in tunneling appear as the result of nonzero coupling of the two degrees of freedom.

Before we describe the choice of $U_N(\rho)$ in our model, let us introduce the center of mass ($R$) and relative ($\rho$) coordinates $$\begin{aligned}
\label{SimpleCMandRelativeCoord}
R = (x_1 + x_2)/2, \qquad \rho = x_2 -x_1.\end{aligned}$$ The Hamiltonian (\[HamiltonianCartesianDef\]) in these new coordinates reads $$\begin{aligned}
\label{Hamiltonian_NewCoord}
\hat{H}_N(\alpha) &=& \frac{-1}{2M}\frac{\partial^2}{\partial R^2} + \frac{-1}{2\mu}\frac{\partial^2}{\partial \rho^2} +  \Omega_N (\alpha; \rho, R), \nonumber\\
\Omega_N (\alpha; R, \rho) &=& \alpha V\left( R - \rho/2\right) + 3V\left( R +  \rho/2\right) \nonumber\\
&&+ U_N(\rho), \label{PotentialDeff}\end{aligned}$$ where $\mu = 1/2$ and $M = 2$.

Now we can specify the potential that couples the two degrees of freedom and see its role in the problem. If $U_N(\rho)$ is attractive, as it is in our calculations, it may support bound states. These bound states, and their symmetries, play a key role.

Here, we set $U_N$ to describe a short-range attraction, $$\begin{aligned}
\label{Un_potentials}
U_N (\rho) = -A\exp\left( -\rho^2/r_N^2 \right).\end{aligned}$$ Varying the parameter $r_N$, we change the number of bound states supported by the attracting potential. In the calculations, we use $A=2$ and $r_1 = 1$, $r_2 = 1.961$, $r_4 =3.162$ corresponding to one, two, and four bound states 