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Determining bound states for delta function potential

  1. Oct 13, 2014 #1
    I'm working on a problem out of Griffith's Intro to QM 2nd Ed. and it's asking to find the bound states for for the potential ##V(x)=-\alpha[\delta(x+a)+\delta(x-a)]## This is what I'm doing so far:

    $$
    \mbox{for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}\\
    \mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=Be^{-\kappa x}+Ce^{\kappa x}\\
    \mbox{and for $x\gt a$:}\hspace{1cm}\psi=De^{-\kappa a}
    $$

    However, this is what the solution reads:
    $$
    \mbox{for $x\lt a$:}\hspace{1cm}\psi=Ae^{-\kappa a}\\
    \mbox{for $-a\lt x\lt a$}\hspace{1cm}\psi=B(e^{\kappa x}+e^{-\kappa x})\\
    \mbox{and for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}
    $$

    Can someone explain what I'm doing wrong - why they are getting the coefficients they are - and what I should be doing? Maybe go over the general way of approaching these kind of problems? I'm also wondering why they don't have the case for $x>a$?

    Thanks
     
  2. jcsd
  3. Oct 13, 2014 #2

    Nugatory

    User Avatar

    Staff: Mentor

    The missing case for ##x>a## has to be a typo - the first line of the solution has a ##<## where ##>## was intended.

    You will make more headway with the constants if you impose the additional condition that ##\psi## has to be symmetric or anti-symmetric. These constraints are OK because the potential is symmetric and the solution for a symmetric potential can always be written as a linear combination of symmetric and antisymmetric solutions.
     
  4. Oct 13, 2014 #3
    Thanks! That makes it much clearer. I figured it was a typo but was kind of confused to begin with.
     
  5. Oct 14, 2014 #4
    After this I am trying to figure out what ##\Delta\left(\frac{d\psi}{dx}\right)## is. Integrating the potential part of SWE and taking the limit as ##\epsilon## approaches ##\pm a## I get:

    $$
    \Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m}{\hbar ^2}\left[\alpha\psi(a)+\alpha\psi(-a)\right]
    $$
    but the solution reads ##\Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m\alpha}{\hbar^2}\psi(a)##
    I think I'm close but not sure how they got their result. Do you see what I'm doing? Can you correct where I made any mistakes?
     
  6. Oct 14, 2014 #5

    Nugatory

    User Avatar

    Staff: Mentor

    Look for solutions to the Schrodinger equation that work everywhere except at the two points where the potential is not zero. By inspection (also called "lucky guess" or "thoughtful selection of ansatz" or "the way most differential equations get solved") these have to be linear combinations of ##e^{kx}## and ##e^{-kx}##. Your boundary conditions tell you that ##\psi## goes to zero as ##x## goes to infinity in either direction; and you allowed to assume that the solution is either symmetric or antisymmetric. Those conditions are sufficient to determine which linear combinations (that is, what the coefficients are) of ##e^{kx}## and ##e^{-kx}## are solutions in the three regions of interest.

    If find yourself tiring of the PhysicsForums rule that we will help you find the solution yourself but won't just give you the answer... Google will find a bunch of solutions. But I hope you'll keep working at it yourself - you are a differential equation away from victory.
     
    Last edited: Oct 14, 2014
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