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Homework Help: Commuting operators

  1. Jun 27, 2005 #1
    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.
     
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  3. Jun 27, 2005 #2

    Galileo

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    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.
     
  4. Jun 27, 2005 #3

    dextercioby

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    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: Jun 27, 2005
  5. Jun 28, 2005 #4

    CarlB

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    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
     
  6. Jun 28, 2005 #5

    dextercioby

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    I have no idea why i didn't think of linear combinations of eigenstates.:rolleyes:

    Daniel.
     
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