Bound States in a Potential Well *

We will work with the same potential well as in the previous section but assume that \bgroup\color{black}$-V_0<E<0$\egroup, making this a bound state problem. Note that this potential has a Parity symmetry. In the left and right regions the general solution is

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


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

The \bgroup\color{black}$e^{-\kappa x}$\egroup term will not be acceptable at \bgroup\color{black}$-\infty$\egroup and the \bgroup\color{black}$e^{\kappa x}$\egroup term will not be acceptable at \bgroup\color{black}$+\infty$\egroup since they diverge and we could never normalize to one bound particle.

u_1(x)&=&C_1 e^{\kappa x} \\
u_3(x)&=&C_3 e^{-\kappa x} \\

In the center we'll use the sine and cosine solutions anticipating parity eigenstates.

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

Again we will have 4 equations in 4 unknown coefficients.


The calculation shows that either \bgroup\color{black}$A$\egroup or \bgroup\color{black}$B$\egroup must be zero for a solution. This means that the solutions separate into even parity and odd parity states. We could have guessed this from the potential.

The even states have the (quantization) constraint on the energy that

\kappa=\tan(ka)k \\
...E\over E+V_0}=\tan\left(\sqrt{2m(E+V_0)\over\hbar^2}a\right) \\

and the odd states have the constraint

\kappa=-\cot(ka)k \\
\sqrt{-E\over E+V_0}=-\cot\left(\sqrt{2m(E+V_0)\over\hbar^2}a\right) \\

These are transcendental equations, so we will solve them graphically. The plot below compares the square root on the left hand side of the transcendental equations to the tangent on the right for the event states and to ``-cotangent'' on the right for odd states. Where the curves intersect (not including the asymptote), is an allowed energy. There is always one even solution for the 1D potential well. In the graph shown, there are 2 even and one odd solution. The wider and deeper the well, the more solutions.


Try this 1D Potential Applet. It allows you to vary the potential and see the eigenstates.

Jim Branson 2013-04-22