Quantum Field Theory Course

Total: 440+ pages covering classical field theory through the Standard Model

Part III, Chapter 5 — Page 3 of 3

Worked Examples

Computing correlation functions diagram by diagram in $\phi^4$ theory

5.12 The 2-Point Function at One Loop

At tree level, the 2-point function is simply the free propagator. The first quantum correction comes at order $\lambda$. There is one diagram at this order: the tadpole.

Tadpole Diagram

The tadpole consists of a single vertex with one self-contracted loop and two external legs. Using the momentum-space Feynman rules:

$$-i\Sigma_{\text{tadpole}} = \frac{(-i\lambda)}{2}\int \frac{d^4k}{(2\pi)^4}\frac{i}{k^2 - m^2 + i\epsilon}$$

Let us trace where each factor comes from:

  • Vertex factor: $-i\lambda$
  • Loop propagator: $i/(k^2-m^2+i\epsilon)$ integrated over loop momentum
  • Symmetry factor: $S=2$ (can swap the two ends of the loop), giving the $1/2$

Note this diagram is independent of the external momentum $p$ — it is a constant mass shift. In dimensional regularization ($d = 4-2\varepsilon$):

$$-i\Sigma_{\text{tadpole}} = \frac{-i\lambda}{2}\frac{m^2}{(4\pi)^2}\left(\frac{1}{\varepsilon} - \gamma_E + \ln\frac{4\pi\mu^2}{m^2} + 1\right)$$

💡Tadpole = Mass Renormalization

The tadpole is momentum-independent and simply shifts the mass: $m^2 \to m^2 + \delta m^2$. It is the first sign that bare parameters in the Lagrangian differ from physical (measured) parameters. This is the beginning of renormalization!

Bubble Diagram (Self-Energy at Order $\lambda^2$)

At order $\lambda^2$, the bubble (or sunset-type) diagram contributes to the self-energy. It has two vertices connected by two internal propagators, plus external legs:

$$-i\Sigma_{\text{bubble}}(p^2) = \frac{(-i\lambda)^2}{2}\int \frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(p-k)^2-m^2+i\epsilon}$$

Here the symmetry factor is $S=2$ from exchanging the two internal propagators. Unlike the tadpole, this integral depends on the external momentum $p$, contributing both mass and wavefunction renormalization.

To evaluate, we use the Feynman parametrization:

$$\frac{1}{AB} = \int_0^1 dx\,\frac{1}{[xA + (1-x)B]^2}$$

Applying this with $A = k^2 - m^2$ and $B = (p-k)^2 - m^2$, completing the square, and performing the $d$-dimensional momentum integral:

$$-i\Sigma_{\text{bubble}}(p^2) = \frac{\lambda^2}{2}\int_0^1 dx \frac{1}{(4\pi)^2}\left(\frac{1}{\varepsilon} - \gamma_E + \ln\frac{4\pi\mu^2}{\Delta}\right)$$

where $\Delta = m^2 - x(1-x)p^2$. The $1/\varepsilon$ pole signals the UV divergence.

5.13 The 4-Point Function at Tree Level

The tree-level 4-point function in $\phi^4$ theory is the simplest scattering process. There is exactly one diagram: a single vertex with four external legs.

$$i\mathcal{M}_{\text{tree}} = -i\lambda$$

This is remarkably simple: the tree-level scattering amplitude is just the coupling constant! The symmetry factor is $S=1$ because with four distinguishable external legs, there are no nontrivial automorphisms.

The full 4-point Green function in momentum space includes external propagators:

$$\tilde{G}^{(4)}(p_1,p_2,p_3,p_4)\Big|_{\text{tree}} = \prod_{i=1}^{4}\frac{i}{p_i^2-m^2+i\epsilon}\cdot(-i\lambda)\cdot(2\pi)^4\delta^4(p_1+p_2+p_3+p_4)$$

💡Amputated vs Full Diagrams

The amputated diagram strips off the external propagators, leaving just $-i\lambda$. The LSZ formula tells us the S-matrix element is obtained from the amputated, on-shell Green function. For the 4-point tree, the scattering amplitude is simply $\mathcal{M} = -\lambda$.

