Poisson bracket significance (Classical Mechanics)

1. May 9, 2007

We have to show that [Lx,Ly] = Lz
[Ly,Lx] = -Lz
[Lx,Lx] = 0

and I have done this. We then need to comment on the significance of these results, which I'm not sure of. I know in QM you get similar results for commutators of these quantities, and it means that you can't simultaneously know e.g Lx and Ly, but I'm not sure what happens in CM.

2. May 9, 2007

lalbatros

Good question. I have no good answer yet.
The correspondance principle associates any QM commutator with a CM Poisson bracket. If there is a meaning for the first in QM, what could be the meaning for the second in CM? quite a good question.

As a first naïve strategy to answer it, I imagine to put one of the L's as a pertubation in an Hamiltonian.
Without the perturbation, Lx and Ly and Lz could be constants of motion, all simultaneously.
Now, if Lx perturbs the hamiltonian, we can deduce from the equations of motion that Lx can remain a constant of motion, but not Ly nor Lz. This would be quite natural as the result of a magnetic field along the x axis.

Would that not be a way to give some physical meaning to the bracket commutation relations?
I would also draw your attention on the fact that measurements in QM always imply an interaction, even if this does not appear explicitely in the infamous QM postulates. This tends to indicate that the analogy is actually more than a simple analogy.

It would be very nice and exciting if you could help me clarify these raw ideas.

Thanks for this question.

3. May 9, 2007

nrqed

Good question. Someone else will probably do better than what I can say here, but let me mention two applications. One is that one can test whether a transformation between two parametrizations of phase space is a canonical transformation is to check if the Poisson brackets between the phase space coordinates obey the fundamental brackets, $\{ q_i , p_j \} = \delta_{ij}$ and the other being zero.

But another, maybe more relevant application is that the time evolution of a quantity "A" is given by the Poisson bracket of A with the Hamiltonian (plus a partial derivative of A if it has explicit time dependence), viz.

$$\frac{ dA }{dt} = \{ A,H \}_{class} + \frac{\partial A}{\partial t}$$

The connection with qm is therefore the most direct in the Heisenberg picture where the same equation appears with quantum commutators.

Going back to CM, your result shows, for example, that if a Hamiltonian contains only one angular momentum component, the other two components will not be conserved in time.

Patrick

4. May 9, 2007