We looked at the divergences in the one-loop diagrams in the four-dimensional $\phi^4$ theory, and how they are renormalized. This requires mass, coupling constant, and field renormalization.

I concluded with pictures of the running coupling constants in the Standard Model:

- Gauge coupling constants without supersymmetry
- Gauge coupling constants with supersymmetry meet in a point. Does it mean something?
- Quartic Higgs coupling runs to smaller and possibly negative values at high energies.

TODO: type my lecture notes

]]>\begin{equation} V_\text{eff}(\varphi) = V(\varphi) – \frac{i\hbar}{2} \int \frac{d^4k}{(2\pi)^4} log \left( \frac{k^2 – V”(\varphi)}{k^2} \right) + O(\hbar^2) \end{equation}

TODO: type my lecture notes

]]>Let us start by taking a step back and reformulate some of the
combinatorics of Feynman diagrams in terms of algebra of the
generating function. First, consider a Feynman diagram
\begin{equation}
D = \cup_{j=1}^k n_J C_j
\end{equation}
consisting of connected components $C_j$ with multiplicity $n_j$. The
Feynman rules for $D$ contain a symmetry factor
\begin{equation}
\frac{1}{|\mathop{Aut} D|} =
\prod_{j=1}^k \frac{1}{n_j! ~ |\mathop{Aut} C_j|}
\end{equation}
To evaluate the path integral we have to sum over all diagrams,
connected or not. Writing $C_j$ for both the Feynman diagram and the
expression obtained by applying the Feynman rules, we find
\begin{equation}
Z = \int \mathcal{D}\phi e^{iS} =
\sum_k D_k =
\prod_j \sum_{n_j} \frac{1}{n_j!} C_j =
\prod_j e^{C_j} = e^{\sum_j C_j}.
\end{equation}
This suggests to define a functional $W[J]$ as
\begin{equation}
Z[J] = e^{i W[J]}
,\quad
Z_E[J] = e^{W_E[J]}
\end{equation}
in Minkowski and Euclidean signature, respectively. The $W[J]$ then
has an expansion into a sum over connected Feynman diagrams. We can
define “connected” correlation functions
\begin{equation}
\langle \phi(x_1) \cdots \phi(x_n) \rangle_C =
\tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \;
\cdots
\tfrac{1}{i} \frac{\delta}{\delta J(x_n)} \;
W[J] \Big|_{J=0},
\end{equation}
and, in particular,
\begin{equation}
\langle \phi(x)\rangle_C =
\tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \;
W[J] \Big|_{J=0} =
\frac{1}{i^2} \frac{1}{Z}
\frac{\delta Z}{\delta J(x)} \Big|_{J=0} =
\frac{1}{i^2} \langle 0|\phi(x) |0\rangle.
\end{equation}
The vev of $\phi$ is, of course, $0$ in $\phi^4$-theory because of the
$\phi\mapsto -\phi$ symmetry. To get something non-trivial, we should
*not* set $J=0$, this amounts to measuring the response of the
field to an external source. Hence, we define the *mean field* as
\begin{equation}
\label{eq:meanfield}
\varphi(J) = \frac{\delta W}{\delta J}
\end{equation}
It is perhaps suggestive to think of $W$ in
\begin{equation}
e^{iW[J]} = \int\mathcal{D}\phi e^{i\int d^4x (\mathcal{L}+J\phi)}
\end{equation}
as an “effective action” which incorporates all the quantum effects
from the path integral. However, it depends on the source $J$ and not
on the field. Hence, to define something that deserves the name, we
have to replace the $J$-dependence with a dependence on the mean field
$\varphi$. The right way to do this is to use the Legendre transform,
just like how we change between $(q, \dot q)$ and $(q,p)$-dependence
in classical mechanics. Hence, we invert $\varphi=\varphi(J)$ to
obtain $J=J(\varphi)$ and define the *1PI* effective action} as
\begin{equation}
\Gamma[\varphi] =
W[J(\varphi)] – \int d^dx \; \varphi J(\varphi).
\end{equation}
Functionally differentiating with respect to the mean field yields
\begin{equation}
\frac{\delta \Gamma[\varphi]}{\delta \varphi} + J = 0
\quad \Leftrightarrow \quad
J = -\frac{\delta \Gamma[\varphi]}{\delta \varphi}
\end{equation}
We interpret this in the following way: The stationary point of the
effective action $\Gamma[\varphi]$ is where the mean field satisfies
$J(\varphi)=0$. Hence
\begin{equation}
\frac{\delta \Gamma[\varphi]}{\delta \varphi}
\quad\Rightarrow\quad
J=0
\quad\Rightarrow\quad
\varphi =
\frac{\delta W}{\delta J} \;
\Big|_{J=0} =
\tfrac{1}{i} \langle 0|\phi(x) |0\rangle,
\end{equation}
that is, the stationary points of the effective action determine the
quantum one-point function. A similar calculation for higher
derivatives shows that all correlation functions are determined by the
higher derivatives of $\Gamma[\phi]$.

