_{t_0}^t dt_1\cdots\int_{t_0}^t dt_n\, T\{H_I(t_1)\cdots H_I(t_n)\}$

Verification

Check that Dyson’s formula satisfies the differential equation. Differentiate:

$i\frac{\partial}{\partial t}U(t, t_0) = i\frac{\partial}{\partial t}\sum_n \frac{(-i)^n}{n!}\int dt_1\cdots dt_n T\{H_I(t_1)\cdots H_I(t_n)\}$

The derivative hits the upper limit of integration. Each integral contributes a term where $t_i = t$ and we’ve lost one integration. After some algebra (cleaner in the nested-integral form), this gives $H_I(t) U(t, t_0)$. ✓

Why It’s Beautiful

Dyson’s formula is remarkable: it rewrites time evolution in an interacting theory as a time-ordered exponential of free-theory operators. No nested integrals; just unordered integrations with the time-ordering handled by the $T$ symbol.

Every perturbative QFT calculation starts with Dyson’s formula. Feynman diagrams (next document) are a graphical representation of its expansion.

5. The S-Matrix

Definition

The S-matrix (scattering matrix) is the interaction-picture time evolution from the far past to the far future:

$S = \lim_{t_\pm \to \pm\infty} U(t_+, t_-)$

Using Dyson’s formula:

$S = T\exp\left[-i\int_{-\infty}^\infty dt\, H_I(t)\right] = T\exp\left[-i\int d^4x\, \mathcal{H}_I(x)\right]$

where $\mathcal{H}_I$ is the interaction Hamiltonian density.

For QED in the interaction picture:

$\mathcal{H}_I(x) = e\bar\psi_I(x)\gamma^\mu\psi_I(x)A_{I\mu}(x)$

So:

$S = T\exp\left[-ie\int d^4x\, \bar\psi_I\gamma^\mu\psi_I A_{I\mu}\right]$

S-Matrix Elements

A scattering process $|i\rangle \to |f\rangle$ has amplitude:

$S_{fi} = \langle f | S | i\rangle$

where $|i\rangle$ and $|f\rangle$ are multi-particle states (wave packets, strictly; we’ll use plane-wave idealization).

The probability for the transition:

$P_{i \to f} = |S_{fi}|^2$

which, after factoring out momentum-conservation delta functions and proper normalization, becomes a cross section or decay rate.

The Trivial Part of $S$

$S = 1 + \text{(interaction part)}$. The “1” represents non-interacting evolution (particles fly past without interacting). We extract the nontrivial part:

$S = 1 + iT$

where $T$ is the transition matrix. For scattering amplitudes, we often write:

$\langle f | S | i \rangle = \delta_{fi} + (2\pi)^4\delta^4(p_f - p_i)\, i\mathcal{M}(i \to f)$

The delta function enforces total 4-momentum conservation; $\mathcal{M}$ is the invariant amplitude, which is what perturbation theory actually computes.

Unitarity

$S$ is a unitary operator: $S^\dagger S = S S^\dagger = 1$. This is required because time evolution must conserve probability. Unitarity constrains $\mathcal{M}$ through the optical theorem:

$2\,\text{Im}\,\mathcal{M}(p \to p) = \sum_f |\mathcal{M}(p \to f)|^2$

(for forward scattering $p \to p$, with the sum over all possible final states $f$). This is how bound-state poles and resonances show up in QFT calculations.

6. Correlation Functions and the Gell-Mann–Low Theorem

The Road to Amplitudes

To actually compute $S$-matrix elements, we need to evaluate vacuum expectation values of time-ordered products of fields:

$\langle 0 | T\{\phi(x_1)\phi(x_2)\cdots\phi(x_n)\}|0\rangle$

These are called n-point correlation functions or Green’s functions, and they’re the central objects of QFT.

Why correlation functions? Because the LSZ reduction formula (next section) extracts scattering amplitudes from them.

The Problem with the Vacuum

A subtlety: the "$|0\rangle$" in the interacting theory is not the same as the free-theory vacuum. The free vacuum is annihilated by free-field annihilation operators; the interacting vacuum is annihilated by the full Hamiltonian.

