This is the second post about fermions on the lattice. In this post we're going look at how the 16 extra "tastes" of lattice fermions show up, and one way that we can correct for them. The nice thing about this calculation is we don't really need to worry about the gauge fields in all this. We'll work in the free quark case, everything will apply to the interacting case as well.

In Euclidean space we can write the Dirac Action as

$$

S = \int dx \bar{\psi} [ \sum_{mu} \gamma_{mu} d_{mu} + m ] \psi

$$

If we transform to momentum space using

$$

\psi(x) = \int dk/(2\pi)^4 exp(ipx) \psi(p)

$$

we'll find the fermion propagator

$$

S = (-i \sum{mu} \gamma_{\mu} p_{\mu + m)/(p^2 + m^2)

$$

Of crucial importance is the presence of a single pole at p^{2} = - m^{2}. This tells us that the propagator is describing a single particle. This pole only fixes one componant of p, for example we could take p_{4} = im.

Now, lets put this theory on the lattice, discretizing the derivative in the most naive way we can,

$$

d_{\mu} \psi(x) = (\psi(x + \hat{a\mu}) - \psi(x-a\hat{\mu}))/(2a)

$$

$a$ is the lattice spacing. Our conversion to momentum space is now

$$

\psi(x) = \sum_{k} exp[ i px ] \psi(p)

$$

We're in a finite box, so the integration over all momenta has become a sum. With this expansion it's easy to read off the propagator

$$

S_{L} = (-i \sum_{\mu} \gamma_{\mu} pbar_{mu} + ma)/(pbar^2 + (ma)^2)

$$

where

$$

pbar_{mu} = sin(p_{\mu} a )

$$

and

$$

pbar^2 = \sum_{\mu} p_{\mu}^{2}

$$

This *looks* okay. We've just replaced the momenta in the continuum propagator with some trig functions. In the small $a$ limit, they go over into the correct factors, so we should be fine. But we're not.

We have poles in our lattice propagators at

$$

pbar^{2} = -(ma)^2

$$

Let's try p_{4} being the only non-zero momenta, then our condition is

$$

sin(p_{4}a) = i ma

$$

or

$$

p_{4}a = i arcsinh(ma)

$$

Again, this looks okay, for small $a$ $arcsinh(ma) ~ ma$ so we recover our continuum condition. But, we *could* have just as easly taken

$$

p_{4}a = i arcsinh(ma) + pi

$$

Since $sin(x+pi) = sin(x)$ this will be equivalent to my original situation. We've found two poles in the fermion propagator. We can carry this further, noting that for the spatial momenta we can take $p_{i} = 0$ or $p_{i} = \pi$. So overall there are $2^{4}=16$ different combinations of momenta that produce poles in the propagator. So this propagator describes not one, but 16 different types (tastes) of quarks.

This is the famous fermion doubling problem. As we discussed in the last post, there are a three primary ways around this. The first (and oldest) is due to Wilson. The idea here is simple, add some term to the action to change the propagator, so it has only one pole. Wilson added the following term

$$

- a/2 \laplacian

$$

Acting on a lattice field we have

$$

\laplacian \psi(x) = \sum_{\mu} [ \psi(x + a\hat{\mu}) + \psi(x - a\hat{\mu}) - 2 \psi(x) ]

$$

Notice that this term is explictly supressed by a power of the spacing, so it shouldn't cause problems in the continuum limit.

This new term does have a big effect on the fermion propagator. The denominator of the propagator becomes

$$

pbar^2 + (ma + a/2 phat^2)^2

$$

where

$$

phat^2 = \sum_{\mu} 4(sin(0.5 p_{\mu} a))^2

$$

Setting the spatial momenta to zero, and solving for $denominator = 0$ we find

$$

(sin(p_{4} a))^2 = -[ma + 2(sin(p_{4}a/2))^2]^2

$$

If we put $p_{4}a = i arcsinh(ma)$ we'll find that as $a \to 0$ this equation is satisfied (ie we'll find m = m), but for $p_{4}a = i arcsinh(ma) + \pi$ the new term well produce a term which destroys the solution. So this equation has only *one* solution in the continuum limit, not two. The same is true for the spatial momenta.

In more concrete terms what is happening is that the Wilson term gives the doubler fermions a mass proportional to the inverse lattice spacing

$$

M_{d} ~ m/a

$$

so in the continuum limit they become infinitly massive and decouple.

The Wilson solution to the doubling problem is easy to understand, implement and compute with. But it suffers from a number of flaws. The first is that the Wilson term breaks Chiral symmetry (CS). Without CS the quark mass is not protected from additive mass renormalizations. This makes it difficult to simulate at small quark masses. You need to preform large subtractions to keep the quark mass you get out (the renormalized mass) close to the mass you put in (the bare mass). Furthermore, these large additive renormalizations can push you into a region where the renormalized mass is negative. When this happens you have to throw out the run. This problem (jargon: "exceptional configurations") slows Wilson quark simulations down by a large factor. Another problem with Wilson quarks is that the Wilson term introduces a linear error. This can be very large.

