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During analysis the constraint from a theory, suppose my canonical Hamiltonian is $$H_c=P^A\dot{A}+P^B\dot{B}-L$$ where $P^A=\frac{\partial L}{\partial \dot A}$ and $P^B=\frac{\partial L}{\partial \dot B}$ In this case does the commutation of $[P^A,P^B]=0$?

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Comment to the question formulation (v1): This question (i) seems to mix Lagrangian and Hamiltonian definition of momentum, (ii) is not clear about whether it speaks of Poisson brackets or operator commutators, (iii) is not clear about what types of constraints are present. –  Qmechanic Apr 4 '13 at 23:29

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If there are not second class constraints, then yes. In general, see http://en.wikipedia.org/wiki/Dirac_bracket

The result is $$ \left[P^A,P^B \right]=i\hbar\{P^A,P^B\}_{DB}=-i\hbar\sum _{j,k}\{P^A,\phi_j\}_{PB}\,\left( \{\phi_j,\phi_k\}_{PB}\right)^{-1}\,\{\phi _k,P^B\}_{PB} $$ where $\phi _i$ is a second class constraint, and DB and PB stand for Dirac and Poisson brackets, respectively.

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What if one of them is second class and other is not? –  aries0152 Sep 6 '12 at 11:29
The number of second class constraints is always even. –  drake Sep 6 '12 at 16:58
@aries0152 And the reason why is an even number is the following: by definition of Poisson bracket (it has to be antisymmetric), the matrix $\{\phi _j,\phi _k\}_{PB}$ is antisymmetric and it is easy to prove that an antisymmetric matrix has even dimension if and only if its determinant is different from zero. And the determinant must be different from zero because otherwise the constraints would not be linearly independents. –  drake Sep 7 '12 at 18:08

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