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Simultaneous diagonalization and replacement of operators with eigenvalues ?

  1. Jul 21, 2009 #1
    Apparently, if I have a Hamiltonian that contains an operator, and that operator commutes with the Hamiltonian, not only can we "simultaneously diagonalize" the Hamiltonian and the operator, but I can go through the Hamiltonian and replace the operator with its eigenvalue everywhere I see it show up. Can someone please explain why this is? Thanks!
     
  2. jcsd
  3. Jul 21, 2009 #2
    The proof that you can simultaneously diagonalize commuting operators should be available in any elementary quantum mechanics text. IIRC it goes something like this:

    1. We examine an eigenvector of operator A with eigenvalue a: [tex]A|a> = a|a>[/tex]
    2. Now act with B (remember numbers commute with operators: [tex]BA|a> = a B |a>[/tex]
    3. We assume B commutes with A: [tex] BA |a> = AB |a>[/tex]
    4. Use 2 to replace in 3: [tex] A(B |a> )= a (B |a>) [/tex]
    5. Now we see that B|a> is an eigenvector of A with eigenvalue a. Thus, B can only scale |a> by a constant which means |a> is an eigenvector of B. I've assumed A has no degenerate eigenvectors; the situation gets a bit trickier under conditions of degeneracy, but you can still prove that commuting operators can be simultaneously diagonalized.

    You can only replace the operator with its eigenvalue when the operator is operating on one of its eigenvectors, and then you have to use the eigenvalue associated with that eigenvector. And that's because of the definition of eigenvectors/eigenvalues.
     
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