Wilson Lattice Action and the Strong-Coupling Expansion

[crackjack]

Consider the Wilson lattice action for a Yang-Mills theory with two parameters - color $N$ and coupling $g$.

1) The strong coupling expansion on the lattice is given in terms of $\beta = N/g^2$.
But what is the other parameter of the lattice theory? Is it $N$? In that case, does the $\beta$-expansion break down at large-$N$ and large-$g$?2) Is there a direct continuum limit (at a physicist's rigor) of the strong-coupling lattice theory ie. a continuum theory constructed from Wilson line variables with coupling $~1/g$, rather than from fields coupling at $g$?

 

[torquil]

Consider the Wilson lattice action for a Yang-Mills theory with two parameters - color $N$ and coupling $g$.

1) The strong coupling expansion on the lattice is given in terms of $\beta = N/g^2$.
But what is the other parameter of the lattice theory? Is it $N$? In that case, does the $\beta$-expansion break down at large-$N$ and large-$g$?

I'm not sure I understand the question, but I'll try anyway. You select the value of N that you are interested in, e.g. N=3 for QCD. The value of g has to do with the coupling strength and continuum limit. Simulation results are determined in units of "a", which is initially unknown. One determines "a" by matching a numerical and experimental observable quantity.

Are large N studies relevant in lattice gauge theory? Are you talking about working algebraically with the lattice theory, or doing computer simulations?

 

[crackjack]

Thanks for the reply Torquil!

My questions were regarding working algebraically in some generic theory. In that sense, we are starting with some usual local, continuum field theory with arbitrary $N$ and $g$. When such a theory is put on a lattice, we get a Wilson action in terms of [itex] [/itex] $\beta = N/g^2$, with the new "fields" being non-local Wilson loops.

1) I would like to know the other parameter in this lattice action (since we started with two parameters - $N$ and $g$ - in the original field theory).

2) Why is it that when we take the continuum limit of lattice Wilson action (with non-local loops as basic constituents), we invariably lead to a local field theory (the usual field theories in QFT)? Is there an intuitive or hand-waving picture that can show how non-locality transforms into locality in the continuum limit?

 

[torquil]

Thanks for the reply Torquil!

My questions were regarding working algebraically in some generic theory. In that sense, we are starting with some usual local, continuum field theory with arbitrary $N$ and $g$. When such a theory is put on a lattice, we get a Wilson action in terms of [itex] [/itex] $\beta = N/g^2$, with the new "fields" being non-local Wilson loops.

The 1x1 Wilson loop quantities are gauge group valued quantities that are related to the non-commutative gauge algebra valued field strength. The degrees of freedom in the usual lattice gauge theory are the gauge group elements on each edge, which are related to the gauge algebra valued gauge field along each edge.

1) I would like to know the other parameter in this lattice action (since we started with two parameters - $N$ and $g$ - in the original field theory).

There are two parameters, N and g, in the continuum theory. The value N determines the size of the gauge group and g is the coupling constant. In the lattice theory, N has exactly the same meaning, and beta is used instead of g just by convention. But you can use N and g if you want in the lattice theory.

2) Why is it that when we take the continuum limit of lattice Wilson action (with non-local loops as basic constituents), we invariably lead to a local field theory (the usual field theories in QFT)? Is there an intuitive or hand-waving picture that can show how non-locality transforms into locality in the continuum limit?

In the continuum limit, the physical size of the 1x1 Wilson loop becomes infinitesimal, and in this infinitesimal loop size limit, the WIlson loop quantity approaches a value that involves the differential of the field strength, i.e. it approaches the continuum Yang-Mills Lagrangian. Derivatives are also local quantities.

This is similar to the way that finite difference equations approach differential equations in the limit of infinitesimal lattices.

 

[crackjack]

The 1x1 Wilson loop quantities are gauge group valued quantities that are related to the non-commutative gauge algebra valued field strength. The degrees of freedom in the usual lattice gauge theory are the gauge group elements on each edge, which are related to the gauge algebra valued gauge field along each edge.

