I have a problem with this finite square potential well:
$$V(x) = \begin{cases} 0, & \mbox{if } x<|a| \\ V_0, & \mbox{if } x>|a| \end{cases}$$
with $V_0>0$.
The TISE of this potential well are:
\begin{cases} \psi''(x)=\frac{2mE}{\hbar^2}\psi(x), & \mbox{if } x<|a| \\ \psi''(x)=\frac{2m(V_0-E)}{\hbar^2}\psi(x), & \mbox{if } x>|a| \end{cases}
Since we want the solutions to be $L^2$ and the potential is symmetric we can consider only the problem with $x>0$ and only the solutions with definite parity so that the even solutions of the TISE for positive $x$ are:
\begin{cases} A\cos(kx), & \mbox{if } 0<x<a \\ Be^{-cx}, & \mbox{if } x>a \end{cases}
and the odd solutions are:
\begin{cases} A\sin(kx), & \mbox{if } 0<x<a \\ Be^{-cx}, & \mbox{if } x>a \end{cases}
with $k=\frac{\sqrt{2mE}}{\hbar}$ and $c=\frac{\sqrt{2m(V_0-E)}}{\hbar}$
Now from the continuity of the wave function and the first derivative for $x=a$ I get the Cauchy problem:
\begin{cases} A\cos(ka)=Be^{-ca}, & \mbox{if } 0<x<a \\ kA\sin(ka)=cBe^{-ca}, & \mbox{if } x>a \end{cases}
for even functions and:
\begin{cases} A\sin(ka)=Be^{-ca}, & \mbox{if } 0<x<a \\ kA\cos(ka)=-cBe^{-ca}, & \mbox{if } x>a \end{cases}
for odd functions.
The main problem at this point is:
Finding $A$ and $B$ and finding the eigenvalues of $H$ of the bound states.
What I don't understand is:
Once I have written the total wave function as combinations of $L^2$ functions (at infinity the wave function so constructed behaves like an exponential),
why do I find the "acceptable" values of $E$ from the continuity of the wave function and its first derivative?
I would have expected to find the constants $A$ and $B$ from Cauchy's problem and the values of $E$ from the definition of a function belonging to the class $L^2$.
Instead, I find the values of $E$ from Cauchy's problem. How are the conditions on the continuity of the wave function and its prime derivative in $x=a$ related to the membership of the wave function in $L^2$?
Best Answer
As you seem to be assuming that there is only an $e^{-cx}$ for $x>a$ and that $c>0$. Thus you have already assumed that you are in $L^2$. This condition on $c$ requires that $E<V_0$ and you have bound states. The pair of equations for the even case, for example, can by simplified by dividing one by the other to get $$ k\tan ka = c $$ which needs to be solved numerically for the allowed $E$'s.
You can satisfy continuity of $\psi'(x)/\psi(x)$ at $x=a$ with any $E$ --- but to do this you will need to include an $C e^{+cx}$ in $x>a$ as well as the decaying $B e^{-cx}$. These enlarged solutions are now not in $L^2$ because they blow up at large $x$. So yes: being in $L^2$ selects the allowed values of $E$.
There are also scattering eigenfunctions with $E>V_0$. These are in the continuous spectrum and any $E>V_0$ is an allowed eigenvalue. The continuous spectrum eigenstates are not in $L^2$ but need to be included if you are to get a complete set of eigenfunctions. The scattering eigenstates will have delta function normalization and so you will need a rigged Hilbert space to accommodate them.