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In the renormalization of $\phi^4$ theory the integral: $$I=\int \frac{d^Dk}{(2\pi)^D} \frac{1}{(k^2+M_0^2)((k-p)^2+M_0^2)}$$ appears during renormalization, where I am taking $M_0$ to be the bare mass. In the massless case the renormalized mass is zero, $M=0$. If I want to find the bare vertex function in terms of renormalized parameters to lowest order - I can see two possible ways to treat $I$:

  1. Expand $I$ to lowest order initially which since $M_0 =0+\mathcal{O}(\lambda)$ becomes: $$I=\int \frac{d^Dk}{(2\pi)^D} \frac{1}{k^2(k-p)^2}$$ This gives me a non-zero value.
  2. Alternatively I could calculate $I$ with $M_0$ and then expand. This gives me (in $D=4-2\varepsilon$ dimensions): $$I\propto M_0^{-\varepsilon}$$ $$ \sim \lambda^{-\varepsilon}$$ Now from what I remember we can treat $\lambda$ as arbitrarily small and as such I would say that this should go to $\infty$.

With my interpretation of $\lambda^{-\varepsilon}$ the two methods therefore do not agree. Which method is the correct way to approach this and how should I interpret the quantity $\lambda^{-\varepsilon}$

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  • $\begingroup$ @marmot Actually surly this is more worrying since not only do we no longer have a perturbation expansion in this last case we get terms of the form $\lambda^{-a}$ for arbitrarily large $a$ if we carry the expansion in terms of $\lambda$ to higher order. $\endgroup$
    – Quantum spaghettification
    Commented Jan 1, 2018 at 15:20
  • $\begingroup$ Sorry, I cannot understand the very long sentence in your last comment. But your question reads now that you get infinity in both computations, and you're wondering why the results do not agree. (I agree of course that the results are not the same, but I am wondering what the question is. The first integral diverges also if you resurrect $M_0$.) $\endgroup$
    – user178876
    Commented Jan 1, 2018 at 15:26
  • $\begingroup$ @marmot (ignore my last comment is wasn't very important). The problem is I think that we should expect the same result for the coefficient of $\lambda^0$ in both cases. This is not the case. Further-more if we put $\lambda=0$ in the latter case (equivalent to putting $M_0=0$) we do not return to the result of the first case. $\endgroup$
    – Quantum spaghettification
    Commented Jan 1, 2018 at 15:31
  • $\begingroup$ I am still struggling to understand the question. But let me tell you this. You're computing the correction to a mass squared, and surely this should have mass dimension two. On the other hand, you're dropping all quantities with mass dimension (such that your theory becomes scale invariant at the classical level). If you were to use cut-off regularization, the correction would be proportional to $\Lambda^2$. In dimensional regularization it becomes proportional to $\mu^2$. But you are right that it is nontrivial to see this, but well known (c.f. Coleman-Weinberg potential). $\endgroup$
    – user178876
    Commented Jan 1, 2018 at 15:38
  • $\begingroup$ @marmot I will try and make my question more explicit; Why do the two methods not agree and which one gives the correct coefficient at order $\lambda^0$. $\endgroup$
    – Quantum spaghettification
    Commented Jan 1, 2018 at 15:48

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OK so coming back to this after a short brake from work I realize I have made an error when writing the post. The expression I wrote for the case (2) is wrong and infact we should have that: $$I \propto \int^1_0 (x(1-x)p^2+M_0^2)^{D/2-2}dx$$ This essentially makes my question void and clears up the confusion I had about it. Sorry.

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