Let’s use notation: $|\Omega\rangle$ for the interacting vacuum, $|0\rangle$ for the free vacuum. The correlation functions we want are:

$\langle\Omega|T\{\phi_H(x_1)\cdots\phi_H(x_n)\}|\Omega\rangle$

where $\phi_H$ is the Heisenberg-picture field (evolving with the full $H$). But we only know how to compute with interaction-picture fields and the free vacuum.

The Gell-Mann–Low Theorem

The bridge:

$\boxed{\langle\Omega|T\{\phi_H(x_1)\cdots\phi_H(x_n)\}|\Omega\rangle = \frac{\langle 0|T\{\phi_I(x_1)\cdots\phi_I(x_n)S\}|0\rangle}{\langle 0|T\{S\}|0\rangle}}$

where $S$ is the S-matrix as an operator (Dyson-expanded).

Key features:

• Left side: what we want (physical correlation functions in the interacting theory)
• Right side: computable, involving only the free vacuum and interaction-picture fields
• The denominator $\langle 0|S|0\rangle$ accounts for the “vacuum bubbles”; disconnected diagrams; which factor out

Why This Works (Sketch)

The interacting vacuum $|\Omega\rangle$ can be obtained from the free vacuum $|0\rangle$ by adiabatic switching: slowly turn on the interaction, and the free vacuum evolves into the true interacting vacuum. Mathematically:

$|\Omega\rangle \propto \lim_{T \to \infty(1-i\epsilon)}U(0, -T)|0\rangle$

Inserting this into the correlation function definition and manipulating (this is the proper statement of the Gell-Mann–Low theorem) gives the result above. The $i\epsilon$ prescription projects onto the actual ground state.

What This Means Practically

We’ve reduced the problem:

1. Expand $S$ using Dyson’s formula in powers of the coupling
2. Compute each term; a time-ordered product of free fields; using Wick’s theorem (next section)
3. Sum up the contributions (order by order)

Every QFT calculation reduces to computing correlation functions of free fields. That’s what we’ll learn to do next.

7. Wick’s Theorem

The Problem

Dyson’s formula expresses $S$ as a sum of terms like $T\{H_I(x_1)\cdots H_I(x_n)\}$, each a time-ordered product of free fields. To compute matrix elements, we need to evaluate these products.

Direct evaluation (using commutation relations to reorder) is a combinatorial nightmare for more than a few fields. Wick’s theorem automates the procedure.

Normal-Ordered Products

Recall normal ordering ($:\!\cdots\!:$) puts creation operators to the left of annihilation operators:

$:\!a^\dagger a\!: = a^\dagger a, \qquad :\!a a^\dagger\!: = a^\dagger a$

(For fermions, include a sign for each anticommutation needed.)

Key fact: $\langle 0|:\!\cdots\!:|0\rangle = 0$ unless the expression is itself the identity (all operators cancel).

Contractions

For two fields, the contraction is defined as:

$\overbrace{A(x_1)A(x_2)}^{\text{contraction}} \equiv \langle 0|T\{A(x_1)A(x_2)\}|0\rangle$

For scalar fields: the Feynman propagator $D_F(x_1 - x_2)$.

For Dirac fields: the fermion propagator $S_F(x_1 - x_2)$.

For photons: the photon propagator $D_F^{\mu\nu}(x_1 - x_2)$.

All of these are c-numbers (not operators); just functions of the coordinates.

Wick’s Theorem: The Statement

For any product of free fields:

$T\{A_1 A_2 \cdots A_n\} = :\!A_1 A_2 \cdots A_n\!: + \text{(all possible contractions)}$

where the sum includes:

• All single contractions: pair up two fields, contract them, leave the rest in normal order
• All double contractions: pair up four fields into two pairs, contract each pair, leave rest in normal order
• …
• Full contractions: pair up all fields (only possible for even $n$)

For a product of $2n$ scalar fields, the full contraction has $(2n-1)!!$ terms.

