The Faddeev-Popov determinant in case of $U(1)$ turns out to be ${\rm Det}(\partial^2)$.

My question is: what is the determinant of $\partial^2$?

  • 1
    $\begingroup$ Every text that introduces the notion of the FP determinant should also "compute" it e.g. by the standard procedure of expressing it as a fermionic Gaussian integral, thereby introducing ghost fields. What exactly is your question? $\endgroup$
    – ACuriousMind
    Nov 10, 2016 at 16:56
  • $\begingroup$ I thought by introducing the ghost fields we bypass the calculation of the determinant. $\endgroup$
    – Razor
    Nov 11, 2016 at 3:24

1 Answer 1


It is a constant number. The $U(1)$ is indeed a quite trivial case.


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