1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:))
  2. jcsd
  3. Nov 5, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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:))
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Commuting operators