Angular momentum operators in quantum mechanics

Click For Summary
The discussion revolves around a Hamiltonian in quantum mechanics expressed as H=(J1^2+J2^2)2A+J3^2/2B, where J1, J2, and J3 are angular momentum operators. The user successfully rewrote the Hamiltonian and derived the eigenvalues as (h^2L(l+1)-h^2m^2)/2A+h^2m^2/2B. However, the user is uncertain about how to find the eigenstates of the Hamiltonian, expressing confusion over calculating eigenvalues without utilizing the eigenstates |l,m>. The thread highlights the challenge of transitioning from eigenvalue calculations to determining the corresponding eigenstates in quantum mechanics. Understanding the relationship between eigenvalues and eigenstates is crucial for solving the problem.
myb821
Messages
1
Reaction score
0

Homework Statement


H=(J1^2+J2^2)2A+J3^2/2B where J1,2,3 are the angular momentum operators and A and B are just numbers


Homework Equations





The Attempt at a Solution


I rewrote the Hamiltonian as (J^2-Jz^2)/2A + J3^2/2B and got the eigenvalues to be (h^2L(l+1)-h^2m^2)/2A+h^2m^2/2B but i don't know how to find the eigenstates of this hamiltonian help
 
Physics news on Phys.org


I don't see how you could calculate eigenvalues without using the eigenstates |l,m>
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
46
Views
1K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K