is [*not*]{} a gauge transformation (i.e. its not of the form ).

The end result of Eqs. (\[Psitransform\])–(\[Vdef2\]) is that when we rotate the Hilbert space by ${\cal F}$ or, equivalently, conjugate the fermionic part of the Hamiltonian by ${\cal F}$ we indeed get : $$\begin{aligned}
H_F &\to & H'_F = {\cal F} \, H_F \, {\cal F}^{\dag} = -i\sum_{x}\psi^{\dag a}_x \sigma_3 \, e^{i\varphi_a } \, \psi^{a}_{x+1}+h.c. + m\sum_x \psi^{\dag a}_x \sigma_1 \, \psi^{a}_x.\label{H'}\end{aligned}$$ This change of quantum basis is how the process of axial gauge fixing works in the Hamiltonian formalism.

We now proceed to find the transformed version of $H_E$, of the commutation relations, and of the Gauss-law constraint. We begin by defining the transformed electric fields $E'$ via $$\begin{aligned}
E^{'i}_x&\equiv & {\cal F} \, E^{i}_x \, {\cal F}^\dag,\label{E'}\end{aligned}$$ which allows us to write the transformed electric field Hamiltonian as $$\begin{aligned}
H_E &\to & H'_E={\cal F} \, H_E \, {\cal F}^\dag = \frac{g^2}{2}\sum_{x,i} \left(E^{'i}_x\right)^2.\end{aligned}$$ It is clear that the commutation relation between the $E'$ fields and the $U$ fields are the same as in and (this is so because of ). Next, we turn to transform the Gauss law constraint. For that we conjugate by ${\cal F}$ $${\cal 