# 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 $$D = dL – 2I$$

# Week 5, Wednesday

The possible Wick contractions of the $4$ external positions $x_1$, $\dots$, $x_4$ are just products of the leading $2$-point functions, $$\begin{split} \langle \phi_1 \phi_2\phi_3 \phi_4 \rangle =&\; \langle \phi_1 \phi_2\phi_3 \phi_4 \rangle_0 + O(\lambda) \\ =&\; \langle \phi_1 \phi_2 \rangle \langle \phi_3 \phi_4 \rangle + \langle \phi_1 \phi_3 \rangle \langle \phi_2 \phi_4 \rangle + \langle \phi_1 \phi_4 \rangle \langle \phi_2 \phi_3 \rangle + O(\lambda) \end{split}$$

## First-Order 4-Point Function

The order-$\lambda$ contribution to the 4-point correlator is $$\langle \phi_1 \phi_2 \phi_3 \phi_4 \rangle^{(1)} = – \frac{i}{24} \int d^4y \Big( \langle \phi_1 \phi_2 \phi_3\phi_4 \phi_y^4 \rangle_0 – \langle \phi_1 \phi_2 \phi_3 \phi_4 \rangle_0 \langle \phi_y^4 \rangle_0 \Big)$$ The Wick contractions of the first term are $$\begin{split} \langle \phi_1 \phi_2 \phi_3 \phi_4 \phi_y^4 \rangle_0 =&\; 24~ \overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_3 \phi_y}~ \overline{\phi_4 \phi_y} \\ &\; + 12\Big( \overline{\phi_1 \phi_2}~ \overline{\phi_3 \phi_y}~ \overline{\phi_4 \phi_y}~ \overline{\phi_y \phi_y} + \text{perm.} \Big) \\ &\; + 3\Big( \overline{\phi_1 \phi_2}~ \overline{\phi_3 \phi_4}~ \overline{\phi_y \phi_y}~ \overline{\phi_y \phi_y} + \text{perm.} \Big) \end{split}$$ The Wick contractions of the second term just cancel the last summand above (the terms multiplied by $3$). If you draw the corresponding diagram, you see that they are “vacuum bubbles”, that is, contain a subdiagram that is disconnected from all external positions. Such vacuum bubbles are generally divergent, though fortunately they end up being subtracted off by the second term.

We also notice that many of the terms are just products of 2-point functions, for example $$\begin{split} \langle \phi_1 \phi_2 \rangle \langle \phi_3 \phi_4 \rangle =&\; \overline{\phi_1\phi_2}~ \overline{\phi_3\phi_4} \\ &\; – \frac{i\lambda}{2} \int d^4y\; \Big( \overline{\phi_1 \phi_2}~ \overline{\phi_3 \phi_y}~ \overline{\phi_4 \phi_y}~ \overline{\phi_y \phi_y} + \overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_3 \phi_4}~ \overline{\phi_y \phi_y} \Big) \\ &\; + O(\lambda^2) \end{split}$$ This should also not be surprising, part of amplitude for the scattering process of two ingoing and two outgoing particles is just two particles not interacting at all. In fact, only the single connected diagram at $O(\lambda)$ remains after collecting everything we can into products of two-point functions: $$\begin{split} \langle \phi_1 \phi_2\phi_3 \phi_4 \rangle =&\; \langle \phi_1 \phi_2 \rangle \langle \phi_3 \phi_4 \rangle + \langle \phi_1 \phi_3 \rangle \langle \phi_2 \phi_4 \rangle + \langle \phi_1 \phi_4 \rangle \langle \phi_2 \phi_3 \rangle \\ &\; – i \lambda \int d^4y \overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_3 \phi_y}~ \overline{\phi_4 \phi_y} + O(\lambda^2) \end{split}$$

In fact, this is true in general:

• Vacuum bubbles are cancelled by the expansion of the denominator $$\frac{1}{\int \mathcal{D}\phi e^{iS_\text{int}}},$$ which we got because we are generally not able to normalize the path integral measure.

• $n$-point correlators are sums of products of disconnected correlators plus the connected diagrams.

• Each internal node is accompanied by a factor of $-i\lambda$ and an integral over its position. We will make these also part of the graphical notation.

• Each term comes is multiplied with a combinatorial symmetry factor $\frac{1}{|\mathop{Aut} G|}$, where $\mathop{Aut}(G)$ is the automorphism group of the diagram. That is, count all ways to map the vertices to vertices and lines to lines. Note that we divide by the number of automorphisms because each symmetry reduces the number of distinct Wick contractions we can make. In the diagrams so far, we had

• $\overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_3 \phi_y}~ \overline{\phi_4 \phi_y}$, $\mathop{Aut} G = 1$, coefficient $\tfrac{24}{24} = 1$,

• $\overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_y \phi_y}$, $\mathop{Aut} G = \mathbb{Z}_2$, coefficient $\tfrac{12}{24} = \tfrac{1}{2}$,

• $\overline{\phi_1 \phi_2}~ \overline{\phi_y \phi_y}~ \overline{\phi_y \phi_y},$ $\mathop{Aut} G = D_8$, that is, the dihedral group with $8$ elements, coefficient $\tfrac{3}{24} = \tfrac{1}{8}$.

## Feynman Rules in Position Space

Definition: A Feynman graph in position space, for the $\phi^4$-theory, is a graph (undirected, without edge labels) with

• 1-valent vertices labelled by $x_i$, called “external”,
• 4-valent vertices labelled by $y_j$, called “internal”, and
• without vacuum bubbles.

