# Week 5, Thursday

## Feynman Rules in Momentum Space

It turns out that the Feynman rules take a nicer for in momentum space. Partly, this is because it is just more convenient for accelerators where we scatter particles with fixed momentum. More technically, recall that the Feynman propagator in position space is quite complicated involving Bessel functions. And we have to compute convolution integrals of the Feynman propagator because of the $y_j$-integrals.

Hence, we apply Fourier transformation to each external position $x_i$ to obtain the $n$-point correlator in momentum space as $$F(k_1, \dots, k_n) = \int d^4x_1 \cdots \int d^4x_n \; e^{i \sum_j k_j x_j} \langle \phi(x_1) \cdots \phi(x_n) \rangle.$$ The Fourier transformation of just $\frac{1}{i}$ times the Feynman propagator (i.e.\ the free field 2-point function) is $$\begin{split} G(k_1, k_2) =&\; \frac{1}{i} \int d^4x_1 \int d^4x_2\; e^{i( k_1 x_2 + k_2 x_2)} \int \frac{d^4k}{(2\pi)^4} \frac{e^{i k (x_1-x_2)}}{k^2 + m^2 -i\epsilon} \\ =&\; \frac{1}{i} (2\pi)^4 \delta^4(k_1+k_2) \frac{1}{k_1^2 + m^2 – i\epsilon}. \end{split}$$ For the vertex, note that its position $y$ occurs in the exponent of the $4$ propagators that it is connected to. If we let $k_1$, $\dots$, $k_4$ be the momenta in the $4$ propagators then the contribution of the vertex boils down to $$-i \lambda \int d^4y \; e^{iy(k_1+k_2 + k_3 + k_4)} = -i \lambda (2\pi)^4 \delta^4(k_1+k_2 + k_3 + k_4).$$ We notice that the delta functions just implement momentum conservation on each line and each internal vertex. We make this part of our graphical notation, and define

Definition: A Feynman graph in momentum space for the $\phi^4$-theory is a graph with

• 1-valent external vertices labeled with inflowing $4$-momentum $k_j$,

• edges labelled by directed $4$-momenta $\ell_j$,
• 4-valent internal vertices, and
• without vacuum bubbles.

Each unique graph again can be translated into a particular summand in the series expansion of the $n$-point correlator using

Definition: The Feynman rules for $\phi^4$-theory in momentum space are

• for each connected component with inflowing momenta $k_j$, multiply with $(2\pi)^4 \delta^4(\sum k_j)$,

• use momentum conservation along lines and vertices to replace internal momenta $\ell_j$ as far as possible,

• for each remaining internal momentum, integrate $\int \frac{d^4\ell}{(2\pi)^4}$,

• for each edge carrying momentum $k$, multiply the integrand with a factor $\frac{1}{i} \frac{1}{k^2+m^2-i \epsilon}$,

• for each 4-valent vertex multiply with $-i\lambda$,
• multiply with the symmetry factor $\frac{1}{|\mathop{Aut} G|}$.

For example, for the tree graph $4$-point function with a single internal vertex we get $$F^{(1)}_\text{conn} = -i \lambda (2\pi)^4 \delta^4(k_1+k_2+k_3+k_4) \prod_{j=1}^4 \frac{1}{k_j^2+m^2-i \epsilon}.$$ A more complicated diagram is the “fish”, which has a $\mathbb{Z}_2$ symmetry: \begin{multline} F = \frac{1}{2} (-i\lambda)^2 (2\pi)^4 \delta^4(k_1+k_2+k_3+k_4) \prod_{j=1}^4 \frac{1}{k_j^2+m^2-i \epsilon} \\ \times \int \frac{d^4\ell}{(2\pi)^4}~ \frac{1}{\ell^2 + m^2 -i\epsilon}~ \frac{1}{(k_1+k_2-\ell)^2 + m^2 -i\epsilon} \end{multline} There is a $4$-dimensional momentum integration, but the integrand only falls off like $\ell^{-4}$. This results in a logarithmic divergence if you try to do the momentum integral. Because it appears at high energies, this is called a UV divergence. We will have to understand how to systematically deal with these.

If $m^2=0$ and, $k_1=-k_2$, then there is in addition a divergence as $\ell\to 0$. Such a divergence is called an IR divergence. It typically appears for special values of the external momenta, whereas the UV divergence is independent of the external momenta. In the following, we will only consider $m^2\not=0$, so at least we do not have any problems with IR divergences.

## Power Counting

Counting the powers of the momenta in the integral gives us a first hint at whether there is a UV divergence, so we want to do it more systematically for all diagrams. So let us define the counts of the constituents of a Feynman graph as

• $V$ = number of (internal) vertices,
• $I$ = number of internal lines,
• $E$ = number of external lines.

Each internal line ends at two vertices, each external line ends at one vertex. Since the vertices are 4-valent, we get $$4V = 2I + E.$$ The number of loops $L$ equals the number of unconstrained internal momenta. We get one from each internal line, but can eliminate $V-1$ using all momentum conservation rules except for the overall momentum conservation. Hence, $$L = I – (V-1) = I – V + 1.$$ We now define the superficial (apparent) degree of divergence $D$ as the overall power of the momentum in the integral. For reasons that will be clear later, we want to do it in arbitrary space-time dimension $d$. Then we get $Ld$ powers of the momentum from the measure of the $L$ integrals, and we get a inverse momentum squared from each internal propagator. Hence, the degree of divergence is $$D = dL – 2I$$