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Wave function at infinity

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data
    If I have a wave function that goes to infinity can I assume that the derivative also goes to 0 at infinity?


    2. Relevant equations



    3. The attempt at a solution
    The reason I think it does is because the wavefunction and its derivative must be continuous everywhere except at potentials that go to infinity. Is this the correct logic?
     
  2. jcsd
  3. Apr 28, 2014 #2
    Well, this is a little too little information. What is the equation for your wavefunction?
     
  4. Apr 28, 2014 #3
    We aren't given any. Some other information which might be helpful though, the wave function is normalized, it is a solution to the time dependent schrodinger equation and the potential function is real. And again it approaches zero as x goes to +- infinity. Is that enough?
     
  5. Apr 28, 2014 #4
    Just that in general the most sure way to decide is to calculate. So, what you should do is write down a wavefunction that satisfies the conditions and see what happens if you differentiate it under your stated conditions.

    For example, try to solve this one, you will find it interesting:

    [itex]\psi(x) = \frac{Asin^{2}x}{\sqrt{x^{2}+1}}[/itex]
     
  6. Apr 28, 2014 #5
    I think I see what you are saying. I will give a little more information into what I am trying to do.

    The question is:

    Consider two normalizable wave functions Ψ1(x, t) and Ψ2(x, t), both of which are solutions
    of the time-dependent Schrodinger equation. Assume that the potential function is real.
    The functions are normalized, and the functions both approach zero as x goes to ±∞.

    Show that these propeties can be used to prove that

    [itex]\frac{d}{dt}[/itex] [itex]\int_{-∞}^{∞}[/itex] Ψ1*(x, t)Ψ2(x, t)dx = 0

    We get a hint to conert the temporal derivative to the spatial dertivative using the time-dependent schrodinger equation and you get:

    [itex]\frac{i\bar{h}}{2m}[/itex] [itex]\int_{-∞}^{∞}[/itex] Ψ1*(x, t)[itex]\frac{d^2}{dx^2}[/itex]Ψ2(x, t) - Ψ2(x, t)[itex]\frac{d^2}{dx^2}[/itex]Ψ1*(x, t)

    Then using integration by parts you entually get a term (for one part of the integral) that is

    Ψ1*(x, t)[itex]\frac{d}{dx}[/itex]Ψ2(x, t) [itex]|^{∞}_{-∞}[/itex]

    and I need that to go to 0 or else the rest doesn't really follow through correctly. I justified it by saying that since Ψ2(x, t) goes to 0 and +- infinity then its derivative will go to 0 and +- infinity.
     
  7. Apr 28, 2014 #6
    Well, as far as I know, in general it is assumed that

    [itex]Lim(x \rightarrow ∞) \frac{∂^{n}ψ}{∂x^{n}} = 0 [/itex]

    where n is some positive integer

    but it does not necessarily have to be the case. As far as I know it is only an assumption.
     
  8. Apr 28, 2014 #7
    Okay thank you. I think it is sufficient for my course. I appreciate your time!
     
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