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Special cases of the Schroedinger Eq?

  1. Apr 20, 2010 #1
    According to the lecture I'm hearing the SE can be simplified for the cases of a timeinvariant potential and a constant potential:

    Time invariant

    [itex]\frac{\partial^2\psi}{\partial x^2} = 2\frac m{\hbar^2}(V(x)-E)\psi[/itex].

    Then, the lecture states that for the case of the potential being not just time but also spatially invariant it can be simplified to

    [itex]\frac{\partial^2\psi}{\partial x^2} = -k^2\psi[/itex]

    My question is, why this is said to be possible, only if the potential is constant. Given the term [itex]V(x) - E[/itex] we can conclude that [itex]V(x) - E = E_{kin}[/itex] and hence [itex]\frac12\hbar^2\frac{k^2}m[/itex]

  2. jcsd
  3. Apr 20, 2010 #2
    If the potential is non-constant then the kinetic energy V-E is once again a function of position. Your last result is not a function of position.
  4. Apr 21, 2010 #3
    Call it k(x) Then it is a function of position.

    nvm replying, I got the "idea" and I've to blame it on the lecture that this wasn't clear.
    Last edited: Apr 21, 2010
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