Simple Example: Two Fields

$T\{\phi(x)\phi(y)\} = :\!\phi(x)\phi(y)\!: + \overbrace{\phi(x)\phi(y)}^{\text{contraction}}$

$= :\!\phi(x)\phi(y)\!: + D_F(x - y)$

Taking the vacuum expectation value (normal-ordered part vanishes):

$\langle 0|T\{\phi(x)\phi(y)\}|0\rangle = D_F(x - y)$

Which we already knew; but this is the simplest case of the theorem.

Four Fields

$T\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\}$

Single contractions (three choices of pair × three remaining pairs):

• $\phi_1 \phi_2$, leaving $:\phi_3\phi_4:$, times $D_F(x_1 - x_2)$
• $\phi_1 \phi_3$, leaving $:\phi_2\phi_4:$, times $D_F(x_1 - x_3)$
• $\phi_1 \phi_4$, leaving $:\phi_2\phi_3:$, times $D_F(x_1 - x_4)$
• $\phi_2 \phi_3$, leaving $:\phi_1\phi_4:$, times $D_F(x_2 - x_3)$
• …etc (6 single contractions total)

Double contractions (3 pairings):

• $D_F(x_1 - x_2)D_F(x_3 - x_4)$
• $D_F(x_1 - x_3)D_F(x_2 - x_4)$
• $D_F(x_1 - x_4)D_F(x_2 - x_3)$

Vacuum expectation (only double contractions survive):

$\langle 0|T\{\phi_1\phi_2\phi_3\phi_4\}|0\rangle = D_F(x_1 - x_2)D_F(x_3 - x_4) + D_F(x_1 - x_3)D_F(x_2 - x_4) + D_F(x_1 - x_4)D_F(x_2 - x_3)$

Three propagator products. Each corresponds to a different way of pairing up the four points. These will become the three “topologies” of a tree-level 4-point Feynman diagram.

Fermions: Signs Matter

For fermion fields, swapping two fields generates a minus sign. Wick’s theorem statements must account for this:

$T\{\psi_a \psi_b \psi_c \psi_d\} = :\!\psi\psi\psi\psi\!: + S_F(a - b):\!\psi_c\psi_d\!: - S_F(a - c):\!\psi_b\psi_d\!: + \cdots$

The sign for each contraction is $(-1)^{\text{number of fermion anticommutations to bring contracted fields together}}$. Getting these signs right is tedious but mechanical.

Why Wick’s Theorem Matters

It converts a scary operator product into a finite sum of c-number propagator products. Each term can be evaluated as a simple integral.

The theorem also directly motivates Feynman diagrams: each pairing of fields corresponds to a “line” connecting two spacetime points, and each line corresponds to a propagator.

For a general $n$-point function of free scalars:

• Odd $n$: vacuum expectation vanishes (no way to fully contract an odd number of fields)
• Even $n$: the VEV is a sum over all ways of pairing the $n$ fields, with each pair contributing a propagator

This is why Feynman diagrams for scalar field theories have only even-legged vertices; reflecting Wick’s theorem applied to interaction terms.

Generalization: Wick for Interacting Fields

When we’re computing $\langle 0|T\{\phi(x_1)\cdots\phi(x_n)S\}|0\rangle$, the $S$-operator brings additional field products (from expanding the interaction Lagrangian). Wick’s theorem applies to the whole expression, giving contractions between the “external” $\phi(x_i)$ fields and the “internal” fields from $S$.

This is how interactions enter the calculation. Each interaction term in $S$ contributes fields that get contracted with external fields and with each other; producing Feynman diagrams with vertices.

Proof Idea

Wick’s theorem is proved by induction. The base case (two fields) is checked directly. The inductive step uses the fact that:

$T\{A\, B\} - :\!AB\!: = \text{c-number}$

for any two free fields $A$, $B$ (since moving all annihilation operators to the right produces only c-number contributions from the commutators). Extending to $n$ fields gives the combinatorial sum over pairings.

