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Analytical solutions of S.E. for unusual potential

  1. Mar 7, 2007 #1
    Analytical solutions of S.E. for "unusual" potential

    Hi,

    Given the potential

    V(x)=0, when x<0 (region I)
    V(x)=V_0, when 0<=x<=a, V_0=real constant (region II)
    V(x)=infinite when x>a (region III)

    What would be the general form on the solutions for each region?

    I would think the solution for region I would be on the form A*exp(ikx) rather than A*exp(ikx)+B*exp(-ikx), since the exp(-ikx) term would "blow up" when x -> -infinite ?? But in the problem text I'm asked later "name the coefficient for the right-moving term A, and the left-moving term B", implying that I should use two terms for the solution ... ?
     
  2. jcsd
  3. Mar 7, 2007 #2

    dextercioby

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    Science Advisor
    Homework Helper

    No term would "blow up" in the I region simply because both the sine and the cosine are bounded functions.
     
    Last edited: Mar 8, 2007
  4. Mar 7, 2007 #3

    jtbell

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    Staff: Mentor

    [itex]e^{ikx}[/itex] and [itex]e^{-ikx}[/itex] both oscillate. Their magnitude (absolute value) never gets larger than 1.
     
  5. Mar 7, 2007 #4
    Ofcourse, my mistake.

    So I end up with solutions:

    psi_I(x) = A*exp(-ikx) + B*exp(ikx)
    psi_II(x) = C*exp(-iqx) + D*exp(iqx)
    psi_III(x) = 0

    with boundary conditions

    psi_I(0) = psi_II(0)
    (d/dx)psi_I(0) = (d/dx)psi_II(0)
    psi_II(a) = 0 (loss of continuinity here, right?)

    Which gives the relations:

    A+B=C+D
    k(-A+B)=q(-C+D)
    C*exp(-iqa)+D*exp(iqa)=0

    Somehow I suspect I've missed something. A possibility of eliminating the exponentials in that last formula, or missed a boundary condition, or ... ???
     
  6. Mar 7, 2007 #5

    Mentz114

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    Gold Member

    Devoboy,

    I assume you know that

    exp(ia) = cos(a) + isin(a)

    .
     
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