1D Model of a Molecule Derivation *

\begin{displaymath}\bgroup\color{black} \psi(x)=\left\{ \matrix{e^{\kappa x} & x...
...x}\right) & -d<x<d\cr
e^{-\kappa x} & x>d\cr} \right. \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black} \kappa=\sqrt{-2mE\over\hbar^2} .\egroup\end{displaymath}

Since the solution is designed to be symmetric about \bgroup\color{black}$x=0$\egroup, the boundary conditions at \bgroup\color{black}$-d$\egroup are the same as at \bgroup\color{black}$d$\egroup. The boundary conditions determine the constant \bgroup\color{black}$A$\egroup and constrain \bgroup\color{black}$\kappa$\egroup.

Continuity of \bgroup\color{black}$\psi$\egroup gives.

e^{-\kappa d}=A\left(e^{\kappa d}+e^{-\kappa d}\right) \\
A={e^{-\kappa d}\over e^{\kappa d}+e^{-\kappa d}} \\

The discontinuity in the first derivative of \bgroup\color{black}$\psi$\egroup at \bgroup\color{black}$x=d$\egroup is

-\kappa e^{-\kappa d} - A\kappa \left(e^{\kappa d}-e^{-\kappa ...
...-\kappa d}} \\
{2maV_0\over\kappa\hbar^2}=1+\tanh(\kappa d) \\

We'll need to study this transcendental equation to see what the allowed energies are.

Jim Branson 2013-04-22