# 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.