• Support PF! Buy your school textbooks, materials and every day products Here!

Addition of Angular Momentum for identical particles

  • #1
This is the problem I'm trying to understand:

Consider two particles with spin 1 without orbital angular momentum. If they are distinguishable, from the rule of addition of angular momentum applied to spin, we'll have states of total spin [itex]j=0,1,2[/itex]. If we have, however, identical particles which are the possible states?

In textbooks, the addition of angular momentum is never treated in terms of distinguishable and identical particles, at least I don't recall it. The way I would approach this problem is to acknowledge that the possible total spin would be [itex]j=0,1,2[/itex] and then, from the state [itex]|-1\rangle |-1\rangle[/itex], I would use the ladder operator to build all the other four states compatible with [itex]j=2[/itex].


How do I build the states for [itex]j=0,1[/itex]? The state [itex]|j=1,m_{j}=-1\rangle[/itex] must be built from the same states as [itex]|j=2,m_{j}=-1\rangle[/itex], that is, [itex]|-1\rangle |0\rangle[/itex] and [itex]|0\rangle |-1\rangle[/itex]. So how are they any different?

Thank you very much.


The Attempt at a Solution

 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,817
6,625
What is important for distinguishable particles is that the total state is (anti-)symmetric (depending on Bose/Fermi statistics). You need to take the spin state into account as well as the spatial wave function.
 

Related Threads on Addition of Angular Momentum for identical particles

Replies
5
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
858
  • Last Post
Replies
2
Views
1K
Replies
0
Views
933
Replies
2
Views
1K
Replies
5
Views
520
Replies
1
Views
10K
  • Last Post
Replies
5
Views
3K
Top