In Dirac's book Principles of quantum mechanics (4th ed., pgs 87-88), he seems to give a very elementary argument as to how the commutator $[X,P]$ reduces to the Poisson brackets ${x,p}$ in the limit $\hbar\to 0$. However, I don't understand the argument that he is making. Could someone please explain this?

Dirac P.A.M. Principles of quantum mechanics (4ed., Oxford, 1958)


Dirac's argument is on pp.85-86, and it goes like this:

the classical Poisson bracket obeys the rules

$$\{A,B\} = -\{B,A\}$$ $$\{aA + bB ,C\} = a\{ A,C\} + b\{B,C\} $$ $$\{AB,C\} = A{B,C} + \{B,C\}A $$ $$\{\{A,B\},C\} = \{B,\{A,C\}\} - \{A,\{B,C\}\}$$

Where I rewrote the Jacobi identity in the way that makes sense. Now Dirac asks whether you can define such a thing for noncommuting quantum objects, and he notes that you can, if

$$ i\hbar [A,B] = (AB - BA) $$

Where $\hbar$ is a proportionality context, fixed by dimensional analysis, while i is there to make the Poisson Bracket analog Hermitian, as observables should be by convention (the anticommutator is anti-Hermitian).

He deduces this by expanding the commutator: $[AB,CD]$ using the formal rules above as axioms, in two different ways. From this, he finds that

$$ [A,C](BD - DB) = (AC-CA)[B,D]$$

In the classical theory, this gives $0=0$, but in QM, the observables don't commute, so you learn you should identify commutators as the quantum analogs of Poisson brackets. He then argues that $\hbar$ commutes with everything, and therefore that the commutation relations should hold, and from there deduces that the Schrodinger picture is available.

The argument is streamlined, and not historically accurate. For Heisenberg's original argument (or something very close to it) see Wikipedia's Matrix Mechanics page.

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On page 87, Dirac writes that the commutation relation between two observables u and v is given by $$uv-vu=i \hbar [u,v]$$ Now, see Eqs. 8: $$[q_{r},q_{s}] = 0$$ $$[p_{r},p_{s}] = 0$$ $$[q_{r},p_{s}] = \delta _{rs}$$ In order to find the corresponding quantum commutators for the above relations, we just plug these into the first equation, giving the quantum versions:$$q_{r}q_{s} - q_{s}q_{r} = 0$$ $$p_{r}p_{s} - p_{s}p_{r} = 0$$ $$q_{r}p_{s} - p_{s}q_{r} = i \hbar\delta _{rs}$$ You can see that setting $\hbar = 0$ gives 0 for the third equation - this is the classical limit, since any two observables will commute in classical physics (In classical physics, observables aren't operators, and there is no uncertainty principle). So, as $\hbar \rightarrow 0$, quantum mechanics reduces to classical physics.

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  • 1
    $\begingroup$ This isn't what OP is asking--- he wants the first nontrivial correction to zero, namely $[A,B] = i\hbar \{A,B\}$ semiclassicaly. $\endgroup$ – Ron Maimon Aug 9 '12 at 3:29
  • $\begingroup$ For completeness, both this leading and all subleasing corrections to the commutator, which in phase space presents as the celebrated Moyal bracket, are provided in this Reference: Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. The PDF file is available here. $\endgroup$ – Cosmas Zachos Feb 8 '16 at 14:04

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