Field-strength formulation of gauge theories: Transformation of the functional integral

Abstract

We present a formulation of gauge field theories in which the gauge potentials ${A}_{\ensuremath{\mu}}(x)$ are eliminated in a simple way in terms of the field strengths ${F}_{\ensuremath{\mu}\ensuremath{\nu}}(x)$. Our results are closely related to, but are much simpler than, Halpern's dual variable formulation of gauge theories in the axial gauge. We work in the coordinate gauge ${x}^{\ensuremath{\mu}}{A}_{\ensuremath{\mu}}(x)=0$, and show both analytically and geometrically that the potential ${A}_{\ensuremath{\mu}}$ can be determined uniquely from the field strengths ${F}_{\ensuremath{\mu}\ensuremath{\nu}}$ for a suitable class of $F'\mathrm{s}$, $A\ensuremath{\rightarrow}A[F]$. We show, furthermore, that a tensor ${F}_{\ensuremath{\mu}\ensuremath{\nu}}(x)$ is a coordinate-gauge field tensor if and only if it satisfies the restricted set of Bianchi identities ${\ensuremath{\epsilon}}^{\ensuremath{\mu}\ensuremath{\nu}\ensuremath{\sigma}\ensuremath{\lambda}}{x}_{\ensuremath{\sigma}}{D}^{\ensuremath{\rho}}[F]^{*}F_{\ensuremath{\rho}\ensuremath{\lambda}}=0$, $D=\ensuremath{\partial}+[A[F], \ifmmode\cdot\else\textperiodcentered\fi{}]$. These results permit us to transform the functional integral for the vacuum-to-vacuum amplitude $Z[J]$ for the gauge theory to a form in which the potentials are completely eliminated in terms of the field strengths. When the Bianchi constraints are eliminated using a set of Lagrange multiplier fields ${\ensuremath{\lambda}}_{\ensuremath{\sigma}}(x)$, the $F'\mathrm{s}$ can be integrated out completely to obtain a form of the theory which appears to be useful for strong coupling.