7
$\begingroup$

There is an well-known infamous DICTUM:

-Second Quantization is a functor, First Quantization is a mystery-.

Indeed, second quantization is the "Fock functor", which builds the Fock space in a canonical way out of the Hilbert space of a single particle.

But, what about first quantization? There is probably no hope to canonically associate an Hilbert space to the manifold of states of a classical particle (mathematically, there seem to be an inherent element of choice as far as turning functions into operators).

However, there is (I suspect) some functorial description for going the other way around, FROM the quantum scenario INTO the classical one (corresponding to the limit $h\rightarrow 0$). If this is true, there maybe a "fiber" of candidate quantum descriptions, all collapsing into the same classical one.

Any place where this has been worked out clearly?

$\endgroup$
1

1 Answer 1

5
$\begingroup$

You may be interested in the answers to a question on physics.SE, "Quantum Mechanics on Manifold".

The gist of it is that there is a plethora of different quantization schemes (canonical quantization, geometric quantization, ...). As you note, they all have the same classical theory as a limit. Overview:

Quantization Methods: A Guide for Physicists and Analysts


That said, speaking as a physicist (caveat emptor), I think the paper

http://link.aps.org/doi/10.1103/PhysRevA.23.1982

pretty much covers everything that is of practical relevance. Unsurprisingly, the Schrödinger equation on an embedded manifold $M \subseteq \mathbb{R}^3$ depends not only on intrinsic data, like the metric or curvature, but also on extrinsic data about the embedding of the manifold inside 3D space. That's to be expected because an electron can tunnel through the full 3D space and hence take "extrinsic short-cuts". I think that's also the reason why there is so much ambiguity in quantization schemes: it's simply not something that you can do completely intrinsically.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .