# Week 6, Thursday

## Two-Point Correlators at Second Order

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.

## Renormalization Prescription

To summarize, we can combine the Lagrangian and counterterms into a renormalized Lagrangian $$\mathcal{L}_\text{Ren} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi -\frac{1}{2} m_\text{bare}^2 \phi^2 – \frac{1}{4!} \lambda_\text{bare} \phi^4$$ with “bare” parameters $$\begin{split} \lambda_\text{bare} =&\; \lambda_0 M^2 = \lambda \\ m_\text{bare}^2 =&\; m^2 – \frac{\lambda_0 M^2}{4\pi} \left( \frac{2}{\omega} + F\right) \end{split}$$ 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 $$\Gamma_{\mathcal{L}_\text{Ren}} (k_1, \dots, k_n; m_\text{bare}^2, \lambda_\text{bare}, \omega) = \Gamma_{\mathcal{L}+\mathcal{L}_\text{ct}} (k_1, \dots, k_n; m^2, \lambda_0, M, \omega).$$ 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 $$\frac{1}{\Gamma^{(2)}(k_1,k_2)} \Big|_{k_1=k_2=0} = m^2.$$ 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 $$\frac{1}{\Gamma^{(2)}(k_1,k_2)} \Big|_{k_1=k_2=0} = m^2 \left[ 1 + \frac{\lambda_0 M^2}{4\pi m^2} \left( -\gamma + \ln\left(\frac{4\pi M^2}{m^2}\right) – F \right) + O(\lambda^2) \right]$$ which we can solve by setting $$F = -\gamma + \ln\left(\frac{4\pi M^2}{m^2}\right)$$

• 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.

# Week 6, Wednesday

## Regularization

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 $$A(\Lambda) = – \frac{\lambda}{2} \int_{|\ell| \leq \Lambda} \frac{d^4 \ell}{\ell^2+m^2}.$$ 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 $$A(z) = – \frac{\lambda}{2} \int \frac{d^4 \ell}{\big(\ell^2+m^2\big)^z}, \quad z \gg 1.$$ 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 $S_d$, $$D_d = \frac{1}{d} S_{d-1} ,\quad S_d = 2\pi D_{d-1},$$ which have a simple geometric origin. We can solve them in closed form as $$D_d = \frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+2)} ,\quad S_{d-1} = \frac{2 \pi^{d/2}}{\Gamma(\frac{d}{2})}.$$ 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 $$A(\omega) = -\frac{\lambda}{2} \int \frac{d^{d-\omega}\ell}{(2\pi)^{d-\omega}} ~ \frac{1}{\ell^2 + m^2} = -\frac{\lambda}{2} S_{d-1-\omega} \int_0^\infty \frac{d\bar\ell ~ \bar\ell^{d-1-\omega}}{\bar\ell^2 + m^2}.$$ The remaining one-dimensional integral can be solved by the formula $$\int_0^\infty \frac{x^k dx}{\big(x^n + a^n\big)^r} = \frac{ (-1)^{r-1}\; \pi \; a^{k+1-nr} \; \Gamma\big(\tfrac{k+1}{n}\big) }{ n \; \sin\big(\tfrac{k+1}{n} \pi\big) \; \Gamma\big(\tfrac{k+1}{n}-r+1\big) \; (r-1)! }$$ and using Euler’s reflection formula $$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}.$$ The result is $$A(\omega) = -\frac{\lambda}{4\pi} \Gamma\big(\tfrac{\omega}{2}\big) \left(\frac{2 \sqrt{\pi}}{m} \right)^\omega,$$ 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 $$\lambda = \lambda_0 M^{2+\omega}$$ into a dimensionless coupling constant $\lambda_0$ times a mass scale $M$. Then, $$A(\omega) = -\frac{\lambda_0 M^2}{4\pi} \Gamma\big(\tfrac{\omega}{2}\big) \left(\frac{2 \sqrt{\pi}M}{m} \right)^\omega,$$ 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) $$\Gamma(\omega) = \frac{1}{\omega} – \gamma + O(\omega) ,\quad x^\omega = 1 + \omega \ln(x) + O(\omega^2)$$ we obtain $$A(\omega) = -\frac{\lambda_0 M^2}{4\pi} \left[ \frac{2}{\omega} – \gamma + \ln\left( \frac{4\pi M^2}{m^2} \right) + O(\omega) \right]$$ 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.

## Renormalization

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 $$A(\omega) \delta(k_1-k_2) \frac{1}{k_1^2+m^2} ~ \frac{1}{k_2^2+m^2}.$$ 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 $$—\times— ~= \frac{\lambda_0 M^2}{4\pi} \left(\frac{2}{\omega} + F \right) \delta(k_1-k_2) \frac{1}{k_1^2+m^2} ~ \frac{1}{k_2^2+m^2},$$ 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 $$\mathcal{L}_\text{ct} = \frac{1}{2} \frac{\lambda_0 M^2}{4\pi} \left(\frac{2}{\omega} + F \right) \phi^2$$

# Week 6, Monday

We simplify the superficial degree of divergence to $$\begin{split} D =&\; dL – 2I \\ =&\; d – \frac{d-2}{2}E + V(d-4) \\ =&\; \begin{cases} 4-E & \text{in 4d} \\ 2-2V & \text{in 2d}. \end{cases} \end{split}$$ A superficial divergence (by power counting) does not necessarily translate into an actual divergence, for example the tree level interaction (E=4, V=1) has superficial degree of divergence $D=0$. But, since there is no loop integral, it obviously does not diverge logarithmically. Neither does a superficial convergence $D<0$ guarantee convergence, for example attach a tadpole to any convergent diagram. The tadpole integral remains divergent. But in that case the divergence just comes from a sub-diagram, it is not inherently due to the larger diagram. This suggests that we should only look at diagrams without superficially divergent sub-diagrams; If there is a divergent sub-diagram then you really only have to worry about that sub-diagram. This suggests the following:

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.

## Overlapping Divergences

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 $$\begin{split} I =&\; \int d^4p \; d^4\ell \; \frac{1}{p^2+m^2-i\epsilon} \frac{1}{\ell^2+m^2-i\epsilon} \\ &\; \qquad \times \frac{1}{(p+\ell)^2+m^2-i\epsilon} \frac{1}{(p-k_1)^2+m^2-i\epsilon} \frac{1}{(\ell-k_2)^2+m^2-i\epsilon} \end{split}$$ 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 $$I \sim \int \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4}.$$ 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, $$U_p = \big\{ p \big| p^2 \geq \ell^2,~ (p+\ell)^2 \geq \ell^2 \big\},$$ and combine it with the necessary IR cutoff $$U_\ell = \big\{ \ell \big| \ell^2 \geq 1 \big\}.$$ 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 $$I \sim \int_{U_p} \int_{U_\ell} \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4} = \int_0^\infty d\bar\ell \bar\ell^3 \frac{1}{\bar\ell^6} \int_{S^3} de_\ell \int_{U_{p’}} \frac{d^4 p’}{(p’)^4 (p’+e_\ell)^2}$$ where we rescaled $p’ = p \bar\ell$ and integrate over the rescaled region $$U_{p’} = \big\{ p’ \big| (p’)^2 \geq 1,~ (p’+\ell)^2 \geq 1 \big\}.$$ 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.