Without <gauge fixing>, every <gauge orbit> is integrated infinitely many times and the quadratic gauge-field operator has a <zero mode in field theory> for every pure-gauge direction. It therefore has no inverse and no propagator. A gauge condition selects one representative per orbit, up to global transformations and the possible nonperturbative <Gribov ambiguity>, and makes perturbative Gaussian integration well defined.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 4 / d / i / Solution

 Solution (source code)