Today I was musing over the following problem. Consider a non-relativistic particle confined in an one-dimensional double well potential of the form $V(x)=\kappa(x^2-a^2)^2$ where both $a$ and $\kappa$ are positive constants. As far as I know, it is not possible to analytically determine the position space wavefunction $\psi_0(x)$ corresponding to the exact ground state. However, the reflection symmetry of the Hamiltonian dictates that the ground state wavefunction $\psi_0(x)$ be either an even or an odd function of $x$ (there is no ground state degeneracy in this problem). Suppose the particle is prepared in the ground state at $t=0$. What will happen to the ground state wavefunction as the barrier between the wells (and therefore the Hamiltonian) is adiabatically (slowly) allowed to approach infinity by letting $\kappa\to\infty$ and $a$ fixed?
1 Answer
The behavior is known already. One study that does a numerical treatment of a double well problem is (http://iopscience.iop.org/article/10.1088/0143-0807/33/6/1651). Roughly speaking the ground state has a peak around each minimum of the potential. They deal with three approximations, square double-well, double well pasted with a cusp, and the "Sombrero" one you ask. They all share the same qualitative behaviour for the ground-state. Making $\kappa$ large will just change dimensions of the peaks. Infinity per se is unphysical, since you must have a potential which is defined in the region in which you intend to solve the Schrödinger equation. Making $\kappa$ infinity is equivalent to isolating the two wells completely.
P.D. I believe in one dimension the ground-states have to be even. (See Ground_state_has_no_nodes_in_one-dimension)