By this argument, the classical solutions of the effective action
$\Gamma$ are equivalent to full path integral with the action $S=\int
d^4x \mathcal{L}$. In terms of Feynman diagrams, this means that the
sum over all tree level $\Gamma$-diagrams equals the sum over all
connected $S$-diagrams. It remains to explain why $\Gamma$ is called
the *one-particle irreducible* (1PI) effective action. This is
because it can be expanded
\begin{equation}
\Gamma[\varphi] = S[\varphi] + \sum_{1PI} A(G)
\end{equation}
where the sum runs over all 1PI graphs $G$ and $A(G)$ is the
corresponding amputated diagram. A mathematical proof is quite
involved, but we can understand it by the following heuristic
argument. According to this equation, the $\Gamma$-Feynman rules have
a different type of vertex for each 1PI $S$-Feynman graph. The
correspondence between all $\Gamma$ and $S$-Feynman graphs is obtained
by shrinking each 1PI $S$-subgraph to the corresponding
$\Gamma$-vertex.

Let us go back to the 2-dimensional $\phi^4$ theory on the lattice. In perturbation theory and after Wick rotation, we have the two-point function \begin{equation} \begin{split} \Gamma^{(2)} =&\; \frac{1}{k^2+m^2} + \frac{1}{(k^2+m^2)^2} \left( -\frac{\lambda}{2} T + C \right) + O(\lambda^2) \\ \frac{1}{\Gamma^{(2)}} =&\; k^2 + m^2 + \frac{\lambda}{2}T – C + O(\lambda^2) \end{split} \end{equation} where \begin{equation} T = \int \frac{d^2p}{(2\pi)^2} \frac{1}{p^2+m^2} \end{equation} is the divergent tadpole integral and $C$ is a counterterm which cancels the divergence. On the lattice every quantity is a sum over the finitely many lattice points, so is necessarily finite. However, in the continuum limit we have to reproduce the divergence. In renormalization terms, the lattice discretion acts as a regularization just like dimensional regularization. The lattice regularized version of the divergent integral $T$ is \begin{equation} T_L = \frac{1}{N^2} \sum_{p_L^2} \frac{1}{p_L^2 + m_L^2} \end{equation} where $p_L$ are the lattice momenta and $N$ is the number of lattice points in each of the 2 directions. Also, recall the dimensionless lattice parameters (in 2d) are \begin{equation} m^2_L = m^2 a^2 ,\quad \lambda_L = \lambda^2 a^2 . \end{equation} By definition, the $p_L^2$ are the eigenvalues of the lattice Laplacian $\Delta_L$, that is, the discretized Laplacian.

In one dimension we can think of the discretized function as a vector with $N$ entries. The Laplacian, with periodic boundary conditions, is then the matrix \begin{equation} \Delta_L = \left( \begin{smallmatrix} 2 & -1 & 0 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 2 & -1 & \cdots & 0 & 0 \\ 0 & 0 & -1 & 2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 2 & -1 \\ -1 & 0 & 0 & 0 & \cdots & -1 & 2 \end{smallmatrix} \right). \end{equation} The zero mode is the constant vector $\phi_0 = (1, \dots, 1)$ and the highest frequency mode is the alternating vector $\phi_4 = (1, -1, 1, -1, \dots, -1)$ whose eigenvector is $4$. Hence there are $N$ eigenvalues in the range $p_j^2 \in [0,4]$. A computation yields \begin{equation} p_j^2 = 4 \sin^2\left(\tfrac{\pi j}{N}\right) ,\quad j \in \{0, \dots, N-1\}. \end{equation}

