Why Don't L2 and Ly Commute When L2 and Lx Do?

Click For Summary

Discussion Overview

The discussion revolves around the commutation relations between angular momentum operators, specifically why the operators L2 and Ly do not commute while L2 and Lx do. The scope includes theoretical exploration of quantum mechanics and the implications of these commutation relations on eigenstates.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that since [L2, Lx] = 0, L2 and Lx have simultaneous eigenstates, leading to the question of whether L2 and Ly should also commute.
  • One participant argues that even if A and B commute (having a common eigenbasis) and B and C commute (having a different common eigenbasis), it does not imply that A and C commute.
  • Another participant suggests that L2 and Lx have a common eigenbasis due to their commutation, while L2 and Ly have a different common eigenbasis, raising the point that since Lx and Ly do not commute, the two sets of eigenbases cannot be the same.
  • A later reply introduces the context of a spin-1/2 system, noting that the operators can be represented as simple 2x2 matrices and encourages working through this example for better understanding.
  • It is mentioned that the identity operator commutes with everything, and every vector is an eigenvector of the identity operator, which has implications for shared eigenbases with other operators.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the commutation relations, particularly regarding the relationship between the eigenbases of L2, Lx, and Ly. The discussion remains unresolved with multiple competing interpretations of the commutation properties.

Contextual Notes

The discussion highlights the complexity of commutation relations and their implications for eigenstates, with participants noting that the relationships depend on the specific properties of the operators involved. There are unresolved assumptions regarding the nature of the eigenbases and the implications of different commutation relations.

dyn
Messages
774
Reaction score
63
Hi.
To show that [ L2 , L+ ] uses the following commutators [ L2 , Lx ] = 0 and [ L2 , Ly ] = 0 . But if [ L2 , Lx ] = 0 this shows that L2 and Lx have simultaneous eigenstates ; but then should L2 and Ly not commute ?
Thanks
 
Physics news on Phys.org
dyn said:
Hi.
To show that [ L2 , L+ ] uses the following commutators [ L2 , Lx ] = 0 and [ L2 , Ly ] = 0 . But if [ L2 , Lx ] = 0 this shows that L2 and Lx have simultaneous eigenstates ; but then should L2 and Ly not commute ?
Thanks
No. If A and B commute so have a common eigenbasis and B and C commute so have a different common eigenbasis, it does not follow that A and C commute and have a common eigenbasis.
 
  • Like
Likes   Reactions: dyn
Thanks. So L2 and Lx have a common eigenbasis as the 2 operators commute and L2 and Ly have a different common eigenbasis. But as Lx and Ly do not commute these 2 sets of eigenbases can never be the same ?
 
dyn said:
Thanks. So L2 and Lx have a common eigenbasis as the 2 operators commute and L2 and Ly have a different common eigenbasis. But as Lx and Ly do not commute these 2 sets of eigenbases can never be the same ?

A spin-1/2 system is useful because the operators are simple 2x2 matrices. I suggest you work through this example for these AM operators.

Note that the identity operator, ##I##, commutes with everything (and every vector is an eigenvector of ##I##). So, ##I## has a shared eigenbasis with all operators that have an eigenbasis (such as Hermitian operators).
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K