Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

For my thesis, I have calculated the constraints for a system using Dirac method of constraint analysis. The problem is I got odd number of second class constraints (!), which gives me unusual numbers of degrees of freedom in phase space. I might have made some mistakes.

Is there any other method beside Dirac, to analyze the constraints of the system?

share|cite|improve this question

3 Answers 3

up vote 2 down vote accepted

If the constrained Hamiltonian system has a finite number of real degrees of freedom$^1$, and if all the constraints are regular, then it is mathematically impossible to have an odd number of second-class constraints. (The proof is very similar to the reason why a symplectic manifold or vector space must be even-dimensional.)

Perhaps OP is actually considering a constrained Hamiltonian field theory with an infinite number of degree of freedom and an infinite number of second-class constraints? (Typically this happens because all the fields, say a position field $\phi(\vec{x},t)$ and a momentum field $\pi(\vec{x},t)$, are labeled by a continuous index, namely the space point $\vec{x}$). In that case, it does not make sense to label $\infty$ as an odd number.

Example$^2$: A typical example of a second-class constraints in 1+1 dimension field theory with canonical equal-time Poisson brackets $$\tag{1} \{\phi(x,t),\pi(y,t)\}~=~ \delta(x-y), $$ $$\tag{2} \{\phi(x,t),\phi(y,t)\}~=~0, $$ $$\tag{3} \{\pi(x,t),\pi(y,t)\}~=~0, $$ is

$$\tag{4} \chi(x,t)~:=~\pi(x,t) -\partial_x\phi(x,t). $$

Naively one may think of (4) as a single (i.e. odd!) second-class constraint, but it is really infinitely many second-class constraints labeled by the position $x$. Their equal-time Poisson brackets are

$$\tag{5} \Delta(x,y)~:=~\{\chi(x,t),\chi(y,t)\}~=~ 2 \delta^{\prime}(x-y) $$

with a formal$^3$ inverse

$$\tag{6} \Delta^{-1}(x,y) ~=~ \frac{1}{4}{\rm sgn}(x-y).$$

For another related example of second-class constraints in Hamiltonian field theory, see also e.g. this Phys.SE answer.

$^1$ The definition of degrees of freedom (d.o.f.) is e.g. discussed in this Phys.SE post. (Note that there is also a field-theoretic notion of d.o.f., which is different. E.g. in pure QED in 3+1 dimensions, the photon has 2 physical polarizations, so one would say that pure QED has 2 physical d.o.f., etc. This is not the notion of d.o.f, that I'm considering here. If OP is counting field-theoretic d.o.f., there is no reason to be surprised to meet an odd number, cf. the Example.)

$^2$ This example is sometimes referred to as a chiral/self-dual boson in 1+1 dimensions.

$^3$ One should impose appropriate boundary conditions at $|x| \to \infty$.

share|cite|improve this answer
The fields (I am working with) have finite number of DOF of themselves separately. My work was to analyze how they react if they all interact in the same field. In this case, is it possible to have infinite number of DOF? – aries0152 Oct 2 '12 at 8:59
@aries0152: If you would like a more focused response about your system of second-class constraints, you would have to display the explicit form in the question. – Qmechanic Oct 2 '12 at 9:53
My work has a similarity with this ( paper. Here the Fields/massive terms has their own number of DOF. But if we couple them together, the pseudomass term eliminates all other DOF except the two traverse DOF. Now let us consider a different coupling with other fields or a group of fields. Instead of decreasing, is it possible that it could result infinite numbers of DOF? – aries0152 Oct 2 '12 at 12:30
Right, that paper considers field theory, i.e. infinitely many d.o.f. – Qmechanic Oct 2 '12 at 13:37
So with infinite number of dof, does the commutation chain breaks in a point (Like $\dot \phi=0$) ? Or it just continue producing new constraints (second class) ? – aries0152 Oct 2 '12 at 17:32

There is an elegant constraint quantization method by Jackiw; see

share|cite|improve this answer

The matrix of the Poisson bracket of the secondary constraints is anti-symmetry, so the number of the secondary constraint is even. If it is odd, there maybe have some new first-class constraints which can be combined by those secondary constraints.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.