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.(adsbygoogle = window.adsbygoogle || []).push({});

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)

[tex]\psi_I(x) = A\sin{kx} + B\cos{kx}[/tex]

where [tex] k^2 = \frac{2mE}{\hbar^2}[/tex]

and we know that the infinite wall forces B=0

and for region II (x>L)

[tex]\psi_{II}(x) = De^{\alpha{x}} +Ce^{-\alpha{x}} [/tex]

where[tex]\alpha^2 = \frac{2m(U-E)}{\hbar^2}[/tex]

and since [tex]\int_L^\infty |\psi(x)|^2 dx[/tex] must be finite, this D=0

So I'm left with

[tex]\psi_I(x) = A\sin{kx}[/tex]

[tex]\psi_{II}(x) = Ce^{-\alpha{x}} [/tex]

The matching conditions give me:

[tex]\psi_I(0) = 0[/tex] (I already used this to make B=0)

[tex]\psi_I(L) = \psi_{II}(L) [/tex] therefore [tex] A\sin{kL} = Ce^{-\alpha{L}}[/tex]

[tex]\frac{d\psi_I}{dx}(x=L) = \frac{d\psi_{II}}{dx}(x=L)[/tex] therefore [tex]kA\cos{kL} = -\alpha{C}e^{-\alpha{L}}[/tex]

To normalize I did this:

[tex]\int_0^\infty |\psi(x)|^2 dx = \int_0^L |\psi_I(x)|^2 dx + \int_L^\infty |\psi_{II}(x)|^2 dx = 1[/tex]

[tex]\int_0^L A^2\sin^2{kx}\; dx + \int_L^\infty C^2e^{-2\alpha{x}} dx = 1[/tex]

[tex]\frac{A^2}{2}\int_0^L (1-\cos{2kx}\; dx + C^2\int_L^\infty e^{-2\alpha{x}} dx = 1[/tex]

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

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)[tex] A\sin{kL} = Ce^{-\alpha{L}}[/tex]

2)[tex]kA\cos{kL} = -\alpha{C}e^{-\alpha{L}}[/tex]

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

4)[tex] k^2 = \frac{2mE}{\hbar^2}[/tex]

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

It's very easy to divide (1) by (2) to get

[tex]\tan{kL} = -\frac{k}{\alpha}[/tex]

and then using the equations for k and α this becomes

[tex]\tan{kL} = - \sqrt{\frac{E}{U-E}}[/tex]

but I don't know what, if anything, this tells me.

The solution in the book is that allowed energies satisfy:

[tex]\frac{kL}{\sin{kL}} = \left[\frac{2mUL^2}{\hbar^2}\right]^{\frac{1}{2}}[/tex],

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

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.

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# Homework Help: Infinitely annoying square well

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