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I am now reading the quantum mechanics textbook by Dirac (chap. 4, $\S21$, p. 88).

He says that his quantization procedure does not include all possible systems in quantum mechanics and there are systems without a classical analogue.

How is this possible?

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    The spin. As $s$ is fixed (and not variable as in the case of the angular momentum), $\hbar^2 S^2 \to 0$ as $\hbar \to 0$. – Valter Moretti Mar 24 '16 at 13:36
  • Classical mechanics is not the limiting case to $\hbar \to 0$. That's an old meme that should have bitten the dust a long time ago but simply doesn't want to die. It is not clear to me why there has to be a limiting case, to begin with. Maybe the early authors felt that theories needed to have clean mathematical relationships? That is certainly not the case. Th dividing line between quantum behavior and classical behavior is complicated and sometimes simply non-existent. One can't for instance, do classical mechanics without matter and there simply is no classical version of matter. – CuriousOne Mar 25 '16 at 01:37
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    This appears to be a list-based question – Kyle Kanos Mar 25 '16 at 10:10

1 Answers1

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1) It is true that not all quantum systems have classical analogues. E.g. if we have a quantum algebra $({\cal A},\ast)$ of Laurent polynomials in an indeterminate $\hbar$, and endowed with an associative star product $\ast$, it might not be meaningful to take the classical limit $\hbar\to 0$.

2a) However on the linked page Dirac makes a slightly different point that involves canonical position and momentum coordinates, cf. below boldfaced excerpts.

[...] dynamical systems which have a classical analogue and which are describable in terms of canonical coordinates and momenta. This does not include all possible systems in quantum mechanics. [...]

[...] we may have a system in quantum mechanics for which canonical coordinates and momenta do not exist and we can still give a meaning to P.B.s. Such a system would be one without a classical analogue [...]

2b) Recall that the first few entries in the dictionary between

$$\tag{0} \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}$$

read

$$\tag{1} \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,$$

$$\tag{2} \text{Commutator}\quad \frac{1}{i\hbar}[\hat{f},\hat{g}] \quad\longleftrightarrow\quad\text{Poisson bracket}\quad \{f,g\}_{PB} , $$

$$ \text{Heisenberg's EOMs}\quad\quad\quad\quad\quad\quad \text{Hamilton's EOMs} $$ $$\tag{3} \frac{d\hat{f}}{dt}~=~\frac{1}{i\hbar}[\hat{f},\hat{H}]+ \frac{\partial\hat{f}}{\partial t}\quad\longleftrightarrow\quad \frac{df}{dt}~=~\{f,H\}_{PB}+ \frac{\partial f}{\partial t} .$$ See also this Phys.SE post.

2c) In a modern language, Dirac is essentially saying (in the above quotes) that if we dequantize a quantum theory, then on the classical side the resulting (possibly degenerate) Poisson manifold may not be a symplectic manifold, or even a regular$^1$ Poisson manifold.

Recall that the Darboux's theorem, which guarantee the existence of Darboux/canonical coordinates (in each sufficiently small neighborhoods) does not hold for (singular$^2$) Poisson manifolds.

2d) One simple counterexample is the $su(2)$ Lie algebra

$$\tag{4} [\hat{J}^I,\hat{J}^J] ~=~i\hbar\varepsilon_{IJK}\hat{J}^K $$

of angular momentum operators, which leads to a singular Poisson manifold $\mathbb{R}^3$ with Poisson bracket

$$\tag{5} \{J^I,J^J\}_{PB} ~=~\varepsilon_{IJK}J^K. $$

The symplectic leaves are concentric 2-spheres. The symplectic leaf $\{0\}$ at the origin is singular. There does not exist Darboux/canonical coordinates in a neighborhood of the origin.

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$^1$ A regular Poisson tensor has constant rank. A singular Poisson tensor can have rank jumps. A Poisson structure of maximal rank is a symplectic structure.

$^2$ There exists a generalized version of Darboux's theorem for regular Poisson manifold where Darboux neihborhoods contains position, momentum and casimir coordinates.

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