The Key Payoff

With Dyson + Wick, we have an algorithm:

1. Write $S$ as a time-ordered exponential of $H_I$
2. Expand to desired order
3. Each term is a time-ordered product of free fields
4. Apply Wick’s theorem to reduce to propagators
5. Integrate

This algorithm is exactly what Feynman diagrams encode graphically. Document 5 will show the translation.

8. The LSZ Reduction Formula

The Problem

Correlation functions $\langle\Omega|T\{\phi(x_1)\cdots\phi(x_n)\}|\Omega\rangle$ are the natural objects of QFT; they’re what Wick + Dyson compute. But they’re not directly scattering amplitudes. Experimentally, we measure amplitudes like $\langle p_1', p_2'|S|p_1, p_2\rangle$; matrix elements between definite-momentum states.

The LSZ reduction formula (Lehmann-Symanzik-Zimmermann, 1955) bridges the gap.

The Idea

An amplitude for $n$ in-going particles to become $m$ out-going particles can be extracted from the $(n+m)$-point correlation function by:

1. Fourier-transforming the correlation function to momentum space
2. Amputating external propagators (dividing out the $1/(p^2 - m^2)$ factors)
3. Going on-shell ($p^2 = m^2$ for each external particle)
4. Multiplying by appropriate wave function factors

The poles of the correlation function at $p_i^2 = m^2$ contain the physical amplitude information. LSZ systematically extracts them.

The Formula (Scalar Fields)

For scattering $p_1 p_2 \to p_3 \cdots p_{n+2}$:

$\langle p_3, \ldots, p_{n+2}|S|p_1, p_2\rangle = \prod_i \int d^4x_i e^{\pm ip_i \cdot x_i}\left(i\sqrt{Z}(\partial^2 + m^2)_i\right)\langle\Omega|T\{\phi(x_1)\cdots\phi(x_{n+2})\}|\Omega\rangle$

Each incoming particle gets an $e^{-ip\cdot x}$ factor; outgoing gets $e^{+ip\cdot x}$. The $(\partial^2 + m^2)$ factor applies the Klein-Gordon operator to each external leg, which produces inverse propagators in momentum space; i.e., cancels (amputates) the external propagator. The $\sqrt{Z}$ is the field-strength renormalization, which we’ll discuss below.

The Ingredients

External Klein-Gordon operators: For a free field, $(\Box + m^2)\phi = 0$. But in an interacting theory, $(\Box + m^2)\phi = \text{source}$ where the source involves the interaction. When Fourier-transformed, this gives $(-p^2 + m^2)\tilde\phi =$ something, meaning $\tilde\phi$ has poles at $p^2 = m^2$. LSZ says: extract the residue at that pole.

Amputation: An $(n+m)$-point function has external propagators attached to each vertex. These external propagators are just “free legs” that don’t contribute to the scattering itself; they just propagate in/out. Amputating them isolates the interaction kernel.

On-shell condition: Physical particles have $p^2 = m^2$. External legs must be on-shell to represent real particles.

Field-strength renormalization $Z$: In an interacting theory, $\langle\Omega|\phi(0)|p\rangle \neq 1$ as it would be in the free theory; instead, it equals $\sqrt{Z}$. This factor accounts for the fact that the physical “particle” associated with $\phi$ differs from the bare free-theory particle. $Z$ is generally computed perturbatively.

A Crucial Simplification

At lowest order in perturbation theory, $Z = 1$ and the complications vanish. At higher orders, $Z$ receives corrections. But at tree level, the LSZ formula simplifies dramatically; you just compute the correlation function using Dyson + Wick, identify the appropriate $n$-point function, Fourier transform, and read off the amplitude.

The story: in momentum space, the amputation and on-shell procedure amounts to not including external-leg propagators in Feynman diagrams. External lines are just wave function factors; only internal lines carry propagators.

LSZ for Fermions

Similar structure, but with $(i\slashed\partial - m)$ instead of $(\Box + m^2)$, and external-line factors $u^s(p)$, $\bar u^s(p)$, $v^s(p)$, $\bar v^s(p)$ for incoming/outgoing particles/antiparticles.