5.14 The 4-Point Function at One Loop

At order $\lambda^2$, there are three distinct one-loop diagrams for 4-point scattering, corresponding to the three Mandelstam channels:

s-Channel Diagram

Particles 1 and 2 enter a vertex, propagate through a loop, and exit at particles 3 and 4. Define $s = (p_1+p_2)^2$:

$$i\mathcal{M}_s = \frac{(-i\lambda)^2}{2}\int\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(k-p_1-p_2)^2-m^2+i\epsilon}$$

t-Channel Diagram

Particles 1 and 3 exchange momentum. Define $t = (p_1-p_3)^2$:

$$i\mathcal{M}_t = \frac{(-i\lambda)^2}{2}\int\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(k-p_1+p_3)^2-m^2+i\epsilon}$$

u-Channel Diagram

Particles 1 and 4 exchange momentum. Define $u = (p_1-p_4)^2$:

$$i\mathcal{M}_u = \frac{(-i\lambda)^2}{2}\int\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(k-p_1+p_4)^2-m^2+i\epsilon}$$

All three integrals have the same structure. Each has symmetry factor $S=2$. The total one-loop amplitude is:

$$i\mathcal{M}^{(1)} = i\mathcal{M}_s + i\mathcal{M}_t + i\mathcal{M}_u$$

Evaluating via Feynman parametrization and dimensional regularization:

$$i\mathcal{M}_s = \frac{(-i\lambda)^2}{2(4\pi)^2}\int_0^1 dx\left(\frac{1}{\varepsilon} - \gamma_E + \ln\frac{4\pi\mu^2}{m^2 - x(1-x)s}\right)$$

with analogous expressions for the t- and u-channels replacing $s \to t$ and $s \to u$. The Mandelstam variables satisfy the constraint:

$$s + t + u = 4m^2$$

💡Crossing Symmetry

The three channels are related by crossing symmetry: exchanging incoming and outgoing particles. The s-channel describes direct scattering, the t-channel describes forward exchange, and the u-channel describes backward exchange. The full amplitude must be symmetric under permutations of identical external particles.

5.15 Symmetry Factor Worked Examples

Let us verify several symmetry factors by explicit counting from the Wick expansion.

Example 1: Tadpole ($S=2$)

At order $\lambda$ in $\langle\phi(x)\phi(y)\rangle$, we contract $\phi(x)\phi(y)\phi^4(z)$. The 6 fields must be fully contracted. One contraction pairs $\phi(x)$ with one $\phi(z)$, $\phi(y)$ with another $\phi(z)$, and the remaining two $\phi(z)$'s with each other (the loop). Count: choose which of 4 z-legs connects to x ($4$ ways), which of the remaining 3 connects to y ($3$ ways), and the last 2 form the loop ($1$ way). Total: $4 \times 3 \times 1 = 12$. From the Lagrangian: $1/4! = 1/24$. So the combinatorial factor is $12/24 = 1/2$, giving $S = 2$.

Example 2: 4-Point Tree ($S=1$)

For $\langle\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle$ at order $\lambda$, we contract 8 fields: 4 external plus $\phi^4(z)$. Each external field connects to one z-leg. Number of ways: $4 \times 3 \times 2 \times 1 = 4! = 24$. From the Lagrangian: $1/4! = 1/24$. Combinatorial factor: $24/24 = 1$, so $S = 1$.

Example 3: Figure-Eight Vacuum Bubble ($S=8$)

The figure-eight is a vacuum diagram at order $\lambda$ with one vertex and two self-loops. Starting from $\phi^4(z)/(4!)$, we pair up 4 fields into 2 loops. Number of pairings: choose 2 of 4 for the first loop ($\binom{4}{2}=6$ ways), the other 2 form the second loop ($1$ way), but the two loops are indistinguishable so divide by 2: $6/2 = 3$. From the Lagrangian: $1/4! = 1/24$. Combinatorial factor: $3/24 = 1/8$, giving $S = 8$. The automorphisms are: swap the two ends of loop 1 ($\times 2$), swap the two ends of loop 2 ($\times 2$), swap the two loops ($\times 2$), total $2\times 2\times 2 = 8$.

Example 4: s-Channel Bubble ($S=2$)

Two vertices connected by two internal propagators with 2 external legs at each vertex. At order $\lambda^2$, the $1/(2!\cdot (4!)^2)$ prefactor combines with the number of Wick contractions. After accounting for all combinatorics, the only remaining symmetry is swapping the two internal propagators, giving $S = 2$.

Key Concepts — Worked Examples

  • Tadpole: momentum-independent, contributes mass renormalization at $O(\lambda)$
  • Bubble self-energy: momentum-dependent at $O(\lambda^2)$, requires Feynman parametrization
  • Tree-level 4-point: $\mathcal{M} = -\lambda$, the simplest scattering amplitude
  • One-loop 4-point: three Mandelstam channels (s, t, u) with $s+t+u=4m^2$
  • Symmetry factors verified by explicit Wick contraction counting
  • UV divergences appear as $1/\varepsilon$ poles in dimensional regularization
  • Crossing symmetry relates the s, t, and u channels
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