3D Problems Separable in Cartesian Coordinates

We will now look at the case of potentials that separate in Cartesian coordinates. These will be of the form.

\begin{displaymath}\bgroup\color{black} V(\vec{r})=V_1(x)+V_2(y)+V_3(z) \egroup\end{displaymath}

In this case, we can solve the problem by separation of variables.

H=H_x+H_y+H_z \\
(H_x+H_y+H_z)u(x)v(y)w(z)=Eu(x)v(y)w(z) \\
...\over u(x)}=E-{(H_y+H_z)v(y)w(z)\over v(y)w(z)}&=&\epsilon_x \\

The left hand side of this equation depends only on \bgroup\color{black}$x$\egroup, while the right side depends on \bgroup\color{black}$y$\egroup and \bgroup\color{black}$z$\egroup. In order for the two sides to be equal everywhere, they must both be equal to a constantwhich we call \bgroup\color{black}$\epsilon_x$\egroup.

The \bgroup\color{black}$x$\egroup part of the solution satisfies the equation

\begin{displaymath}\bgroup\color{black} H_xu(x)=\epsilon_xu(x) .\egroup\end{displaymath}

Treating the other components similarly we get

H_yv(y)=\epsilon_yv(y) \\
H_zw(z)=\epsilon_zw(z) \\

and the total energy is

\begin{displaymath}\bgroup\color{black} E=\epsilon_x+\epsilon_y+\epsilon_z .\egroup\end{displaymath}

There are only a few problems which can be worked this way but they are important.

Jim Branson 2013-04-22