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Physical meaning of probability density

  1. Feb 21, 2014 #1
    Hi guys. I'm trying to get the idea of probability density for 1s hydrogen atom.

    I just don't understand that probability density reaches maximum at nucleus (r → 0) if the most probable radius where electron can be found is at Bohr radius according to radial probability (Which also states probablity of finding electron is 0 at r → 0.

    Could you please give me a more physical, visual kind of interpretation? Is there a physical meaning anyways? I've seen a lot of 'that's how the math works' but it doesn't quite work for me. Thanks!
  2. jcsd
  3. Feb 22, 2014 #2


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    Staff: Mentor

    The probability density P(r) = |ψ|2 is the probability per unit volume.

    The radial probability (which I think most books call R(r)) is the probability per unit radius (distance from center).

    If you have a spherical shell of radius r and thickness dr, the probability that the electron can be found in the shell, i.e. in the range dr can be calculated two ways:

    1. using the probability density and the volume of the shell: Pshell = P(r)4πr2dr

    (note this is the volume of the shell itself, not the volume inside the shell!)

    2. using the radial probability and the thickness of the shell: Pshell = R(r)dr

    The two probabilities have to be equal, so

    R(r) = P(r)4πr2

    At the center (r = 0), R(0) must be zero even though P(0) is not zero (provided P(0) is not infinite, of course).

    Think of this as due to the volume of the shell (itself) going to zero as r goes to zero, for a fixed dr.
  4. Feb 22, 2014 #3
    Thanks for the answer, but I just can't wrap my head behind PRACTICAL meaning of this. If I said that most probable location where I would find the electron would be at Bohr radius and there is practically no chance of finding it at radius 0 (as the radial probability is 0 at that location) how can I even think about other probability that states the probabilty is infinite at radius 0? What is PRACTICAL meaning of this maximum at r = 0? Is there an analogy?
    Last edited: Feb 22, 2014
  5. Feb 22, 2014 #4


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    Staff: Mentor

    Consider a small volume V, the size of an atomic nucleus, around the r = 0 point. The probability of the electron being in that volume is nonzero. This is what allows electron capture to happen:


    Using the volume probability density P, this probability is P(0)*V. Approximately, of course. The approximation becomes better as V becomes smaller and smaller.

    The fact that R(0) = 0 is a purely geometrical artifact. Any volume probability distribution P that isn't infinite at r = 0 will give you R(0) = 0.

    As an example, consider a uniform volume probability distribution, P = (some constant) everywhere in some region around r = 0. Using the equation in my previous post, R(r) = (that constant) * 4πr2, which is zero at r = 0.
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