= Even bound state in every attractive square well
{title2=$V_0>0\Longrightarrow\text{an even bound state exists}$}
= Existence of a bound state in a one-dimensional square well
{synonym}
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 $\rho=a\sqrt{2mV_0}/\hbar$, the dimensionless decay constant $\alpha=a\sqrt{-2mE}/\hbar$, and interior oscillation constant $\beta=a\sqrt{2m(E+V_0)}/\hbar$. <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>.
Back to article page