\subsubsection{Lattice Laplacian in 2D}

The 2-dimensional eigenfunctions on the $N\times N$ square lattice are \begin{equation} \phi_{ij}(x,y) = \phi_i(x) \phi_j(y) \end{equation} so there are $N^2$ eigenvalues \begin{equation} p^2_{i} = 4 \sin\left(\tfrac{\pi i}{N}\right) + 4 \sin\left(\tfrac{\pi j}{N}\right) ,\quad i, j \in \{0, \dots, N-1\}. \end{equation} Hence, the lattice approximation to the divergent tadpole integral is \begin{equation} T_L = \frac{1}{N^2} \sum_{i,j = 0}^{N-1} \frac{1}{ m_L^2 + 4 \sin\left(\tfrac{\pi i}{N}\right) + 4 \sin\left(\tfrac{\pi j}{N}\right) } \end{equation} A bit of complex analysis shows that the large volume limit $N\to \infty$ exists. With some more work one can show that \begin{equation} \lim_{N\to \infty} T_L = \int_0^\infty e^{-m_L^2 t} \left( e^{-2t} I_0(2t) \right)^2 = \frac{2}{(4+m_L^2)\pi} K\big(\tfrac{4}{4+m_L^2}\big) \end{equation} where $I_0$ is the Bessel function and $K$ is the elliptic integral of the first kind. In the lattice literature you can usually find only the semidefinite integral, though it is not as convenient as the closed form for actually evaluating $T_L$. Also, note that we haven’t done the continuum limit $a\to 0$ yet. In that limit, $m_L^2 = m^2 a^2 \to 0$ which diverges because $K$ has a simple pole at $K(1)$. This is of course necessary to reproduce the divergence in the field theory. In other words, this is the lattice analog to the divergence of $T$ in the dimension $d\to 2$ limit in dimensional regularization.

To renormalize the lattice values we need to pick a suitable renormalization prescription. We also want it to satisfy two physical requirements:

- The renormalization prescription should distinguish the two phases, because that is what we are interested in. But just looking at the $1/\Gamma^{(2)}$ limit as $k\to 0$ does not, in the zero momentum limit the field just sits around the minimum. Both the symmetry-preserving and symmetry-breaking mimimum look locally the same, so our renormalization prescription should include non-zero momenta.
- The renormalization prescription should be convenient for lattice calculations.

The solution to both of these requirements is to take $T_L$ as the
counterterm,
\begin{equation}
\label{eq:2dphi4bare}
\begin{split}
m^2_{L, \text{bare}} =&\; m_L^2 – \frac{\lambda_L}{2} T_L(m^2_L)
\\
\lambda_{L, \text{bare}} =&\; \lambda_L
\end{split}
\end{equation}
In the lattice simulation, we necessarily picked *bare* values
for the parameters by hand. Also, note that the lattice code used
$\frac{1}{4}\lambda \phi^4$ instead of $\frac{1}{4!}\lambda \phi^4$ as
interaction term, this leads to extra factors of $6$ below. Using the
$\frac{}{4!}$ convention, the transition line was numerically around
\begin{equation}
m^2_{L,\text{bare}} \approx -1.27
,\quad
\lambda_{L, \text{bare}} \approx 6.0.
\end{equation}
To get the physical parameters, we have to numerically invert
eq.~\eqref{eq:2dphi4bare}, which yields
\begin{equation}
m_L^2 \approx 0.0980
,\quad
\lambda_L \approx 6.0
.
\end{equation}
We now take the continuum limit while staying on the transition
line. In the $m_L^2$ vs. $\lambda_L$ plane, the transition is almost
on a line. The physically interesting values is the slope at the
origin, which defines the *critical value*
\begin{equation}
f = \lim_{a=0}
\frac{\lambda_L}{6 m^2_L}
\approx
10.8
\end{equation}

Up to $O(\lambda^2)$, the Feynman rules including the counterterm yield

- Tree level propagator $\frac{1}{k^2+m^2}$,
- At $O(\lambda)$: Tadpole + counterterm.
- Four one-particle reducible diagrams, and
- Three one-particle irreducible diagrams.

