infinitely annoying square well

[gnome]

Edit: I corrected an error in the "normalizing" (forgot to square the functions). But since I wasn't really using it anyway it doesn't seem to matter.

This square well has an infinite wall at x=0 and a wall of height U at x=L. For the case E < U, obtain solutions to the Schrodinger equation, satisfy the appropriate boundary conditions ...etc,etc... enforce the proper matching conditions at x=L to find an equation of the allowed energies of this system.

Starting with the general solution for region I inside the box (0<x≤L)
\psi_I(x) = A\sin{kx} + B\cos{kx}
where k^2 = \frac{2mE}{\hbar^2}
and we know that the infinite wall forces B=0

and for region II (x>L)
\psi_{II}(x) = De^{\alpha{x}} +Ce^{-\alpha{x}}
where\alpha^2 = \frac{2m(U-E)}{\hbar^2}
and since \int_L^\infty |\psi(x)|^2 dx must be finite, this D=0

So I'm left with
\psi_I(x) = A\sin{kx}
\psi_{II}(x) = Ce^{-\alpha{x}}

The matching conditions give me:
\psi_I(0) = 0 (I already used this to make B=0)
\psi_I(L) = \psi_{II}(L) therefore A\sin{kL} = Ce^{-\alpha{L}}
\frac{d\psi_I}{dx}(x=L) = \frac{d\psi_{II}}{dx}(x=L) therefore kA\cos{kL} = -\alpha{C}e^{-\alpha{L}}

To normalize I did this:
\int_0^\infty |\psi(x)|^2 dx = \int_0^L |\psi_I(x)|^2 dx + \int_L^\infty |\psi_{II}(x)|^2 dx = 1
\int_0^L A^2\sin^2{kx}\; dx + \int_L^\infty C^2e^{-2\alpha{x}} dx = 1
\frac{A^2}{2}\int_0^L (1-\cos{2kx}\; dx + C^2\int_L^\infty e^{-2\alpha{x}} dx = 1
\frac{A^2}{2}\left(L - \frac{\sin{2kL}}{2k}\right) + \frac{C^2}{2\alpha}e^{-2\alpha{L}} = 1
although I'm not sure that added any information. I don't see how, or why, I would use that ugly expression.

So to summarize, I have:
1) A\sin{kL} = Ce^{-\alpha{L}}

2)kA\cos{kL} = -\alpha{C}e^{-\alpha{L}}

3)\frac{A^2}{2}\left(L - \frac{\sin{2kL}}{2k}\right) + \frac{C^2}{2\alpha}e^{-2\alpha{L}} = 1

4) k^2 = \frac{2mE}{\hbar^2}

5)\alpha^2 = \frac{2m(U-E)}{\hbar^2}

It's very easy to divide (1) by (2) to get
\tan{kL} = -\frac{k}{\alpha}
and then using the equations for k and α this becomes

\tan{kL} = - \sqrt{\frac{E}{U-E}}
but I don't know what, if anything, this tells me.

The solution in the book is that allowed energies satisfy:
\frac{kL}{\sin{kL}} = \left[\frac{2mUL^2}{\hbar^2}\right]^{\frac{1}{2}},

which has solutions only if \frac{2mUL^2}{\hbar^2} > 1.

I see why that statement is true, and apparently more useful than my answer. But I don't see how he got that expression in terms of sin kL, and more importantly, I don't see how I would even know to try and find a solution in that form if I didn't already have the published answer.

Any ideas? Do you see any mistakes in what I did? Many thanks.

[gnome]

Well, I just realized that I can get to

(1): \frac{kL}{\sin{kL}} = \left[\frac{2mUL^2}{\hbar^2}\right]^{\frac{1}{2}}

from

\tan{kL} = -\frac{k}{\alpha}

Simply using a right triangle, \tan{kL} = -\frac{k}{\alpha} gives \sin{kL} = -\frac{k}{\sqrt(\alpha^2 + k^2)} which can be rearranged using the expressions for k2 and α2 to give the expression (1) shown in the text.

But I still don't see the usefulness of this result.

It shows that solutions (allowed energies) only exist where
\frac{2mUL^2}{\hbar^2} > 1
but it does nothing to show what those solutions are, does it?

On the other hand, I think one could take the "easy" solution of
\tan{kL} = -\frac{k}{\alpha}
and graph curves of tan(kL) vs kL and
-k/α vs kL on the same set of axes
and find solutions for the actual allowed energies where those curves intersect.

What do you think?