What is the exact connection between Poisson brackets and commutators

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SUMMARY

The connection between Poisson brackets in classical mechanics and commutators in quantum mechanics is established through their shared algebraic properties and underlying mathematical structures. Specifically, the Poisson bracket corresponds to the commutator divided by \(i \hbar\) in quantum mechanics. This relationship is rooted in the geometric properties of phase space, where both classical and quantum systems are described by Poisson manifolds and symplectic manifolds, respectively. Advanced quantization methods, such as geometric quantization and deformation quantization, are essential for mapping classical symplectic geometry to quantum Poisson geometry.

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  • Understanding of Poisson brackets in classical mechanics
  • Knowledge of commutators in quantum mechanics
  • Familiarity with Lie groups and Lie algebras
  • Basic concepts of symplectic geometry and phase space
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Physicists, mathematicians, and students interested in the foundations of quantum mechanics and classical mechanics, particularly those exploring the mathematical structures that connect these two fields.

snoopies622
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I'm not perfectly clear on the connection between Poisson brackets in classical mechanics and commutators in quantum mechanics.
For any classical mechanical system, if I can find the Poisson bracket between two physical observables, is that always the value of the corresponding commutator in the quantum mechanical formulation of the same system (divided by i \hbar )?
If so, there must be some mathematical reason for this. Some kind of homomorphism between the two systems, perhaps?
I know that Poisson brackets and commutators share some algebraic properties and that
Hamilton's <br /> <br /> \frac{df}{dt}= \{ f,H \} + \frac { \partial {f} }{\partial {t}}<br /> <br />

looks similar to Heisenberg's <br /> <br /> \frac{d \hat{A} }{dt}= \frac {i}{\hbar} [ \hat{A},\hat{ H} ] + \frac { \partial { \hat{A} } }{\partial {t}}

but for that method to work every time, it feels to me like there must be something more.
 
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There are two main mathematical structures that you should look into. The first one is that of transformations of the system described by Lie groups and the corresponding Lie algebras. The Poisson brackets and commutators form the product of the Lie algebra, the so called Lie bracket.

The second structure comes from the geometric property of phase space that assures that geometric transformations need to preserve the phase space volume locally. This constraint makes the classical and quantum phase spaces poisson manifolds (and the classical one even a symplectic manifold), whose structure are characterised by a preserved canonical poisson bracket or commutator respectively.

The two structure are not independent. The latter implies that valid transformations of a system need to be symplectomorphisms. This partly determines the common mathematical structure of classical mechanics and quantum theory. However it is not a simple correspondence, because the geometry of quantum phase spaces is even more restrictive than that of classical ones. The additional structure in quantum theory is of course the non-commutativity of observables.

Methods for quantising classical systems need to map the classical symplectic geometry to the poisson geometry of the quantum system. This is simple only for very basic systems that are described by cartesian spatial coordinates and the conjugate momenta. Anything beyond that needs advanced quantisation method that address exactly the kind of correspondence you ask for. They are usually termed "geometric quantisation methods" and of particular interest to you might be the so called deformation quantisation. It takes the algebraic structure of classical mechanics and translates it to a deformed algebra with a continuous deformation parameter ##\hbar##.

I hope you have enough keywords now to start your own research.

Cheers,

Jazz
 
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Wow, thanks Jazz - much to look into here! In the meantime I will assume that the answer to my first question is, "yes".
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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