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How to get the results like that in Finite Square Barrier?

  1. Sep 28, 2016 #1
    [this thread was moved from the Quantum Physics subforum, hence no template]
    In this page :
    http://www.physicspages.com/2012/08/06/finite-square-barrier-scattering/

    When the E<V
    The boundary condition tells us the equation (9) (10) and (11) (12).
    I tried to get the results from those equation ,but still cannot get the results like (13)(14)(15)(16)(17).
    I being watching a lot video about the Finite Square Barrier, they always skip the solving,only show the final results of this 4 unknown constants.....
    Still don't know how to solving those unknown constants like that.

    If any one can show me how to solving or some of the crucial step in the solving,I will really appreciate that !
     
    Last edited by a moderator: Sep 28, 2016
  2. jcsd
  3. Sep 28, 2016 #2

    jtbell

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    Show us how you tried to do it, and someone here can help you find your mistake(s) and continue towards the solution.
     
  4. Sep 28, 2016 #3
    In the boundary conditions x=-a ,
    IMG_1475115225.157844.jpg
    But if i try to set the D=0 ,like the Step Potential ,and going to solving what B is , it is the same result with the Step Potential Problem.Seem to be not right,when E<V the particle still have probability tunneling through the barrier.

    If I not setting D equal to 0,keep that in equation, I always cannot solving the unknown constant.
     
  5. Sep 29, 2016 #4

    vela

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    It looks like you didn't understand why you set D=0 in the step-potential problem. That reasoning doesn't apply to this problem.

    Of course you can't solve it. You have two equations and four unknowns. At best, you could solve for two of the constants in terms of the other two. You need more equations.
     
  6. Sep 29, 2016 #5
    Yes , there have another two equation,but for the different boundary conditions x=a.
    IMG_1475201454.381195.jpg
    Seems putting four equation together in different boundary conditions its not right....

    And each of this results have same denominator with A.
    IMG_1475202038.287318.jpg
    Look like those results solved with out considered the different of the boundary conditions.

    Do I really need putting different boundary conditions equation together then try to solving it?
     
  7. Sep 30, 2016 #6

    vela

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    Why? The wave function is going to depend on the width of the barrier, right? It's easier to get past a narrow barrier than a wide one. The reflection coefficient depends on the ratio B/A. How can B/A depend on the width of the barrier if it's determined solely by the boundary conditions at x=-a?

    I don't see how you're concluding that.

    Yes.
     
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