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Delta force

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data
    In delta potential barrier problem Schrodinger equation we get

    [tex]\psi(x)=Ae^{kx}, x<0[/tex]

    [tex]\psi(x)=Ae^{-kx}, x>0[/tex]

    We must get solution of

    [tex]lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx[/tex]




    2. Relevant equations



    3. The attempt at a solution

    [tex]lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \frac{d\psi}{dx}|^{\epsilon}_{-\epsilon}[/tex] and get the solution

    I can say that the whole function is

    [tex]\psi(x)=Ae^{-k|x|}[/tex]

    I don't have first derivative in 0.

    [tex]lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=0[/tex]


    Why I don't get same solution different then zero like in case

    [tex]lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \frac{d\psi}{dx}|^{\epsilon}_{-\epsilon}[/tex]

    ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 28, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    First, I'd like to point out how disappointed I was when I clicked on this thread and found that it wasn't a Chuck Norris movie:biggrin:


    No you cant. The wavefunction [tex]\psi(x)=\left\{\begin{array}{lr}Ae^{kx}, & x<0 \\ Ae^{-kx}, & x>0\end{array}\right.[/itex] is undefined at [itex]x=0[/itex] (as it should be for a delta function potential). The wavefunction [itex]\psi(x)=Ae^{-k|x|}[/itex] is defined at [itex]x=0[/itex]; the two wavefunctions are not equivalent.

    I don't see how you are getting zero for that limit. Show the rest of your steps.
     
    Last edited: Apr 28, 2010
  4. Apr 29, 2010 #3
    [tex]
    lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=0
    [/tex]

    [tex]lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0}\frac{d\psi}{dx}|^{0}_{-\epsilon}+lim_{\epsilon \rightarrow 0}\frac{d\psi}{dx}|^{\epsilon}_{0}
    [/tex]

    [tex]\frac{d\psi}{dx}=kAe^{kx}[/tex] for [tex]x<0[/tex]

    [tex]\frac{d\psi}{dx}=-kAe^{-kx}[/tex] for [tex]x>0[/tex]

    [tex]lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=
    kA-lim_{\epsilon \rightarrow 0}kAe^{k\epsilon}-lim_{\epsilon \rightarrow 0}kAe^{-k\epsilon}+kA=2kA-2kA=0[/tex]
     
  5. Apr 29, 2010 #4
    Double check your exponentials (hint hint) and maybe expand them out ignoring terms higher than [itex]\epsilon^1[/itex]
     
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