From what I understand (which is mainly from http://arxiv.org/abs/0906.4487v1 ), in lattice gauge theories, the matter fields sit at the lattice points, while Wilson lines sit on a link. The Wilson lattice action has non-local Wilson loops (that sit on plaquettes) as fundamental excitations, summed over all possible lattice loops (ie. plaquettes).

In the lattice theory, N has exactly the same meaning, and beta is used instead of g just by convention. But you can use N and g if you want in the lattice theory.

I agree. But, since the whole point of lattice/wilson action is to do a strong-coupling expansion, let us stick to $\beta$ as one parameter. If $N$ is taken to be the other independent parameter, one should be able to do a large-$N$ expansion over and above the [itex] [/itex] $\beta$-expansion in lattice/wilson gauge theories? (something similar to doing a doubled expansion in large-$N$ and 't Hooft coupling, $\lambda = g^2 N$, in the 't Hooft limit of quantum field theories)
(PS: The paper I referenced above says $g^2 \sim 1/N$ in eq-97. I don't really understand why, given that 't Hooft coupling can be fixed at any value!)

In the continuum limit, the physical size of the 1x1 Wilson loop becomes infinitesimal, and in this infinitesimal loop size limit, the WIlson loop quantity approaches a value that involves the differential of the field strength, i.e. it approaches the continuum Yang-Mills Lagrangian. Derivatives are also local quantities.

This is similar to the way that finite difference equations approach differential equations in the limit of infinitesimal lattices.

What is not clear to me is the following: The original lattice wilson action sums over all possible loops on the lattice. So, even if the lattice spacing is taken to zero, there would still be loops of all sizes (including those almost the size of the entire lattice). This is the reason why the continuum limit derived in eq-57 of above review paper is unconvincing - because, it the final continuum limit still retains a sum over all possible $\mu$ and $\nu$.

 

[torquil]

From what I understand (which is mainly from http://arxiv.org/abs/0906.4487v1 ), in lattice gauge theories, the matter fields sit at the lattice points, while Wilson lines sit on a link. The Wilson lattice action has non-local Wilson loops (that sit on plaquettes) as fundamental excitations, summed over all possible lattice loops (ie. plaquettes).

No, the Wilson action only includes 1x1 loops, not loops of all sizes. That's why it becomes local in the continuum limit. In the paper you linked to it is stated after eq. (52) that it is a sum over all "elementary plaquettes", i.e. the 1x1 plaquettes.

I agree. But, since the whole point of lattice/wilson action is to do a strong-coupling expansion, let us stick to $\beta$ as one parameter. If $N$ is taken to be the other independent parameter, one should be able to do a large-$N$ expansion over and above the [itex] [/itex] $\beta$-expansion in lattice/wilson gauge theories? (something similar to doing a doubled expansion in large-$N$ and 't Hooft coupling, $\lambda = g^2 N$, in the 't Hooft limit of quantum field theories)
(PS: The paper I referenced above says $g^2 \sim 1/N$ in eq-97. I don't really understand why, given that 't Hooft coupling can be fixed at any value!)

The relationship you refer to is determined by demanding that the continuum limit of the Wilson action converge to the Yang-Mills continuum action. It is simply stated for a general value of N. The reason that they are considering large N limits in continuum theories is that they are unable to calculate strong coupling properties otherwise. In the lattice gauge theory this problem disappears, so you should only consider N=3. The strong coupling expansion in lattice QCD is valid for N=3.

What is not clear to me is the following: The original lattice wilson action sums over all possible loops on the lattice. So, even if the lattice spacing is taken to zero, there would still be loops of all sizes (including those almost the size of the entire lattice). This is the reason why the continuum limit derived in eq-57 of above review paper is unconvincing - because, it the final continuum limit still retains a sum over all possible $\mu$ and $\nu$.

It has only a sum over elementary 1x1 plaquettes. The mu,nu indices refer to spacetime directions, not the size of the Wilson loops, which are always 1x1 in the Wilson action.

