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Finite Square Well Solutions

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Already defined that for a 1D well with one finite wall the eigenvalue solutions are given by
    k cot(kl) = -α

    Show the eigenvalue solutions to well with both walls finite is given by

    tan(kl) = 2αk / (k^2 - α^2)

    Well is width L (goes from 0 to L) with height V_0

    2. Relevant equations

    k cot(kl) = -α

    tan(kl) = 2αk / (k^2 - α^2)

    Time Independent Schrödinger Equation.

    3. The attempt at a solution

    I will explain briefly as there is a lot of equations and it will look a mess, I can always upload a picture of my work if needs be.

    General Solution:
    u(x) = Ce^-αx + De^αx

    For x < 0
    C = 0, so exponential doesn't go to ∞.
    For x > L
    D = 0 so exponential doesn't go to ∞.

    Using boundary conditions where the function and it's derivative must be continuous
    With the solution between 0 and L being

    u(x) = A sin(kx)

    Substituting in L for x

    A sin(kL) = Ce^-αL
    Ak cos(kL) = -αCe^-αL

    Dividing the second by the first

    k cot(kL) = -α (As given in question)

    This breaks down when substituting in 0 for x.

    A sin(k0) = De^α0
    Ak cos(k0) = αDe^α0
    0 = αD

    I thought about changing the coordinates so the well runs from -L/2 to L/2, but that just appears to give

    k cot(kL/2) = -α

    And

    k cot(kl/2) = α

    I was/am looking in one text book, and that manages to get it down to

    k tan (kL/2)= α
    k cot (kL/2)= -α

    Still not what I need, but might be closer?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2012 #2
    Guess nobody can help then. :(
     
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