In particular, disconnected graphs are allowed but each connected component must be attached to at least one external vertex. As we have seen, such a graph translates directly into a summand in the series expansion of correlation functions. Explicitly, the rules are

Definition: Feynman rules for $\phi^4$-theory in position space

• For each line joining two vertices $u$, $v$ multiply the integrand with a free propagator $\overline{\phi(u)\phi(v)} = \frac{1}{i} \Delta(u-v)$.

• For each vertex, integrate $-i\lambda \int d^4y_j$
• Multiply with the symmetry factor $\frac{1}{|\mathop{Aut} G|}$.

# Week 5, Monday

## Correlators in the Interacting Theory

In quantum mechanics, perturbation theory is generally done in the interaction picture where we split the Hamiltonian $H=H_0 + H_\text{int}$ into a free Hamiltonian (harmonic oscillator of in some disguise) and the interaction part $H_\text{int}$. The path integral analog is to split the action $$S = S_0 + S_\text{int}$$ into the action of a free particle and interactions. We use this split to write correlators as $$\label{eq:corrSint} \langle 0 | T\mathcal{O} |0\rangle = \frac{ \int \mathcal{D}\phi \; \mathcal{O} e^{i(S_0+S_\text{int})} }{ \int \mathcal{D}\phi \; e^{i(S_0+S_\text{int})} } = \frac{ \langle0|T \mathcal{O} e^{iS_\text{int}} |0\rangle_0 }{ \langle0|T e^{iS_\text{int}} |0\rangle_0 }$$ where the subscript zero on $\langle \cdots \rangle_0$ means that we use the free action $S_0$ only to compute the correlator. In particular, this means that we can use Wick contractions to evaluate the right hand side. For $\phi^4$ theory, the free and interacting actions are the integrals over the Lagrange densities $$\mathcal{L}_0 = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi – \frac{1}{2} m^2 \phi^2 , \quad \mathcal{L}_\text{int} = – \frac{1}{4!} \lambda \phi^4$$ The basic idea behind perturbation theory is to use the series expansion $$e^{i S_\text{int}} = 1 + \int d^4y \big(-\tfrac{i\lambda}{24} \phi^4 \big) + \frac{1}{2!} \left[\int d^4y \big(-\tfrac{i\lambda}{24} \phi^4 \big)\right]^2 + \cdots$$ and compute correlators order-by-order in the coupling constant $\lambda$. Note that the coupling constant is in both numerator and denominator of the correlator, so to first order we obtain \begin{multline} \begin{split} \langle 0|T\mathcal{O}|0\rangle = &\; \sum_{n=0}^\infty \langle 0|T\mathcal{O}|0\rangle^{(n)} \lambda^n \\ &\; \langle 0|T\mathcal{O}|0\rangle_0 \\ &\; – \frac{i\lambda}{24} \int d^4y \Big( \langle 0|T\mathcal{O} \phi(y)^4 |0\rangle_0 – \langle 0|T\mathcal{O} |0\rangle_0 \langle 0|T \phi(y)^4 |0\rangle_0 \Big) \\ &\; + O(\lambda^2) \end{split} \end{multline}

Let us start with the correlator of two fields. For simplicity, we define the short-hand notation $$\phi_i = \phi(x_i) ,\quad \phi_y = \phi(y) ,\quad \langle \mathcal{O} \rangle = \langle 0|T \mathcal{O} |0\rangle$$ Then the leading term in the expansions $$\langle \phi_1 \phi_2 \rangle = \sum_{n=0}^\infty \lambda^n \langle \phi_1 \phi_2 \rangle^{(n)} \lambda^n$$ is just the free field result $$\langle \phi_1 \phi_2 \rangle^{(0)} = \langle \phi_1 \phi_2 \rangle_0 = \overline{\phi_1 \phi_2} = \tfrac{1}{i} \Delta(x_1 – x_1)$$ The following graphical notation will be useful in the future for Wick contractions:
• Draw a point for each $x_i$, and
The first order contribution is $$\langle \phi_1 \phi_2 \rangle^{(1)} = – \frac{i}{24} \int d^4y \Big( \langle \phi_1 \phi_2 \phi_y^4 \rangle_0 – \langle \phi_1 \phi_2 \rangle_0 \langle \phi_y^4 \rangle_0 \Big)$$ The Wick contractions of the first term are $$\langle \phi_1 \phi_2 \phi_y^4 \rangle_0 = 3~ \overline{\phi_1 \phi_2}~ \overline{\phi_y \phi_y}~ \overline{\phi_y \phi_y} + 12~ \overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_y \phi_y}$$ and the Wick contractions of the second term are $$\langle \phi_1 \phi_2 \rangle_0 \langle \phi_y^4 \rangle_0 = 3~ \overline{\phi_1 \phi_2}~ \overline{\phi_y \phi_y}~ \overline{\phi_y \phi_y}.$$ The result is that $$\langle \phi_1 \phi_2 \rangle = \overline{\phi_1\phi_2} – \frac{i\lambda}{2} \int d^4y\; \overline{\phi_1 \phi_y}~ \overline{\phi_2 \phi_y}~ \overline{\phi_y \phi_y} + O(\lambda^2)$$ The picture for the order-$\lambda$ term is what is called a tadpole: A part of a diagram that is attached only through a single vertex. The tadpole is divergent: $$\overline{\phi_y \phi_y} = \tfrac{1}{i} \Delta(0)$$ Going back to the path integral, we see that the problem arises when translating the $\phi(y)^4$ in the action into a product of coincident operators. Had we used normal ordering $:\phi_y^4:$, for example, then the result would be finite. Equivalently, we could add $\frac{i\lambda}{4}\Delta(0)\phi^2$ to the Lagrangian which also turns out to cancel the divergent term. We will see how to deal with this divergence systematically later.