1D Model of a Molecule Derivation *

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

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

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

Continuity of $\psi$ gives.

$$\begin{eqnarray*} e^{-\kappa d}=A\left(e^{\kappa d}+e^{-\kappa d}\right) \\ A={e^{-\kappa d}\over e^{\kappa d}+e^{-\kappa d}} \\ \end{eqnarray*}$$

The discontinuity in the first derivative of $\psi$ at $x=d$ is

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

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