Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability density versus radial distribution function

  1. May 18, 2010 #1
    Okay, this is a really basic question. I'm just learning the basics of QM now.

    I can't wrap my head around the idea that the radial distribution function goes to zero as r-->0 but that the probability density as at a maximum as r-->zero. How can this be?

    Thanks!
     
  2. jcsd
  3. May 18, 2010 #2

    phyzguy

    User Avatar
    Science Advisor

    What are you referring to exactly? The hydrogen ground state wave function? As you say, the probability density is a maximum at r=0. Perhaps the point you've missed is that the volume element (r^2 sin(theta))dr dtheta dphi goes to zero at r=0?
     
  4. May 18, 2010 #3

    alxm

    User Avatar
    Science Advisor

    Ah, the 1s orbital. It's a common source of confusion, that.
    Well, the radial wavefunction has a maximum at r=0. This is the point, i.e. infinitesimal volume element that has the highest probability. It's the volume of space the electron's most likely to be in.

    The radial distribution function, on the other hand, is the sum of probabilities over a given r. It's not the probability of just one spot, but of every spot on the surface of an infinitesimally thin sphere of that radius. Since a sphere with zero radius has zero area, it means the radial distribution is 0 at r=0 no matter what the wave function's value there.

    An analogy I once thought up: Imagine you paint a bunch of different-sized spheres, and where the thickness of the paint is [tex]e^{-r}[/tex].. The sphere with the thickest coat of paint is not the same thing as the sphere with the most paint on it!
     
  5. May 18, 2010 #4
    Thank you. Yes, I'm referring to the 1s H-atom.

    Mathematically it makes perfect sense: if r=0 then the radial distribution function is zero. So let's say we take a concentric sphere with the inner surface being the surface of the nucleus, and the outer surface at a distance dr from the surface of the nucleus. Saying that the radial distribution function is zero means that electrons will never be found within that volume. On the other hand, probability density (psi^2) tells us that there is some finite probability that an electron will be found in that very same region of space. Doesn't it?
     
  6. May 18, 2010 #5

    phyzguy

    User Avatar
    Science Advisor

    It's only right at the origin that the radial distribution function goes to zero. Since the 1s wave function is spherically symmetric, the probability of finding the electron between r and r+dr is give by the expresssion: [tex]4\pi r^2 \psi(r) \psi^*(r) dr[/tex]. This will be nonzero (but very small) when r=r(nucleus), as you asked. The reason it will be very small is that the volume of space at r=r(nucleus) is very small. Try plotting the functions [tex]4\pi r^2 \psi(r) \psi^*(r) [/tex] and [tex] \psi(r) \psi^*(r) [/tex], and I think you will see why it is this way.
     
  7. May 18, 2010 #6
    So r=0 is defined as the center of the nucleus?
     
  8. May 18, 2010 #7

    phyzguy

    User Avatar
    Science Advisor

    Yes. Where else would it be?
     
  9. May 18, 2010 #8
    Okay, but then how can there be any electron density inside the nucleus?
     
  10. May 18, 2010 #9

    phyzguy

    User Avatar
    Science Advisor

    Why can't the electron be inside the nucleus? Elementary particles appear to be pointlike as far down as we can measure. So even inside the nucleus it appears to be mainly empty space between the quarks and gluons that make it up. We know the electron definitely penetrates inside the nucleus because there are types of radioactive decay (called electron capture) where the nucleus basically captures one of the atomic electrons and transmutes into a nucleus with atomic number reduced by one.
     
  11. May 18, 2010 #10
    Electron capture occurs in large nuclei; however, the proton of the hydrogen atom cannot capture an electron yielding a free neutron.
     
  12. May 18, 2010 #11

    phyzguy

    User Avatar
    Science Advisor

    No, of course not. I didn't mean to say that it could. I was just saying that this is evidence that the electron wave function does penetrate the nucleus. Do you think the electron wave function does not penetrate the nucleus of a hydrogen atom?
     
  13. Jun 2, 2012 #12
    If the radial wave function represents the probability that an electron will be contained within an infinitesimal volume at some specified radius, how can it be then that some plots of the radial wave functions have negative values, would this not imply a negative probability at these distances?

    For example, what does the radial wave function tell us at the distances where minima occur in the 2s, 3s and 3p plots?

    Thanks.:uhh:
     
  14. Jun 3, 2012 #13

    phyzguy

    User Avatar
    Science Advisor

    It's not the wave function itself that represents the probability of finding the electron at a given location. It is the square of the wave function ψψ* . This quantity is always non-negative, although it can be zero.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook