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.