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.

  • $\begingroup$ 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 $\endgroup$ – Andrew Oct 19 '13 at 17:06
  • $\begingroup$ 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. $\endgroup$ – Andrew Oct 19 '13 at 17:11

The problem lies in what we learn about good old constrained dynamics from traditional Dirac approach is not complete and is somehow inconsistent, and the above is one example of this. This was the message of Pitts' paper mentioned in the question above, who reviewed a bunch of previous work on this very matter. I will mention couple of references from that paper which should clarify the problem and solve it.

What is wrong is that the generator of gauge transformation in a certain constrained system is not an isolated first class constraint, but a certain linear combination of first class constraints. This is what the mentioned Pitts' paper tries to emphasize. Somehow, people worked by inertia ever since Dirac's lecture notes, and never got to notice that a single first class constraint does not generate a gauge transformation, but a ''bad change'' (as Pitts calls it) as you have stated in your question. Then came Castellani in 1982, and in his paper "Symmetries in constrained systems" in Annals Phys. 143, p. 357 (1982) formulated a generator of gauge symmetries as a well-defined linear combination of first class constraints. This paper is very insightful and I recommend it as a starting point when one starts with constrained systems. There he derives an algorithm which determines the form of a gauge generator, and then he derives gauge generators for a simple toy model, for General Relativity, and for Yang-Mills theories (of which Electromagnetism is a special case, so the results also apply to the question above). All of them are a linear combination of first class constraints.

There is also a nice discussion and possibly a very detailed answer to the exact question posted above, in paper by Pons, as well as in Sundermeyer's book "Symmetries in Fundamental Physics".

The bottom line is: isolated first class constraints (primary or secondary or tertiary...) do not, in general, generate gauge transformations. But they are each a part of a gauge generator, which is defined as a linear combination of these constraints.

  • $\begingroup$ Welcome to Physics SE and thank you for the answer. Could you please add the full reference for the work by Pitt you are quoting? $\endgroup$ – Sanya Jul 26 '16 at 16:03

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