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Characteristic for such an operator is that the equation (Hf, 9 ) = (f,9 H ) holds for every f and g in 5. n. system is used to obtain the matrix representation. Bounded, self-adjoint operators possess a number of interesting properties. We shall establish a few of these. (1) If (HAY)=0 for everyf, then Proof. From the relation * (H(f * H = 0. * (H. 9)k (Hg,f)+ (Hg,9) 91, (f 9))= (HAf) and (Hg,f)= (9,Hf)= (HA g)* it follows that 4 Re(H. 9)= ( H ( f + g), (f+ 9))- ( H ( f - g), (f- 9)). 2) On replacing g by ig, we get 4 WHf, 9)= (H(f+ig), (f+i d ) - Mfid,(f- is)).

Thus the discovery of a subspace which reduces H already brings us one step toward the diagonal representation. The important matter is to find an entire sequence of reducing subspaces ordered in a very special way. We return temporarily to the finite, Hermitean matrices, but we will discuss them in terms of the Hilbert space formalism. 6). Let us now define E,= Pi. 6b). It commutes with H , therefore the range of E, reduces H . 1 1) Thus the matrix H has been split into a positive and a negative part; H(Z - E,) and HE,.

4) If A is a closed operator with domain everywhere dense, the operator At has the same properties. Beta A. Lengyel 48 C. 9) ( A f ,g ) = (f,A d holds for everyfand g in D. This definition implies that the adjoint operator A t exists and is an extension of A . Thus D' 2 D is everywhere dense and the existence of A" is also assured. Since Att is a closed extension of A whose adjoint is again A', it is sufficient to consider closed symmetric operators only. A closed symmetric operator that has no further symmetric extensions is called a maximal symmetric operator.