 

[crackjack]

No, the Wilson action only includes 1x1 loops, not loops of all sizes. That's why it becomes local in the continuum limit. In the paper you linked to it is stated after eq. (52) that it is a sum over all "elementary plaquettes", i.e. the 1x1 plaquettes.

True, but I have also read that, while the simplest Wilson action (like in the paper I quoted) consists of 1x1 Wilson loops as fundamental excitations, there are more complicated Wilson actions (~ Symanzik's improvement) with longer Wilson loops that further reduce the error when taking the continuum limit.

The relationship you refer to is determined by demanding that the continuum limit of the Wilson action converge to the Yang-Mills continuum action. It is simply stated for a general value of N. The reason that they are considering large N limits in continuum theories is that they are unable to calculate strong coupling properties otherwise. In the lattice gauge theory this problem disappears, so you should only consider N=3. The strong coupling expansion in lattice QCD is valid for N=3.

I think large-N doesn't necessarily lead to easier calculation in large-(t' Hooft-)coupling - eg. N=4 SUSY-YM and the whole industry of AdS/CFT. I wanted to know what it means to formulate a large-N version of a complete Wilson action at strong coupling.

 

[torquil]

True, but I have also read that, while the simplest Wilson action (like in the paper I quoted) consists of 1x1 Wilson loops as fundamental excitations, there are more complicated Wilson actions (~ Symanzik's improvement) with longer Wilson loops that further reduce the error when taking the continuum limit.

Yes, those exist, but only use loops with limited area in units of the elementary plaquette area. So in the continuum limit the physical size of these will also approach zero, so these also reduce to local continuum actions.

I wanted to know what it means to formulate a large-N version of a complete Wilson action (my motivation was the large-N version of N=4 SUSY-YM).

Lattice QCD can easily be formulated for a general value of N, although it would not be numerically tractable in the usual way of LQCD simulations since the computer would be given the task of multiplying and sampling over a space of extrelemy large matrices. It simply means that the edge matrices and plaquette matrices are elements of the Lie algebra su(N) instead of su(3). Beta would be the "coupling constant" of the lattice theory, and N would be baked into the theory by way of the size of the edge matrices. This matrix size would affect the coefficient values in a strong coupling polynomial expansion in the beta parameter.

EDIT: I'm assuming you meant "large-Nc version of N=4 SUSY-YM with gauge group SU(Nc)" ? I.e. Nc to indicate that the limit has to do with number of colours? Because one cannot take a large-N limit for N="number of supersymmetries".

 

[crackjack]

Thanks for the fast reply! (I will keep this in mind when editing. Sorry about the editing mess in my last post)

Yes, those exist, but only use loops with limited area in units of the elementary plaquette area. So in the continuum limit the physical size of these will also approach zero, so these also reduce to local continuum actions.

Is it true that, theoretically speaking, the lattice Wilson action will exactly coincide with the Yang-Mills action in the continuum limit only when the Wilson action includes sum over all possible (area and configuration) Wilson loops with suitable coefficients? One possible discrepancy would be that of the SO(4) symmetry that is present in euclidean Yang-Mills - without the above "counter terms", there would be some higher-dimensional terms that might break SO(4) in the continuum limit (ref: ch 2.3 & 4.1 in http://www.physics.mcgill.ca/~guymoore/latt_lectures.pdf )

Algebraically, since large-N expansion simplifes YM theory (at leading orders), what is the corresponding simplification in lattice Wilson action?
(Can this simplification be a reduction in the large/infinite number of terms in the completely improved Wilson action?)

And yes, I denoted number of colors by N.

 

[torquil]

Thanks for the fast reply! (I will keep this in mind when editing. Sorry about the editing mess in my last post)

Is it true that, theoretically speaking, the lattice Wilson action will exactly coincide with the Yang-Mills action in the continuum limit only when the Wilson action includes sum over all possible (area and configuration) Wilson loops with suitable coefficients?

