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Self-Adjoint Extensions

  1. Mar 28, 2009 #1
    Let A be a closed symmetric operator with deficiency indices n_+/-.

    A has self-adjoint extension iff n_+=n_-. In this case the set of self-adjoint extensions is in natural correspondence with the set of isomorphisms of L_+ onto L_-.

    Isn't this just von Neumann's theorem?

    Let A be a symmetric operator and suppose that there exists a conjugation C with C: dom A-> dom A and AC = CA. Then A has equal deficiency indices and therefore has self-adjoint extensions.
     
  2. jcsd
  3. Apr 5, 2009 #2
    I only know of a VN thm involving one-parameter unitary groups of operators. Your problem looks as if you have to prove closure and to look at the associated resolvent operator.
     
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