Hund's rule and angular momentum coupling

302
0
Hi.

In Hund's second rule, it seems that we calculate the value of L simply by summing the [itex]L_z[/itex] components of the individual electrons. But L has to do with the eigenvalue of the L^2 operator, i.e. the eigenvalue is L(L+1). So how can this be correct?
 

Meir Achuz

Science Advisor
Homework Helper
Gold Member
2,169
63
L is the maximum eigenvalue of L_z.
L(L+1) is the eigenvalue of a different operator, L^2.
 
302
0
L is the maximum eigenvalue of L_z.
L(L+1) is the eigenvalue of a different operator, L^2.
Yeah that is exactly my point. We know that that L_z has some particular value. Now why is L=L_z ?
 

Related Threads for: Hund's rule and angular momentum coupling

Replies
6
Views
254
  • Posted
Replies
1
Views
2K
Replies
3
Views
2K
  • Posted
Replies
7
Views
17K
  • Posted
Replies
1
Views
2K
Replies
1
Views
2K
Replies
5
Views
7K
Replies
0
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top