For these reasons, people have looked for other ways of simulating quarks. In the next post we'll look at Kogut-Susskind (or Staggered) fermions.

## Monday, November 29, 2004

## Monday, November 22, 2004

### In case you have nothing to do

Here [1.5 Mb pdf] are the slides for a talk I gave Friday at Syracuse University. It went pretty well, lots of questions, which is always good (since I could answer them). The talk ranged pretty wide, I tried to cover our whole lattice field theory program, rather than bore people with technical details of lattice peturbation theory. I think it was the right approach for a non-lattice field theory crowd.

I like to go give talks to other groups. It is a good way of seeing how your ideas play with people who don't work with you on a day by day basis.

I like to go give talks to other groups. It is a good way of seeing how your ideas play with people who don't work with you on a day by day basis.

Labels:
publicity

## Thursday, November 18, 2004

### Fermions on the lattice, part one of four

This is a "fessing up" post, before I talk about all the great things that lattice QCD can do. I'm going to talk in general about fermions on the lattice, and discuss the ways in which you can simulate them. This will take us down three roads, and at the end we'll find pros and cons to each approach. I'll focus on an outstanding issue with staggered fermions at the end.

The formulation of fermion fields on the lattice has been a huge problem. Compared to Wilson's elegant formulation of gauge fields on the lattice fermions look downright ugly. To top it off, they're slow, so it's hard to deal with them. We'll look briefly at the latter problem, the speed first, and then talk about why lattice fermions are ugly.

Fermion fields in quantum field theory are represented by anti-commuting numbers, known as Grassman numbers. This is fine when you're working with a pencil and paper, but it's impossible to put a Grassman number into a computer. Computers are good for reals, integers and complex numbers (really two reals), but they can't do Grassman numbers.

This is not quite as big a problem as it sounds. The reason is every fermion action that people use can be written as

$$

S_{F} = \bar{\psi} M[U] \psi

$$

where $M[U]$ is a complex matrix functional of the gauge fields. This is happy, since we can do path integrals of this form,

$$

\int D\bar{\psi} D\psi exp(-S_{F}) = \det(M[U])

$$

and we're done! We now have fermion fields on the lattice. The trouble is that we need to evaluate the determinant. And that is mad slow. So slow, in fact, that until the mid-nineties it was essentially impossible. This led people to just drop it, using something called the quenched approximation. This causes uncontrollable ten or twenty percent errors, but it makes things faster. Physically, it amounts to neglecting dynamical fermions in your simulation.

As people started to remove the quenched approximation another problem came up. Evaluating the determinant gets slower as you go to smaller and smaller quark masses. This forces you to work at unphysical values for the up and down quark masses. That's the situation today. Simulations are forced to work at masses that are larger than the physical up and down quark masses. Then you have to extrapolate to the physical limit (which is near m=0).

That's the present situation. It is possible to simulate with dynamical fermions, but, for the most part, you have to simulate at unphysical masses.

Now why are lattice fermions ugly? Well, it turns out that if you take the Dirac action

$$

S = \bar{\psi} ( \gamma \cdot D + m) \psi

$$

and naively discretize it (by replacing D with a covariant difference operator) you get an action which describes 16 degenerate types of fermions. These degenerate types are called "tastes". So the naive discritization of lattice fermions induces 16 tastes. Obviously we don't want to simulate 16 degenerate quarks, we want to simulate one quark, so we need to figure out some way of getting rid of the other 15 tastes. In the next installment we'll see how the extra tastes show up, and how to remove them by adding a Wilson term to the action.

The formulation of fermion fields on the lattice has been a huge problem. Compared to Wilson's elegant formulation of gauge fields on the lattice fermions look downright ugly. To top it off, they're slow, so it's hard to deal with them. We'll look briefly at the latter problem, the speed first, and then talk about why lattice fermions are ugly.

Fermion fields in quantum field theory are represented by anti-commuting numbers, known as Grassman numbers. This is fine when you're working with a pencil and paper, but it's impossible to put a Grassman number into a computer. Computers are good for reals, integers and complex numbers (really two reals), but they can't do Grassman numbers.

