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Schroedinger boundry equation: Smooth?

  1. Apr 29, 2012 #1
    I tried to solve a time independent schroedinger equation with a finite potential well today by solving it in 3 pieces, one for in the box and 2 for the outsides. By setting the equations equal to each other where they met at the edges of the box, by setting the integral of everything squared = 1, and by setting the integral of the exponential functions to the right and left equal to each other (I figured it made sense for them to by symmetrical), I was able to write all the constants equal to 1 other constant, which I could set equal to a function involving energy. I don't like this because it looks as though the constants and energy could vary infinitely, but energy is supposed to be quantized. It seems intuitive to make it a smooth function, but I don't ever remember being taught that one is supposed to do that. I tried making it smooth and it looks as though I get convenient answers.

    1. Is it appropriate to solve the Schrodinger equation in pieces like this?

    2. Is it good to make it smooth?

    3. An unrelated question: Is there also a corresponding time dependent equation? I'm familiar with separation of variables, and the full Schrodinger equation makes it look as if there should always be a time dependent solution, and if there's not, then the total energy E is equal to 0. When, if ever, is there a wave-function that is independent of time?

    4. Is there anything else I missed?
     
  2. jcsd
  3. Apr 29, 2012 #2

    mfb

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    Staff: Mentor

    The equation is true at every point in space. If it helps to solve different regions separately (which is usually the case in these examples), just do that.

    It should have a second derivative everywhere.

    The time-dependence of your energy eigenfunctions is quite easy to calculate, it is just a rotation in the complex plane. Every state can be written as a superposition of them, and therefore solved with this system.
    You can take every state and just let it evolve in time according to the Schrödinger equation, if you want.
     
  4. Apr 29, 2012 #3
    Thanks so much mfb
     
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