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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?

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  • $\begingroup$ 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 $\endgroup$
    – Invenietis
    Aug 2 at 9:44
  • $\begingroup$ 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. $\endgroup$
    – Invenietis
    Aug 2 at 9:53
  • $\begingroup$ It is the square of the linear momentum, $p^2=\boldsymbol{p}·\boldsymbol{p}$ $\endgroup$
    – Invenietis
    Aug 2 at 10:39
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Invenietis
    Aug 2 at 11:17
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    $\begingroup$ as it stands the statement is false. the function $f$ must be a scalar function constructed out of $p$, $q$ only. $\endgroup$ Aug 6 at 14:01
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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.

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  • $\begingroup$ 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, $L_i$? $\endgroup$
    – Invenietis
    Aug 3 at 15:50
  • $\begingroup$ @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"? $\endgroup$ Aug 3 at 19:55
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    $\begingroup$ (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.) $\endgroup$ Aug 3 at 19:56
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    $\begingroup$ 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. $\endgroup$ Aug 6 at 15:49
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    $\begingroup$ yes, if interpreting in that way the term "scalar", the answer is almost a tautology. $\endgroup$ Aug 8 at 5:52

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