# Infinitely annoying square well

1. Apr 30, 2004

### 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.

Last edited: Apr 30, 2004
2. May 2, 2004

### 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?