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Apologies if this is a really basic question, but what is the physical interpretation of the Poisson bracket in classical mechanics? In particular, how should one interpret the relation between the canonical phase space coordinates, $$\lbrace q^{i}, p_{j} \rbrace_{PB}~=~\delta^{i}_{j} $$ I understand that there is a 1-to-1 correspondence between these and the commutation relations in quantum mechanics in the classical limit, but in classical mechanics all observables, such as position and momentum commute, so I'm confused as to how to interpret the above relation?

Qmechanic
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Will
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  • Related: http://physics.stackexchange.com/q/32738/2451 , http://physics.stackexchange.com/q/133952/2451 , http://physics.stackexchange.com/q/130800/2451 Related mathoverflow post: http://mathoverflow.net/q/19932/13917 , http://mathoverflow.net/q/157633/13917 – Qmechanic Apr 14 '15 at 22:27
  • If I understand correctly, the fundamental bracket you show indicates that the phase space variables are truly independent of each other, and it is easy to see why from the maths why you get the delta. You can then use these in some canonical quantisation scheme. Both quantum and classical mechanics (KvN theory) can be converted to an operator form and we only allow for commuting algebras in classical mechanics while quantum mechanics has non-commuting observables too! I understand the Poisson bracket to exemplify the change in some quantity with respect to the phase space variables e.g. .. – AngusTheMan Apr 14 '15 at 22:35
  • (cont) for some function $F$ of phase space, the PB with the Hamiltonian is actually the time derivative of the function (add a $\partial F/\partial t$). Since time evolution can be viewed as an infinitesimal canonical transformation we see a relationship between the generators of the CTs and the effect on the function. Therefore, the Poisson bracket is most useful in the Hamiltonian version of Noether's theorem, such that if a function is invariant under the action of a generator its PB with the generator is zero. – AngusTheMan Apr 14 '15 at 22:38

1 Answers1

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In a rather general approach you can consider the Poisson bracket $\{g,f\}$ as expressing the rate of change of $g$ as a consequence of a flow induced by $f$. As mentioned by AngusTheMan in the comments, you get the time variation of $g$ if $f=H$ (assuming that quantities are not explicitly time-dependent). Here $g$ and $f$ are any (smooth) functions on the phase space, i.e. observables. When $g=q$ and $f=p$, since the momenta are the generators of translations, the flow generated by $f$ can be interpreted as translations, so that the canonical bracket $$\{q,p\} = 1$$ implies a variation of $\delta q = \{q,p\}\epsilon = \epsilon$. Generalising this to many dimensions you get $$\delta q_i = \{q_i,p_j\}\epsilon = \delta_{ij}\epsilon,$$ which is expressing the fact that $p_j$ generates the translations along the $j$-th coordinate (indeed $q_i$ changes by $\epsilon > 0$ only if $j=i$).

Phoenix87
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  • So is it that the Poisson bracket has nothing to do with commutativity of conjugate variables? If so, how does one obtain commutation relations from these in QM? Also, is the Poisson bracket a derived quantity, or is it just defined as such? – Will Apr 15 '15 at 07:12
  • Classical mechanics has a commutative structure. In terms of operator algebras it is described by the algebra of smooth functions on the phase space, and the real value functions are the observables. The fact that Heisenberg relations are linked to the canonical brackets is just a quantisation procedure. Poisson brackets have a precise definition and the canonical brackets follow from it. – Phoenix87 Apr 15 '15 at 07:49
  • So is the point that there is a relation between them, but that the Poisson bracket is not describing whether observables commute or not, where as the "quantised" version, the commutator is? Do you know of any lecture notes (or other sources) that give a good description of the link between the two? – Will Apr 15 '15 at 09:18
  • I would try Dirac's Lectures on Quantum Mechanics perhaps, although the description is in a sense "advanced". However I think it gives a good feeling of what the PB are when you have constraints, and how they then relate to quantum systems. – Phoenix87 Apr 15 '15 at 09:24
  • Ok, thanks I'll take a look. Referring to your original answer, would it be correct at all to say that (at least intuitively) the Poisson bracket in CM is related to the commutator in QM in the sense that the PB $\lbrace q^{i},p_{j}\rbrace$ measures the translation along $q^{i}$ generated by $p_{j}$ ... – Will Apr 15 '15 at 10:24
  • ... and hence if we measure momentum of a particle in QM we can not simultaneously specify its position (as the measured momentum will have generated a change in position). Similarly, if we measure its position, then we cannot simultaneously determine the momentum that generated its translation to that position?! – Will Apr 15 '15 at 10:25