Probability of finding electron.

Click For Summary
The discussion focuses on the probability density of finding electrons in hydrogen atom orbitals, particularly the s orbital. It is clarified that while s orbital electrons have the highest probability density at the nucleus (r=0), the expectation value of the electron's distance from the nucleus is not zero. The radial distribution function indicates that the likelihood of finding an electron increases at certain distances from the nucleus due to the spherical nature of s orbitals. As the radius increases, the surface area of the sphere grows, affecting the probability distribution. Understanding these concepts is essential for grasping quantum mechanics.
nil1996
Messages
301
Reaction score
7
HAtomOrbitals.png


The picture shows the first few hydrogen atom orbitals (energy eigenfunctions). These are cross-sections of the probability density that are color-coded (black=zero density, white=highest density).according to the picture -Do s orbital electrons have highest probability density in the nucleus?

(i am totally new to quantum mechanics.So please tell if i am wrong.)
 
Last edited:
Physics news on Phys.org
If you take the radial part of the wave function only, then yes, the probability density ##|\psi|^2## is maximum at ##r=0## for s electrons. But that doesn't mean that the expectation value ##\langle r \rangle## is 0. A measurement will most likely find the electron at a certain distance from the nucleus.

Another way to see it is to consider the radial distribution too. As s orbitals are spherical, the surface of the sphere increases as ##r## increases, so the probability of finding the electron at a certain distance of the nucleus goes through a maximum away from the nucleus.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 43 ·
2
Replies
43
Views
5K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
8K
  • · Replies 24 ·
Replies
24
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K