I am studying the Faddeev-Popov procedure to quantize a non-Abelian gauge theory, and I got confused by the status of the time reversal symmetry. People have different definitions of the time reversal symmetry, and I will define what I mean by it below.

For concreteness, consider some fermions coupled to an SU(2) gauge field in the fundamental representation. Suppose we want to impose the Lorenz gauge, $\partial_\mu A^\mu=0$, the path integral of this theory takes the form: $$ \mathcal{Z}=\int D\bar\psi D\psi DA\det\left(\partial_\mu D^\mu_{\rm adj}\right)\exp\left(iS[\psi, A]\right) $$ where $\psi$ and $\bar\psi$ represent the fermionic matter fields, $A$ represents the SU(2) gauge field, $D^\mu_{\rm adj}=\partial^\mu\delta^{ab}-f^{abc}A^{c\mu}$ is the covariant derivative in the adjoint representation, where $f^{abc}=\epsilon^{abc}$ is the anti-symmetric tensor (structure constant for SU(2)). $S$ is the action that describes the fermionic matter coupled to the SU(2) gauge field.

Usually there is a time reversal symmetry that takes $A^\mu$ to be $$ A^0\rightarrow A^0, \vec A\rightarrow-\vec A $$ I am leaving the symmetry action on the fermionic matter implicit, which is irrelevant for this discussion. The action $S$ is invariant under this symmetry action. Now let us look at the symmetry action on the Faddeev-Popov determinant, $\det\left(\partial_\mu D^\mu_{\rm adj}\right)$. The covariant derivative transforms in a "bad" way: $$ \begin{split} D^0_{\rm adj}&=\partial^0\delta^{ab}-f^{abc}A^{c0}\rightarrow -\partial^0\delta^{ab}-f^{abc}A^{c0}\\ D^i_{\rm adj}&=\partial^i\delta^{ab}-f^{abc}A^{ci}\rightarrow \partial^i\delta^{ab}+f^{abc}A^{ci} \end{split} $$ where that $f^{abc}$ is real has been used. Namely, within the covariant derivative the ordinary derivative and the gauge field term pick up opposite signs.

My question is: how can the Faddeev-Popov determinant, $\det\left(\partial_\mu D^\mu_{\rm adj}\right)$, be invariant under this time reversal symmetry? Or in other words, how can this theory have a time reversal symmetry of the form defined above?

  • $\begingroup$ May be this points in the right direction, all we need to check is that the $\det$ gives the same, so it is enough if the transformed operator inside turns out to have the same spectrum as the original. So note as well that T transforms the $t$ argument within $A^0$ and $\vec{A}$, so my idea would be to find a mappping between eigenfunctions. Something like if $h(t,\vec{x})$ is an eigenfunction for the original operator $\partial_\mu D^\mu$, perhaps $h(-t,\vec{x})$ will be an eigenfunction for the transformed one with the same eigenvalue (however such a simple mapping doesn't seem to work) $\endgroup$ – ohneVal Aug 20 at 9:53

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