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**Example text**

The question immediately arises, what symmetric operators can be extended so that they become self-adjoint? In order to discuss this question and related other questions we have to digress and introduce the concept of closure. The next section is devoted to this topic. Exercises 1. n. set {q,}in &,define a linear operator H formally by the equations H q , = n q , ,n = 1,2,3, ... Now specify the domain of H first so that H is symmetric, but not self-adjoint; then extend the domain of it so as to obtain a self-adjoint operator.

Moreover, we found that this decomposition is unique. 35) defines the projection operator P associated with M. The domain of P is 5, its range is M. 34) by complex constants and arriving at af+ Pg = (afo + Pgo) + (4+ PSI), (TI. 36) where the first expression in parenthesis belongs to M, the second to MI. Every projection is a bounded operator with the bound 1. Moreover P 2 = P,because Pfo =fo. For arbitrary f and g we have (PAg) = (Yo, g) Conversely, given a bounded, self-adjoint operator P,such that P2 = P, 27 Functional Analysis for Quantum Theorists there is always a closed linear manifold M of which P is the projection operator.

Thus the operator U has the entire space as its domain, it is permutable with Q and therefore with H, it is symmetric because it is the sum of such operators. 55) it follows that ( I - Q)U= HU=Z. Since U is permutable, with H it is equal to H-'. 61) Functional Analysis for Quantum Theorists 59 The bound of U = H - l cannot exceed a-'. ij. Note: It is instructive to construct the inverse of the operator H by making use of the spectral theorem for bounded self-adjoint operators. f. Therefore H-' may be represented in the form jobf dE, .