Register to reply

Spherical harmonics and wavefunctions

by Chemist20
Tags: harmonics, spherical, wavefunction
Share this thread:
Chemist20
#1
Feb14-12, 05:52 AM
P: 87
What's the difference in the representation of spherical harmonics and the orbitals themselves? they look exactly the same to me... unlike the radial part of the wavefunction though.
Phys.Org News Partner Physics news on Phys.org
Scientists uncover clues to role of magnetism in iron-based superconductors
Researchers find first direct evidence of 'spin symmetry' in atoms
X-ray laser probes tiny quantum tornadoes in superfluid droplets
jtbell
#2
Feb14-12, 08:40 AM
Mentor
jtbell's Avatar
P: 11,748
The angular part of an orbital wave function for hydrogen (or any other spherically symmetric potential) is a spherical harmonic:

[tex]\Psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)[/tex]
Chemist20
#3
Feb14-12, 08:41 AM
P: 87
Quote Quote by jtbell View Post
The angular part of an orbital wave function for hydrogen (or any other spherically symmetric potential) is a spherical harmonic:

[tex]\Psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)[/tex]
yes but for a 2p for example, what's the difference in representation between the orbital itself and the spherical harmonic? they look the same to me.

cgk
#4
Feb14-12, 10:21 AM
P: 426
Spherical harmonics and wavefunctions

Quote Quote by Chemist20 View Post
yes but for a 2p for example, what's the difference in representation between the orbital itself and the spherical harmonic? they look the same to me.
The angular parts look the same, because they are identical (see jtbell's comment). Unlike spherical harmonics, orbital wave functions, however, do not consist only of an angular part. They also have a radial part. And this radial part is non-trival and comes from solving the Schroedinger equation for some potential (e.g., in hydrogen the nuclear attraction of the proton, in higher spherical atoms from nuclear attraction and the mean field of the other electrons (Fock potential)). But this has no influence on the angular part. E.g., 2p and 3p orbitals have the same angular part, not only in a single atom, but across all atoms (in the nonrelativistic case etc.).
Chemist20
#5
Feb14-12, 12:09 PM
P: 87
Quote Quote by cgk View Post
The angular parts look the same, because they are identical (see jtbell's comment). Unlike spherical harmonics, orbital wave functions, however, do not consist only of an angular part. They also have a radial part. And this radial part is non-trival and comes from solving the Schroedinger equation for some potential (e.g., in hydrogen the nuclear attraction of the proton, in higher spherical atoms from nuclear attraction and the mean field of the other electrons (Fock potential)). But this has no influence on the angular part. E.g., 2p and 3p orbitals have the same angular part, not only in a single atom, but across all atoms (in the nonrelativistic case etc.).
yes, but when drawing the 2p orbital and the 2p spherical harmonic what's the difference? THEY ARE THE SAME!
questionpost
#6
Feb14-12, 09:36 PM
P: 198
If they are the same, then by the property of sharing the same identity they are not different.
lugita15
#7
Feb16-12, 10:03 AM
P: 1,583
Quote Quote by Chemist20 View Post
yes, but when drawing the 2p orbital and the 2p spherical harmonic what's the difference? THEY ARE THE SAME!
Note that multiplying the radial wave function by constant factor changes the size, not the shape, of the "drawing" of the orbital, which is really just a drawing of the surface of maximum probability density. By looking at how more general changes in the radial wave function affects this surface, you can see why spherical harmonics look so much like these surfaces.


Register to reply

Related Discussions
Vector Spherical harmonics/spherical coordinates question General Physics 0
Spherical harmonics Calculus & Beyond Homework 0
Spherical Harmonics Science & Math Textbooks 1
Spherical Harmonics Calculus & Beyond Homework 1