6
$\begingroup$

Context:

The coupling action of a particle of charge $q$ to a $U(1)$ gauge field is given by \begin{equation} S = q \int d \tau A_\mu \left( X \right) \frac{dX^\mu(\tau)}{d \tau} = -i \ln W_q, \tag{1} \end{equation} where \begin{equation} W^{\text{abelian}}_q = \exp{ \left(iq \int A_\mu dX^\mu\right) } \end{equation} 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: \begin{equation} 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). \end{equation} 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 \begin{equation} 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), \end{equation} 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 \begin{equation} \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}}}, \end{equation} but I was not able to check this, due to the complicated nature of the operator inside the determinant.

$\endgroup$

1 Answer 1

0
+50
$\begingroup$

The evaluation of determinants like this one can be found in various references, including, for example,

The idea is to work in the Cartan subalgebra of the gauge group where everything is nice and diagonal, and then express the result in terms of group invariants. Ultimately this reproduces the non-Abelian Wilson line you started with (in a chosen representation), so the real advantage of this first quantised representation would be to find an alternative, more convenient way of evaluating the one-dimensional path integral, perhaps in perturbation theory.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.