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In BRST Symmetry in the Classical and Quantum Theories of Gauge Systems, Henneaux says the Fock representation is not applicable to an odd number of constraints. Then he goes on to say that the Kugo-Ojima quartet requires the constraints to be in pairs. For BRST theories, when are they not in pairs?

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  • $\begingroup$ Full reference: M. Henneaux, BRST Symmetry in the Classical and Quantum Theories of Gauge Systems. Published in Quantum Mechanics of Fundamental Systems, (Editor C. Teitelboim), 1988, chapter 10, p. 138. $\endgroup$ – Qmechanic Feb 8 '14 at 13:39
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A stupid answer is to take the action $$L = \frac{(\dot{q}_1-\dot{q}_2)^2}{2} + \frac{(q_1-q_2)^2}{2}.$$ It has only one constraint, $p_1+p_2=0$.

I don't know any more physical example, but I wouldn't be surprised if one exists.

There are at least two less stupid answers.

The first is (3+1)-d Maxwell theory in $A_0 = 0$ gauge, because then the only constraint is Gauss' law $\nabla \cdot E = 0$.
There are corresponsing ghosts that have no dynamics because this only consists of time-independent gauge transformations.
As a result, no one really talks about them.

The second is (0+1)-d Maxwell theory, where the only gauge-invariance is that generated by $\Pi^0$, the momentum conjugate to $A_0$.
This is basically the converse of the previous example, freezing space instead of time.

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  • $\begingroup$ How does this answer the question "For BRST theories, when are they not in pairs?"? $\endgroup$ – AccidentalFourierTransform Jul 21 '17 at 17:22
  • $\begingroup$ Well, you can write a BRST charge $Q = \eta (p_1+p_2)$ and all the machinery of BRST works. Is there a more involved meaning of "BRST theory?" $\endgroup$ – Ronak M Soni Jul 21 '17 at 18:14

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