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Homework Help: Commuting operators

  1. Nov 5, 2006 #1
    hello....this might look very stupid but I am totally confused...........

    Let have operators A, B, C. Let [A,B]=[A,C]=0 and [B,C] not 0...

    When two operators commute, they have the same base in (Hilbert) space.
    So base in A representation is the same as in the B representation and also
    the basis in A and C representations are the same.

    But if B and C do not commute, their basis are different.



    There has to be a fatal error in here:))
    thanx:tongue:
     
  2. jcsd
  3. Nov 5, 2006 #2

    nrqed

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    As far as I know, this happens only when there is degeneracy.

    Let's say that the eigenstates of A are degenerate. Then if [A,B]=0, it means that the eigenstates of B can be written as some linear combination of the eigenstates of A. If [A,C]= 0, the same thing applies for the eigenstates of C (they can be written as a linear combination of the eigenstates of A).

    Consider a set of eigenstates of A all corresponding to the same eigenvalue of A, say [itex] A \psi_n(x) = a \psi(x) [/itex] . Then it will be possible to write some of the eigenstates of B (corresponding possibly to different eigenvalues of B) as some linear combinations of these [itex] \psi_n(x)[/itex]. And it will be possible to write some of the eigenstates of C as a different linear combination of the eigenstates of A (still corresponding to the same degenerate eigenvalue [itex] a [/itex]). So A and B share a common set of basis states, A and C share a common basis of states but B and C don't.

    An exception is if there is a state for which applying A, B or C all gives zero. Then all three operators may share the same eigenstate even if they don't all commute.


    This is what happens, say, with [itex] {\vec L}^2[/itex] and [itex] L_x[/itex] and [itex] L_z [/itex] for example. We have [itex] [ {\vec L}^2,L_x] = [{\vec L}^2,L_z]=0 [/itex] but [itex] [L_x,L_y] \neq 0 [/itex] .
     
    Last edited: Nov 5, 2006
  4. Nov 5, 2006 #3
    :!!)

    YES, thats it!!!!!!! :!!)

    so......there must be degeneracy in all eigenvalues of A, yes???

    thankx a lot.........



    .........and this was not any homework, someone replaced it from QT:))
     
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