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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 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

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