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Sch. eq. Delta potential

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data
    ##\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-\alpha\delta(x))\psi(x)=0##
    Show that ##\psi(x)## is continuous and that first derivative has discontinuity ##\frac{2m\alpha}{\hbar^2}\psi(0)##.


    2. Relevant equations



    3. The attempt at a solution
    I'm not sure how to show that function ##\psi(x)## is continuous. Or that ##\psi(0^+)=\psi(0^-)##. ##\alpha\delta(x)## is even potential so ##\psi(x)## could be even function or odd function. Right. And because potential in zero is ##\infty##, ##\psi(0)=0##. Question. Could I say that
    ## \int^{\epsilon}_{-\epsilon}\psi(x)dx \leq 2\epsilon \psi(y)##
    where ##\psi(y)## is maximum of the function ##\psi(x)## at ##(-\epsilon,\epsilon)##.
    and that
    ## \int^{\epsilon}_{-\epsilon}\psi(x)dx \geq 0##?
    and why ## \int^{\epsilon}_{-\epsilon}\psi(x)dx \geq 0##?
     
  2. jcsd
  3. May 25, 2014 #2

    vanhees71

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    The trick is to integrate the Schrödinger equation over an interval [itex](-\epsilon,\epsilon)[/itex] and then to consider the limit [itex]\epsilon \rightarrow 0^+[/itex]. Think about what this procedure tells you about [itex]\mathrm{d}\psi/\mathrm{d} x[/itex] around [itex]x=0[/itex]!
     
  4. May 25, 2014 #3
    I know how to solve this if I suppose that ##\psi(0^+)=\psi(0^-)##. My problem is how to solve that
    ##\lim_{\epsilon \to 0}\int^{\epsilon}_{-\epsilon}\psi(x)dx=0##?
    I try to do that in my last post but I'm not sure why
    ##\lim_{\epsilon \to 0}\int^{\epsilon}_{-\epsilon}\psi(x)dx=0## can not be negative!
     
  5. May 26, 2014 #4
    I try to do that in my last post but I'm not sure why
    ##\lim_{\epsilon \to 0}\int^{\epsilon}_{-\epsilon}\psi(x)dx## can not be negative! Thanks in advance.
     
  6. May 27, 2014 #5
    Start with TDSE:

    [tex]\int _{-\epsilon}^{+\epsilon}\frac{d^2\psi}{dx^2}= - \int _{-\epsilon}^{+\epsilon} \frac{2m}{\hbar^2}(E-\alpha\delta(x))\psi(x)[/tex]

    Integrate both sides from ##-\epsilon## to ##+\epsilon##, then take limit as ##\epsilon \rightarrow 0##. (Hint: What happens to the delta function on the right hand side?)?



    (Interesting fact: If you sketch a delta function, you will see that it is symmetrical about x=0, then you see that the gradients at slight less and slightly more than x=0 should have the same magnitude, but opposite signs. This gives the discontinuity.)
     
    Last edited: May 27, 2014
  7. May 27, 2014 #6

    vela

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    You could look at the Dirac delta function as the limit of a sequence of functions that get increasingly taller and narrower.

    This isn't true. If the potential was infinite over some interval, then the wave function would vanish on that interval. However, the potential here is infinite only at one point, which only results in a discontinuity in ##\psi'(x)##.

    Sure.

    No. What if ##\psi(x)<0## on that interval?

     
  8. May 28, 2014 #7
    Well I think if potential is even function as in this case wave funtions could be even or odd. If the wave function is odd then
    ## \int^{\epsilon}_{-\epsilon}\psi(x)dx=0##
    odd function in symmetric boundaries. However, if function is even I could use that ##\psi(x)<0## and ##\psi(x)>0## has the same norm ##|\psi(x)|^2##. And because wave function does not have physical sence if state function is symmetric and negative I could just take ##-\psi(x)## in that interval. Right?
     
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