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quasar987

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Thanks.

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- Thread starter quasar987
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- #1

quasar987

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Thanks.

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fzero

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Thanks.

Non-Hamiltonian systems are already known from classical mechanics (which are probably what a text on geometrical quantization is referring to). These are typically open systems, like a system connected to a heat bath. Another term for some of these is "dissipative system." In some of these systems, energy is not conserved, however there can be non-Hamiltonian systems where an energy-like quantity is conserved.

It might also be illuminating to note the condition for a classical Hamiltonian to exist, namely if [itex] \mathbf{X} = (q_i,p_i)[/itex] are the phase space variables, then the equations of motion can be expressed as

[tex] \mathbf{\dot{X}} = \mathbf{M(X)}.[/tex]

If

[tex]\nabla_\mathbf{X}\cdot \mathbf{M} = 0,[/tex]

then the Hamiltonian exists as a potential function such that

[tex] \mathbf{M} = (-\partial_{q_i}H , \partial_{p_i} H),[/tex]

whereas if [itex]\nabla_\mathbf{X} \mathbf{M} \neq 0,[/itex] then the Hamiltonian does not exist globally.

A somewhat simple example of a non-Hamiltonian system is the Lorenz system http://en.wikipedia.org/wiki/Lorenz_system.

Edit: Perhaps the Lorenz system isn't a great example because it has an odd number of variables. It's obviously easy to construct examples, but I'm not familiar enough with the topic to give a well-motivated example.

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quasar987

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Therefor, when you say

Non-Hamiltonian systems are already known from classical mechanics (which are probably what a text on geometrical quantization is referring to).

I would think that the opposite is true: since this is coming from a text in geometric quantization, the hint about non quantizable systems refer to classical systems wich are hamiltonian to begin with, but we don't know how to chose X and p for it.

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strangerep

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Your question might be clearer if you included a larger quote from the textbook you mentioned. (I didn't see it on French Wiki, but my French is very minimal.)I thought geometric quantization was concerned with the process of associating a quantum theory to a classical system. This means: to a symplectic manifold Mand an hamiltonianH:M-->R, find a hilbert space X and a map p:R-->L(X) where R is some subset of functions on Mcontaining H.

I'm wondering whether you're talking about the old ordering ambiguity problem (Groenewold-van Hove theorem) that arises in quantization when higher products of the basic observables are involved.

Cf. http://en.wikipedia.org/wiki/Geometric_quantization

and http://en.wikipedia.org/wiki/Moyal_bracket

- #5

quasar987

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The quote I'm refering to is on page 2-08 (and not 2-06 like I said in post #1 - sorry!)

the author just introduced the definition of geometric quantization, and shows how to quantize the easiest case when M is the cotangent bundle of

the author then says, right before section (d) begins:

The use of the word "many" there suggests to me that the physicists don't know how to do it always. I ask what is the simplest case when they don't know how to do it.

Thx!

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fzero

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If you want specific conditions for geometric quantization to fail, you can start with a list of what's required, such as http://math.ucr.edu/home/baez/quantization.html and try to find an example where something goes wrong. This is not my area of expertise, so the only example I can come up with at the moment is the case where phase space is a 2-sphere. The volume form is symplectic and [itex]\partial/\partial\theta[/itex] is Hamiltonian. However, there is no Lagrangian submanifold, so there is no polarization. Namely [itex]\omega[/itex] vanishes at the poles, but these points are not maximally isotropic.

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Matterwave

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For example, in a large system, it might be too difficult to construct the full Hamiltonian, so we tend to "trace out" the excess degrees of freedom.

Or, in canonical GR, the Hamiltonian turns out to be 0 (being composed entirely of constraints) with the usual boundary conditions (closed universe), in which case you get no Schroedinger equation, since you simply get Hψ=0.

Probably this is not what you're looking for XD.

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tom.stoer

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I think a good starting point are complicated manifolds with non-trivial geometry and metric g

At the moment I can think of no system which could be relevant in QM and for which H is not known or for which H is known not to exist as a QM operator.

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quasar987

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If you want specific conditions for geometric quantization to fail, you can start with a list of what's required, such as http://math.ucr.edu/home/baez/quantization.html and try to find an example where something goes wrong. This is not my area of expertise, so the only example I can come up with at the moment is the case where phase space is a 2-sphere. The volume form is symplectic and [itex]\partial/\partial\theta[/itex] is Hamiltonian. However, there is no Lagrangian submanifold, so there is no polarization. Namely [itex]\omega[/itex] vanishes at the poles, but these points are not maximally isotropic.

In fact, the lagrangian situation is the opposite of this! :) On any 2-dimensional symplectic manifold,

- #10

fzero

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Thanks to everyone for the answers.

In fact, the lagrangian situation is the opposite of this! :) On any 2-dimensional symplectic manifold,any1-dimensional submanifold is lagrangian! More generally, on any symplectic manifold, any 1-dimensional submanifold is isotropic. This is because the symplectic form ω is skew, and hence ω(v,v)=0 for any tangent vector v. In particular, if v,w are in T_{p}L for some 1-submanifold L of M, then it is that w=cv for some constanc c. And hence ω(v,w)=cω(v,v)=0 and L is isotropic.

I think that you're right. I was mis-assuming additional structure (special Lagrangian) that would have required the submanifold to be an integral homology class in [itex]H_1(S^2,\mathbb{Z})=\{0\}[/itex].

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