Tuesday, October 26, 2004

Good coupling, bad coupling

Let's pick up where we left off. We saw in the last post that lattice perturbation theory, when applied naively, didn't work. That is, one and two loop estimates of short distance quantities didn't agree with the non-perturbativly measured values. There are three reasons for this: The first is the suitability of the bare lattice coupling $\alpha_0$ as a perturbative expansion parameter.

Forget about lattice perturbation theory for a minute. Just imagine we've computed something in standard QCD perturbation theory, as a series in the MSbar coupling
$$
F = f_0 + f_1 \alpha + f_2 \alpha^2 + \cdots
$$
We'll assume that this series is perfectly convergent, with each coefficient of order 1. If we assume they are one, and use $\alpha = 0.1$ we find
$$
F ~ 1.11 + O(0.001)
$$
We have a good series. Notice the one loop estimate would have been 1.1.

Now, what if we were perverse, and did the following
$$
\alpha = \alpha_b ( 1 + 1000 \alpha_b)
$$
Remember, in QCD the coupling constant is scheme dependant. So there's some scheme in which this is true. Solving this with $\alpha= 0.1$ shows that $\alpha_b ~ 0.0095$ so we've made the coupling really small.

In terms of our new coupling we have
$$
F = f_0 + f_1 \alpha_b + (f2 + 1000 f_1) \alpha_b^2 + \cdots
$$
The series is still formally correct, but we've made the two loop contribution huge, and the one loop contribution tiny. So if we had only had the one loop contribution at hand, we'd get
$$
F ~ 1.0095
$$
Likewise, even with the two loop result, we'll still get the wrong answer because the three loop term is going to be messed up. What all this illustrates is that it's important to pick a coupling that's not abnormally small. If it is, the convergence of the perturbation series is very slow.

This is exactly what happens in lattice perturbation theory. The bare lattice coupling is quite small (the 1000 in the relationship above is more like 10, but it's still enough to screw things up), so perturbation theory really doesn't work well. The solution is to find a different coupling constant, and work with that.

There are any number of choices one could make. For example, you could use the MSbar coupling (which is good), but it makes more sense to use a coupling defined in some "latticey" sense. The coupling Lepage and Mackenzie proposed using is one defined from the short distance static quark potential $\alpha_V$. This coupling is defined such that the short distance part of the potential (the Coulomb part) is given in momentum space by
$$
V(q) = -4/3 \alpha_V(q)/q
$$
With this definition you can derive the connection between the bare coupling $\alpha_0$ and $\alpha_V$ by computing V(q) as an expansion in the former, then demanding it has the form given above. Quantities expressed in $\alpha_V$ tend to agree with non-perturbative data much better than those expressed in terms of the bare coupling.

One thing that we'd like to understand is the origin of the large numbers in the connection between the bare coupling, and the good (V) coupling. That is, what could give rise to the factor of 1000 in the example above, or the factor of 10 that occurs in practice? That'll
be the subject of the next post on tadpole diagrams.

UPDATE: Well, the cool "post by email" thing is nice, except the formatting gets all messed up. Here's hoping this is better.

5 comments:

Anonymous said...

Juest found this nice blog.

"The bare lattice coupling is quite small (the 1000 in the relationship above is more like 10, but it's still enough to screw things up), so perturbation theory really doesn't work well."

Any simple explanation why it is 10? Persumably a function of discretization currently used.

Some other questions (maybe future blog entries?):

* How does lattice QCD deal with asymptotic series, i.e., how does it know when to stop?

* How does one find in lattice that chiral symmetry is broken? And confinement (Wilson'a area law, but I am forgetting details).

* Can lattice QCD currently tell us what bound states (mesons, baryons etc) can arise? How could that be done in principle?

Many thanks!

Matthew said...

Any simple explanation why it is 10?
Yes, it'll be in the next post on tadpole factors. I'm writing it now.

How does lattice QCD deal with asymptotic series, i.e., how does it know when to stop?
It doesn't use a series expansion. The perturbation theory is only for matching actions. The actual computations are done using Monte-Carlo methods, so there's no series, Asymtotic or otherwise.

How does one find in lattice that chiral symmetry is broken?
This is tricky since most lattice discretizations of fermions screw up the fermions. But, in prinicple, you can dial the quark masses to be whatever you want, and observe in the massless limit that you get a trio of massless mesons (the pions). You can also measure things like the chiral condensate.

And confinement
You can do it in the most naive way if you have the computing. Namely put two quarks on the lattice, and measure the potential between them as a function of their separation. If you do this, you see that the potential is linear, as you'd expect for confinement. This is the same as Wilson's area law (that is if the expectation value of the Wilson loop grows as the area, you have a linearly rising static quark potential).

Can lattice QCD currently tell us what bound states (mesons, baryons etc) can arise? How could that be done in principle?
To measure a bound state you create a state of the appropriat quantum numbers and propagate it across the lattice. If it's a bound state of mass M you'll find that the propagator goes like (up to constant coefficients)

exp(-M0 * t) + exp(-M1 * t) + ...

Where M0 < M1 < M2 ... are the masses of the ground state the 1st excited state and so one. If you take t large enough you can ignore all but the ground state. If you want the excited states, you can (in principle) get those as well.

If the state is unstable there will be an oscillatory componant, which (again in principle) you can use to get the decay width.

I hop that answers you questions, please ask more if it doesn't.

Anonymous said...

Many thanks. You have answered most of my questions very well! I greatly appreciate them.

Just one question on them:

"You can also measure things like the chiral condensate."

I understand this part of your answer on CSB: simply calculate the VEV of $\bar{\psi}\psi$.

But I do not understand this:

"But, in prinicple, you can dial the quark masses to be whatever you want, and observe in the massless limit that you get a trio of massless mesons (the pions)."

How does one "see" that? That is, what is being computed on the lattice that leads to that conclusion?
In theory, one looks at \langle vac|SU(2) current| pi\rangle...

Also, another question:

Does one see U(1) anomaly violation effects caused by
instantons in the lattice? Or they are too small for current computational resources?

Thanks! Have a great weekend.
DMS

PS: What happened to your 'Current Mood' and 'Current Music'?

Matthew said...

But I do not understand this:

"But, in prinicple, you can dial the quark masses to be whatever you want, and observe in the massless limit that you get a trio of massless mesons (the pions)."

How does one "see" that? That is, what is being computed on the lattice that leads to that conclusion?
In theory, one looks at \langle vac|SU(2) current| pi\rangle...
You do it in the same way as you do in the continuum, measure \langel vac | SU(2) | \pi \rangle on the lattice. In Euclidean space this will give you the exponential form I mentioned before, and you'll find $M_{\pi} -> 0$ as $m_{up,down) -> 0$.

Now there's problems with this, one is that it's hard to simulate with small quark masses. The other is that as $M_{\pi$ -> 0$ the pion's compton wavelength is getting large, so you have to worry about finite volume effects.

The other thing you can do is measure $M_{\pi}$ as a function of quark masses and see if it fits onto predicitions of Chiral Perturbation theory.

Does one see U(1) anomaly violation effects caused by
instantons in the lattice? Or they are too small for current computational resources?
You can observe mass differences between the \eta' and the \eta if that's what you're getting at.

PS: What happened to your 'Current Mood' and 'Current Music'?
That was a feature of livejournal, which blogger doesn't have. Imagine "busy" and "gangsta rap" as reasonable stand ins :)

Anonymous said...

Excellent! Many thanks for your answers.

DMS