Week 6, Wednesday


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}