LSZ for Photons

$(-\Box \delta^{\mu\nu} + \partial^\mu\partial^\nu)/\xi$ type factors (depending on gauge), external factors $\epsilon^\mu(k)$ or $\epsilon^{\mu*}(k)$ for polarization.

Summary of the Full Recipe

To compute $\langle f|S|i\rangle$:

1. Determine the external particles (number, types, momenta) and Fock states $|i\rangle$, $|f\rangle$
2. Identify the needed correlation function
3. Use Dyson’s formula: $S = T\exp(-i\int H_I)$
4. Expand to desired order in the coupling
5. Use Wick’s theorem: convert the operator product to propagator products
6. Fourier transform and amputate external legs (LSZ)
7. Result: the amplitude $\mathcal{M}$

This procedure will be automated in the next document via Feynman diagrams; but conceptually, this is what’s happening.

9. Cross Sections and Decay Rates

To connect amplitudes $\mathcal{M}$ to experimentally measurable quantities, we need formulas for cross sections and decay rates.

The Setup

Transition probability per unit time per unit flux:

$d\sigma = \frac{|\mathcal{M}|^2}{\text{flux}}d\Pi$

where $d\Pi$ is the Lorentz-invariant phase space for the final state, and the flux involves the initial-state momenta.

Lorentz-Invariant Phase Space

For $n$ final-state particles with momenta $p_i$:

$d\Pi = (2\pi)^4\delta^4\left(\sum p_i - P_i\right)\prod_{i=1}^n \frac{d^3 p_i}{(2\pi)^3 2E_i}$

The delta function enforces 4-momentum conservation. The $d^3p/(2E)$ factor is the Lorentz-invariant momentum-space measure.

Cross Section for 2 → $n$ Scattering

For two incoming particles with momenta $p_1$, $p_2$ and masses $m_1$, $m_2$:

$d\sigma = \frac{1}{2E_1 \cdot 2E_2 \cdot |\vec v_1 - \vec v_2|}|\mathcal{M}|^2 \, d\Pi$

The factor $2E_1 \cdot 2E_2 \cdot |\vec v_1 - \vec v_2|$ is the flux (particle density times relative velocity).

In terms of Mandelstam variable $s = (p_1 + p_2)^2$, this simplifies in the high-energy limit:

$d\sigma \approx \frac{|\mathcal{M}|^2}{2s}d\Pi$

Decay Rate

For a particle of mass $M$ decaying into $n$ final-state particles:

$d\Gamma = \frac{1}{2M}|\mathcal{M}|^2\, d\Pi$

The lifetime is $\tau = 1/\Gamma_{\text{total}}$ where $\Gamma_{\text{total}}$ is the total rate summed over all final states.

Averaging and Summing over Spins

For unpolarized scattering, you average over initial spin configurations and sum over final ones:

$\overline{|\mathcal{M}|^2} = \frac{1}{2s_1 + 1}\cdot\frac{1}{2s_2 + 1}\sum_{\text{spins}}|\mathcal{M}|^2$

For fermion scattering (spin 1/2 each), this gives $\tfrac{1}{4}\sum_{\text{spins}}|\mathcal{M}|^2$.

The spin-summed squared amplitude can be computed using the completeness relations $\sum_s u^s\bar u^s = \slashed p + m$ from the workbook. The calculation reduces to traces of gamma matrix products; which is why the trace identities (workbook II.3) are so important.

Example: 2 → 2 Scattering, Center-of-Mass Frame

For 2 → 2 scattering in the CM frame, with all particles of the same mass for simplicity:

$\frac{d\sigma}{d\Omega} = \frac{|\mathcal{M}|^2}{64\pi^2 s}$

where $d\Omega$ is the solid angle of one outgoing particle. For specific processes like $e^+e^- \to \mu^+\mu^-$, computing $|\mathcal{M}|^2$ and doing the trace algebra gives the explicit differential cross section.

We’ll do this computation in document 5.

