# Poisson bracket significance (Classical Mechanics)

1. May 9, 2007

### deadringer

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.

I hope someone will add more information.

Patrick

4. May 9, 2007

### Physics Monkey

Hi deadringer,

The connection between the Poisson bracket and the quantum commutator is really quite astounding. As you know, the interpretation of [Lx,Ly] = i Lz in quantum theory is that one cannot label physical states by simultaneous eigenvalues of Lx and Ly. The interpretation of {Lx,Ly} = Lz in classical mechanics is that one cannot use both Lx and Ly as canonical variables. In other words, any set of phase space variables containing both Lx and Ly cannot be a set of canonical variables. So we have a sort of correspondence where a maximal set of commuting operators (QM) is related to a canonical set of phase space variables (CM). In both cases there is a notion of the "correct" description of a system. Of course, in the classical case one can still compute the value of the Lx and Ly from a good canonical set of phase space variables, but there is a real sense in which using both Lx and Ly at the same time is the "wrong" way to describe the system. This is one of many possible interpretations of the Poisson bracket. Does this make sense to you?

Last edited: May 9, 2007
5. May 9, 2007

### deadringer

Yes this does make sense, since for a set of canonical coordinates we should get [qi, qj] = zero
Thanks to everybody for helping.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook