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Simplest example where quantization is unknown

  1. Mar 24, 2012 #1

    quasar987

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    As I understand (from reading p. 2-06 of Marle's 1975 text on geometric quantization available on the french wiki page on "quantification géométrique") , there are physical situations where we do not know how to write the Schrodinger equation. Namely, we do not know what operator to take as the hamiltonian H in the equation dψ/dt = Hψ. Is this correct? What is the simplest physical system where this situation occurs?

    Thanks.
     
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  3. Mar 24, 2012 #2

    fzero

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    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.
     
    Last edited: Mar 24, 2012
  4. Mar 24, 2012 #3

    quasar987

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    I thought geometric quantization was concerned with the process of associating a quantum theory to a classical system. This means: to a symplectic manifold M and an hamiltonian H:M-->R, find a hilbert space X and a map p:R-->L(X) where R is some subset of functions on M containing H.

    Therefor, when you say

    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.
     
  5. Mar 24, 2012 #4

    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'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
     
  6. Mar 24, 2012 #5

    quasar987

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    Here is the link: http://archive.numdam.org/ARCHIVE/S...1975-1976__2__A2_0/SPK_1975-1976__2__A2_0.pdf

    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 R^n (or R^2n with the standard symplectic form if you prefer). This yields the familiar quantization X=L²(R^n) where q becomes the operator "multiplication by q" and p becomes i[itex]\hbar[/itex]∂/∂q.

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

    "It remains, in order to write the Schrodinger equation, to say what the operator corresponding to the hamiltonian H should be. This is something that physicists know how to do in many practical situations."

    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!
     
  7. Mar 24, 2012 #6

    fzero

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    OK, in my classical definition [itex]\mathbf{M}[/itex] is not a Hamiltonian vector field, so we would never get those equations of motion from a Poisson bracket. Perhaps part of the problem is that geometric quantization starts from a symplectic manifold and this is not really an entire physical system. A physical system is a phase space plus equations of motion. The symplectic approach requires that the equations of motion follow from a Hamiltonian flow on the phase space, which is great when it happens, but is obviously not the most general case we might want to consider. Your last quote seems to fall into this category, where we don't know the Hamiltonian for the system for some reason, either it doesn't exist, or we're not clever enough to deduce it.

    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.
     
  8. Mar 24, 2012 #7

    Matterwave

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    I don't really know anything about geometric quantization, but there are many cases where we don't know the exact Hamiltonian, I don't know if this is what you're looking for?

    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.
     
  9. Mar 25, 2012 #8

    tom.stoer

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    In QG having Hψ=0 is not really problematic. The problem is to find a different time variable which describes a 'physical flow of time'. In relativistic mechanics you start with H²=papa-m² and H~0 but you can find a different time variable for which po plays the role of the Hamiltonian.

    I think a good starting point are complicated manifolds with non-trivial geometry and metric gab for wich the kinetic energy ~Δg is problematic (Δg is the Laplace-Beltrami operator).

    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.
     
  10. Mar 25, 2012 #9

    quasar987

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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, any 1-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 TpL 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.
     
  11. Mar 25, 2012 #10

    fzero

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    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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