By one-particle irreducible diagram (1PI) we mean any diagram that can not be disconnected by cutting a single internal line. Like the disconnected diagrams, the one-particle reducible diagrams are again a type of diagram that we have to add up to get the correlation function, but that really is just a combination of lower-order diagrams and does not present any real difficulty. In particular, up to a factor of the external propagator $\frac{1}{k^2+m^2}$, the sum of the four one-particle reducible diagrams is just the square of tadpole and counterterm.

The really interesting part are the three 1PI diagrams. The “double scoop” (first) diagram is divergent only because the top loop integral is a tadpole, so it yields the same $A(\omega)$ as the tadpole times whatever the convergent integral over the lower loop is. To it, we have to add the tadpole-with-counterterm (second) diagram. Its loop integral is just the same as the lower loop integral of the previous diagram, now multiplied by the counterterm. Hence, the first two diagrams just cancel the $\frac{1}{\omega}$ pole just like tadpole and counterterm alone. Finally, the “sunset” diagram (third) is convergent in two dimensions.

By as similar reasoning, the addition of the counterterm to the Feynman rules always cancels the $\frac{1}{\omega}$ pole from any tadpole sub-diagram, rendering every correlation function finite. As a word of warning, however, this is not typical of quantum field theories. For example in 4-d $\phi^4$-theory, there are new divergences at each order in $\lambda$, for which we have to add more and more counterterms. In particular, the leading $\frac{1}{\omega^2}$-divergence will cancel between the first two diagrams as above, but only to leave a $\frac{1}{\omega}$-pole that still diverges. However, as we will see, the $\phi$-dependence of all of the counterterms is just like one of the terms that is already in the Lagrangian, and this is what makes the theory renormalizable.

To summarize, we can combine the Lagrangian and counterterms into a renormalized Lagrangian \begin{equation} \mathcal{L}_\text{Ren} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi -\frac{1}{2} m_\text{bare}^2 \phi^2 – \frac{1}{4!} \lambda_\text{bare} \phi^4 \end{equation} with “bare” parameters \begin{equation} \begin{split} \lambda_\text{bare} =&\; \lambda_0 M^2 = \lambda \\ m_\text{bare}^2 =&\; m^2 – \frac{\lambda_0 M^2}{4\pi} \left( \frac{2}{\omega} + F\right) \end{split} \end{equation} The bare parameters in the Lagrangian are unphysical (and usually infinite in the $\omega\to 0$ limit). We can now compute correlation functions in two ways, either using the renormalized Lagrangian and bare parameters or using the original Lagrangian with (divergent) counterterms. Either way, the divergences in the $\omega\to 0$ limit cancel to give finite correlation functions \begin{equation} \Gamma_{\mathcal{L}_\text{Ren}} (k_1, \dots, k_n; m_\text{bare}^2, \lambda_\text{bare}, \omega) = \Gamma_{\mathcal{L}+\mathcal{L}_\text{ct}} (k_1, \dots, k_n; m^2, \lambda_0, M, \omega). \end{equation} Analyzing the $M$-dependence of this equation will lead us to the renormalization group later, but before we get there we need to understand how to relate the parameters to the physical mass and coupling strength.

Really, the mass and coupling constant are experimental input. Because of the ambiguities introduced in the renormalization procedure, these are not directly related to any set of parameters in the Lagrangian. Instead, we have to fix a certain number of observables and match the computed value (as a power series in the coupling constant, most likely) to the experimental input. The ambiguity in choosing a particular set of observables is called the “renormalization prescription”, and reflects the ambiguity in the finite part of the counterterms.

In quantum field theory, the observables are simply the correlation functions. The renormalization prescription for the mass are always linked to the two-point functions in some way. Popular choices are

