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 Schrödinger 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?

 

[M98Ranger]

First of all, I appreciate your dedication to correcting your mistake and seeking to understand the solution in the book. It shows a great level of commitment and determination in your studies.

As for your question, I believe the key to understanding the solution in the book lies in the boundary conditions and the concept of standing waves. Let me try to explain it in simpler terms.

In a square well potential, the particle can only exist within the bounds of the well, meaning between x=0 and x=L. This is represented by the first term in your general solution, A sin(kx), as it satisfies the boundary conditions at x=0 and x=L. However, we also know that the particle cannot penetrate the infinite wall at x=0, which is why we set B=0.

Now, for the particle to exist in the well, its wavefunction must also be continuous at x=L. This is where the concept of standing waves comes in. Standing waves are formed when two waves with the same frequency and amplitude travel in opposite directions and interfere with each other. This results in a wave that appears to be standing still. In the case of the square well, the wave traveling from left to right (represented by A sin(kx)) interferes with the wave traveling from right to left (represented by C e^(-αx)), creating a standing wave.

To find the allowed energies, we need to find the values of k and α that satisfy the boundary conditions and result in a standing wave. This is where the equation in the book comes in. It essentially represents the condition for a standing wave to exist, where the wavelength of the wave (kL) is equal to the circumference of the well (2L). This results in a relationship between k and L, and when substituted into the equations for k and α, gives us the allowed energies.

So to answer your question, I believe the key to understanding the solution lies in understanding the concept of standing waves and how they relate to the boundary conditions in a square well potential. I hope this helps clarify things for you. Keep up the good work!