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H atom electron in combined spin/position state

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    An electron in a H atom occupies the combined spin and position state: R21{(sqrt(1/3)Y10χ+) + (sqrt(2/3)Y11χ-)} If you measured both the z component of spin and the distance from the origin, what is the probability density for finding the particle with spin up and at radius r?

    2. Relevant equations

    3. The attempt at a solution

    The answer should just be |R21|^2*(1/3)*|Y10|^2*|χ+|^2 = (r^2)/(96πa^5) * exp(-r/a) * (cosθ)^2, right? Or do I need to do an integral? The theta dependence of my answer is bugging me, but I'm not entirely sure if I need to integrate over theta and phi to just get an r dependent answer? Could somebody please help me think through this? Thanks very much.
  2. jcsd
  3. Oct 22, 2011 #2


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    If the question just asks for "radius r", then, yes, you do have to integrate out the theta and phi dependence.
  4. Oct 22, 2011 #3
    Sorta makes sense, but if the probability density is not isotropic in theta, it still seems wrong to me to integrate that dependence away.
  5. Oct 22, 2011 #4


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    If you don't integrate over the angles, the expression you have doesn't give the probability density that the electron is at a distance r; it's the probability density that the electron is at a distance r, angle θ, and angle φ.
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