For conventions used in this note, see my blog here. I understand that my choice of the metric, the mostly negative one, will intrigue some strong emotional reaction among certain groups of physicist, but I will keep using it anyway, partly because it is the most widely used in the context of quantum field theory.

One-dimensional gauge invariant objects in pure YM

This chapter is meant for people who are not very familiar with Wilson loops and ‘t Hooft loop, if you are not the target reader, feel free to skip this chapter all together.

I assume the reader has some basic notion of what quantum Yang-Mills theory is, if not, there exist an ocean of textbooks and introductory readings on quantum field theory, all necessarily include Yang-Mills theory. Personally I recommend the books written by Matthew D. Schwartz, A. Zee and Mark Srednicki. It’s not that the all-time classic, Peskin&Schroeder’s book is not good, just that it is not for beginners and is heavily focused on high energy physics calculations, you practically learn no geometry of QFT. I also recommend David Tong’s lecture notes, which provides a pretty good balance between calculation detail and profound understanding.

The gauge group is assumed to be $SU(N)$ if not mentioned otherwise, sometimes when we talk about electric and magnetic field we will use $U(1)$ which is a abelian version of $SU(N)$. The Yang-Mills field is normalized such that the action reads

\[S = - \frac{1}{g^{2}} \int d^{4}x \, \frac{1}{2}\mathrm{Tr}\,F_ {\mu \nu}F^{\mu \nu}\]

Thus the coupling $g^{2}$ plays the rule of $\hbar$, at the limit $g^{2}\to 0$ it is the same as $\hbar\to 0$ and the classical contribution will dominate.

Wilson line

In this chapter we will assume the gauge field $A_ {\mu}$ is a fixed background field, in which we place a test particle. The test particle is under our control and its own contribution to the potential is neglected, in other word it tests the field but sources nothing. Suppose the particle moves along a curve $C$, which starts at $x_ {i}$ (i for initial) and ends at $x_ {f}$ (f for final), we define the corresponding Wilson line as

\[U[C] := \mathcal{P} \exp\left( i \int_{x_ {i}}^{x_ {f}} dx^{\mu} \, A_ {\mu} \right) = \mathcal{P}\exp\left( i\int_{x_ {i}}^{x_ {f}} A \right)\]

where $\mathcal{P}$ stands for the path-ordering and in the last term we adopted the 1-form expression for the gauge field (or connection, if you prefer to treat is as a fiber bundle). The Most important property of Wilson line is how it transforms under a gauge transformation. Turns out it only depends on the end point,

\[U[x_ {i},x_ {f};C]\to \Omega(x_ {i})\; U[x_ {i},x_ {f};C] \; \Omega ^{\dagger}(x_ {f}).\]

If we compactify $C$ to a closed curve, namely a loop, then we get a Wilson loop,

\[W[C] = tr \mathcal{P} \exp\left( i \int A \right),\]

Note we have also put a trace in the definition for future convenience. Mathematically it is known as a gauge holonomy. The important thing is that, Wilson loop is gauge invariant!

Remarks.

  1. Wilson loop is a one-dimensional thing,
  2. Wilson line is not gauge invariant but Wilson loop is,
  3. It will play the most important rule in the study of deconfinement phase transition.

The Wilson loop can also be regarded as operators acting on the Hilbert space of the quantum system. However, remember that in quantum theories, states are defined as living on a spacelike slice of the spacetime, thus it is important to first rotate our Wilson loop so that it lives in a time slice, namely in a space at fixed time. It turns out that when $W[C]$ acts on a state, it adds to the state a loop of electric flux along $C$. Roughly speaking, in the language of canonical quantization, let $\pi$ be the canonical momentum, similar to $p = -i \frac{\partial}{\partial x}$, we have the following functional expression

\[\pi_ {i} = -i \frac{\delta \mathcal{L}}{\delta {A}^{} {i}} = E_ {i}+\frac{\theta}{8\pi^{2}}B_ {i}\]

thus when the $\theta$-angle is zero, we simply get

\[E_ {i} = -i \frac{\delta}{\delta A^{i}}\]

Thus when acting on a state $W[C]\left\lvert{\psi}\right\rangle$, we have

\[E^{i}W[C]\left\lvert{\psi}\right\rangle = \mathrm{Tr}\, \mathcal{P} \left\{ \left( \frac{\delta}{\delta A_ {i}}\int_ {C} A \right) W[C] \right\} \left\lvert{\psi}\right\rangle\]

whose support is $C$, meaning the electric field is zero at any place but $C$. For more detail, see David Tong’s note.

‘t Hooft line

We have talked about test particles with charge, they are electric probes. Then what about magnetic probe? We run into problems immediately, since we don’t know what is the operator that inserts a magnetic monopole in the spacetime. For electric probe, a charged particle is inserted by the source term $j^{\mu}$, there is no magnetic counterpart of it. It turns out there is still a way to do it, and the operator that is responsible for the insertion of a magnetic field is called a 't Hooft line.

Take the simplest gauge group, $U(1)$ for example. We know the magnetic monopole solution, although they are singular at the origin, but it doesn’t matter for our discussion. Suppose that a monopole of magnetic charge $m$ traces out a world line $C$ in our Minkowski spacetime. For any $\mathbb{S}^{2}$ that surrounds $C$ in space (not 4D spacetime but 3D space), we have

\[\int _ {\mathbb{S}^{2}} \, B\cdot dS =m.\]

We now define the ‘t Hooft line by requiring that we take the path integral only over fields that satisfy the above equation on $C$.

The logic is really simple, since we can’t simply find the operator that sources the monopole, then let’s start from the path integral, whenever we want to calculate the expectation value of something in the background of a monopole (with worldline $C$), we require that only the field configurations satisfy this condition to be included into the path integral! Then, at least in principal, we can calculate the correlation between a ‘t Hooft line $T[C]$ with other stuff.

