Groenewold-van Hove theorem

  • Thread starter paweld
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I've found recently that there is very interesting theorem about quantization
procedure which states that the Dirac procedure of exchanging Poisson bracket
to commutator can't work for observables which are of order grater then 2 in momentum
or position (no matter which ordering we chose). This is Groenewold-van Hove theorem.
Unfortunately I cannot find the precise formulation of it. I don't know what assumption
are needed there. This theorem probably comes from the paper
http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X42-4FW6JXD-4&_user=3934946&_coverDate=10/31/1946&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_searchStrId=1630632579&_rerunOrigin=scholar.google&_acct=C000054122&_version=1&_urlVersion=0&_userid=3934946&md5=03eadd9bb38467a28554de78206452a5&searchtype=a"
(but I don't have access to papers on ScienceDirect ...).
Could anyone give me a reference to diferent paper or state the
assumptions needed in this theorem.
 
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strangerep

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Try looking up Gotay's papers on Google scholar. E.g., math-ph/9809015 has
some info, and there's more in later papers. Also try Rieffel.
 

tom.stoer

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... the Dirac procedure of exchanging Poisson bracket
to commutator can't work for observables which are of order grater then 2 in momentum
or position (no matter which ordering we chose).
Is this really what the quantization procedure says? Of course one uses {x,p}=1 and replaces it by [x,p]=i, but for other observables A(x,p) and B(x,p) there is no reason why the replacement {A,B}=C with [A,B]=iC should work with the same function C(x,p); one has to calculate the commutator and perhaps one gets adifferent Cq - of course depending on the ordering.

For me there is absolutely no reason why the "quantization" should work via this kind of replacement. It is clear that every quantum theory contains more information and structure than the classical theory. So why should there be one unique quantum theory for a given classical theory? I always had the impression that in simple cases it works fine, whereas for difficult cases one as to check different quantum theories (w/o rigorous constructions; having all the same classical limit) and to select the right one using physical arguments beyond this simple "quantization".
 

A. Neumaier

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Of course one uses {x,p}=1 and replaces it by [x,p]=i, but for other observables A(x,p) and B(x,p) there is no reason why the replacement {A,B}=C with [A,B]=iC should work with the same function C(x,p);
{A,B}=C promotes only to [A,B]=iC+O(hbar), and what you get in the correction term depends on the ordering.

So why should there be one unique quantum theory for a given classical theory?
The goal of geometric quantization (the modern name for what the Grunewald-vanHove theorem lead to) is to get a unique quantum kinematics from the classical kinematics. This works essentially whenever one has a classical Lie algebra that generates the classical observables. Representations on coadjoined orbits then lift to unitary representations, provided a Bohr-Sommerfeld type of quantization condition is satisfied for the parameters defining the classical representation. (Well, this is a bit simplified, but it works like that for simple Lie algebras. The general case is complicated.)

Since quadratics in p and q are closed under the Poisson bracket, one can lift them to the quantum case (using symmetric ordering) without getting a correction term. But higher order polynomiials don't close, whence the Grunewald-vanHove theorem.

This still leaves open how to determine the correct generalization of the classical Hamiltonian. This is very nonunique, and is typically handled empirically or based on simplicity arguments.
 
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The goal of geometric quantization (the modern name for what the Grunewald-vanHove theorem lead to) is to get a unique quantum kinematics from the classical kinematics. This works essentially whenever one has a classical Lie algebra that generates the classical observables. Representations on coadjoined orbits then lift to unitary representations, provided a Bohr-Sommerfeld type of quantization condition is satisfied for the parameters defining the classical representation. (Well, this is a bit simplified, but it works like that for simple Lie algebras. The general case is complicated.)

Since quadratics in p and q are closed under the Poisson bracket, one can lift them to the quantum case (using symmetric ordering) without getting a correction term. But higher order polynomiials don't close, whence the Grunewald-vanHove theorem.

This still leaves open how to determine the correct generalization of the classical Hamiltonian. This is very nonunique, and is typically handled empirically or based on simplicity arguments.
I think all your arguments only lead to the conclusion that we have to get rid of classical thought and quantization procedures. Btw. what you call geometric quantization, is that the same as deformation quantization based upon the Moyal product (because this program avoids the Van Hove result too)?
 

tom.stoer

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{A,B}=C promotes only to [A,B]=iC+O(hbar), and what you get in the correction term depends on the ordering.
Exactly. Thanks for a simple formula instead of my lengthy blahblahblah.

Since quadratics in p and q are closed under the Poisson bracket, one can lift them to the quantum case (using symmetric ordering) without getting a correction term. But higher order polynomiials don't close, whence the Grunewald-vanHove theorem.

This still leaves open how to determine the correct generalization of the classical Hamiltonian. This is very nonunique, and is typically handled empirically or based on simplicity arguments.
Exactly. This is well known e.g. for quantization on (arbitrarily) curved manifolds where a natural choice for the kinetic energy is the Laplace Beltrami operator. However - it's natural, but not unique!
 
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Exactly. This is well known e.g. for quantization on (arbitrarily) curved manifolds where a natural choice for the kinetic energy is the Laplace Beltrami operator. However - it's natural, but not unique!
Ordinary physicists simply say that the Feynman propagator is not uniquely defined (for the simple reason there is no Killing time), the Laplace-Beltrami operator itself is of course unique (since it has a local prescription). Again, the conclusion you shoud draw from this, is the same: free QFT is not understood (and as it stands, wrong).
 
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tom.stoer

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I am talking about QM on curved manifolds, not QFT.

The Laplace-Beltrami operator is unique mathematically, but it's not unique that it should be the kinetic energy; this is "only natural".
 
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I am talking about QM on curved manifolds, not QFT.

