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Finite potential well

  1. Mar 30, 2013 #1
    Lets say we have a finite square well symetric around ##y## axis (picture below).

    screenshot-from-2013-03-.png

    I know how and why general solutions to the second order ODE (stationary Schrödinger equation) are as follows for the regions I, II and III.

    \begin{align}
    \text{I:}& & \psi_{\text{I}}&= Ae^{\kappa x} \\
    \text{III:}& & \psi_{\text{III}}&= Be^{-\kappa x} \\
    \text{II:}& & \psi_{\text{II}}&= C \cos(k x) + D\sin(kx)
    \end{align}

    But now i got to a point where i have to start applying a boundary conditions to get a speciffic solution. So i start with the 1st boundary condition which is ##\psi_{\text{I}}\left(-\frac{d}{2}\right)=\psi_{\text{II}}\left(-\frac{d}{2}\right)## for the left potential shift and ##\psi_{\text{II}}\left(\frac{d}{2}\right)=\psi_{\text{III}}\left(\frac{d}{2}\right)## for the right potential shift. These leave me with a system of 2 equations (one for left and one for right potential shift):

    \begin{align}
    {\scriptsize\text{left potential shift:}}& & Ae^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) - D\sin\left(k \tfrac{d}{2}\right)\\
    {\scriptsize \text{right potential shift:}}& & Be^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) + D\sin\left(k \tfrac{d}{2}\right)\\
    \end{align}

    Question 1:
    From here on authors of most books don't seem to explain much. Most of them only say that we must use ##\boxed{D\!=\!0}## to solve for even solutions and ##\boxed{C\!=\!0}## to solve for *odd solutions*. What is this argument based on?
     
    Last edited: Mar 30, 2013
  2. jcsd
  3. Mar 30, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    How is cos(x) related to cos(-x) and how is sin(x) related to sin(-x)?

    (apologies in advance if I've missed the point of your question)
     
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