10. Example: First-Order $\phi^4$ Theory

Let’s concretize everything by computing at first order in a simple scalar theory.

The Setup

$\mathcal{L} = \tfrac{1}{2}(\partial\phi)^2 - \tfrac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4$

The interaction Hamiltonian density is $\mathcal{H}_I = \frac{\lambda}{4!}\phi^4$. (The $1/4!$ is conventional; it cancels symmetry factors in Feynman diagrams.)

Scattering: $\phi\phi \to \phi\phi$

Consider two incoming scalars with momenta $p_1, p_2$ going to two outgoing scalars with $p_3, p_4$.

Zeroth Order

$S = 1$. Contribution to the amplitude: $\delta^4(p_1 - p_3)\delta^4(p_2 - p_4) + \delta^4(p_1 - p_4)\delta^4(p_2 - p_3)$. These represent “no scattering”; particles fly past each other. Not what we want.

First Order

$S^{(1)} = -i\int d^4x\, \mathcal{H}_I(x) = -i\frac{\lambda}{4!}\int d^4x\, \phi^4(x)$

The first-order amplitude for $\phi\phi \to \phi\phi$:

$\langle p_3 p_4|S^{(1)}|p_1 p_2\rangle = -i\frac{\lambda}{4!}\int d^4x\, \langle p_3 p_4|\phi^4(x)|p_1 p_2\rangle$

Applying Wick’s Theorem

We need $\langle p_3 p_4|\phi^4(x)|p_1 p_2\rangle$. The four fields at spacetime point $x$ need to be contracted with the external particle states. Each $|p_i\rangle$ requires a $a^\dagger$ operator (in the expansion $\phi = a + a^\dagger + \ldots$); similarly each $\langle p_j|$ requires an $a$.

There are $4! = 24$ ways to match up the four external particles with the four $\phi$‘s at the interaction vertex. Each gives a factor of $e^{\pm i p_i \cdot x}$ (from the mode expansion) times the appropriate $1/\sqrt{2E_i}$ normalization.

But we should be careful: some of these 24 ways are equivalent; they differ only by a permutation of the outgoing or incoming particles. After accounting for symmetry, the distinct contributions are 3 (from matching $\{1,2\}$ with $\{3,4\}$, or $\{1,3\}$ with $\{2,4\}$, etc.) times $4! \cdot 2 / 3 = 8$ each.

Actually, the standard way to count: there are exactly $4!$ matchings, and the $1/4!$ in the Lagrangian cancels this, leaving an overall factor of 1.

The Result

Integrating over $x$ gives a delta function $(2\pi)^4\delta^4(p_1 + p_2 - p_3 - p_4)$ from momentum conservation.

The amplitude is:

$i\mathcal{M} = -i\lambda$

Just a constant. Remarkable: at first order, the $\phi^4$ scattering amplitude is just the coupling constant.

The Feynman Diagram

Graphically, this is represented by a single vertex where four lines meet:

• Two incoming lines (particles $p_1, p_2$)
• Two outgoing lines (particles $p_3, p_4$)
• One vertex at position $x$ (integrated over)

This is the simplest possible Feynman diagram. The “Feynman rule” extracted is: a $\phi^4$ interaction vertex contributes a factor of $-i\lambda$.

Cross Section

For identical scalars scattering in the CM frame:

$\frac{d\sigma}{d\Omega} = \frac{|\mathcal{M}|^2}{64\pi^2 s} = \frac{\lambda^2}{64\pi^2 s}$

(Constant in angle; the $\phi^4$ theory scatters isotropically at tree level.)

Total cross section (integrated over half the sphere, to account for identical final particles):

$\sigma = \frac{\lambda^2}{32\pi s}$

This is as simple as a scattering calculation gets. More complicated theories (QED, QCD) introduce propagators, vertices with tensor structure, and so on; but the conceptual structure is identical.

11. Preview: Where Feynman Diagrams Come From

Everything we’ve done in this document can be organized graphically. Feynman diagrams are the graphical representation of the perturbative expansion.

