pronounced if $R_1\ll R_2$. The difference from the Coulomb case is that, for the square well, when a bound state collapses from the “atomic” to the “nuclear” energy range, a new state is created from the continuum.

Even states in a double well {#sub:evensq}
----------------------------

The even spectrum of the potential (\[double-well\]) is given by $w(x)=w_1(x)=\cos(x\sqrt{V_1-k^2})$ for $0\le x<R_1$, and $w(x)=w_2(x)=w_1(R_1) \cos[k'(x-R_1)]+ w'_1(R_1) \sin[k' (x-R_1)]/k'$ if $R_1<x<R_2$ with $k'^2=V_2-k^2$, and suitable changes $\sin\to\sinh$ and $\cos\to\cosh$ if $k^2> V_2$. Then the matching to $w(x)=w_3(x)=\exp(- k x)$ gives the eigenenergies.

The results are shown in Fig. \[OddEvenSq\], with the same parameters as for the odd part. The same pattern of “plateaux” is seen as for the odd parts, with, however, some noticeable differences:

-   In quantum mechanics with space dimension $d=1$ (actually for any $d\le2$), any attractive potential supports at least one bound state. In particular, a nuclear state develops in the narrow potential of width $2R_1$ even for arbitrarily small values of its depth $V_1$. Hence the ground-state level starts immediately falling down as $V_1$ increases from zero,

-   The first even excitation does not stabilise near the value of the unperturbed even ground state, it reaches a plateau corresponding to 