Does the path of a Wilson line in a quark-correlator matter?

Question

Consider a gauge-invariant quark correlation function nested inside an arbitrary state $|p\rangle$

$$\langle p |\bar \psi(z)_{\alpha,a}\left( W_{\Gamma}(z,0)\right)_{ab}\psi(0)_{\beta,b}|p\rangle \tag{1}$$

where $\alpha,\beta$ are spinor indices, and $a,b$ are the quarks color/gauge indices. Here $W_{\Gamma}(z,0)$ is a Wilson line connecting the points $0$ and $z$ along the path $\Gamma$.

My question is, does the path $\Gamma$ matter here? This question is quite important because when $|p\rangle$ is the state of some hadron (e.g. pion, proton, neutron), this kind of correlator is exactly what enters into parton distribution functions!

If it didn't matter, then we would have that

$$\langle p |\bar \psi(z)_{\alpha,a}\left( W_{\Gamma_1}(z,0)-W_{\Gamma_2}(z,0)\right)_{ab}\psi(0)_{\beta,b}|p\rangle=0\tag{2}$$

But to leading order in the gauge coupling, this equation tells me

$$\langle p |\bar \psi(z)_{\alpha,a}\left( \oint_{\Gamma_1+\Gamma_2}A^\mu dx_\mu\right)_{ab}\psi(0)_{\beta,b}|p\rangle=0\tag{3}$$

Now, if $A^\mu(x)$ was an external field and $|p\rangle$ some neutral hadron (or even the vacuum state!), then this would turn into

$$\left[\Phi_{\Gamma_1+\Gamma_2}\right]_{ab}\langle p |\bar \psi(z)_{\alpha,a}\psi(0)_{\beta,b}|p\rangle=0\tag{4}$$

where $\Phi_{\Gamma_1+\Gamma_2}$ is the "flux" passing through the area spanned by the paths $\Gamma_1+\Gamma_2$. Because I believe in general we cannot say that $\Phi_{\Gamma_1+\Gamma_2}=0$, we cannot say that (2) vanishes and therefore we must conclude that in general, eq. (1) depends on $\Gamma$.

But what the heck?? This doesn't seem so natural, because then the definition of a parton distribution function will depend on the path of the gauge-link/Wilson-line.

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Some possible rebuttals that I see:

1. Parton distributions, as well as my general gauge-invariant correlator in eq. (1), are not physical objects. You cannot directly measure them. For example, a PDF is already regularization-scheme dependent. Therefore, who cares if eq. (1) depends on $\Gamma$.

2. More to come...

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I've raised this question with some smart guys, and they told me that in general yes, these kinds of correlators (which appear in parton distribution functions) do indeed depend on the choice of path $\Gamma$, up to a certain extent. Supposedly there are certain equivalence classes of paths. The equivalence classes of paths are something like:

1. Paths where the tangent vector to $n^{\mu}$ remains spacelike for the entire path.
2. Paths where the tangent vector remains lightlike for the entire path.
3. Paths where the tangent vector remains timelike for the entire path.

However I was not given any rigorous definitions of these classes, nor was I given any simple/comprehensible intuition for why these might indeed be equivalence classes.

Does anybody know anything about this? Or maybe it wouldn't be so hard to prove such claims?

Answer 1

Wilson lines for general gauge theories are definitely not topological, their expectation values really do depend on the path.

You can sometimes make them topological by hand. For example doing some sort of supersymmetric twisting that makes the full theory topological, so in particular the line operators. Or, if you begin with a theory that is topological from the beginning, such as Chern-Simons theories in 3d, the line operators are topological too. The classical example is Witten's seminal paper re Jone polynomials 1, where the expectation value of two Wilson loops along paths $\gamma_1,\gamma_2$ on a three-sphere was shown to be $$ \langle W_{R_1}(\gamma_1)W_{R_2}(\gamma_2)\rangle\sim S_{R_1R_2}^{\text{Link}(\gamma_1,\gamma_2)} $$ where $\text{Link}$ denotes the linking number of the paths, and $S$ is the modular matrix of an associated Kac-Moody algebra. Note that this expectation value depends on $\gamma_1,\gamma_2$ only through their topology, namely the number of times they cross each other. Wilson lines in Chern-Simons theories are topological: their expectation values only care about the global features of the paths, and not on their actual geometry.

For generic gauge theories, this is not the case. Line operators are not topological, they depend on the full path and not just its topology.

That being said, it is often the case that we care about Wilson lines over very large paths. In that limit, the leading contribution is indeed quasi-topological: it depends only on its area (or length, depending on whether the theory confines).

(Regarding classes of paths. The physical interpretation of the line changes depending on its causal structure: if you have a spacelike line connecting two quarks, it represents a glue string that connects them, and this allows you to measure the force between them; if you have a timelike line connecting two quarks, it represents the worldline of a single quark, propagating in spacetime. Not sure if this is relevant at all though.)

References.

1. E. Witten - Quantum field theory and the Jones polynomial. https://doi.org/10.1007/BF01217730

Answer 2

I would say the expectation value you write down depends on the path of the Wilson loop you write down, the reason is perhaps best appreciated by considering what happens in for classical gauge fields, and then looking at the Hilbert space formulation of the quantum theory. The Hilbert space formulation of the quantum theory lacks manifest Lorentz invariance, and is therefore not oftend considered explicitly in QFT, but in this case, explains the situation quite well, I would say.

In the classical theory, consider the system state to be the field configuration $A$, which is an $\mathfrak{su}_3$ valued one form. The equation $\nabla = d + A$ defines a covariant derivative, and $W_\Gamma$ measures the parallel transport along a curve $\Gamma$. Basically, if you take a quark field configuration where the derivative term vanishes, the Wilson line measures how the colour charge of the field changes along the line. This parallel transport is independent of perturbations in the path only if the curvature $F$ of the connection, which we interpret as the field strength, vanishes. This also makes sense in the colour charge picture, the presence of a gluon field will cause the colour charge to change. For a $U(1)$ theory, this is the Aharonov-Bohm effect. This means that, for a general field configuration where the field strength is not zero, the Wilson loop will be path dependent in the classical theory.

Now, in the Hilbert space formulation of the QFT, a state $|\psi\rangle$ will correspond to a Gauge invariant functional $\psi(A)$, so that $|psi(A)|^2$ gives the probability density of finding the field configuration $A$ when the system is measured. That the wave function is a Gauge invariant functional means that these probabilities are gauge independent.

The expectation value $\langle \psi |W_\Gamma|\psi\rangle$ is then an average of the parallel transports in the classical theory over those configurations specified by the wave function. Every field configuration with non-zero $F$ will yield a non-zero contribution to the average, and it would be quite miraculous for them to cancel out. As mentioned in another answer here, for topological theories, one expects the Wilson lines to be path independent, but this gives quite a constraint on the possible wave-functions, meaning they have small phase spaces in some sense.

I am not that familiar with QCD, so I will take the word of the other answers to your question that the Wilson loops in QCD are path dependent. However, what I've outlined above is my reason for not finding this surprising.