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Schrodinger eqn. b.c.

  1. Dec 17, 2009 #1
    When solving Schrodinger's eqn. one comes across the expression:

    [tex]\frac{d^2 \psi}{dx^2}=(V-E)\psi[/tex]

    where the mass has been chosen to make [tex]\frac{\hbar^2}{2m}=1[/tex]

    If V is infinity at some x, then it is said that [tex]\frac{d \psi}{dx} [/tex] can have a finite jump at that x, since [tex]\frac{d^2 \psi}{dx^2}=\infty [/tex]

    But doesn't this assume that [tex]\psi[/tex] doesn't go to zero? Then you would get: [tex] \frac{d^2 \psi}{dx^2}=(V-E)\psi=\infty*0[/tex] and it is not necessarily true that [tex]\frac{d^2 \psi}{dx^2}=\infty [/tex], hence no longer necessarily true that [tex]\frac{d \psi}{dx} [/tex] can have a finite jump.

    The easiest case is the infinite well. The solution at the boundary of the well does have a finite jump in its slope. However, the function [tex]\psi[/tex] itself is zero at the boundaries of the well.

    So in general, if you have a differential equation [tex]y'=f(x)y [/tex], where f(x) is singular at a point [tex]x_0 [/tex], can you assume a finite jump in the solution y at [tex]x=x_0 [/tex]?
  2. jcsd
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