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Is commutativity transitive for non degenerate eigenvalues?

  1. Oct 31, 2012 #1
    Suppose we have three hermitian operators A,B,C each with only non degenerate eigenvalues.

    If A and B commute, then for each eigenvector of A we can find an eigenvector of B, and because the eigenvalues are non degenerate the mapping is one to one. If B and C commute we can do the same. This implies that A and C also have a one to one mapping of their eigenbases, and so A and C commute.

    But in general we know that the commutation relation is not transitive for some arbitrary set of operators {X,Y,Z}. Is this because some of them might have degenerate eigenvalues?
     
  2. jcsd
  3. Oct 31, 2012 #2

    Bill_K

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    Suppose [X,Y] = 0 and [Y,Z] = 0, and suppose transitivity fails. That means that within some eigenspace of Y, [X,Z] ≠ 0, implying that the eigenspace is at least two-dimensional.
     
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