- Pole mass: Let the mass be the location of the pole in the two-point function $\Gamma^{(2)}$ in Minkowski space. This is a very physical prescription, but not convenient in Euclidean space after Wick rotation.
- More convenient for calculations (as long as there is no IR divergence) is to demand that \begin{equation} \frac{1}{\Gamma^{(2)}(k_1,k_2)} \Big|_{k_1=k_2=0} = m^2. \end{equation} This is just a little bit unphysical as $k_1=k_2=0$ is not really allowed for massive external particles, it violates the mass shell condition. But nothing stops us from evaluating the $2$-point function and use it in our prescription. Inverting the two-point function to first order in $\lambda$ is easy enough, and by summing the tree-level, tadpole, and counterterm we obtain \begin{equation} \frac{1}{\Gamma^{(2)}(k_1,k_2)} \Big|_{k_1=k_2=0} = m^2 \left[ 1 + \frac{\lambda_0 M^2}{4\pi m^2} \left( -\gamma + \ln\left(\frac{4\pi M^2}{m^2}\right) – F \right) + O(\lambda^2) \right] \end{equation} which we can solve by setting \begin{equation} F = -\gamma + \ln\left(\frac{4\pi M^2}{m^2}\right) \end{equation}
- Even more convenient for calculations is the “minimal subtraction”, where we set $F=0$ as our choice of renormalization prescription. In other words, we only use the minimial counterterm necessary to precisely cancel the $\frac{1}{\omega}$-pole.

As we have seen, it is unavoidable that some of the Feynman diagrams contain divergent momentum integrals. It turns out that this can be dealt with, but we must be careful when handling the infinities. Naively, $1+\infty = \infty$ so how can we make any sensible computation? The key is to introduce a suitable parameter (called “regulator”) that makes all loop integrals finite. Perhaps the simplest way to do that is to cut off momentum integrals at some scale $\Lambda$, for example for the tadpole integral \begin{equation} A(\Lambda) = – \frac{\lambda}{2} \int_{|\ell| \leq \Lambda} \frac{d^4 \ell}{\ell^2+m^2}. \end{equation} Only at the end we then let $\Lambda \to \infty$ to recover the integration over the entire momentum space. This is actually very similar to the lattice computation that we have seen before, there the lattice spacing $a$ also ensures that we can only sample waves with momentum up to some upper limit $\approx \frac{1}{a}$.

Both the direct momentum cutoff and the lattice regularization break Lorentz invariance badly: Cutting off momenta in some special frame breaks boost invariance. And the lattice even breaks rotational invariance down to a discrete subgroup, namely rotations by $\tfrac{2\pi}{4}$. Since symmetries are one of the key guiding principles in physics, a lot of attention was spent on finding regularization schemes that do not break a particular symmetry that one is interested in. One such example is analytic regularization \begin{equation} A(z) = – \frac{\lambda}{2} \int \frac{d^4 \ell}{\big(\ell^2+m^2\big)^z}, \quad z \gg 1. \end{equation} However, the most common regularization scheme is dimensional regularization where we compute the momentum integral in $d-\omega$ dimensions for some $\omega > 0$. This is the regularization scheme that we will always be using in the following. For example, the tadpole integrand is actually rotationally symmetric so we can just split it into a one-dimensional integral times the area of the $(d-\omega)$-sphere.

In each integral dimension $d\in \mathbb{Z}$, the volume of the $d$-dimensional disk and the area of the $d$-dimensional sphere are related by the recurrence relations for the area of the $d$-dimensional disc $D_d$ and sphere $S_d$, \begin{equation} D_d = \frac{1}{d} S_{d-1} ,\quad S_d = 2\pi D_{d-1}, \end{equation} which have a simple geometric origin. We can solve them in closed form as \begin{equation} D_d = \frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+2)} ,\quad S_{d-1} = \frac{2 \pi^{d/2}}{\Gamma(\frac{d}{2})}. \end{equation} The result is actually an analytic function of the dimension, and we will use it to define what me mean by the area of the $(d-1)$-sphere for arbitrary $d\in\mathbb{C}$. It is by no means the unique analytic function that equals $S_{d-1}$ when restricted to the integers, but it is certainly the most convenient choice. Any other choice would just be a different regularization scheme.

