Why does the anisotropic harmonic oscillator has no coupling between different directions? The hamiltonian of the anisotropic HO e.g. in 2d is typically written as 
$$H=\frac{1}{2m}\left(p_x^2+p_y^2\right)+\frac{1}{2}m(\omega_x^2 x^2+\omega_y^2y^2)$$
What I wonder is why there is no coupling between different directions, i.e. a term proportional to ~xy in the potential part of the Hamiltonian. Wouldn't the most general form of e.g. Hook's law be a tensor law where one also has cross terms that couple the different directions? How would a potential of this form look like?
Any help is highly appreciated.
 A: If you have 
$$
V(x,y) = \frac 12(ax^2+2bxy+cy^2), \quad a,c>0,\quad ac-b^2>0
$$
you can rotates the axes $(x,y)\to (x',y')$ so that they line up with the principle diameters  of the ellipse whose equation is 
$$
1= ax^2+2bxy+cy^2. 
$$ (The condition $ac-b^2>0$ ensures that the curve is an ellipse, so the system is stable.)
Then, in the new axes,
$$
V\to \frac 12(\omega_1^2x'^2 +\omega_2^2 y'^2)
$$
for some positive  numbers $\omega_{1,2}^2$,
and the cross term has disappeared. Consequently   nothing of any physical consequnce is gained by including a cross term in the potential. 
A: The Hamiltonian
$$
H=\frac{1}{2m}\left(p_x^2+p_y^2\right)+\frac{1}{2}m(\omega_x^2 x^2+\omega_y^2y^2)
$$
has equations of motion given by the Hamilton equations with the Poisson bracket $\dot p_i~=~\{H,~p_i\}$ or
$$
\dot p_i~=~\frac{\partial H}{\partial p_j}\frac{\partial p_j}{\partial q_i}~-~\frac{\partial H}{\partial q_j}\frac{\partial p_j}{\partial p_i}
$$
$$
=-\frac{\partial H}{\partial x_j}\delta_{ij}~=~-\frac{\partial H}{\partial x_i}.
$$
Here $q_i$ refers to $x$ and $y$ for $i~=~1$ or $2$ and momentum and position are independent with $\frac{\partial p_j}{\partial q_i}~=~0$. Now putting in the potential function $V~=~\frac{1}{2}m(\omega_x^2 x^2$ $+~\omega_y^2y^2)$ gives
$$
\dot p_i~=~-\omega^2_i x_i.
$$
These are then two independent differential equations with no coupling terms such as $q_1q_2$ or $p_1q_2$ and so forth.
