Consider the following 2-loop 1PI diagram for the $\lambda \phi^4$ theory.


This is given by the following integral:

$$-i\ \Sigma(p^2) = \frac{(i\lambda_0)^2}{\#} \int\frac{\mathrm d^4s}{(2\pi)^4}\frac{\mathrm d^4t}{(2\pi)^4} \frac{i}{s^2 - \mu_0^²+i\varepsilon} \frac{i}{t^2 - \mu_0^²+i\varepsilon}\frac{i}{(p-s-t)^2 - \mu_0^²+i\varepsilon} \,.$$

We can see that it is quadratically divergent. Note that a derivative with respect to $p^2\,,$

$$ \frac{\partial}{\partial p^2} = \frac{1}{2p^2} p_\mu \frac{\partial}{\partial p_\mu}\,,$$

decreases the degree of divergence by $1$.

That is to say, $$\frac{\partial \Sigma}{\partial p^2}\Big|_{p^2=0} \text{ is linearly divergent,} \\ \quad \text{ and } \frac{\partial^2 \Sigma}{\partial^2 p^2}\Big|_{p^2=0} \text{ is logarithmically divergent.} $$

Therefore, a Taylor series expansion of this function about $p^2 = 0$ should contain three divergent terms.

$$ \Sigma(p^2) = \color{red}{\Sigma(0)} + \color{red}{\frac{\partial \Sigma}{\partial p^2}\Big|_{p^2=0}} p^2 + \color{red}{\frac{\partial^2 \Sigma}{\partial^2 p^2}\Big|_{p^2=0}} (p^2)^2 + \tilde{\Sigma}(p^2) \,,$$

where $\tilde{\Sigma}(p^2)$ is UV-finite and $\tilde{\Sigma}(0) = 0\,, \tilde{\Sigma}'(0) = 0 \,.$

That is my understanding. However, the general consensus is that there is no linearly divergent term in the above series. In fact, the coefficient of $p^2$ should be logarithmically divergent. I do not understand why.

Cheng and Li (p.34) say that it is because the coefficient should in fact be $\frac1{8} (\partial/\partial p_\mu) (\partial/\partial p^\mu) \Sigma(p^2)|_{p^2 =0}\,.$ Furthermore, the linearly divergent term is absent due to Lorentz invariance.

First of all, I do not understand why Cheng and Li have a different coefficient than mine. Secondly, what has Lorentz invariance got anything to do with Taylor expansion?

To summarise, why are there no linearly divergent terms in the sunset diagram?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.