Fall 2007 15 Qm

A particle of mass m moves in a spherically symmetric potential well V(r) = -V0 < 0 at r < a and V(r) = 0 at r > a. Find the smallest V0 at which a bound state exists at zero angular momentum.




The Schrodinger equation for zero angular momentum reads

\begin{align} -\frac{\hbar^2}{2m}\left[\psi''(r)+\frac{2}{r}\psi'(r) \right] = E\psi, \;\; r>a, \end{align}
\begin{align} -\frac{\hbar^2}{2m}\left[\psi''(r)+\frac{2}{r}\psi'(r) \right] = (V_0+E)\psi, \;\; r<a, \end{align}

The sought solution for E < 0 is

\begin{align} \psi(r)=A \exp{(-\alpha r)}/r, \;\; \alpha= \sqrt{-2mE}/\hbar, \;\; r>a, \end{align}
\begin{align} \psi(r)=B \sin{(-\beta r)}/r, \;\; \beta= \sqrt{-2m(V_0+E)}/\hbar, \;\; r<a, \end{align}

The continuity of $\psi'(r)/\psi(r)$ at r = a demands

As $E \rightarrow 0^{-},\; \alpha\rightarrow 0^{+}$, and so ,$\beta a\rightarrow \pi/2$. Thus,

\begin{align} \mbox{min}V_0 = \frac{\pi^2\hbar^2}{8 m a^2}. \end{align}
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