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Probability of finding particle in half of a box for a given state

  1. Dec 14, 2009 #1
    Basically this is easy stuff i'm doing but i'n not sure whether i made a mistake or my answer is actually correct physically.

    I'm solving the 1-dimensional TISE for a particle confined to a box -a<x<a.

    I found that the eigenfunctions for this particle in a box with periodic boundary conditions are:[tex]u _{n}(x)=Ae ^{ \frac{in \pi x}{a} } [/tex] I'm pretty sure that this is correct.
    I normalised A and got [tex]A= \frac{1}{ \sqrt{2a} } [/tex], which I'm also confident about.
    And now I compute probability of finding particle in region 0<x<a when it is in state:
    [tex]\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)] [/tex]
    I get that it is 1/2-1/pi which is about 0.2... Is that correct? Or should it be rather 1/2 simply.
  2. jcsd
  3. Dec 14, 2009 #2
    Be careful that you are meeting the boundary conditions. I assume you are talking about an infinite square well. If this is the case, the wave functions that you use have to be 0 at x=a, and x=-a. I believe this gives the solutions: [tex]u _{n}(x)=\sqrt{\frac{2}{a}}sin ( \frac{n \pi}{2a} (x+a)) [/tex]

    Now whatever [tex]\psi(x)[/tex] you use, the probability of finding the particle in region 0<x<a is [tex]\int_0^a |\psi(x)|^2 dx[/tex]
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