The Translation

• External line: Incoming or outgoing particle. Corresponds to a factor from the wave function (momentum eigenstate) in the external state.
• Internal line: A propagator. Comes from contracting two field operators.
• Vertex: An interaction term in $H_I$. Corresponds to a factor of $-i$ times the coupling and any tensor structure.

Rules for QED (from the Lagrangian)

The Feynman rules I listed in document 3’s appendix are derived from this machinery:

1. Each photon propagator: $-i\eta^{\mu\nu}/(k^2 + i\epsilon)$
2. Each fermion propagator: $i(\slashed{p} + m)/(p^2 - m^2 + i\epsilon)$
3. Each vertex: $-ieQ\gamma^\mu$
4. External photons: $\epsilon^\mu$ or $\epsilon^{\mu*}$
5. External electrons/positrons: $u$, $\bar u$, $v$, $\bar v$
6. Momentum conservation at every vertex
7. Integrate over loop momenta
8. Symmetry factors and fermion loop signs

Each rule has a specific origin in Wick’s theorem plus the QED Lagrangian.

The Big Picture

The flow:

QED Lagrangian → H_I in interaction picture → Dyson's formula
                                                  ↓
                               n-th order term in S (Wick's theorem)
                                                  ↓
                               Feynman diagrams of order e^n
                                                  ↓
                               Integrals over propagators
                                                  ↓
                               Amplitudes M
                                                  ↓
                               Cross sections via LSZ + phase space

Document 5 will:

1. Codify the Feynman rules explicitly
2. Compute specific QED processes: $e^+e^- \to \mu^+\mu^-$, Compton scattering, electron-electron scattering
3. Extract differential cross sections
4. Compare to experimental data

This is where QFT becomes calculational physics.

12. Appendix: Formulas and Identities

The Three Pictures of Quantum Mechanics

Key Formulas

Interaction-picture state evolution: $i\partial_t|\psi\rangle_I = H_I(t)|\psi\rangle_I$

Dyson’s formula: $U(t, t_0) = T\exp\left[-i\int_{t_0}^t dt'\, H_I(t')\right]$

S-matrix: $S = T\exp\left[-i\int d^4x\, \mathcal{H}_I(x)\right]$

Gell-Mann–Low theorem: $\langle\Omega|T\{\phi_H(x_1)\cdots\}|\Omega\rangle = \frac{\langle 0|T\{\phi_I(x_1)\cdots S\}|0\rangle}{\langle 0|T\{S\}|0\rangle}$

Contractions (Propagators)

• Scalar: $\overbrace{\phi(x)\phi(y)} = D_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2 - m^2 + i\epsilon}e^{-ip(x-y)}$

• Dirac: $\overbrace{\psi(x)\bar\psi(y)} = S_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\frac{i(\slashed{p} + m)}{p^2 - m^2 + i\epsilon}e^{-ip(x-y)}$

• Photon (Feynman gauge): $\overbrace{A^\mu(x)A^\nu(y)} = D_F^{\mu\nu}(x - y) = \int\frac{d^4k}{(2\pi)^4}\frac{-i\eta^{\mu\nu}}{k^2 + i\epsilon}e^{-ik(x-y)}$

Wick’s Theorem

$T\{A_1 A_2 \cdots A_n\} = \sum_{\text{all pairings}} (\text{contractions})\cdot\, :\!(\text{remaining normal-ordered fields})\!:$

For fermions, pairings pick up a sign $(-1)^P$ from the permutation required.

