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How can I define the Hamiltonian independent of the Lagrangian? For instance, let's assume that i have a set of field equations that cannot be integrated to an action. Is there any prescription to construct the Hamiltonian of a such system starting from the field equations?

Qmechanic
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John Doe
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    It sort of depends on what you mean by a hamiltonian. If you mean 'is there a system whose dynamics can be rephrased as Hamilton's equations, but for which there's no lagrangian that reproduces the dynamics and transforms into the correct hamiltonian?', you should express that in the question. (If not, use a similarly precise statement.) That case is answered in the negative by linear hamiltonians such as $H=p$. – Emilio Pisanty Jan 21 '16 at 17:54
  • ok, let me provide the simplest example that i can think of. Consider that I have the following field equation for a symmetric rank-2 tensor $h_{\mu\nu}: \eta^{\rho \sigma} \partial_\mu \partial_\nu h_{\rho\sigma} + \eta_{\mu\nu} \partial_{\rho} \partial_{\sigma} h^{\rho\sigma}$. Obviously, such a field equation can come from $L = \eta^{\rho\sigma} \partial_\mu h^{\mu\nu} \partial_\nu h_{\rho \sigma}$. But if I change the relative coefficients between two terms in the field equation, then you can no longer find a Lagrangian for such a field equation. – John Doe Jan 21 '16 at 18:09
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    I think what you are asking about comes under the general heading of the 'Inverse problem for Lagrangian/Hamiltonian mechanics'. There's a theorem due to Douglas (1941) that gives a set of necessary and sufficient conditions under which a there exists a Lagrangian for any given ODE. I'm sure you can then extract the conditions on the existence of a Hamiltonian from this. The problem is much harder for general PDE systems -- but I think this is still quite an active field of research. – Arthur Suvorov Jan 25 '16 at 07:20
  • @JohnDoe Your example is in principle interesting, but is the field language actually necessary? Surely the issue is the fundamentals of the mechanics, and there should be at least a discretized version - hopefully a single-dof one - with the same core issue. Also, you mentioned "equation" but you didn't actually display any equals signs. – Emilio Pisanty Jan 25 '16 at 22:48

2 Answers2

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Comments to the question (v2):

  1. First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein.

  2. On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2.

  3. On the other hand, if we have a (possible constrained) Hamiltonian formulation, of the type discussed in Refs. 1 and 2, then it is possible to give a Hamiltonian action formulation, which in itself can be interpreted as a Lagrangian formulation, e.g. after integration out momentum variables along the lines indicated in my Phys.SE answer here.

  4. In other words, Lagrangian and Hamiltonian formulations traditionally go hand in hands. Thus it is unclear what precisely OP is looking for.

References:

  1. P.A.M. Dirac, Lectures on QM, (1964).

  2. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

Community
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Qmechanic
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    (-1) Any set of equations can be put into a Lagrangian setting by adding additional unobservable degrees of freedom. For $F(x)=0$ add new variables $y$ and take $L=\langle y|F(x)|\rangle>$. Variation of $y$ gives $F(x)=0$, and variation of $x$ gives $F'(x)^Ty=0$ which in many cases is equivalent $y=0$ - which makes it unobservable since it never changes. – Arnold Neumaier Jan 26 '16 at 15:57
  • The Lagrangian side (and its general feasibility) has been discussed extensively in several threads here: http://physics.stackexchange.com/q/20188/ http://physics.stackexchange.com/q/4749/ http://physics.stackexchange.com/q/167027/ http://physics.stackexchange.com/q/147341/ – Arnold Neumaier Jan 26 '16 at 16:08
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The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense.

Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any Lagrangian, in a purely Hamiltonian form. Both formalisms can be made fully covariant.

This holds for classical fields. How to quantize a theory in one of these formalisms is very poorly understood.

Danu
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Arnold Neumaier
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