The General Solution for a Constant Potential

We have found the general solution of the Schrödinger Equation in a region in which the potential is constant. Assume the potential is equal to \bgroup\color{black}$V_0$\egroup and the total energy is equal to \bgroup\color{black}$E$\egroup. Assume further that we are solving the time independent equation.

\begin{displaymath}\bgroup\color{black} {-\hbar^2\over 2m}{d^2u(x)\over dx^2}+V_0u(x)=Eu(x) \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black} {d^2u(x)\over dx^2}=-{2m(E-V_0)\over\hbar^2}u(x) \egroup\end{displaymath}

For \bgroup\color{black}$E>V_0$\egroup, the general solution is

\begin{displaymath}\bgroup\color{black} u(x)=Ae^{+ikx}+Be^{-ikx} \egroup\end{displaymath}

with \bgroup\color{black}$k=\sqrt{2m(E-V_0)\over\hbar^2}$\egroup positive and real. We could also use the linear combination of the above two solutions.


We should use one set of solutions or the other in a region, not both. There are only two linearly independent solutions.

The solutions are also technically correct for \bgroup\color{black}$E<V_0$\egroup but \bgroup\color{black}$k$\egroup becomes imaginary. For simplicity, lets write the solutions in terms of \bgroup\color{black}$\kappa=\sqrt{2m(V_0-E)\over\hbar^2} $\egroup, which again is real and positive. The general solution is

\begin{displaymath}\bgroup\color{black} u(x)=Ae^{+\kappa x}+Be^{-\kappa x} .\egroup\end{displaymath}

These are not waves at all, but real exponentials. Note that these are solutions for regions where the particle is not allowed classically, due to energy conservation; the total energy is less than the potential energy. Nevertheless, we will need these solutions in Quantum Mechanics.

Jim Branson 2013-04-22