Hence, the dimensionally-regulated tadpole is \begin{equation} A(\omega) = -\frac{\lambda}{2} \int \frac{d^{d-\omega}\ell}{(2\pi)^{d-\omega}} ~ \frac{1}{\ell^2 + m^2} = -\frac{\lambda}{2} S_{d-1-\omega} \int_0^\infty \frac{d\bar\ell ~ \bar\ell^{d-1-\omega}}{\bar\ell^2 + m^2}. \end{equation} The remaining one-dimensional integral can be solved by the formula \begin{equation} \int_0^\infty \frac{x^k dx}{\big(x^n + a^n\big)^r} = \frac{ (-1)^{r-1}\; \pi \; a^{k+1-nr} \; \Gamma\big(\tfrac{k+1}{n}\big) }{ n \; \sin\big(\tfrac{k+1}{n} \pi\big) \; \Gamma\big(\tfrac{k+1}{n}-r+1\big) \; (r-1)! } \end{equation} and using Euler’s reflection formula \begin{equation} \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}. \end{equation} The result is \begin{equation} A(\omega) = -\frac{\lambda}{4\pi} \Gamma\big(\tfrac{\omega}{2}\big) \left(\frac{2 \sqrt{\pi}}{m} \right)^\omega, \end{equation} but there is something troubling about the $m^{-\omega}$: what kind of units (or mass dimension) does such a quantity have? The answer is that we forgot that the mass dimension of $\lambda$ is also non-trivial and dependent on the ambient space dimension. In particular, in $d-\omega$ dimensions we have $[\lambda] = 2+\omega$. To better understand how the units work in the equation, it is convenient to split the dimensionful coupling constant \begin{equation} \lambda = \lambda_0 M^{2+\omega} \end{equation} into a dimensionless coupling constant $\lambda_0$ times a mass scale $M$. Then, \begin{equation} A(\omega) = -\frac{\lambda_0 M^2}{4\pi} \Gamma\big(\tfrac{\omega}{2}\big) \left(\frac{2 \sqrt{\pi}M}{m} \right)^\omega, \end{equation} At the end of the day, we will of course be interested in the limit $\omega\to 0$, and we can understand the behavior of $A(\omega)$ in this limit by expanding it in a Laurent series. Using ($\gamma\approx 0.577\dots$ is the Euler-Mascheroni constant) \begin{equation} \Gamma(\omega) = \frac{1}{\omega} – \gamma + O(\omega) ,\quad x^\omega = 1 + \omega \ln(x) + O(\omega^2) \end{equation} we obtain \begin{equation} A(\omega) = -\frac{\lambda_0 M^2}{4\pi} \left[ \frac{2}{\omega} – \gamma + \ln\left( \frac{4\pi M^2}{m^2} \right) + O(\omega) \right] \end{equation} Clearly this is still infinite in the limit $\omega\to 0$, as it must. But the infinity just comes from the simple pole at $\omega=0$, so we can easily handle it.

We did not write the external propagators in the previous section for brevity, this is also called the amputated Feynman diagram. The actual contribution to the two-point function is \begin{equation} A(\omega) \delta(k_1-k_2) \frac{1}{k_1^2+m^2} ~ \frac{1}{k_2^2+m^2}. \end{equation} To get rid of the $\frac{1}{\omega}$ pole, we now add a new Feynman rule that cancels it. The easiest way is to add a new 2-valent vertex with just the right interaction strength such that \begin{equation} —\times— ~= \frac{\lambda_0 M^2}{4\pi} \left(\frac{2}{\omega} + F \right) \delta(k_1-k_2) \frac{1}{k_1^2+m^2} ~ \frac{1}{k_2^2+m^2}, \end{equation} where $F$ is some arbitrary finite quantity. Then the $\frac{1}{\omega}$-pole cancels in sum of the tadpole and the counterterm, rendering the 2-point correlator finite at order $O(\lambda)$. Of course we cannot just add new Feynman rules at will, they must come from an interaction term in the Lagrangian. Since the counterterm vertex is 2-valent, it corresponds to \begin{equation} \mathcal{L}_\text{ct} = \frac{1}{2} \frac{\lambda_0 M^2}{4\pi} \left(\frac{2}{\omega} + F \right) \phi^2 \end{equation}

]]>Theorem:[Dyson-Weinberg convergence theorem] A superficially convergent diagram such that all of its sub-diagrams are also superficially convergent is actually convergent.

The theorem might seem obvious but is actually rather tricky due to the issue of overlapping divergences. We will explain what this in a second, but first let us classify the primitively divergent diagrams in $\phi^4$-theory. That is, look for the divergent diagrams that do not have a divergent sub-diagram. These are the real troublemakers, and we really only have to deal with their divergences.

