This is the final installment in this series. We've been seeing how to make perturbation theory work, using a lattice cutoff. The main point so far has been that the bare lattice coupling gets large renormalizations from lattice tadpole diagrams. This makes it a very poor expansion parameter for perturbative series'. We can correct this by using a "natural" coupling, in practice we use the "V" coupling, define through the static quark potential. We could also use the standard MSbar coupling, but that isn't really a "lattice thing".

There's one final thing we need to do to make the perturbation series as accurate as possible. That is to set the scale. Remember, the quantities we compute perturbativly on the lattice are all short distance, so we expect that they will be dominated by momentum scales of the order of the cutoff \pi/a. But they're not *zero*distance* quantities. A 1x1 square is still a finite sized object, so the scale won't be exactly \pi/a. How do we determine the scale?

Let's consider what we do when we calculate something to one loop. We'll get a loop integral over some integrand

$$

\int dq f(q)

$$

times the coupling constant at some scale

$$

\alpha_{V}(q*)

$$

I've used the "V" scheme here, and called the optimal scale we're looking for $q*$. Remember the QCD coupling varies with $q$. This suggests that the ideal scale might be found by averaging the coupling, weighted by the integrand $f(q)$. That is

$$

\alpha_{V}(q*) \int dq f(q) = \int dq \alpha_{V}(q) f(q)

$$

This would seem to be a natural definition.

The trouble is that this definition, if naively applied, doesn't work. The reason is that the running coupling is usually written as

$$

\alpha_{V}(q) = \frac{\alpha_{V}(\mu)}{1 - \beta\log(q^2/\mu^2) \alpha_{V}(\mu)}

$$

and this function has a pole in it. A couple of points, $\beta$ is the one loop beta function. For those of you who don't know what that is, just imagine that it's a negative constant. $\mu$ is the renormalization scale, it'll go away soon. If I jam this into the integral on the right hand side it'll blow up. There's a simple, and sensible, fix to this, expand the running coupling to the correct order in $\alpha_{V}(\mu)$ on both sides, and solve. I'll spare you the algebra, if you take

$$

\alpha_{V}(q) = \alpha_{V}(\mu) + \beta \log(q^2/\mu^{2}) \alpha_{V}^{2}(\mu)

$$

and the same for $\alpha_{V}(q*) and jam them into the defining equation for $q*$ you'll find (trust me!)

$$

\log(q*^{2}) = (\int dq \log(q) f(q)) / \int dq f(q)

$$

Viola! This is a sensible, convergent definition. And that's how you set the scale.

So let's review how we do lattice perturbation theory.

1) Use tadpole improved actions and operators. This will remove large perturbative coefficients that come from lattice tadpole diagrams

2) Use a sensible coupling $\alpha_{V}$. This coupling is not too small, so the convergence of the perturbation series is pretty good

3) Use a correctly set scale. The quantities we're going to compute are short distance, not zero distance, so the scale is not \pi/a

Just to be clear, none of this is new. It's all in a famous paper by Lepage and Mackenzie (PRD48, 2250 1993) so if you want to learn more, and see examples of how it all works in practice, go look there.

Next post (which I promise will be sooner than this one came in) I'll talk a little bit about some of the non-perturbative results that our group has calculated.