Even bound state in every attractive square well ($V_0>0⟹\text{an even bound state exists}$)

For the finite square well $V=-V_0$ on $|x|<a$ and $V=0$ outside, with $a,V_0>0$, there is always an even bound state, however small $V_0$ is. Define $\rho =a\sqrt{2m{V}_0}/\hslash$, the dimensionless decay constant $\alpha =a\sqrt{-2mE}/\hslash$, and interior oscillation constant $\beta =a\sqrt{2m(E+V_0)}/\hslash$. Wavefunction matching at a finite potential step gives $\alpha =\beta \tan \beta$ and ${\alpha }^2+{\beta }^2={\rho }^2$.

On $0<\beta <\pi /2$, these equations reduce to $\beta /\cos \beta =\rho$. The left side is continuous, strictly increasing from zero to infinity, with derivative $\sec \beta (1+\beta \tan \beta )>0$. The intermediate value theorem gives exactly one root on this first branch for every $\rho >0$. It has $0<\beta <\rho$, hence $\alpha >0$ and $-V_0<E<0$. The matched cosine interior and exponential tails therefore give a nonzero even square-integrable function with continuous first derivative, which can be normalized. As the well becomes shallow the binding energy approaches zero but remains negative; see shallow finite-square-well energy.