case of an infinite square well in which an additional well is digged: it can be proposed as an exercise.

The starting point $V_1=0$ with the model (\[double-well\]) is an one-dimensional square well of depth $V_2$ and radius $R_2$. Its intrinsic spectral properties depends only on the product $R_2^2V_2$. With a value 80, which is realised in the following examples with $R_2=1$ and $V_2=80$, there are six bound states, three even levels and three odd ones. See, e.g., [@bonfim:43] for solving the square well problem.

Odd states in a symmetric double well {#sub:oddsq}
-------------------------------------

Besides a normalisation factor $\sqrt2$, the odd sector is equivalent to the S-wave sector in a central potential $V(r)$. The radial wave function $u(r)$ is thus $u(r)=u_1(r)=\sin(r\sqrt{V_1-k^2})$ for $r<R_1$, and $u(r)=u_2(r)=u_1(R_1) \cos[k'(r-R_1)]+ u'_1(R_1) \sin[k' (r-R_1)]/k'$ if $R_1<r<R_2$ with $k'^2=V_2-k^2$, and suitable changes $\sin\to\sinh$ and $\cos\to\cosh$ if $k^2> V_2$. The eigenenergies can be obtained by matching this intermediate solution $u_2$ to the external solution $u_3(r)=\exp(-k r)$ at $r=R_2$ , i.e., imposing $u_2(R_2)u'_3(R_2)-u'_2(R_2)u_3(R_2)=0$. The calculation involves only elementary trigonometric functions, and the spectrum can be computed easily.

The energy levels as functions of $V_1$ are displayed in Fig. \[OddEvenSq\]. The rearrangement pattern is clearly seen, and is especially 