Poisson bracket of the angular momentum and a scalar function

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?

• It appears in one of my class exercises. Indeed, the statement is supposed to be correct, although I also find the part corresponding to the scalar function a bit strange Commented Aug 2, 2021 at 9:44
• I have seen for instance that this statement is used to argue that, in the case of a particle of magnetic moment $\boldsymbol{\mu} \equiv \gamma \boldsymbol{L}$ in a magnetic field $\boldsymbol{B}$, the Poisson Bracket $\left[\boldsymbol{L}, \frac{p^{2}}{2 m}\right]$ is zero. Commented Aug 2, 2021 at 9:53
• It is the square of the linear momentum, $p^2=\boldsymbol{p}·\boldsymbol{p}$ Commented Aug 2, 2021 at 10:39
• Commented Aug 2, 2021 at 11:17
• as it stands the statement is false. the function $f$ must be a scalar function constructed out of $p$, $q$ only. Commented Aug 6, 2021 at 14:01

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.
• So, for every non-vectorial function (say the square of the linear momentum, $p^2$, or a combination with the $q$ coordinate such as $q^3p^5$), we can assure that it will automatically Poisson-Commute with the components of the angular momentum, $L_i$? Commented Aug 3, 2021 at 15:50
• @Invenietis I don't know what "non-vectorial function" means, and I'm sure that if you try to come up with a precise definition, you'll be able to convince yourself that there is no good definition other than "a function that Poisson-commutes with $L$". Can I invite you to try to find a precise definition of "non-vectorial function"? Commented Aug 3, 2021 at 19:55
• (But anyway, more to your question: any function of $q_i,p_i$ that depends on these variables only through the combinations $\sum_i q_i^2, \sum_i p_i^2,\sum_i q_ip_i$, will Poisson-commute with $L$. Any (analytic) function that depends on $q_i,p_i$ in a way that does not involve these combinations only, will not Poisson-commute.) Commented Aug 3, 2021 at 19:56
• If $a\neq 0$ is a constant vector, $f(x,p) = a \cdot p$ is a scalar but it is not invariant under rotations...I understand that you are considerinf functions of $x$ and $p$ only, but in the original question this constraint is not assumed. I think that the original question is ambiguous. Your correct answer is the previous comment actually. Commented Aug 6, 2021 at 15:49