In the context of the Hamiltonian mechanics, I am trying to demonstrate the following statement:
For any scalar function $f$, just as the dot product $\boldsymbol{q}·\boldsymbol{p}$, the Poisson Brackets with the components of the angular momentum vanishes: $
\left[L_{i}, f\right] = 0$
My attempt
For the scalar product $\boldsymbol{q}·\boldsymbol{p}$, making use of the expression of the angular momentum in terms of the Levi-Civita symbol, $L_i =\epsilon_{i r s} q_{r} p_{S}$, I can see that the statement is true:
$$
\left[L_{i}, q_{j} p_{j}\right]=\frac{\partial L_{i}}{\partial q_{k}} q_{j} \frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial L_{i}}{\partial p_{k}} p_{j} \frac{\partial q_{j}}{\partial q_{k}}=q_{j} \frac{\partial L_{i}}{\partial q_{j}}-p_{j} \frac{\partial L_{i}}{\partial p_{j}}
$$
$$
=q_{j} \frac{\partial \epsilon_{i r s} q_{r} p_{s}}{\partial q_{j}}-p_{j} \frac{\partial \epsilon_{i r s} q_{r} p_{s}}{\partial p_{j}}=\epsilon_{i j s} q_{j} p_{s}-\epsilon_{i r s} q_{r} p_{s} = 0
$$
However, I cannot find a way to prove this for the case of a general scalar function $f$. How could the statement be reasoned?
Best Answer
In the context of Hamiltonian mechanics, Poisson-commuting with $L_i$ is the definition of being a scalar function.
In general, a tensor $T$ of rank $2j$ is by definition a function that satisfies $$ \{L^i,T^a\}=t^i{}^a{}_b T^b $$ where $t^i$ are the matrices that generate $SO(3)$ in the $2j+1$ dimensional representation. A scalar is $j=0$ so the one-dimensional representation, whose generators vanish, $t^i=0$. So a scalar Poisson-commutes with $L_i$, by very definition of being a scalar.
If you don't like this, you will need to come up with an alternative definition of being a scalar. If you try, you'll convince yourself that all attempts lead to the condition $\{L_i,T\}=0$ one way or another, so there is no good alternative definition, really. The underlying reason is that, scalar means invariant under rotations, and $L_i$ are precisely the generators of rotations, implemented via $\{L_i,\cdot\}$, so being a scalar is quite literally being annihilated by this operator.