This is not quite as big a problem as it sounds. The reason is every fermion action that people use can be written as

$$

S_{F} = \bar{\psi} M[U] \psi

$$

where $M[U]$ is a complex matrix functional of the gauge fields. This is happy, since we can do path integrals of this form,

$$

\int D\bar{\psi} D\psi exp(-S_{F}) = \det(M[U])

$$

and we're done! We now have fermion fields on the lattice. The trouble is that we need to evaluate the determinant. And that is mad slow. So slow, in fact, that until the mid-nineties it was essentially impossible. This led people to just drop it, using something called the quenched approximation. This causes uncontrollable ten or twenty percent errors, but it makes things faster. Physically, it amounts to neglecting dynamical fermions in your simulation.

As people started to remove the quenched approximation another problem came up. Evaluating the determinant gets slower as you go to smaller and smaller quark masses. This forces you to work at unphysical values for the up and down quark masses. That's the situation today. Simulations are forced to work at masses that are larger than the physical up and down quark masses. Then you have to extrapolate to the physical limit (which is near m=0).

That's the present situation. It is possible to simulate with dynamical fermions, but, for the most part, you have to simulate at unphysical masses.

Now why are lattice fermions ugly? Well, it turns out that if you take the Dirac action

$$

S = \bar{\psi} ( \gamma \cdot D + m) \psi

$$

and naively discretize it (by replacing D with a covariant difference operator) you get an action which describes 16 degenerate types of fermions. These degenerate types are called "tastes". So the naive discritization of lattice fermions induces 16 tastes. Obviously we don't want to simulate 16 degenerate quarks, we want to simulate one quark, so we need to figure out some way of getting rid of the other 15 tastes. In the next installment we'll see how the extra tastes show up, and how to remove them by adding a Wilson term to the action.

Labels:
lattice fermions

## Monday, November 08, 2004

### The scale of things

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.

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.

Labels:
perturbation theory,
strong coupling

## Monday, November 01, 2004

### The life of a lattice tadpole

At the end of the last post we saw that the bare coupling in lattice gauge theory is a bad coupling to do a perturbative expansion in. If we use a good coupling, like $\alpha_{V}$ we will find a much more convergent perturbative series. This is crucially important if we want to determine things like improvement coefficients to good accuracy. An open question at the end of the last post was *why* the bare lattice coupling is bad, that's the subject of this post.

To understand this, we need to understand how gauge fields are represented in the lattice. In lattice QCD, the matter fields are assigned to the sites of the lattice, and the gauge fields "live" on the links. This is very much in keeping with the geometric picture of gauge theory, where you have a parallel transporter ($U$) which allows you to compare matter fields at two different points. That is,

$$

\psi(x+\hat{\mu}) = U(x+\mu,x) \psi(x)

$$

In continuum gauge theory we have

$$

U(x+\mu,x) = P exp(ig \int_{x}^{x+\mu} dz^{\nu} A_{\nu}(z))

$$

Here P stands for path ordering.

This is actually much simpler on the lattice. On the lattice, there is a "shortest path" namely one link. So the parallel transporter simplifies to

$$

U(x+\mu,x) = U_{\mu}(x) = exp(i g a A_{\mu}(x))

$$

These parallel transporters (or link fields) are what one builds lattice actions out of. For example, a simple forward covariant derivative on a fermion field might be

$$

D_{\mu} \psi(x) = U_{\mu}(x)\psi(x+\mu) - \psi(x).

$$

Notice that the connection between the link field $U$ and the gauge field $A$ is non-linear. This is where the problems come in. Consider the expectation value of a single link field $(U)$. Expanding the expression for $U$ we find

$$

(U) = 1 - 2 \pi \alpha(\pi/a) a^{2} (A_{\mu}(x) A_{\mu}(x)) + ...

$$

I've used (A) = 0, and evaluated the strong coupling at the cutoff $\pi/a$. This looks fine, in the formal continuum limit (a->0) the lattice corrections vanish. But all is not well! There's a dangerous beast lurking in the loop integral (AA). Let's look at what it is.

The expectation value (A A) can be written as the integral over the momentum space gluon propagator

$$

(A A) = \int_{0}^{2\pi/a} d^{4}q / qhat^{2}

$$

where

$$

qhat^{2} = 4/a^2 \sum_{1}^{4} sin^2 (q_{\mu} a/2).

$$

In the limit where $a$ is small, we can replace $qhat^{2}$ by $q^{2}$, and we find that the integral is

$$

(A A) ~ 1/a^{2} + O(1)

$$

This is the disaster, because when we substitute this back into our expression for the expectation value of $U$ we find

$$

(U) = 1 - 2\pi \alpha(\pi/a) a^{2} (1/a^{2} + O(1)) + ...

$$

The $1/a^{2}$ from the loop integral cancels the $a^{2}$ from the expansion of the exponential leaving a lattice artifact suppressed only by the coupling constant, rather than the lattice spacing. The coupling constant is small (0.2) but not really small (like $\alpha a^{2} ~ 0.01).

