Physical meaning of probability density

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SUMMARY

The discussion centers on the physical interpretation of probability density for a 1s hydrogen atom, specifically addressing the apparent contradiction between maximum probability density at the nucleus (r → 0) and the radial probability being zero at that point. The probability density, defined as P(r) = |ψ|², indicates a non-zero value at the nucleus, while the radial probability R(r) = P(r)4πr² equals zero at r = 0 due to the volume of the spherical shell approaching zero. This discrepancy highlights the importance of understanding the distinction between probability density and radial probability in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wave functions and probability densities.
  • Familiarity with the concept of radial probability and its mathematical formulation.
  • Knowledge of the Bohr model of the hydrogen atom and its implications for electron positioning.
  • Basic grasp of spherical coordinates and volume calculations in three-dimensional space.
NEXT STEPS
  • Explore the mathematical derivation of wave functions for hydrogen atoms, focusing on the 1s orbital.
  • Study the implications of electron capture in quantum mechanics and its relation to probability density.
  • Investigate the geometric interpretation of probability distributions in quantum systems.
  • Learn about the differences between probability density and radial probability in quantum mechanics.
USEFUL FOR

Students and professionals in physics, particularly those studying quantum mechanics, atomic structure, and the behavior of electrons in atoms. This discussion is beneficial for anyone seeking to deepen their understanding of probability concepts in quantum theory.

Absentee
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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 probability 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!
 
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Absentee said:
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 probability of finding electron is 0 at r → 0.

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.
 
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 probability is infinite at radius 0? What is PRACTICAL meaning of this maximum at r = 0? Is there an analogy?
 
Last edited:
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:

http://en.wikipedia.org/wiki/Electron_capture

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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