1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
    Last edited: Nov 21, 2004
  2. jcsd
  3. Nov 21, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper

    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
    \frac {d^2 y}{dx^2} + \beta x y = E_0 y

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Dreaded WKB problem
  1. Problems with problems (Replies: 1)

  2. Dreaded Atwood Problem (Replies: 2)

  3. A problem (Replies: 2)