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Schrödinger equation where V = |x|

  1. Oct 15, 2008 #1
    Hi there!

    I'm looking for the solutions of the stationary Schrödinger equation for a potential of the type

    [tex]V = |x|[/tex]

    I know that the Airy functions are the solutions to the SE where [itex]V \sim x[/itex] but for the above mentioned potential ... I can't find it -- neither in books nor on the net. Do you have some hints?

    Best,
    Dave
     
  2. jcsd
  3. Oct 15, 2008 #2

    malawi_glenn

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    I can Imagine it to be really hard to find analytical solutions to that one since |x| is not continous at x = 0.

    I would personally have done numerical solutions.

    Can you tells us more why you are looking for solutions, what is the orignal problem etc, maybe someone can help you more.
     
  4. Oct 15, 2008 #3

    Vanadium 50

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    Well, it is, but it's derivative is not.

    Odd solutions wouldn't be so bad, but even ones might be a bit tricky to handle, since there's no second derivative at that point.
     
  5. Oct 15, 2008 #4
    Actually, I read that some of our professors ask this question in the theoretical physics exam and I've begun to wonder when I didn't find this problem in the literature.
     
  6. Oct 15, 2008 #5

    malawi_glenn

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    Oh that was not good news =/
     
  7. Oct 15, 2008 #6
    i must admit i am not aware of the properties of airy , but have you tried spliting the line when x is greater than or equal to zero and when x is less than zero and then solving the TISE in both part seperatly (with equal energies). Finally as you require that the wavefunction and its first derivative must be continuous at zero you will have to use the properties of the two airy functions at zero to either get quatised energy ar some other property.
     
  8. Oct 15, 2008 #7

    Avodyne

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    For [itex]x>0[/itex], after some rescaling of [itex]x[/itex] and [itex]E[/itex], the time-indpendent Schrodinger equation is
    [tex]-\psi''(x) + x\psi(x) = E\psi(x),[/tex]
    and the solution that does not blow up as [itex]x\to +\infty[/itex] is
    [tex]\psi(x) = {\rm Ai}(x{-}E),[/tex]
    up to overall normalization. (See http://en.wikipedia.org/wiki/Airy_function for Airy function info.) Then, since [itex]V(x)[/itex] is even, [itex]\psi(x)[/itex] must be even or odd. If odd, [itex]\psi(0)=0[/itex], and so [itex]-E[/itex] must be a zero of [itex]{\rm Ai}(x)[/itex]. This is the energy-eigenvalue condition for odd eigenfunctions. If even, [itex]\psi'(0)=0[/itex], and so [itex]-E[/itex] must be a zero of [itex]{\rm Ai}'(x)[/itex]. This is the energy-eigenvalue condition for even eigenfunctions.
     
  9. Oct 15, 2008 #8

    olgranpappy

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    The literature? I'd try looking around throught some textbooks. For example, I'm almost absolutely positive that you will find the "quantum bouncing ball" potential problem in various textbooks (the quantum bouncing ball is the same potential you are interested in for x>0 but infinite at x=<0)... try Grifffiths maybe?
     
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