Proof that commuting operators have a shared base of eigenfunctions

In summary, the conversation discusses the equivalence between two operators, A and B, having a common base of eigenfunctions if AB = BA. The proof given assumes non-degenerate spectrum, but it is questioned if this assumption is necessary. Some suggest using spectral decomposition to prove the equivalence more generally.
  • #1
jakotaco
2
0
I have been told that if we have two operators, A and B, such that AB = BA then this is equivalent with that A and B have a common base of eigenfunctions.

However, the proof given was made under the assumption that the operators had a non-degenerate spectrum. Now I understand that one rather wants to give a short proof under some simplified condition and hope people will take your word that it applies without those simplifications. But in quantum mechanics, it is not as if degeneracy is uncommon.

So, does anyone know how to prove it more generally? Or maybe an explanation why the assumption isn't as bad as I though?
 
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  • #2
jakotaco said:
I have been told that if we have two operators, A and B, such that AB = BA then this is equivalent with that A and B have a common base of eigenfunctions.

However, the proof given was made under the assumption that the operators had a non-degenerate spectrum. Now I understand that one rather wants to give a short proof under some simplified condition and hope people will take your word that it applies without those simplifications. But in quantum mechanics, it is not as if degeneracy is uncommon.

So, does anyone know how to prove it more generally? Or maybe an explanation why the assumption isn't as bad as I though?

If you search back through this forum for recent weeks, you'll find another thread
where I gave a proof - and got the same objection about degeneracy. Someone else
added the extra detail.

Personally, I don't think it's all that bad, since one can regard the degenerate
eigenspaces as vectors if you think more abstractly in terms of equivalence classes.
Others may find this a bit distasteful, however.
 
  • #3
ok, thanks I found this thread by browsing your post history: https://www.physicsforums.com/showthread.php?t=417407

This is the step I am unsure about

strangerep said:
[tex]
H (A |n\rangle) ~=~ \lambda_n (A |n\rangle) ~.
[/tex]

Therefore, [itex](A |n\rangle)[/itex] is an eigenvector of H with eigenvalue [itex]\lambda_n[/itex]
and hence is proportional to [itex]|n\rangle[/itex]. (This follows because all the
eigenvectors are mutually orthogonal.) So we have:

[tex]
A |n\rangle ~\propto~ |n\rangle ~.
[/tex]

But I will try giving it some more thought, reading some of the similar threads on this forum and see if it comes more clearly.
 
  • #4
One way of doing it is by spectral decomposition:

[tex]A=\sum a_n P_n[/tex]

[tex]B=\sum b_n Q_n[/tex]

where [tex]a_n,b_n[/tex] are eigenvalues and [tex]P_n,Q_n[/tex] are projection operators on eigenspaces of A and B resp. Then it is an easy exercise to show that A and B commute if an only if all [tex]P_n[/tex] and [tex]Q_m[/tex] commute. If so, then

[tex]P_nQ_m[/tex] are also projections and we have orthogonal sum

[tex]\sum_{m,n}P_nQ_m=I[/tex].

You then take any orthonormal basis from vectors contained in [tex]P_nQ_m[/tex] subspaces and you are done one way. The other way is easy.
 

Related to Proof that commuting operators have a shared base of eigenfunctions

What is the concept of commuting operators?

The concept of commuting operators is based on the mathematical idea that two operators, A and B, are said to commute if the order in which they are applied does not matter. This means that applying A and then B will give the same result as applying B and then A.

What does it mean for commuting operators to have a shared base of eigenfunctions?

When two operators commute, they have a shared set of eigenfunctions. This means that there exists a set of functions that are common eigenvectors of both operators. These shared eigenvectors have corresponding eigenvalues for each operator.

Why is it important for commuting operators to have a shared base of eigenfunctions?

Having a shared base of eigenfunctions allows for the operators to be simultaneously diagonalized, meaning they can be written in terms of a common set of basis vectors. This simplifies the analysis and makes it easier to solve problems involving these operators.

What are some examples of commuting operators?

Some examples of commuting operators include position and momentum operators, energy and time operators, and angular momentum components along different axes. These operators can be shown to commute using mathematical proofs.

What are the implications of commuting operators in quantum mechanics?

In quantum mechanics, commuting operators play a crucial role in determining the quantized values of physical observables. The shared base of eigenfunctions allows for simultaneous measurements of these observables, and the commutativity ensures that the results are consistent and predictable.

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