It is factors like these (called gluon tadpoles) which spoil lattice perturbation theory in the bare coupling. The $\alpha_{V}$ coupling doesn't suffer from these corrections because tadpole effects come in an 2nd order in $\alpha$. It would be nice to have a way to systematically account for these effects. This method, called tadpole improvement (or mean-field improvement) was proposed by Lepage and Mackenzie, and is now in wide use.

The idea is to use link fields that have the tadpole effects largely canceled out of them. To do this you take all the link fields you have $U$ and multiple and divide by some average link field $u$. That is

$$

U \to u U/u = u \tilde{U}

$$

The new fields $\tidle{U}$ are much closer to their continuum values, since the average link field division cancels out most of the tadpole effects. The factors in the numerator $u$ can be absorbed into couplings and masses. When combined with a good coupling (such as $\alpha_{V}$) the agreement between perturbation theory and non-perturbative simulations (of short distance quantities) is much better.

There is one last thing one can do to improve the perturbative series, namely set the scale of the coupling in a more sensible way. That will be the subject of my next post.

To understand this, we need to understand how gauge fields are represented in the lattice. In lattice QCD, the matter fields are assigned to the sites of the lattice, and the gauge fields "live" on the links. This is very much in keeping with the geometric picture of gauge theory, where you have a parallel transporter ($U$) which allows you to compare matter fields at two different points. That is,

$$

\psi(x+\hat{\mu}) = U(x+\mu,x) \psi(x)

$$

In continuum gauge theory we have

$$

U(x+\mu,x) = P exp(ig \int_{x}^{x+\mu} dz^{\nu} A_{\nu}(z))

$$

Here P stands for path ordering.

This is actually much simpler on the lattice. On the lattice, there is a "shortest path" namely one link. So the parallel transporter simplifies to

$$

U(x+\mu,x) = U_{\mu}(x) = exp(i g a A_{\mu}(x))

$$

These parallel transporters (or link fields) are what one builds lattice actions out of. For example, a simple forward covariant derivative on a fermion field might be

$$

D_{\mu} \psi(x) = U_{\mu}(x)\psi(x+\mu) - \psi(x).

$$

Notice that the connection between the link field $U$ and the gauge field $A$ is non-linear. This is where the problems come in. Consider the expectation value of a single link field $(U)$. Expanding the expression for $U$ we find

$$

(U) = 1 - 2 \pi \alpha(\pi/a) a^{2} (A_{\mu}(x) A_{\mu}(x)) + ...

$$

I've used (A) = 0, and evaluated the strong coupling at the cutoff $\pi/a$. This looks fine, in the formal continuum limit (a->0) the lattice corrections vanish. But all is not well! There's a dangerous beast lurking in the loop integral (AA). Let's look at what it is.

The expectation value (A A) can be written as the integral over the momentum space gluon propagator

$$

(A A) = \int_{0}^{2\pi/a} d^{4}q / qhat^{2}

$$

where

$$

qhat^{2} = 4/a^2 \sum_{1}^{4} sin^2 (q_{\mu} a/2).

$$

In the limit where $a$ is small, we can replace $qhat^{2}$ by $q^{2}$, and we find that the integral is

$$

(A A) ~ 1/a^{2} + O(1)

$$

This is the disaster, because when we substitute this back into our expression for the expectation value of $U$ we find

$$

(U) = 1 - 2\pi \alpha(\pi/a) a^{2} (1/a^{2} + O(1)) + ...

$$

The $1/a^{2}$ from the loop integral cancels the $a^{2}$ from the expansion of the exponential leaving a lattice artifact suppressed only by the coupling constant, rather than the lattice spacing. The coupling constant is small (0.2) but not really small (like $\alpha a^{2} ~ 0.01).

It is factors like these (called gluon tadpoles) which spoil lattice perturbation theory in the bare coupling. The $\alpha_{V}$ coupling doesn't suffer from these corrections because tadpole effects come in an 2nd order in $\alpha$. It would be nice to have a way to systematically account for these effects. This method, called tadpole improvement (or mean-field improvement) was proposed by Lepage and Mackenzie, and is now in wide use.

The idea is to use link fields that have the tadpole effects largely canceled out of them. To do this you take all the link fields you have $U$ and multiple and divide by some average link field $u$. That is

$$

U \to u U/u = u \tilde{U}

$$

The new fields $\tidle{U}$ are much closer to their continuum values, since the average link field division cancels out most of the tadpole effects. The factors in the numerator $u$ can be absorbed into couplings and masses. When combined with a good coupling (such as $\alpha_{V}$) the agreement between perturbation theory and non-perturbative simulations (of short distance quantities) is much better.

There is one last thing one can do to improve the perturbative series, namely set the scale of the coupling in a more sensible way. That will be the subject of my next post.

Labels:
improvement,
perturbation theory,
tadpoles

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