• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Commuting operators

  • Thread starter Henk
  • Start date
22
0
I was wondering: is every eigenstate of L^2 also an eigenstate of Lz?

I know that commuting operators have the same eigenfunctions but if [A,B] = 0 and a is a degenerate eigenfunction of A the the corresponding eigenfunctions of A are not always eigenfunctions of B.
 

Galileo

Science Advisor
Homework Helper
1,989
6
Not necessarily. What [A,B]=0 means (for hermitian operators) is that there EXISTS a basis of eigenvectors common to A and B.
If A|a>=a|a> and a is a nondegenerate eigenvalue of A, then |a> is also an eigenvector of B (this is easy to prove). If a is degenerate, then you can find an orthonormal basis in the eigenspace of a consisting of eigenvectors common to A and B.
So not every eigenstate of L^2 is an eigenstate of Lz.
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,927
504
You can make it a nice discussion in the following way.

[tex] \hat{J}^{2} ,\hat{J}_{3} [/tex]

form a complete system of commuting observables.It's not difficult to show that.
Therefore,a spectral pair [itex]\left(\hbar^{2} j(j+1), \hbar m_{j}\right) [/itex] determine,up to a phase factor,a vector from the irreducible space of the irreducible finite dimensional linear representation of the angular momentum algebra* (which is isomorphic to [itex] su(2) [/itex] which is isomorphic to [itex] so(3) [/itex]).This is an eigenvector to both operators.

*One can prove that an eigensubspace of [itex] \hat{J}^{2} [/itex] associated to the nondegenerate spectral value [itex] \hbar^{2} j(j+1) [/itex] is invariant to the action of all the linops of the representation of the ang.mom.algebra and does not admit any nontrivial invariant subspaces,therefore it is a subspace of an irred.representation of the ang.mom.algebra.

So all eigenstates of [itex] \hat{J}^{2} [/itex] are eigenstates of [itex] \hat{J}_{3} [/itex] in the simple case of uniparticle systems.

Daniel.

P.S.For the particular case of orbital ang.mom.ops,the things are a bit easier,as one has the particular realization [itex] \mathcal{H}=L^{2}\left(\mathbb{R}^{3}\right) [/itex] and the basis from spherical harmonics.
 
Last edited:

CarlB

Science Advisor
Homework Helper
1,212
8
Since [tex]L_z[/tex] does not commute with any other [tex]L_u[/tex] for [tex]\hat{u}[/tex] any other direction not parallel or antiparallel to [tex]\hat{z}[/tex], it should be clear that it's pretty easy to arrange for an eigenstate of [tex]L_u[/tex] and [tex]L^2[/tex] to not be an eigenstate of [tex]L_z[/tex].

In addition, the reverse is also possible. That is, there are eigenstates of [tex]L_z[/tex] that are not eigenstates of [tex]L^2[/tex]. To make one, just take two eigenstates of both [tex]L^2[/tex] and [tex]L_z[/tex] that happen to have the same [tex]L_z[/tex] eigenvalue but different [tex]L^2[/tex] eigenvalues, and mix em up.

Carl
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,927
504
I have no idea why i didn't think of linear combinations of eigenstates.:rolleyes:

Daniel.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top