Consider the Hamiltonian H given by$$H=(x,y,z,p_x,p_y,p_z)= \frac{p^2_x}{2m}+\frac{p^2_y}{2m}+\frac{p^2_z}{2m}-\frac{1}{\sqrt{x^2+y^2+z^2}}$$ where x(t), y(t) and z(t) give the location of a particle of constant mass m in Cartesian coordinates. The components of the particle’s angular momentum vector are
$L_x=yp_z-zp_y$, $L_y=zp_x-xp_z$, $L_z=xp_y-yp_x$
Given a function A where A(x, y, z,$p_x, p_y, p_z$)=$p_xL_y - p_yL_x$-c($\frac{z}{\sqrt{x^2+y^2+z^2}})$where c is a constant. Calculate the Poisson bracket {A, H}.
Is there a value of c for which A is a constant of motion?
I have been given that {$L_x,H$}=$0$ and I have calculated that {$L_y,H$}=$0$
Please help in any way you can, added tags are appreciated as well as help with the answers. Sorry there is not more working out from me I have done my best to progress the question but have hit a bit of a roadblock.
Let's calculate $\{p_xL_y,H\}$ first:
$$\{p_xL_y,H\} = p_x\{L_y,H\} + \{p_x,H\}L_y = \{p_x,H\}L_y$$
by Leibniz's rule and conservation of angular momentum which you already noted in your question.
Then,
$$ \{p_x,H\} = -\left\{p_x,\frac{1}{\sqrt{x^2+y^2+z^2}}\right\}$$
because $p_x$ commutes with all momenta. And finally,
$$ \{p_x,H\} = \frac{\partial}{\partial x}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right) = -\frac{x}{r^{3}}$$
Where I introduced $r = \sqrt{x^2+y^2+z^2}$. Likewise, by symmetry you can find out that
$$ \{p_y,H\} = -\frac{y}{r^{3}}$$
Then, we need to find the Poisson bracket $\{z/r,H\}$. Again, this reduces to the commutators with the squared momenta, because the other term gives zero. Then we have
$$\{z/r,p_x^2\} = 2p_x\{z/r,p_x\} = 2p_x\frac{\partial}{\partial x}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) = -\frac{2 x z p_x}{r^{3}}$$
and likewise
$$\{z/r,p_y^2\} = -\frac{2 y z p_y}{r^{3}}$$
however,
$$\{z/r,p_z^2\} = 2p_z\{z/r,p_z\} = 2p_z\frac{\partial}{\partial z}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) = \frac{2 p_z}{r} - \frac{2 z^2 p_z}{r^{3}}$$
Now, it's just a matter of putting everything together. Which takes some work and I'll leave to you. However, you should find out that for $c=m$, you have a Poisson bracket equal to zero. This function A is the z-component of a vector known as the Hamilton vector.