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Mathematical conundrum when adding complex exponentials

  1. May 22, 2015 #1
    Hi there,

    Once again I find myself twiddling around with some quantum mechanics, and I bumped into something I find strange. I can't see what the error of my thinking is, so I hope someone could be able to point it out.

    I'm looking at solutions to the infinite square well, and arrive at the simple differential equation

    [tex] \frac{d^2\Psi}{dx^2} = -k^2 \Psi [/tex]

    The solution to this can be written in terms of complex exponentials or sines and cosines. I bumped into the wierd stuff when I use complex exponentials.

    So the general solution in that case would be

    ##\Psi(x) = Ae^{ikx}+Be^{-ikx}##

    Now, what I then started thinking was: "Hmmm... This could be viewed mathematically as a sum of two vectors, and solution is simply another vector."

    So I drew this picture to illustrate the idea:

    ?temp_hash=98f0bfa6626fe3330472390e3bfc5456.png
    So from that perspective it seems that the solution could also be written as

    ##\Psi(x) = Ce^{ik'x}##

    However, using the simple constraints of the infinite square well quickly leads to problems - namely:

    ##|\Psi(0)|^2 = 0##

    ##C = 0##

    So.... Now really what I had hoped for. Where am I going taking a wrong turn? Has it to do with the x? That x should be x' also?

    Thanks in advance :)
     

    Attached Files:

  2. jcsd
  3. May 22, 2015 #2

    Orodruin

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    The problem is that your ##k'## is going to be a function of ##x##. As such, you can not assume it to be constant and apply derivatives to the wave function with that assumption.
     
  4. May 22, 2015 #3

    Strilanc

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    Consider that ##e^{ix} + e^{-ix}## is equal to ##2 cos(x)## and so has no net imaginary component for all x. But a single ##e^{kix}## always has some net imaginary component (for finite ##k## and ##x##).
     
  5. May 22, 2015 #4
    Thanks guys!

    I gave it some more thought and think I nailed it down now.

    I have another question now thought, but I'm gonna make another thread for it.
     
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