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Wave Function Probability

  1. Nov 6, 2008 #1


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    I was reading part of a book which was explaining about the probability of finding a partical on a 1d line.
    [tex]\int^{+\infty}_{-\infty}[/tex]P(x) dx = 1
    This sounds right because if the line was infinitely long then the partical must be on it.
    You can them intergrate between a and b to find the probability of it being in a lenght and if a and b were the same the probability would be 0.
    But when you intergrate P(x) dx you get [tex]\frac{Px^{2}}{2}[/tex]
    by putting the numbers in you get P[tex]\infty[/tex] - -P[tex]\infty[/tex]
    or P[tex]\infty[/tex] + P[tex]\infty[/tex] = P[tex]\infty[/tex]
    A probability can't be more than 1. I must be missing something or dealing with the infinities in the wrong way.
    (Sorry it looks like P^infinity its P x infinity but I couldn't change it.)
  2. jcsd
  3. Nov 6, 2008 #2


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    [tex]\int_\infty^\infty P(x) dx=1[/tex] doesn't mean [tex]\int_\infty^\infty Px dx=1[/tex] P(x) means the probability in function of x not P a constant times x.
  4. Nov 6, 2008 #3


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    If you integrate P(x) dx you get the integral of P(x); I am not quite sure why you think you would get [tex]\frac{Px^{2}}{2}[/tex]?
    P(x) is a FUNCTION, not a constant; there is no way to integrate it unless you know what that function is.
  5. Nov 6, 2008 #4


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    Oh so P(x) is the wave function? I'm going to read the chapter again.
  6. Nov 6, 2008 #5
    No, P(x) is the probability function, which is the wavefunction squared (actually, absolute value squared)... So P(x)dx gives the probability for finding the particle on a bit of length dx at position x.
  7. Nov 6, 2008 #6


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    Yea, I re-read it and think I understand now. W(x) is the wave function and P(x) =|W(x)^2
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