# Representation of nonabelian Wilson line in terms of fermionic fields

Context:

The coupling action of a particle of charge $$q$$ to a $$U(1)$$ gauge field is given by $$$$S = q \int d \tau A_\mu \left( X \right) \frac{dX^\mu(\tau)}{d \tau} = -i \ln W_q, \tag{1}$$$$ where $$$$W^{\text{abelian}}_q = \exp{ \left(iq \int A_\mu dX^\mu\right) }$$$$ is the Wilson line, the integral being over the trajectory. For a particle charged under a nonabelian gauge field, it seems reasonable to try to find the coupling by considering the nonabelian version of the Wison line: $$$$W^{\text{nonabelian}} = \text{Tr} \, \mathcal{P} \exp{ \left(i \int A_\mu dX^\mu\right) } = \text{Tr} \, \prod_{\tau=\tau_i}^{\tau_f} \left( 1 + i \,d \tau A_\mu \left( X \right) \frac{dX^\mu(\tau)}{d \tau}\right).$$$$ My reason for considering this is that such a coupling should appear between the endpoints of open strings to the nonabelian gauge field one finds in the massless spectrum in the presence of D-branes, but nothing in this question depends on the details of string theory.

The question:

How should one deal with the path-ordering $$\mathcal{P}$$ in $$W^{\text{nonabelian}}$$ in order to rewrite this as a simple integral like (1)? The paper Particles with non abelian charges suggests the form $$$$S_{\text{NA}} = \int d\tau \left( \bar{c}^\alpha \frac{dc_\alpha}{d \tau} -i A^a_\mu (X) \frac{d X^\mu}{d \tau} \bar{c}^\alpha \left( T^a \right)_\alpha^{\phantom{a} \beta} c_\beta \right),$$$$ where $$c_\alpha$$ and $$\bar{c}^\alpha$$ are fermionic fields transforming respectively in the fundamental and anti-fundamental representations of $$SU(N)$$, under which the particle is charged, and satisfying a Dirac algebra with respect to these indices. The $$\left( T^a \right)_\alpha^{\phantom{a} \beta}$$ are $$SU(N)$$ generators in the chosen representation. Integrating out these fermions in the generating functional should give $$$$\int \mathcal{D} c \mathcal{D}\bar{c} e^{i S_{\text{NA}}} \sim \det \left( \delta^\beta_\alpha \frac{d}{d \tau} -i A^a_\mu (X) \frac{d X^\mu}{d \tau} \left( T^a \right)_\alpha^{\phantom{a} \beta} \right) = e^{iS_{\text{coupling}}},$$$$ but I was not able to check this, due to the complicated nature of the operator inside the determinant.