The Laplace-Beltrami operator is unique mathematically, but it's not unique that it should be the kinetic energy; this is "only natural".
So you are talking here about the classical Klein Gordon equation ? Why do you say that there might be some ambiguity in the ''kinetic energy'' in this case ? As far as I see, the only ambiguity comes from the choice of foliation (in either the way you split up the Laplace-Beltrami operator) but that was to be expected. In field theory kinetic energy is uniquely defined by considering all quadratic terms in one field only (the only ambiguity may arise in the mass and some cross derivative terms when you perform a field redefinition).

Btw. your comment is even further mysterious to me. From one side you tolerate something exotic and unphysical like the polymer quantization in LQG, while from the other you stick to a theory which should not exist anymore (old QM). Like that, you may be missing out on those things which are actually valuable to describe nature.
 
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strangerep

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To all the other people who responded in this thread,

I'll just make the observation that none of your replies actually
answered the OP's original question, i.e., paweld's request
for online references or a precise statement of the theorem.
Even though I myself found your answers interesting, they
do kinda hijack paweld's thread -- especially since I'm sure some
of you are surely able to suggest better references than the
ones I knew of...
 
1,667
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To all the other people who responded in this thread,

I'll just make the observation that none of your replies actually
answered the OP's original question, i.e., paweld's request
for online references or a precise statement of the theorem.
Even though I myself found your answers interesting, they
do kinda hijack paweld's thread -- especially since I'm sure some
of you are surely able to suggest better references than the
ones I knew of...
The only one who hijacks this thread is you. Regarding the content of the theorem, an answer has clearly been given by Neumaier and Tom Stoer, while I merely picked in on some details. If you think these people did a bad job presenting the content, you may perhaps improve upon it if you can ?
 

tom.stoer

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So you are talking here about the classical Klein Gordon equation ? Why do you say that there might be some ambiguity in the ''kinetic energy'' in this case ?
Even simpler: I am talking about the kinetic energy in the Schrödinger equation of non-rel. QM:

[tex]T=-\frac{\hbar^2}{2m}\Delta[/tex]

Here the symbol refers to the Laplace-Beltrami operator. In flat space, e.g. R³ this is a rather simple expression. For a particle moving on a sphere you get the well-known expression for the Laplacian in spherical coordinates plus fixing r=const. and dropping all r-derivatives. This is a "natural" choice dictated by symmetry arguments. It can somehow be "dervied" using Dirac's constraint quantization (applied to the curved two-dim manifold). But of course one could chose other differential operators having the same classical limit, i.e. differing from the Laplace-Beltrami by ordering. So in that sense Laplace-Beltrami is unique mathematically, but it's not unique to chose the Laplace-Beltrami operator as the kinetic energy; there is another physical argument, namely the symmetry argument.

If you look at more complicated manifolds in phase space (!), there is no such symmetry argument available. If you look at other observables (perhaps not related to a symmetry), again no such symmetry principle may be available.
 

A. Neumaier

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I think all your arguments only lead to the conclusion that we have to get rid of classical thought and quantization procedures. Btw. what you call geometric quantization, is that the same as deformation quantization based upon the Moyal product (because this program avoids the Van Hove result too)?
No. deformation quatization is a perturbative procedure (well, in mosty of the literatire. There is also a rigorous path, trodded by Rieffel and followers), motivated by cohomolcical considerations and the moyal product.

Geometric quantization is from the start rigorous, and based on the insights from the Groenewold-van Hove theorem. it doesn't bypass it but restricts the Homomorphism to a Lie algebra or relevant operators rather than the full Poisson algebra.
 
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Even simpler: I am talking about the kinetic energy in the Schrödinger equation of non-rel. QM:

[tex]T=-\frac{\hbar^2}{2m}\Delta[/tex]

Here the symbol refers to the Laplace-Beltrami operator. In flat space, e.g. R³ this is a rather simple expression. For a particle moving on a sphere you get the well-known expression for the Laplacian in spherical coordinates plus fixing r=const. and dropping all r-derivatives. This is a "natural" choice dictated by symmetry arguments. It can somehow be "dervied" using Dirac's constraint quantization (applied to the curved two-dim manifold). But of course one could chose other differential operators having the same classical limit, i.e. differing from the Laplace-Beltrami by ordering. So in that sense Laplace-Beltrami is unique mathematically, but it's not unique to chose the Laplace-Beltrami operator as the kinetic energy; there is another physical argument, namely the symmetry argument.
Ah, I see. My guideline here is (spacelike) general covariance of the kinetic energy. All different expressions will break it, so it is a very natural assumption given that you have flat coordinates available.

If you look at more complicated manifolds in phase space (!), there is no such symmetry argument available. If you look at other observables (perhaps not related to a symmetry), again no such symmetry principle may be available.
Again, general covariance will save you here. If you let the metric interfere with the partial derivatives, then this is a coordinate dependent expression (since the only appropriate action is by means of the Levi-Civita connection which of course eats away the metric).
 
1,667
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No. deformation quatization is a perturbative procedure (well, in mosty of the literatire. There is also a rigorous path, trodded by Rieffel and followers), motivated by cohomolcical considerations and the moyal product.

Geometric quantization is from the start rigorous, and based on the insights from the Groenewold-van Hove theorem. it doesn't bypass it but restricts the Homomorphism to a Lie algebra or relevant operators rather than the full Poisson algebra.
Oh, I see; that is indeed different then; thanks.
 

A. Neumaier

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Groenewold-van Hove theorem.
Unfortunately I cannot find the precise formulation of it. I don't know what assumption
are needed there.

Could anyone give me a reference to diferent paper or state the
assumptions needed in this theorem.
There are a number of variations of the theorem. Look at
http://arxiv.org/pdf/dg-ga/9605001
(in particular around p.20) where the theorem is embedded in a modern context.
 

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