Variation of the fermion Fock state under small gauge transformation Suppose quantum state of fermions $\Psi_{f}$ in presence of external gauge field $A_{i}^{a}$ with temporary gauge $A_{0}^{a} = 0$:
$$
|\Psi\rangle = \int dA^{a}_{i}|A_{i}^{a}\rangle \otimes|\Psi_{f}(A_{i}^{a})\rangle, \quad i = 1,2,3
$$
Here $a$ is color indice, $|A_{i}^{a}\rangle$ is coherent gauge field state with VEV $A_{i}^{a}$, and $|\Psi_{f}(A_{i}^{a})\rangle$ is Fock state of fermions in presence of external field $A_{i}^{a}$.
The Gauss law is satisfied:
$$
G|\Psi\rangle = 0, \quad G = D_{i}A^{i} - J^{0},
$$
where $J^{0}$ is charge density.
How to prove that under small gauge transformation
$$
\Omega = 1 +\omega, \quad A_{i}^{a} \to (A_{i}^{a})^{\Omega} =  A_{i}^{a} + (D_{i}\omega )^{a}
$$
the fermion state $|\Psi_{f}(A_{i}^{a})\rangle$ is changed as
$$
|\Psi_{f}(A_{i}^{a}+(D_{i}\omega)^{a})\rangle = \left(1+\int d^{3}\mathbf r \text{tr}(\omega J^{0})\right)|\Psi_{f}(A_{i}^{a})\rangle ?
$$
 A: First, let me please correct the Gauss law constraint, whose first term should inlude the electric fields rather than the vector potentials: 
$$G = D_iE^i_a - J^0_a=0$$,
(It is written explicitly in the Lie algebra components).
In temporal gauge, the electric fields are the conjugate momenta of the vector potentials, thua at the quantum level:
$$[E^i_a(x), A_j^b(y)] = \delta^i_j \delta^b_a \delta^3(x-y)$$
Thus they must be represented on the wave functionals by the functional derivatives with respect to the gauge potentials
$$E^i_a(x) = \frac{\delta}{\delta A^{ia}(x)}$$
Therefore:
$$|\Psi_{f}(A_{i}^{a}+(D_{i}\omega)^{a})\rangle = |\Psi_{f}(A_{i}^{a})\rangle + \int d^3x(D_{i}\omega)^{a} \frac{\delta |\Psi_{f}(A_{i}^{a})\rangle}{\delta A^{ia}(x)}$$
$$ = |\Psi_{f}(A_{i}^{a})\rangle + \int d^3x(D_{i}\omega)^{a}  E_i^a(x)|\Psi_{f}(A_{i}^{a})\rangle$$
Perfprming an integration by parts:
$$ = |\Psi_{f}(A_{i}^{a})\rangle - \int d^3x(D_{i}E_i)^{a}  \omega^a(x)|\Psi_{f}(A_{i}^{a})\rangle$$
Then using the Gauss law:
$$ = |\Psi_{f}(A_{i}^{a})\rangle - \int d^3x J^{0a}  \omega^a(x)|\Psi_{f}(A_{i}^{a})\rangle$$
