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\begin{document}

\noindent Solution for assignment 19: Effective Potential.\bigskip
\begin{eqnarray} \nonumber
  U_{\rm eff}(r) &=& -\frac{\alpha}{r} + \frac{M^2}{2mr^2}
\end{eqnarray}
with $M$ angular momentum and $m$ reduced mass. 

\noindent (A) Solve now for $r=r_0$:
\begin{eqnarray} \nonumber
  U'_{\rm eff}(r) &=& +\frac{\alpha}{r} - \frac{2M^2}{2mr^3}\ =\ 0\
  \Rightarrow\ \alpha\,r_0 = \frac{M^2}{m}\ \Rightarrow\ r_0 = 
  \frac{M^2}{\alpha\,m}\, .
\end{eqnarray}
(B) The minimum value of the effective potential:
\begin{eqnarray} \nonumber
  U_{\rm eff}^{\min} = U_{\rm eff}(r_0) &=& -\frac{\alpha^2 m}{M^2}
  + \frac{\alpha^2 m}{2M^2}\ =\ - \frac{\alpha^2 m}{2M^2}\,.
\end{eqnarray}
(C) Calculation of the turning points for $E<0$: We are looking 
of solutions of the quadratic equation
\begin{eqnarray} \nonumber
   E\,r^2 + \alpha\,r - \frac{M^2}{2m} &=& 0
   \\ \nonumber
   r^2 + p\,r - q &=& 0~~{\rm with}~~ p\ =\ \frac{\alpha}{E}\,,~~
                q\ =\ - \frac{M^2}{2m\,E}\,,
   \\ \nonumber
   r_{\min,\max}\ =\ - \frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}
   &=& - \frac{\alpha}{2E} \pm 
   \sqrt{\left(\frac{\alpha}{2E}\right)^2+\frac{M^2}{2mE}}\,.~~
\end{eqnarray}
Recall that $E$ is negative. \smallskip

\noindent (D) When do we have real solutions? The upper bound is $E<0$, 
the lower follows from
\begin{eqnarray} \nonumber
   \left(\frac{\alpha}{2E}\right)^2+\frac{M^2}{2mE} &=& 0\
   \Rightarrow\ \frac{\alpha^2}{4E}+\frac{M^2}{2m}\ =\ 0\
   \Rightarrow\ E\ =\ - \frac{\alpha^2m}{2M^2}\ .
\end{eqnarray}
So, we find
\begin{eqnarray} \nonumber
    - \frac{\alpha^2m}{2M^2}\ <\ E\ <\ 0
\end{eqnarray}
and the lower value is the minimum of the effective potential as
calculated under (B).

\end{document}
