Ladder operators and SU(2) representation

  • I
  • Thread starter kelly0303
  • Start date
  • #1
385
24
Hello! I read in many places the derivation of the representation for SU(2) using ladder operators and in all of the places they say that, due to the fact that we are looking for a finite dimensional representation, the ladder must end at a point, hence why we have an eigenvector of ##L_3## (usually) such that, when acted on with the raising operator gives zero, and the same for the lowering operator. I didn't find an explanation as to why this must be unique. From a physics point of view it makes sense, as you have an irreducible representation of a spin J particle, so in order for the representation to allow for all the spin-z component, all the eigenvectors must have a different eigenvalue, hence why you have just one eigenvector at the top and at the bottom, and in general for any eigenvalue of ##L_3##, but I am not sure I understand mathematically, why can't you have more than one vector with the same eigenvalue in an irrep of SU(2). Can someone help me? Thank you!
 

Answers and Replies

  • #2
15,380
13,405
Hello! I read in many places the derivation of the representation for SU(2) using ladder operators and in all of the places they say that, due to the fact that we are looking for a finite dimensional representation, the ladder must end at a point, hence why we have an eigenvector of ##L_3## (usually) such that, when acted on with the raising operator gives zero, and the same for the lowering operator. I didn't find an explanation as to why this must be unique. From a physics point of view it makes sense, as you have an irreducible representation of a spin J particle, so in order for the representation to allow for all the spin-z component, all the eigenvectors must have a different eigenvalue, hence why you have just one eigenvector at the top and at the bottom, and in general for any eigenvalue of ##L_3##, but I am not sure I understand mathematically, why can't you have more than one vector with the same eigenvalue in an irrep of SU(2). Can someone help me? Thank you!
You can find the underlying theorem (7.1.) here: https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
For the mathematical derivation I recommend [6] in the sources. The main reason is, that all invariant subspaces are one-dimensional, and the upper triangular matrix shifts those subspaces one place higher, the lower triangular one place less.
 

Related Threads on Ladder operators and SU(2) representation

  • Last Post
Replies
2
Views
2K
Replies
2
Views
4K
Replies
6
Views
3K
  • Last Post
Replies
12
Views
9K
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
2
Views
1K
Replies
6
Views
2K
  • Last Post
Replies
4
Views
955
Top