LSZ Reduction (Schematic)

$\langle f|S|i\rangle = \prod_{\text{external}}(\text{wave function factor}) \times (\text{amputated, on-shell, Fourier-transformed correlation function})$

Cross Sections

For 2 → $n$ scattering:

$d\sigma = \frac{|\mathcal{M}|^2}{4|\vec p_1|\sqrt{s}}\,d\Pi$

where:

$d\Pi = (2\pi)^4\delta^4\left(p_i - \sum p_f\right)\prod_f\frac{d^3 p_f}{(2\pi)^3 2E_f}$

Decay Rate

For 1 → $n$ decay of particle of mass $M$:

$d\Gamma = \frac{|\mathcal{M}|^2}{2M}\, d\Pi$

Checklist

By the end of this document, you should be able to:

• Switch between Schrödinger, Heisenberg, and interaction pictures, and know when to use each
• Derive Dyson’s formula from the state evolution equation
• Write the S-matrix as a time-ordered exponential
• Apply Wick’s theorem to a product of 4 or 6 free fields
• Convert a correlation function into an S-matrix element via LSZ
• Compute the tree-level amplitude for $\phi^4$ scattering
• Write the formula for a 2 → 2 cross section
• Recognize how each of these pieces will be encoded graphically in Feynman diagrams

Problems to Work

1. Derive Dyson’s formula by iterating the interaction-picture state equation three times.

2. Using Wick’s theorem, compute $\langle 0|T\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\phi(x_5)\phi(x_6)\}|0\rangle$ for a free real scalar field. How many terms does the answer have?

3. For a Yukawa theory with interaction $\mathcal{L}_I = -g\bar\psi\phi\psi$, write the first-order contribution to the S-matrix. What kind of process does it describe?

4. Show that for the photon propagator in Feynman gauge, $\overbrace{A^\mu(x)A^\nu(y)} = D_F^{\mu\nu}(x-y)$ follows from the photon field commutation relations.

5. Verify by explicit calculation (for 4-point $\phi^4$ scattering) that $\mathcal{M} = -\lambda$ at tree level, including all symmetry factors.

Problems 2 and 5 are especially good for building the algebraic fluency Wick’s theorem demands.

Further Reading

• Peskin & Schroeder, Chapter 4: the standard treatment; careful and complete
• Schwartz, Chapter 7: cleaner modern presentation, especially good on LSZ
• Srednicki, Chapters 5-9: very explicit, includes many worked examples
• Weinberg, Vol. 1, Chapters 6-7: rigorous foundations, heavy reading

Closing Note

This document bridged the gap between “we can quantize free fields” and “we can calculate scattering amplitudes.” The machinery is intricate but the logic is straightforward:

1. Interaction picture lets us work with free-field operators while carrying the interaction in the states
2. Dyson’s formula expands time evolution in powers of the interaction
3. Wick’s theorem reduces products of free fields to propagator combinations
4. LSZ connects correlation functions to physical scattering amplitudes
5. Cross section formulas convert amplitudes to measurable quantities

Every step is mechanical. Every step is reducible to integrals of propagators. That’s why Feynman diagrams work; they encode this mechanical procedure graphically.

What We’ve Built

The full procedure, end to end:

1. Start with a Lagrangian (QED, $\phi^4$, Standard Model, whatever)
2. Extract the free and interaction parts
3. Quantize the free fields (documents 1-3)
4. Write $S = T\exp[-i\int \mathcal{H}_I]$
5. Compute correlation functions order by order using Wick
6. Extract amplitudes via LSZ
7. Square, integrate phase space, get cross sections

The key insight: in perturbative QFT, every physical prediction is a sum of Feynman diagrams. Each diagram is a particular Wick contraction, represented graphically. Each diagram evaluates to an integral over propagators and coupling constants. Summing the diagrams (in a given order) gives the amplitude.

Where We Go Next

Document 5 is where the abstract machinery meets specific calculations:

• Explicit derivation of QED Feynman rules from the Lagrangian
• Tree-level calculation of $e^+e^- \to \mu^+\mu^-$; the classic QED benchmark
• Compton scattering: $e\gamma \to e\gamma$
• Møller scattering: $e^-e^- \to e^-e^-$
• Bhabha scattering: $e^+e^- \to e^+e^-$
• Extracting differential and total cross sections

After that, loops and renormalization (documents 6-7), path integrals (8-9), and Yang-Mills (10-11).

You’re halfway through the core QFT sequence. Take your time, work some problems from this document (especially 2 and 5), and when ready, we’ll compute actual physics.

Nice work getting here.