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Delta Potential Question

  1. Aug 15, 2004 #1

    cepheid

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    Hello,

    While solving the time-independent Schrodinger Equation for a particle in a certain delta potential, I have reached the point where I am trying to satisfy the continuity condition:

    [tex] \psi(a^-) = \psi(a^+) [/tex]

    Which has lead me to:

    [tex] Ce^{\kappa a} + De^{-\kappa a} = Ge^{-\kappa a} [/tex]

    I'm just wondering what (if anything) should I do to simplify this? I have to get a handle on these coefficients in the different regions so that I can normalize later. Note [tex] \inline a>0 [/tex].
     
  2. jcsd
  3. Aug 15, 2004 #2

    Are you sure there is no dependency on x (position) in them exponentials??? Because normally there should be. Then due to the demand that the wave functions are continuous, one states that the relation is valid for x=0 at the boundary of the potentialwell. Then we have C+D=G
    One must do the same for the first derivatives of the wave functions with the same continuity-condition at x=0

    Out of these numbers d,C,... one then calculates the transmission and reflection coefficients

    regards
    marlon
     
  4. Aug 15, 2004 #3
    In order to get rid of those a's, you perform a translation of the well so that a becomes zero


    just an addendum

    marlon
     
  5. Aug 15, 2004 #4

    cepheid

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    It's not a potential well...the potential is given by a dirac delta function at x=a. I'm just trying to simplify that expression so that when I write the overall solution for [tex] \inline \psi (x) [/tex], it's as simple as possible.
     
  6. Aug 15, 2004 #5
    Unfortunately no further simplification can be made to your boundary condition, unless you're willing to perform the translation to a=0. The best you can do is to factor out a term and get something like:

    [tex] Ce^{2 \kappa a} + D = G [/tex]

    Then you can, eg, solve for C in terms of G and D (which is the cleanest way to proceed) and impose your other BC's or normalization equations.
     
  7. Aug 16, 2004 #6
    i know. That is what I meant : inverse potential well or potential barrier

    regards
    marlon

    just perform the translation of the barrier to x=0
     
  8. Aug 16, 2004 #7

    Dr Transport

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    you need two boundary conditions, one for the wave function continuity at x = a and one for the derivative of the wave function at a. To get the continuity equation for the derivative, integrate Schrodingers equation. You should get something like this:

    [tex] \frac{d\Psi(x)}{dx}|_{a+}-\frac{d\Psi(x)}{dx}|_{a-} = \Psi(a) [/tex]

    from there you should be able to find the wave function and then the corresponding energy. also remember that the wave function dies off at infinity, and is exponential in nature, i.e. no factor of [tex] i [/tex].

    dt
     
    Last edited: Aug 16, 2004
  9. Aug 16, 2004 #8

    cepheid

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    ^^^Yes...

    ...that's exactly what dawned upon me today and I am now able to continue with the problem after having integrated Schrodinger's equation. Thank you all for your help!
     
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