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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.

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2 Answers 2

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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.

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  • $\begingroup$ @Lawrence B. Crowell Or, thinking in the reverse: in original Hamiltonian rotate axes $(x,y)$ by an arbitrary angle. In new coordinates $(x',y')$ a cross term $x'y'$ will appear, with no change in physics. $\endgroup$
    – Elio Fabri
    Commented Nov 28, 2018 at 15:46
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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.

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