field operators $\left( U_{x} \right)_{ab}$ create states with higher and higher values of flux and the operators $\vec{E}^2_{x}$ measures the flux on the link $x$. Indeed we shall see shortly that it is proportional to the (kinetic) energy of the gauge fields.

To complete the picture we now discuss gauge invariance. First recall that the starting point of this formalism was to choose the time-like gauge. This leaves only time-independent gauge transformations as a symmetry. The fermion operators belong to the fundamental representation of the gauge symmetry and transform as $$\psi^a_x \rightarrow \left( V_x \right)^{ab} \psi^b_x, \label{gauge_F1}$$ with $V\in SU(N)$ given in general by $$V_x=\exp \left[ i\sum_{i=1}^{N^2-1} \theta^i_x \lambda^i\right].$$ Using the anti-commutation relations (\[anticomm\]), one can show that the quantum operator ${\cal V}_F$ that realizes in Hilbert space as $$\psi^a_x \rightarrow {\cal V}_F \, \psi^a_x \, {\cal V}^\dag_F,$$ is $${\cal V}_F = \exp \left[ -i\sum_x \sum_{i=1}^{N^2-1} \theta^i_x \left( \psi^{\dag a}_x \lambda^i_{ab} \psi^b_x \right) \right]. \label{gauge_F2}$$ The gauge fields transform according to $$\left( U_{x} \right)_{ab} \rightarrow {\cal V}_G\, \left(U_x\right)_{ab}\, {\cal V}^\dag_G = \left( V_x U_{x}  V^\dag_{x+1} \right)_{ab}.$$ Using the commutation relations in , one shows that the quantum operator ${\cal V}_G$ that generates these rotations is given by[^3] $${\cal V}_G = 