Large logarithms and RG

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  • Thread starter Malamala
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Hello! I am reading Schwarz book on QFT and I am at the renormalization group part. I understand the math and the fact that grouping terms perturbatively and requiring them to be 0, when taking the derivatives with respect to a given arbitrary chosen scale, allows perturbative corrections to the coupling constant as a function of momentum transfer. But I am not really sure I understand the large logarithms arguments. Why are they important? I don't really see them used in any of the proofs. You just take the derivative, set it to zero and you get what you want. Can someone explain to me what is the deal with large logarithms and why they seem to be so important in relation to the RG? Thank you!
 

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king vitamin
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I'm not completely clear on your question, so I'll just talk a bit about how I understand the "problem of large logarithms" and hope that it helps out.

In a renormalizable massless quantum field theory, the perturbative expansion for some observable takes the form
[tex]
\Gamma(E) = g + a g^2 \log\left( \frac{E}{\Lambda} \right)
[/tex]
where [itex]E[/itex] is the energy you are observing the observable at, [itex]\Lambda[/itex] is a UV cutoff, and [itex]g[/itex] is the coupling. (And take [itex]a[/itex] to be some numerical constant.)

Now we renormalize. Let's define a renormalized coupling by the value of the above observable at some reference energy [itex]\mu[/itex]. That is,
[tex]
g_R \equiv \Gamma(\mu) = g + a g^2 \log\left( \frac{\mu}{\Lambda} \right)
[/tex]
Then at other energies, the observable takes the form
[tex]
\Gamma(E) = g_R + a g_R^2 \log\left( \frac{E}{\mu} \right)
[/tex]

Here's the issue. Let's say we look at some energy far smaller (or greater) than [itex]\mu[/itex]. Then the second term will eventually dominate over the first, invalidating perturbation theory. The problem is actually worse than it looks like, because if you do higher loops, you'll find higher powers of [itex]g_R \log\left( \frac{E}{\mu} \right)[/itex], and everything breaks down. But on the other hand, this is all kinda stupid because we could have chosen to renormalize the theory at the smaller (or larger) energy scale you're looking at and clearly there wouldn't be a problem. We could just have renormalized at a different scale and no issue would arise (provided the renormalized coupling remains small of course, which I'll comment on later).

So if we consider the scale [itex]\mu[/itex] to be something which can be varied depending on what scale we are interested in, we can solve the problem, since we could choose it to be near whatever energy scale we're looking at and the large logarithm problem will never occur. This leads pretty quickly into the renormalization group. How will the renormalized coupling [itex]g_R[/itex] change with changing [itex]\mu[/itex]? Well that can be parametrized by the beta function:
[tex]
\beta_g \equiv \mu\frac{d g_R}{d \mu} = a g_R^2
[/tex]
Let's assume [itex]a>0[/itex]. Then the renormalized coupling gets larger for larger energies, but as long as the energy scales are small it's still perturbative (which again says that the apparent divergence for small energies we saw in the logarithms above was not physical).

So how do we fix the issue with large logarithms? Let's just solve the differential equation, setting [itex]g_R(\mu_0) = g_{R0}[/itex], which realizes this "sliding scale" renormalized coupling. We get
[tex]
g_R(\mu) = \frac{g_{R0 }}{1 - g_{R0} \log(\mu/\mu_0)}
[/tex]
In this form, we can see that [itex]g_R(\mu)[/itex] remains small as [itex]\mu[/itex] decreases, logarithmically going to zero with no unphysical divergences, so there's no issue anymore. This is the sense in which RG has solved the large logarithm problem.

In contrast, the coupling increases with increasing energy, appearing to diverge at a finite energy [itex]\mu = \mu_0 e^{1/g_{R0}}[/itex]. There are worries about a theory with this structure not being sensible unless it is cutoff at such an energy scale (though this perturbative analysis is not enough to conclude this by itself).
 

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