The continuum action can be approximated by a lattice to any finite order by expanding it in Wilson loops. They will all converge towards the continuum action, for sufficiently smooth gauge fields. The order of accuracy determines how fast they converge towards the continuum action as the grid constant "a" approaches zero. Note that at any finite "a", the order of convergence doesn't really mean much without accompanying regularity assumptions on the gauge field.

A lattice action that includes all possible loops in the lattice (ie arbitrarily large ones) cannot converge towards the YM continuum action, because it won't be local in the continuum limit, if you consider an increasing sequence of finer and finer grids that cover the same physical 4-volume.

The 1x1 Wilson action will coicide with the continuum action in this limit for the same reason that the functions [f(x+a)-f(x)]/a will coincide with df/dx i the same limit.

 

[torquil]

Algebraically, since large-N expansion simplifes YM theory (at leading orders), what is the corresponding simplification in lattice Wilson action?
(Can this simplification be a reduction in the large/infinite number of terms in the completely improved Wilson action?)

There might be some theoretical simplification, but at the same time you cannot expect a computer to work on large-N SU(N) matrices. Of course, it might be possible to prove that in this limit the theory is equivalent to something else which is computationally tractable, who knows?

I had a look in my old Creutz book "Quarks, guons and lattices", in the chapter about the strong coupling expansion. He calculates the IxJ Wilson loop (eq.10.6) and gets

$$W(I,J) = (\frac{\beta}{2N^2})^{IJ}$$

at the lowest order in beta. So in the N -> infinity limit, you get W = 0 for all I,J >= 1. Similary, for the lowest beta-order of the string tension K:

$$K = -\frac1{a^2}\log(\frac{\beta}{2N^2})$$

Therefore, the string tension to lowest order in beta diverges towards infinity in the same limit. Those are some significant simplications for the LQCD with the 1x1 Wilson action.

In any case, the purpose of the lattice is to be able to perform calculations on SU(3) QCD without having to go to such extreme lengths as introducing (probably) largely unphysical theories such as large-N YM or SYM. These are usually introduced because one are not able to calculate using pen and paper in low energy SU(3) QCD. I.e. they are used as "toy models" for QCD, and workers hope to extract information that is relevant also for SU(3) QCD.

In LQCD it is possible to stick with the messy, dirty, real deal, SU(3) QCD :-)

 

[crackjack]

The continuum action can be approximated by a lattice to any finite order by expanding it in Wilson loops. They will all converge towards the continuum action, for sufficiently smooth gauge fields. The order of accuracy determines how fast they converge towards the continuum action as the grid constant "a" approaches zero. Note that at any finite "a", the order of convergence doesn't really mean much without accompanying regularity assumptions on the gauge field.

Lets assume the gauge fields are sufficiently regular. Can we then look at this process of taking lattice spacing to zero, "a" -> 0, as running the theory from IR to UV? In this sense, the higher-dimensional operators (~ those of higher orders in "a") resemble irrelevant operators in renormalization. And the role of the longer Wilson loops in the Wilson action is to suppress these operators?

A lattice action that includes all possible loops in the lattice (ie arbitrarily large ones) cannot converge towards the YM continuum action, because it won't be local in the continuum limit, if you consider an increasing sequence of finer and finer grids that cover the same physical 4-volume.

I understand it this way: the wilson loops themselves may not become local, but the subtle cancellation between the different terms in the sum does lead to a local theory - a better one at that - in the limiting ("a" -> 0) case.

Those are some significant simplications for the LQCD with the 1x1 Wilson action.

Looks like too much simplification to me :-s - since, an important order parameter in the theory is going to zero. Why would such simplifications be desirable in LGT?

My motivation for all this is in AdS/CFT, where the wilson loop on the boundary theory corresponds to the boundary of the string worldsheet stretching into the AdS bulk. But since the limit there is planar 't Hooft limit, I wanted to know what would this limit correspond to in lattice gauge where wilson loops are the fundamental excitations. Add to it the fact that the radial direction in AdS/CFT corresponds to renormalization scale for the boundary theory.