- In the two-dimensional $\phi^4$-theory there is only a single primitively divergent diagram, namely the tadpole. It has superficial degree of divergence $D=0$.
- In the four-dimensional $\phi^4$-theory there are two primitive divergences at $O(\lambda)$, namely the tadpole $D=2$ and the “fish”, the only (up to permutations of the external vertices) connected one-loop diagram contributing to the 4-point function. There are further primitive divergences at all higher powers of $\lambda$, but only with two or four external legs.
- In the four-dimensional $\phi^k$-theory with $k\geq 5$ there are primitive divergences with any number of external legs. This is what makes the theory non-renormalizable, as we will see.

There is one subtlety when dealing with potentially divergent loop integrals: Sometimes you can’t split the integrals into an outer integral times an inner integral and analyze the convergence separately. This happens when an internal line is part of two separate loops, giving rise to a propagator $\frac{1}{(\ell+p)^2+m^2}$ that depends on both the $\ell$ and $p$-loop momentum. More properly, these should be called overlapping integrals.

The simplest overlapping loop diagram would be in $\phi^3$-theory. So, just for this section, let us consider tri-valent interaction vertices and the diagram leading to the loop integral \begin{equation} \begin{split} I =&\; \int d^4p \; d^4\ell \; \frac{1}{p^2+m^2-i\epsilon} \frac{1}{\ell^2+m^2-i\epsilon} \\ &\; \qquad \times \frac{1}{(p+\ell)^2+m^2-i\epsilon} \frac{1}{(p-k_1)^2+m^2-i\epsilon} \frac{1}{(\ell-k_2)^2+m^2-i\epsilon} \end{split} \end{equation} where we left out constants and the external propagators, anything that does not depend on the loop momenta. We now

- Perform Wick rotation: drop the $-i\epsilon$.
- Set $m^2=0$ and $k_1=0=k_2$ since the UV divergence is at $k^2 \gg m^2, k_1^2, k_2^2$.
- Remove a region of small momenta by hand to avoid the IR divergence.

Hence the asymptotic behavior of the loop integral is that of \begin{equation} I \sim \int \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4}. \end{equation} This is an example of an overlapping divergence: by power counting, the diagram superficial degree of divergence $D=2\times 4-5\times 2 =-2$ and no divergent sub-diagram, so we expect convergence. But the actual integral is not so obviously convergent, for example there is a region where $p+\ell$ is constant. In that region, just the $p$-integral seems to have superficial degree of divergence $D=0$, which would indicate a logarithmic divergence. This is actually not true, the whole integral still converges.

To show that the integral does converge, we have to split up the domain according to which momentum factor is the smallest. For example, consider the region where $\ell^2$ is the smallest, \begin{equation} U_p = \big\{ p \big| p^2 \geq \ell^2,~ (p+\ell)^2 \geq \ell^2 \big\}, \end{equation} and combine it with the necessary IR cutoff \begin{equation} U_\ell = \big\{ \ell \big| \ell^2 \geq 1 \big\}. \end{equation} Splitting the 4-momentum $\ell = \bar\ell e_\ell$ into its absolute value $\bar\ell\in\mathbb{R}$ and unit 4-vector $e_\ell$, the loop integral becomes \begin{equation} I \sim \int_{U_p} \int_{U_\ell} \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4} = \int_0^\infty d\bar\ell \bar\ell^3 \frac{1}{\bar\ell^6} \int_{S^3} de_\ell \int_{U_{p’}} \frac{d^4 p’}{(p’)^4 (p’+e_\ell)^2} \end{equation} where we rescaled $p’ = p \bar\ell$ and integrate over the rescaled region \begin{equation} U_{p’} = \big\{ p’ \big| (p’)^2 \geq 1,~ (p’+\ell)^2 \geq 1 \big\}. \end{equation} The $\bar\ell$-integration is, indeed, of superficial degree of divergence $D=-2$. The overlapping diagram does converge as expected by power counting, the potential overlapping divergence does not cause an actual divergence.

]]>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 \begin{equation} 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. \end{equation} The Fourier transformation of just $\frac{1}{i}$ times the Feynman propagator (i.e.\ the free field 2-point function) is \begin{equation} \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} \end{equation} 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 \begin{equation} -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). \end{equation} 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 \begin{equation} 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}. \end{equation} 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.

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 \begin{equation} 4V = 2I + E. \end{equation} 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, \begin{equation} L = I – (V-1) = I – V + 1. \end{equation} 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 \begin{equation} D = dL – 2I \end{equation}

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