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Why does the wave function have to be normalizabled

  1. Dec 3, 2005 #1
    why does the wave function have to be normalizable, and why does it have to go to 0 and x approaches positive/negative infinity and y approaches positive/negative infinity ?
  2. jcsd
  3. Dec 3, 2005 #2
    Think about the physics involved. Here's a simple and intuitive explanation. The particle described by the wave function has to be found somewhere so the wave function must be normalizable. From this it follows that it must go to zero as x->inf.
  4. Dec 3, 2005 #3
    i still don't get it...
  5. Dec 3, 2005 #4
    Could you be more precise about what you don't get?

    [tex]\psi \psi*[/tex] is the probability density. Since the particle is certainly somewhere the probability density integrated over all space must give 1. Or in other words the particle has to be somewhere. If this is to hold psi has to go to zero at infinity.
  6. Dec 3, 2005 #5
    when you normalize the wave function to be 1, do you mean that you do it to make sure that the biggest limit is 1? and what does 0 have to do with infinity? i only see in my textbook that the wave function has no meaning, but its square means probability, so i think i'm confused on what does the boundary limits on the integral mean...

    sorry about that, i'm slow at learning things... :P
  7. Dec 3, 2005 #6
    Ok you normalise the integral of psi^2 between the infinities (+ve and -ve)to be equal to one, i.e. you normalize so that the probability of the particle being somewhere between the origin and the infinities of all spatial dimensions is equal to one. Sensible no? Ok, if the wave function does not go to zero at the infinities then the probability of the particle being somwhere in space is equal to infinity. The reason for that is that if you integrate the probabilty density function over all space from infinity to infinty you get infinity, if the function does not go to zero at the infinities..., hehe, and the maximum possible value for a probabilty is 1... so the wavefuntion has to dissapear at infinity.
  8. Dec 4, 2005 #7
    i see now~ thank you very much for clearing that up!!! :)
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