How is the Poisson bracket $\{\mathbf{c},\mathbf{l}\cdot\hat{n}\}=(\hat{n}\times \mathbf{c})$, for constant $\mathbf{c}$, and not zero?

The starting point is that the $3$-vector $\vec{\bf c}$ transforms in the $3$-dimensional irreducible vector representation of the rotation group $SO(3)$,

$$\tag{1} \{ \vec{\bf c}, \vec{\bf L}\cdot \hat{\bf n} \}_{PB}~=~ \hat{\bf n}\times \vec{\bf c},$$

where $\hat{\bf n}$ is an arbitrary unit vector, whose Poisson bracket (PB) with anything vanishes

$$\tag{2} \{ \hat{\bf n}, \cdot \}_{PB}~=~0.$$

We assume that $\vec{\bf c}$ is not identically zero. Since the PB with $\vec{\bf c}$ does not vanish, the $3$-vector $\vec{\bf c}$ cannot be a constant. It must be a function of the phase space variables $\vec{\bf r}$ and $\vec{\bf p}$. It can be thought of as being of the form

$$\tag{3} \vec{\bf c}~=~\vec{\bf r}f+ \vec{\bf p}g+ \vec{\bf L}h,$$


$$\tag{4} f~=~f(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2), \quad g~=~g(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2), \quad\text{and}\quad h~=~h(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2), $$ are three suitable functions of the phase space $SO(3)$ scalars

$$\tag{5} r^2,\quad p^2,\quad \vec{\bf r}\cdot\vec{\bf p},\quad\text{and}\quad L^2.$$


  1. H. Goldstein, Classical Mechanics; Section 9-6 in 2nd edition or Section 9.7 in 3rd edition.

Let's just do it for a simple example. By $\vec{c}$ I imagine you mean the location of the particle relative to some origin, so $\vec{c}=\vec{r}$. Later on for simplicity we'll suppose further the particle is located on the x-axis (but it is important to do this only after differentiating as we will see).

We'll also suppose we are rotating around the z axis so that $\hat{n}=\hat{z}$.

Then we have \begin{equation} \{\vec{c} , \vec{l}\cdot\vec{n}\}=\{\vec{r},xp_y - y p_x\} =\{x\hat{x}+y\hat{y}+z\hat{z},xp_y - y p_x\}= -y \hat{x}+x\hat{y}. \end{equation}

Now that we have differentiated (meaning, evaluated the brackets) we can set $y=0$ and $x=R$ (that is, we can suppose our particle started on the $x$-axis at the position $R$). Then \begin{equation} \{R \hat{x},\vec{l}\cdot\hat{z}\}=R \hat{y}=\hat{z}\times(R\hat{x}) \end{equation} which is consistent with your formula.

Incidentally, you might be worried that I started off by setting $\vec{c}=\vec{r}$. I think in the framework you are working in--particle mechanics--the vectors should all start from the same origin. If you want to start taking poisson brackets of vectors with different origins, I think you really need to generalize this discussion to field theory (which will complicate the story a bit because in addition to rotating the direction of the vector you need to rotate the origin, so you will end up with an additional term). So I think that may be what you have in mind but that is a more complicated story.