p_1(0) = \left[ \bar{E}_N -\Omega_N (\alpha; \bar{R}, \bar{R}) \right]^{1/2}$. Having calculated classical trajectories with this initial condition, we observe that the classical particle does not reach the first quadrant in Figs. \[Fig\_potentials\](b), \[Fig\_potentials\](d), and \[Fig\_potentials\](f); it is reflected back to the third quadrant. The same conclusion can be reach qualitatively by calculating the force that acts on the classical counterpart, $$\begin{aligned}
F_{1,2}(x_1, x_2) = -\partial \Omega_N (\alpha; x_1, x_2) /\partial x_{1,2}.\end{aligned}$$ Since $$\begin{aligned}
\nonumber
F_1(x,x) =  (2\alpha x^2 -\alpha) e^{-x^2}, \quad
F_2(x,x) = (6x^2 - 3)e^{-x^2},\end{aligned}$$ the classical particle experiences the force that deflects it from moving along the diagonal, $x_1 = x_2$ (which coincides with the axis $\rho=0$), and pushes it toward a knee-like barrier located in the second quadrant (see Fig. \[Fig\_potentials\]); hence, the particle eventually bounces off the barrier back to the third quadrant.

It is noteworthy to mention a peculiarity of numerical calculations. We have found that it is advantageous to employ a (fourth-order) symplectic integrator [@Forest1990; @Yoshida1990; @Candy1991] for solving Hamilton’s equations in this section due to the following reason: A sharp and localized shape of the kneelike barrier leads to an unstable motion of the classical particle. If one employs nonsymplectic integrators (e.g., the 