3
$\begingroup$

This is somewhat related to an earlier question I asked about the following diagram in $\phi^{4}$ theory:

enter image description here

I've been following these lecture notes by H. Kleinert and V. Schulte-Frohlinde.

Saying we're in $D$-dimensions and going to momentum space, the above diagram corresponds to the following: $$ - \lambda \int \frac{d^{D}p}{(2\pi)^{D}} \frac{1}{p^{2}+m^{2}} = - \lambda \frac{(m^{2})^{D/2}}{(4\pi)^{D/2}} \Gamma\left(1 - \tfrac{D}{2}\right) $$

The above is divergent for $D=4$, so we consider small $\epsilon$ for which $4-D=\epsilon$. We consider an arbitrary mass parameter $\mu$, and introduce a dimensionless coupling constant $g =\lambda \mu^{-\epsilon}$. The above then reads: $$ = - m^{2} \frac{g}{(4\pi)^{2}} \left( \frac{4 \pi \mu^{2}}{m^{2}} \right)^{\epsilon/2} \Gamma\left( \tfrac{\epsilon}{2} - 1 \right) $$ And performing a Taylor expansion about small $\epsilon$, we find that the above becomes ($\psi$ is the digamma function): $$ \approx m^{2} \frac{g}{(4\pi)^{2}} \left[ \frac{2}{\epsilon} + \psi(2) + \log\left( \frac{4 \pi \mu^{2}}{m^{2}} \right) + \mathcal{O}(\epsilon) \right] $$

$\ $

I am interested in getting the contribution from the above in position space, in the massless limit $m \to 0$. I have two questions:

  1. In the lecture notes above, it says that the above diagram is IR-divergent in the limit that $m^{2} \to 0$. What does this mean, precisely?

  2. If we have an incoming momentum $k$, and the the above diagram corresponds to a function $\tilde{F}(k)$ in momentum space, then in position space we have a contribution given by $F(x_{1},x_{2}) = \int \frac{d^{4}k}{(2\pi)^{4}} e^{- k \cdot (x_{1} - x_{2})} \tilde{F}(k) $. How do I do this in the framework of dimensional regularization? Can I even do this? Where is the dependence of $k$ in the above that I can even do the integral, and then how do I complete that integral?

At the end of the day, I am trying to understand the nature of the divergence for this diagram in position space (in the massless case).

$\endgroup$
2
$\begingroup$
  1. The IR divergence means that it is divergent at low energies, and you can see that when $m=0$ the integral is divergent at $p =0$.
  2. This loop diagram in phi4 has the particularly not to depend on the outside momentum, so the result should be the same in momentum representation or position representation.
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.