Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 $\sim\lambda \mu^0 = \lambda$.

share|improve this answer
    
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? –  alexarvanitakis Mar 7 '13 at 2:51
    
@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. –  DJBunk Mar 7 '13 at 3:01
    
Yes I am asking precisely that. I edited the question to pinpoint exactly where the problem is. –  alexarvanitakis Mar 7 '13 at 3:13
    
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. –  DJBunk Mar 7 '13 at 3:18
    
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. –  alexarvanitakis Mar 7 '13 at 3:22
show 7 more comments
up vote 0 down vote accepted

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.

share|improve this answer
    
Ugh, sorry I didn't get around to this. Glad it got cleared up. –  DJBunk Mar 13 '13 at 18:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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