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Dreaded WKB problem

  1. Nov 21, 2004 #1
    Consider the potential
    V(x) = [tex]\beta x[/tex] for x [tex]\geq\ 0[/tex]
    V(x) = 0 for x < 0.

    Find the exact and WKB wavefunction for the situation where a particle has
    E = 10 in units where [tex]\beta = \hbar = m = 1[/tex].

    Any suggestions guys?
    James
     
    Last edited: Nov 21, 2004
  2. jcsd
  3. Nov 21, 2004 #2

    Tide

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    For x > 0 you are looking for approximate solutions of an equation having this form:

    [tex]\frac {d^2 y}{dx^2} + k^2\left(1-\frac {\beta x}{E_0}\right) y = 0[/tex]

    which will be of the form

    [tex]e^{\pm i \frac {2}{3} k\left(1-\frac {\beta x}{E_0}\right)^{\frac {3}{2}}[/tex]

    which is valid for x not close to [itex]E_0 / \beta[/itex].
     
  4. Nov 21, 2004 #3
    Ok, what are you calling k because here is what I would think the Shrodinger equation is
    [tex]
    \frac {d^2 y}{dx^2} + \beta x y = E_0 y
    [/tex]

    And is that the WKB solution?
     
    Last edited: Nov 21, 2004
  5. Nov 22, 2004 #4

    Tide

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    Yes, I abbreviated k and just gave the form of the equation. I think your signs are reversed on the [itex]\beta[/itex] and [itex]E_0[/itex] terms.

    Your equation is actually a variant of the Airy differential equation but the point of your problem is to use the WKB approximation which essentially says that the solution of equations like

    [tex]\frac {d^2 y}{dx^2} + k^2(x) y = 0[/tex]

    are approximately of the form

    [tex]e^{\pm i \int^x k(x') dx'}[/itex]

    I'm being a little sloppy here but you can read more here http://www.du.edu/~jcalvert/phys/wkb.htm (and in lots of other places including your textbook!)
     
  6. Nov 23, 2004 #5
    I thought that this is the form of the SE.....[tex]\frac {d^2 y}{dx^2} + V(x) y = E y[/tex]

    So subbing in [tex]\beta x y[/tex] for the potential, I think the signs are rigt.
    I just don' t see how to get
    [tex]\frac {d^2 y}{dx^2} + k^2\left(1-\frac {\beta x}{E_0}\right) y = 0[/tex] :confused:

    Also what is the k that goes in your integral? Shouldn 't it be
    [tex]\beta x y[/tex] - E ?
     
  7. Nov 23, 2004 #6

    Tide

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    No, you have the signs reversed on the E and V terms in the Schrodinger equation. I used [itex]k^2 = \frac {2m}{\hbar^2}(E - V)[/itex]
     
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