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Homework Help: Infinitely annoying square well

  1. Apr 30, 2004 #1
    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)
    [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.
     
    Last edited: Apr 30, 2004
  2. jcsd
  3. May 2, 2004 #2
    Well, I just realized that I can get to

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

    from

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

    Simply using a right triangle, [tex]\tan{kL} = -\frac{k}{\alpha}[/tex] gives [tex]\sin{kL} = -\frac{k}{\sqrt(\alpha^2 + k^2)}[/tex] 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
    [tex]\frac{2mUL^2}{\hbar^2} > 1[/tex]
    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
    [tex]\tan{kL} = -\frac{k}{\alpha}[/tex]
    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?
     
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