The Delta Function Potential *

Take a simple, attractive delta function potential and look for the bound states.

\begin{displaymath}\bgroup\color{black} V(x)=-aV_0\delta(x)\egroup\end{displaymath}

These will have energy less than zero so the solutions are

\begin{displaymath}\bgroup\color{black} \psi(x)=\left\{ \matrix{Ae^{\kappa x} & x<0 \cr Ae^{-\kappa x} & x>0\cr} \right. \egroup\end{displaymath}


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

There are only two regions, above and below the delta function. We don't need to worry about the one point at \bgroup\color{black}$x=0$\egroup - the two solutions will match there. We have already made the wave function continuous at \bgroup\color{black}$x=0$\egroup by using the same coefficient, \bgroup\color{black}$A$\egroup, for the solution in both regions.


We now need to meet the boundary condition on the first derivative at \bgroup\color{black}$x=0$\egroup. Recall that the delta function causes a known discontinuity in the first derivative.

\left.{d\psi\over dx}\right\vert _{+\epsilon}-\left.{d\psi\ove...
...kappa = -{2maV_0\over\hbar^2} \\
\kappa={maV_0\over\hbar^2} \\

Putting in the formula for \bgroup\color{black}$\kappa$\egroup in terms of the energy.

{-2mE\over\hbar^2}={m^2a^2V_0^2\over\hbar^4} \\
E=-{ma^2V_0^2\over 2\hbar^2} \\

There is only one energy for which we can satisfy the boundary conditions. There is only one bound state in an attractive delta function potential.

Jim Branson 2013-04-22