Bound States of a 1D Potential Well *

In the two outer regions we have solutions

u_1(x)=C_1 e^{\kappa x} \\
u_3(x)=C_3 e^{-\kappa x} \\
\kappa=\sqrt{-2mE\over\hbar^2} .\\

In the center we have the same solution as before.

u_2(x)=A\cos(kx)+B\sin(kx) \\
k=\sqrt{2m(E+V_0)\over\hbar^2} \\

(Note that we have switched from \bgroup\color{black}$k'$\egroup to \bgroup\color{black}$k$\egroup for economy.) We will have 4 equations in 4 unknown coefficients.

At \bgroup\color{black}$-a$\egroup we get

C_1e^{-\kappa a}=A\cos(ka)-B\sin(ka) \\
\kappa C_1 e^{-\kappa a}=kA\sin(ka)+kB\cos(ka) .\\

At \bgroup\color{black}$a$\egroup we get

C_3e^{-\kappa a}=A\cos(ka)+B\sin(ka) \\
-\kappa C_3 e^{-\kappa a}=-kA\sin(ka)+kB\cos(ka) .\\

Divide these two pairs of equations to get two expressions for \bgroup\color{black}$\kappa$\egroup.

\kappa&=&{kA\sin(ka)+kB\cos(ka)\over A\cos(ka)-B\sin(ka)} \\
-\kappa&=&{-kA\sin(ka)+kB\cos(ka)\over A\cos(ka)+B\sin(ka)} \\

Factoring out the \bgroup\color{black}$k$\egroup, we have two expressions for the same quantity.

{\kappa\over k}={A\sin(ka)+B\cos(ka)\over A\cos(ka)-B\sin(ka)}...
...kappa\over k}={A\sin(ka)-B\cos(ka)\over A\cos(ka)+B\sin(ka)} \\

If we equate the two expressions,

\begin{displaymath}\bgroup\color{black} {A\sin(ka)+B\cos(ka)\over A\cos(ka)-B\sin(ka)}
={A\sin(ka)-B\cos(ka)\over A\cos(ka)+B\sin(ka)} \egroup\end{displaymath}

and cross multiply, we have


The \bgroup\color{black}$A^2$\egroup and \bgroup\color{black}$B^2$\egroup terms show up on both sides of the equation and cancel. What's left is

AB(\sin^2(ka)+\cos^2(ka))&=&AB(-\cos^2(ka)-\sin^2(ka)) \\
AB&=&-AB \\

Either \bgroup\color{black}$A$\egroup or \bgroup\color{black}$B$\egroup, but not both, must be zero. We have parity eigenstates, again, derived from the solutions and boundary conditions.

This means that the states separate into even parity and odd parity states. We could have guessed this from the potential.

Now lets use one equation.

\begin{displaymath}\bgroup\color{black} \kappa={A\sin(ka)+B\cos(ka)\over A\cos(ka)-B\sin(ka)}k \egroup\end{displaymath}

k If we set \bgroup\color{black}$B=0$\egroup, the even states have the constraint on the energy that

\begin{displaymath}\bgroup\color{black} \kappa=\tan(ka)k \egroup\end{displaymath}

and, if we set \bgroup\color{black}$A=0$\egroup, the odd states have the constraint

\begin{displaymath}\bgroup\color{black} \kappa=-\cot(ka)k .\egroup\end{displaymath}

Jim Branson 2013-04-22