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Unit 28: Self Adjoint Operators




                 T = 0 (zero operator)                                                          Notes
          This completes the proof of the theorem.

          Theorem 6: An operator T on a Hilbert space H is self-adjoint.
                    Tx,x  is real for all x.
          Proof: Let T* =T (i.e. T is self adjoint operator)

          Then for every x H,we have
                  Tx,Tx   x,T * x  x,Tx  Tx,x

              Tx,x equals its own conjugate and is therefore real.

          Conversely, let  Tx,x is real  x H.We claim that T is self adjoint i.e. T*=T.
          since  Tx,x  is real  x H,

                  Tx,x  Tx,x   x,T * x  T * x,x

                  Tx,x  T * x,x  0 x H
                  Tx T * x,x  0 x H

                   T T * x,x  0 x H
                 T – T* = 0                                         [ if (Tx, x) = 0   T = 0]

                 T = T*
                 T is self adjoint.
          This completes the proof of the theorem.
          Cor. If H is real Hilbert space, then A is self adjoint

                    Ax,y   Ay,x  x,y H.

          A is self adjoint   for any x,y H.
                  Ax,y   x,A * y  A * y,x .

          since H is real Hilbert space  A * y,x  A * y,x  so that  Ax,y  Ay,x   A* A

          Theorem 7: The real Banach space of all self-adjoint operators on a Hilbert space H is a partially
          ordered set whose linear and order structures are related by the following properties:

          (a)  If  A  1  A  then A +A A +A for every A S;
                                    2
                      2
                             1
          (b)  If  A  A  and   0, then  A  A .
                  1   2                1    2
          Proof: Let S represent the set of all self-adjoint operators on H. We define a relation  on S as
          follows:
          If  A A  2  S, we write A  1  A  if  A x,x  A ,x   x in H.
                                      1
                                             2
                                 2
              1
          We shall show that  ' ' is a partial order relation on S.  ' '  is reflexive.



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