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.
