# Why don't we add Wilson loops to the SM Lagrangian?

As the title says: why don't we add Wilson loops to common Lagrangians such as the Standard Model? They're gauge invariant and (correct me if I'm wrong, not sure on that) are renormalizable.

Suppose they are indeed renormalizable, one answer I can think of is that they are not per-se a function of spacetime. But then I would say that you can define the Wilson loop with respect to some base-point along the contour of integration, and perhaps there's a clever way of taking care of the redundancy of equivalent loops with different base points. (Maybe similarly to gauge fixing?) You would, of course, also integrate over all possible contours.

What I've described indeed sounds a little shady, but I was wondering if this option has been considered (and perhaps ruled out) a little more rigorously.

The reason Wilson loops are excluded from the lagrangian density (in the continuum limit) of a quantum field theory is simply because it is a nonlocal operator. Many nice properties of a quantum field theory are lost when the lagrangian density is built out of nonlocal operators, the most important being causality. Unitarity is also in jeopardy when nonlocality is introduced (although there are ways of defining nonlocal field theories which get around this).

While you could conceive of a theory which has Wilson loops, it is generically unphysical, and so is thrown out at the beginning of model building.

Now there is reason to study nonlocal quantum field theories, as they are renormalizeable generically. String theory can also be thought of in some sense as a nonlocal quantum field theory. If one wanted to construct a nonlocal gauge theory, Wilson lines would be the primary tool for constructing the action of the theory.

• And finally, a Wilson loop would explicitly break Poincaré symmetry! – Hans Moleman Jun 27 '19 at 0:43
• If one did as OP suggested and summed over all contours, would that restore Poincaré symmetry? – LucashWindowWasher Jun 27 '19 at 0:52
• You could perhaps do that, but it would take you beyond the realm of ordinary QFT. It's pretty clear how to compute observables in the background of a single Wilson loop - but unless you give some explicitly prescription on how you're going to integrate over all contours I don't really know how to make sense of such an idea. – Hans Moleman Jun 27 '19 at 1:12

The kinetic term for a gauge field is a sum of infinitesimal Wilson loops. In this sense, the Lagrangian for a typical gauge theory already has Wilson loops.

In lattice gauge theory, this is completely explicit. In classical lattice gauge theory, a gauge field is described by assigning an element of the gauge group $$G$$ to each pair of neighboring lattice sites. (Note that $$G$$ is the gauge group, not the Lie algebra.) If we write $$U(j,k)$$ for the element of $$G$$ that is assigned to the pair $$j,k$$ of neighboring sites, then the kinetic term in the action is proportional to $$\sum \text{Trace}\big( U(j,k)U(k,l)U(l,m)U(m,j)\big) +\text{complex conjugate}, \tag{1}$$ where $$j,k,l,m$$ are the sites forming the corners of a plaquette (a square's worth of nearest-neighbors) and the sum is over plaquettes. The trace of the product of $$U$$s is a Wilson loop. To relate this to the usual kinetic term for the gauge field in continuous spacetime, think of the lattice sites as a discrete grid of points in continuous spacetime, and write $$U(j,k)\sim\exp\left(i\int_j^k dx\cdot A(x)\right) \tag{2}$$ where $$A_\mu(x)$$ is the usual gauge field (an element of the Lie algebra). Then the product of the four $$U$$s around a plaquette is $$UUUU\sim\exp\left(i\oint dx\cdot A(x)\right) \tag{3}$$ where the integral is around the boundary of the plaquette. Taking the limit of an infinitesimal loop gives $$UUUU\sim\exp(i \epsilon F_{ab})$$ where $$\epsilon$$ is the area of the tiny loop, within which the field-strength tensor $$F_{ab}$$ is essentially constant. Expanding in powers of $$\epsilon$$ and adding this to its complex conjugate gives $$UUUU+\text{c.c.}\sim 1-\frac{\epsilon^2}{2}(F_{ab})^2. \tag{4}$$ After taking the trace, subtracting an appropriate constant term, multiplying by the appropriate normalization factor, and summing over canonical planes (which means summing over index-pairs $$a\neq b$$), we recover the usual gauge-field kinetic term.