Do primary first class constraints change the electric field in the Hamiltonian form of Maxwell's theory?

In my understanding of Dirac's theory of constrained Hamiltonians, the primary (and also the secondary) first class constraints are generators of canonical transformations that do not change the physical state: the electric field is part of the physical state so it has zero response to a primary first class constraint. However, a paper http://arxiv.org/abs/1310.2756 recently appeared which says that the primary first class constraints change the physical state. The paper gives a direct calculation which I'll reproduce below.

Using the notation in Dirac's Lectures on Quantum Mechanics, the p's are $B^{\mu}$ and the q's are the electromagnetic potentials $A_{\mu}$. The primary first class constraints are $B^{0}\approx 0$. So, the generator of the primary first class constraints is, $$G=\int d^{3}x \xi(x) B^{0}(x)$$ The response of the electromagnetic field is given by the PB, $$\frac{dA_{\mu}}{d\epsilon}=[A_{\mu},G]=\delta^{0}_{\mu}\xi(x)\ .$$ The paper defines the electric field as, $$E_{r}=A_{r,0}-A_{0,r}$$ and denies any relation between $E_{r}$ and the canonical momenta $B^{r}$ until the dynamical equation $\dot{q}=[q,H]$ has been used. The paper gets the response of the electric field to the primary first class constraint as, $$\frac{dE_{r}}{d\epsilon}=\frac{\partial}{\partial t}\frac{dA_{r}}{d\epsilon}-\frac{\partial}{\partial x^{r}}\frac{dA_{0}}{d\epsilon}=-\xi_{,r}$$ and this is troubling me because the response should be zero.

I thought I understood constrained Hamiltonians but now I'm not sure, please help.

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Hm. What seems weird to me is denying the connection between E and A until you use the eoms. Ei is the momentum conjugate to Ai. That's normally why things would be consistent ... There is a first class constraint setting the momentum conjugate to A0 to 0, but since A0 only appears as a Lagrange multiplier that constraint generates a trivial gauge symmetry –  Andrew Oct 19 '13 at 17:06
In other words, normally when computing dE/dep i wouldn't write E in terms of A, I would just say it was 0 bc {E,pi0}=0 where pi0 is conjugate to A0. –  Andrew Oct 19 '13 at 17:11