Spherical Harmonic Hydrogen Wavefunction

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
bon
Messages
547
Reaction score
0

Homework Statement



Give a physical explanation of why a spherically symmetric Ylm cannot describe the state of a system with non-zero angular momentum.

Homework Equations





The Attempt at a Solution




I was thinking that if Ylm is spherically symmetric then the particle is equally likely to be found in any direction. But if it has non-zero angular momentum then it is rotating therefore it\'s probability to be found at a particular angle should change with time. But then I thought: assume l=0. Then the particle doesn't have angular momentum. Even so, the spherical harmonic shouldn\'t be spherically symmetric, should it? Surely if it\'s not rotating round the nucleus then there should be a particular direction where it is certain (or v likely) to be found..?

Or am i taking classical ideas too far?

What would you say?
Tanks!
 
Physics news on Phys.org
Look at the form of the l=0 spherical harmonic. It's constant.
 
Thanks but that doesn\'t help. I knew that already, as is clear from my post :)

I\'m trying to understand it physically...
 
Well, think about the idea of "spherically symmetric". Spherically symmetric means that upon rotations, the system's properties do not change. However, if something does have an angular momentum, then one can distinguish a change in a physical system upon rotation. For example, the rotation of an object in a classical orbit has an angular momentum and since the angular momentum is defined by a certain direction, it is no longer spherically symmetric. If you do the typical thing and define the angular momentum in the z-direction with objects orbiting in the xy-plane, the system will change upon any rotation that is not simply in the orbital plane.

So if you have an angular momentum, spherical symmetry no longer holds.
 
Well that\'s basically what i said in my first post..but my question is: why do you have spherical symmetry when l=0. Surely if it is not orbiting, its at a given point - in which case the probability to find it in all directions shouldn't be the same!
 
bon said:
Well that\'s basically what i said in my first post..but my question is: why do you have spherical symmetry when l=0. Surely if it is not orbiting, its at a given point - in which case the probability to find it in all directions shouldn't be the same!

Ahh, I see what you mean. Yes this is where the classical concept of an orbit needs to be abandoned. It doesn't make sense to talk about an electron orbiting a proton, for example. The electron doesn't have a well-defined orbit so that you could say "yes the electron is right there!". Remember, you're not solving for an orbit or equations of motion; you're solving for the wave-function so a lot of classical ideas no longer apply. If you were solving for an orbit, then it would make no sense for something with no angular momentum to be spherically symmetric. However you're solving for probabilities which means the probability distribution for finding the particle is spherically symmetric.