interpretation of $\{H,L^2\}$ In Hamiltonian mechanics, we show $\{H,L_z\}=0$, which can be interpreted as the conservation of angular momentum around $Oz$. Following the same idea, how can we interprete $\{H,L^2\}$?
Is the interpretation the same (or only similar) as in quantum mechanics for $[H,L_z]$ and $[H,L^2]$?
 A: Dear Isaac, yes, $\{H,L^2\}=0$ holds for spherically symmetric Hamiltonians and it means that the magnitude of the angular momentum will be conserved in time. More precisely, the squared magnitude of the angular momentum is conserved but classically, it's the same thing. The time derivative of any observable, whether it's composite or not, is given by its Poisson bracket with the Hamiltonian - all these things are just a classical limit of "commutators" (Poisson bracket) with the corresponding "operators" (observables).
If $\{H,L_x\}=\{H,L_y\}=\{H,L_z\}=0$, then one may also prove $\{H,L^2\}=0$. This is simplest to prove in the quantum mechanical language. If $H$ commutes with $L_z$, it also commutes with $L_z^2$, and similarly it may commute with $L_x^2$ and $L_y^2$ - and with the sum of these three terms which is $L^2$.
The interpretation of classical physics and quantum mechanics is always different but the maths of the Poisson brackets directly reflects the quantum commutators. In quantum mechanics, all these things can only be measured probabilistically and so on; this portion of quantum mechanics is universal.
In the case of the angular momentum, it's useful to talk about $L^2$ and $L_z$ in quantum mechanics - and we usually no longer add $L_x$ or $L_y$. The pair $L^2$ and $L_z$ commutes with each other, $[L^2,L_z]=0$, but commutators such as $[L_x,L_z]$ are not zero but rather $-i\hbar L_y$, and so on. 
Also, $L^2$ has eigenvalues which are always a bit greater than the maximum eigenvalue of $L_z^2$: you can't ever rotate a nonzero angular momentum exactly in the $z$-direction. Equivalently, the uncertainty principle always guarantees that $L_x$ and $L_y$ can't be simultaneously zero when $L_z$ is nonzero. 
The eigenvalues of $L^2$ are $l(l+1)\hbar^2$ for integer values of $l$ while the eigenvalues of $L_z$ are integers between $-l$ and $+l$, where the latter $l$ is the same $l$ used in the formula for the eigenvalue of $L^2$.
