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I am taking a course on quantum field theory where there is some confusion regarding the renormalisation scheme we are using (and a corresponding one in my mind). Apparently the lecturer meant MS-bar when he said MS. Anyway, on with the question.

Consider the* 1-loop contribution to the 1PI 4-point correlation function of phi-4 theory in $d=4-\epsilon$ dimensions, i.e the Lagrangian is $$L=-\frac{1}{2}(\partial \phi)^2 -\frac{1}{2}m^2 \phi^2-\frac{1}{4!} \mu^\epsilon \lambda \phi^4$$ where $\mu$ is a mass scale required to make $\lambda$ dimensionless as we vary $\epsilon$.

The Feynman rules yield an expression of the form $$\frac{(\mu^{ \epsilon}\lambda)^2}{2} \int \frac{d^d k}{(2 \pi)^d}\frac{1}{(k^2 +m^2)((k+p)^2+m^2)}$$where $p$ is a certain sum of external momenta, which eventually reduces to $$\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}}$$I can clearly pull one factor of $\mu^\epsilon$ into $ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}}$ to get $\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)^{-\frac{\epsilon}{2}}$ so that the final, finite result at 1 loop looks like $$ -\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right) +\text{2 other terms*}$$

But where did the other factor of $\mu^\epsilon$ disappear to? Was it implicitly 'countertermed' away along with the factor $ [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$? It looks as if this can be done at this order, but does this generalise to higher loop orders, i.e, is there a prescription telling us how many factors $\mu^\epsilon$ to counterterm away?

Note that in phi-3 theory we get an identical loop integral by considering the 1-loop contribution to the 1PI 2-point correlation function, the difference being that $(\mu^{ \epsilon}\lambda)^2 \to (\mu^{ \frac{\epsilon}{2}} g)^2$ (where $g$ is the phi-3 coupling constant) in front of the integral. So that diagram does not have this problem somehow.

*There are two other contributions obtained by permuting the external momenta. These have identical integral expressions.

EDIT: I'll write this down fully. The term inside the integral can be written as $$(m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}} =\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)^{-\frac{\epsilon}{2}}\Gamma\left(\frac{\epsilon}{2}\right) [4\pi]^{\frac{\epsilon}{2}}=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$$

OR as $$=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}\mu^2]^{\frac{\epsilon}{2}}$$ where the last two equalities are both asymptotic expansions in $\epsilon$ as $\epsilon \to 0$. I can absorb that last square bracket into the counterterm if I want to, at this order, and effectively ignore it.

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2 Answers 2

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Your $\mu^\epsilon$ is still there, it's just that you have expanded in small $\epsilon$ so you got

$$ \mu^\epsilon \approx 1 + \epsilon \log \mu = 1 - \epsilon \log \frac{1}{\mu} $$

The $\log \frac{1}{\mu}$ is there in your expression

$$ -\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)$$

Now if we had instead

$\mu^\epsilon \mu^\epsilon = \mu^{2\epsilon}$

we'd just get

$$ \mu^{2\epsilon} \approx 1 + 2\epsilon \log \mu = 1 - 2 \epsilon \log \frac{1}{\mu} $$

I think what is confusing you is your are writing down a $d = 4+\epsilon$ dimension lagrangian with a $\mu^\epsilon \lambda$ interaction term, and then you want to get a loop correction that is for arbitrary dimensions too. That is fine, I suppose you can calullate that, (although I don't know how far you can take it without specifying the number of dimensions, or equivalently $\epsilon$). At any rate, the calculation you are performing isn't for arbitrary number of dimensions, it is for $d = 4$ since you are expanding in small $\epsilon$ so your corrections are coming out $\sim\lambda \mu^0 = \lambda$.

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  • $\begingroup$ There's a $\mu^{2 \epsilon}=\mu^\epsilon \mu^\epsilon$ in the expression I started with and this accounts only for $\mu^{\epsilon}$. Besides, what about phi-3 as I mention above? $\endgroup$
    – user21299
    Mar 7, 2013 at 2:51
  • $\begingroup$ @alexarvanitakis - I edited as per your $\mu^{2 \epsilon}$ comment, let me know if I still didn't answer it. As for the $\phi^3$ model, I am a little unclear what you are asking - are you asking why you get a factor of $\mu^2$ left over in the 1 loop correction to the 2 point function in this case? Please elaborate. $\endgroup$
    – DJBunk
    Mar 7, 2013 at 3:01
  • $\begingroup$ Yes I am asking precisely that. I edited the question to pinpoint exactly where the problem is. $\endgroup$
    – user21299
    Mar 7, 2013 at 3:13
  • $\begingroup$ If I understand you correctly, there just isn't anything to protect you from getting $\mu^2$ terms in your 2 point function for a scalar. A scalar mass is susceptible to quadratic corrections to any mass scale you have that enters your loops. So, dimensionally we expect them. $\endgroup$
    – DJBunk
    Mar 7, 2013 at 3:18
  • $\begingroup$ I have no idea what that means. Anyway the 2-point function is just fine; it's the 4-point one that's problematic because of the dimensionful logarithm. $\endgroup$
    – user21299
    Mar 7, 2013 at 3:22
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Finally got this cleared up: after a discussion that was somewhat longer than it really should be it was agreed that you pack the remaining $\mu^\epsilon$ into the counterterms like I describe above.

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  • $\begingroup$ Ugh, sorry I didn't get around to this. Glad it got cleared up. $\endgroup$
    – DJBunk
    Mar 13, 2013 at 18:50

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