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Wave functions and probability

  1. May 15, 2008 #1
    1. The problem statement, all variables and given/known data
    I've had lectures on the thoery of this topic, but I've not been given any examples and I'm struggling with how to apply the theory to this homework question:
    A particle is described by the normalised wave function

    Si(x,y,z)=Ae^-h(x^2+y^2+z^2) where A and h are real positive constants.

    Determine the probability of finding the particle at a distance between r and r+dr from the origin.


    2. Relevant equations
    p=integral (mod[Si(x,y,z)])^2 dxdydz


    3. The attempt at a solution
    I thought that the wave function was supposed to involve complex numbers, but there's no i in the wave function? I thought i needed the conjugate of the wave function to find mod[Si(x,y,z)]?
    I think once i find (mod[Si(x,y,z)])^2 i then need to integrate it over the volume of a spherical shell with radius r and thickness dr, so r^2=x^2+y^2+z^2 but I'm confused about how to compute this integral. Do i need to parametrise the variables?

    I'm obviously not asking for someone to give me the solution, but I'm having a hard time figuring out where to start with this problem, so any help would be greatly appreciated.
     
  2. jcsd
  3. May 15, 2008 #2
    Your wavefunction is not normalized! Was there a specific finite domain where it was non-zero and you forgot to include that in your post?

    Anyway you are not evaluating an integral. The probability you are looking for is [tex]|\psi|^2 dV = |\psi|^2 4\pi r^2 dr[/tex].
     
  4. May 15, 2008 #3
    No, i copied the question down word for word, it says that the wave function is normalised! I'm sorry, I'm still very confused because it says in my lecture notes that i have to integrate to find the probability. Can you please tell me where you get |\psi|^2 dV = |\psi|^2 4\pi r^2 dr from?
     
  5. May 15, 2008 #4
    Elementary probability theory--

    The probability of a variable X assuming a value between X=x and X=x+dx is

    [tex]dP=\rho(x)dx[/tex]
    where [tex]\rho[/tex] is the probability density.

    You need that local probability so that you can integrate it to obtain global ones. The integral formula comes from this. The exercise is meant to see if you were paying attention to the derivation.
     
  6. May 15, 2008 #5
    I was paying attention in the derivation, i just found it a bit hard to follow. Thankyou for your help, I'll the question another go.
     
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