Do Commuting Operators Always Share a Common Basis of Eigenvectors?

Click For Summary
SUMMARY

The discussion centers on the relationship between commuting operators and their eigenvector bases in quantum mechanics, specifically regarding angular momentum operators. It is established that if two operators do not commute, they do not necessarily share a common basis of eigenvectors, but additional conditions may apply. Conversely, if two operators do commute, they can have a common basis of eigenvectors, particularly if they are bounded and self-adjoint on a separable Hilbert space. The theorem cited confirms that commuting bounded self-adjoint operators guarantee a shared orthonormal basis of eigenvectors.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly eigenvectors and eigenvalues.
  • Familiarity with quantum mechanics principles, especially angular momentum operators.
  • Knowledge of Hilbert spaces and their properties, including separability and self-adjointness.
  • Basic grasp of operator theory in the context of quantum physics.
NEXT STEPS
  • Study the properties of bounded and self-adjoint operators in quantum mechanics.
  • Explore the implications of the spectral theorem for operators in Hilbert spaces.
  • Learn about the closure of unbounded operators and their role in quantum mechanics.
  • Investigate the concept of common eigenvectors in the context of hermitian operators.
USEFUL FOR

Students and professionals in quantum mechanics, physicists focusing on operator theory, and anyone interested in the mathematical foundations of quantum physics.

Funzies
Messages
20
Reaction score
0
Hey guys,
I'm studying some quantum physics at the moment and I'm having some problems with understanding the principles behind the necessary lineair algebra:

1) If two operators do NOT commutate, is it correct to conclude they don't have a similar basis of eigenvectoren? Or are there more conditions to be verified to conclude this?

2) If two operators DO commutate, you MAY find a similar basis of eigenvectoren? If this is correct, what are the conditions for it to be absolutely true?

The context to which I am asking these question are the ladder operators for angular momentum states and especially the fact that Lx and Ly do not commutate and don't have a similar basis of eigenvectors, whereas L^2 and Lx/Ly/Lz do commutate and do have a basis of eigenvectors.

Thanks in advance!
 
Physics news on Phys.org
In my ancient lecture notes from college I remember having the proof of a strong theorem:

If A and B are 2 bounded and self-adjoint linear operators with discrete spectrum on a separable Hilbert space, they commute if and only if the Hilbert space has an orthonormal basis made up of common eigenvectors of A and B.

This theorem applies to your question, with the minor notice that the orbital angular momentum ops are generally unbounded in the Hilbert space L^2 \left(\mathbb{R}^3, dx\right), so one speaks about the closures of the these operators when referring to self-adjointness and the common dense domain involved is not the domain of the closure, but of the original operator.
 
bigubau said:
If A and B are 2 bounded and self-adjoint linear operators with discrete spectrum on a separable Hilbert space, they commute if and only if the Hilbert space has an orthonormal basis made up of common eigenvectors of A and B.

In the standard QM setup it will be not too dangerous to assume that the following holds:

If A and B are hermitian operators then they commute if and only if there is a common basis of eigenvectors.

This basis can be discrete or continuous or partly discrete and partly continuous - whatever it may mean.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K