Next we go to Yang-Mills theory with gauge group $G$, whose Lie algebra is denoted ${\frak g}$. Recall that ${\frak g}$ include a set of $r$ mutually commute vectors (or generators, anyway if the group is regarded as a manifold then the generators are nothing but vectors at the identity, for more detail refer to Theodore Frankel’s textbook), where $r$ is the rank of the Lie group.

Let ${\frak h}\subset {\frak g}$ be the Cartan subalgebra of ${\frak g}$. Recall that ${\frak h}$ is just the maximal set of $r$ (=rank) mutually commuting generators, or if you prefer the geometric interpretation, substitute generator with vector and commuting with vanishing Lie bracket. ${\frak h}$ can be regarded as a vector of $r$ components.

let $C$ be a timelike (since nothing moves faster than light) curve sitting at the origin. We will require the magnetic field $B^{i}, i=1,2,3$ to take the (spherically symmetric) form

\[B^{i} \to \frac{\hat{r}^{i}}{4\pi r^{2}} {\frak q}(x), \quad r\to \infty,\]

where ${\frak q}(x)$ is some ${\frak g}$-valued things that specifies the magnetic charge of the monopole, we will say more about it later. Just to remind you, the magnetic field $B^{i}$ is just an component of the field strength $F_ {\mu \nu} = F_ {\mu \nu}^{a} T^{a}$, hence is ${\frak g}$-valued.

By a suitable gauge transformation, we can expand ${\frak q}$ in basis of ${\frak h}$ solely. This can be understood from two different perspectives. Algebraically speaking, the operators are certain matrices and the Cartan subalgebra ${\frak h}$ are diagonal matrices, that’s the only way they can be mutually commuting. Then we can choose the suitable gauge transformation that serves as a similar transformation to diagonalize the matrix ${\frak q}$. Geometrically speaking, the gauge transformation can be regarded as a rotation in the vector space of ${\frak g}$, and a suitable rotation can leave ${\frak q}$ in the subspace expanded by ${\frak h}$ solely. Now we can write

\[{\frak q} = \mathbf{m}\cdot {\frak h},\quad {\frak m}=(m^{1},\dots,m^{r})\]

and $\mathbf{m}$ is our magnetic charge. We can think of this as $r$ Dirac monopoles, one for each Cartan generator, embedded in the Cartan algebra.

The Dirac quantization condition now requires

\[\exp(i \mathbf{m}\cdot {\frak h}) = \mathbb{1}.\]

which should hold for all the representations of $G$.

To be able to deal with arbitrary representation, we need to introduce the concept of weight and root. For more details I recommend A. Zee’s textbook on Group Theory (for physicists), here I’ll just list some important definitions and applications, leaving the beauty of the group classification theory behind.

Recall from quantum mechanics that, the weight is given by the eigenvalues of the Cartan generators, which we will treat as operators here. To be specific, given a $d$-dimensional representation, let $\left\lvert{\mu_ {a}}\right\rangle$ be the basis where $a=1,\dots,d$. Then the weight vector is given by

\[{\frak h}\left\lvert{\mu_ {a}}\right\rangle = ({\frak h}^{1},\dots,{\frak h}^{r})\left\lvert{\mu_ {a}}\right\rangle = (\mu_ {a}^{1},\dots,\mu^{r}_ {a}\left\lvert{\mu_ {a}}\right\rangle ) =: \mu_ {a}\left\lvert{\mu}_ {a}\right\rangle.\]

All such weights span the weight lattice $\Lambda_ {w}({\frak g})$, where $w$ is for weight.

The weight of the adjoint representation plays a crucial rule in the classification of Lie groups, and thus deserves its own name, called root. Since the basis of adjoint representation are just generators, and each basis has its own associated weight vector, thus each generator is assigned with a root vector. Let $E_ {\alpha}$ be such a generator label by its root vector $\alpha$, for simplicity we will not use the bold-font convention to denote a vector. In the adjoint representation, the eigenvalue condition translates into a commutation relation (by construction)

\[[H,E_ {\alpha}] = \alpha E_ {\alpha}.\]

The roots also expands a lattice denoted $\Lambda_ {\text{root}}({\frak g})$ which is a subset of $\Lambda_ {w}({\frak g})$, since root is a special kink of weight.

It can be shown with the help of ladder operators that, the weights and roots must satisfy

\[\frac{\alpha \cdot \mu}{\alpha^{2}} = \frac{1}{2} \mathbb{Z}, \quad \mu \in \Lambda_ {w}({\frak g}),\; \alpha \in \Lambda_ {\text{root}}({\frak g})\]

where $\mathbb{Z}$ is the integers. This is exactly what we need to solve the Dirac quantization condition Eq. (2)! In Eq. (2), the operator ${\frak h}$, when acting on the basis, will be replaced by some root vector $\mu$ thus the quantization condition requires that $m\cdot \mu=2\pi \mathbb{Z}$ for all $\mu \in\Lambda_ {w}$. We define the co-root

\[\alpha^{\vee}:= \frac{2\alpha}{\alpha^{2}}\propto \alpha.\]

Obviously we now have

\[\alpha^{\vee}\cdot \mu = \mathbb{Z} .\]

All the co-root vectors also span a lattice, denoted by $\Lambda_ {\text{co-root}}({\frak g})$.

We can define the magnetic charge vector to be the co-root vector, since then the Dirac quantization condition is satisfied. It turns out that for simple group we have

\[\boxed { m = 2\pi \Lambda_ {\text{co-root}}({\frak g}). }\]

This is sometimes referred to as the Goddard-Nuyts-Olive (or GNO) quantization condition.