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I'm trying to calculate the 1 loop correction to the propagator in massless $\phi^4$ theory, in $d = 3$, just for fun. The diagram just looks like a straight line with a circle touching tangently to it.

I'm not sure whether to take $ m = 0$ straightaway or to compute it for $m > 0$ before taking $m \to 0$, because the two methods give different answers.

The diagram is $\sim \int d^3k \frac{1}{k^2 - m^2}$. If I set $m = 0$ straightaway, then I get $\sim \int k^2 dk \frac{1}{k^2} \sim \Lambda$, where $\Lambda$ is the cutoff.

On the other hand, if I do the integral, using the results in the appendix from Peskin and Schroeder \begin{align} \int d^dk \frac{1}{k^2 - m^2} \sim \frac{\Gamma(1-d/2)}{(m^2)^{1-d/2}} =\Gamma(-1/2)(m^2)^{1/2}. \end{align} $\Gamma(-1/2)$ is finite, so taking $m \to 0$ this diagram is $0$.

So one one hand it's UV divergent, on the other hand it's 0.

Which method should I be using? Are the theories where $m=0$ and $m \to 0$ not the same? I also seem to have missed out the $i \varepsilon$ prescription in the propagator in the first case - if I include that in and evaluate the diagram before sending $\varepsilon \to 0$, then the diagram is $0$ too.

Or perhaps it does not matter too much because the mass renormalization term can take care of either divergence and such discrepancies will not show up in calculations of any observables anyway?

Thanks.

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I think that dimensional regularization misses polynomial divergences. Darn, I can't recall a reference right now... Also, $\lambda$ being a dimension 1 coupling, will get renormalized to vanish in the UV; you'll need a counterterm corresponding to that interaction. –  Siva Apr 25 '13 at 6:19
    
@ Siva: thanks. I'll post an answer below, please refer to it :) –  nervxxx Apr 25 '13 at 6:46
    
I'm not sure but it's possible that the DimReg result isn't correct when $m=0$, i.e. that you run into troubl e because of the pole at $k=0.$ But indeed the renormalisation scheme doesn't make a difference. What you should do is calculate the finite part for the 'hard cutoff' method, i.e. coefficient at order $\Lambda^0.$ Since corrections to the renormalisation of the coupling occur at $O(g^2)$, the finite parts of the one-loop corrections should agree. –  Vibert Apr 25 '13 at 7:18
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up vote 3 down vote accepted

Ok I'll answer my own question. I asked my QFT professor, he said different methods of regularization will give different answers. but at the end of the day it doesn't matter because you're going to cancel whatever divergences from that integral from whatever method you use, with the mass counterterm anyway, to impose desired renormalization conditions. so that integral is not physical and all is good.

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