# Computing multiplicity function for self adjoint operator with nonatomic spectral measure

Suppose $$T$$ is a self-adjoint operator in $$B(H)$$ with $$\sigma(T)$$ a spectrum of $$T$$. $$\mu$$ is a spectral measure. For the operators having a generally continuous spectrum how to calculate the multiplicity function? Where multiplicity is usually called spectral multiplicity. Up to compact operators we know how to decompose the spectrum and get the multiplicity function on each bit of disjoint chunks. But how to apply these in continuous spectral measure case? Is related to the direct integral decomposition of the von Neumann algebra generated by $$\pi(C(\sigma(T))$$ or not?

A measurable Hilbert bundle is something of the form $$\bigcup X_n \times H_n$$ where $$(X_n)$$ is a measurable partition of a $$\sigma$$-finite measure space $$X$$ and $$H_n$$ is a Hilbert space of dimension $$n$$, for $$n = 0, 1, 2, \ldots, \infty$$. (I assume we're working with separable Hilbert spaces.) The associated Hilbert space is the $$l^2$$ direct sum of the spaces $$L^2(X_n) \otimes H_n \cong L^2(X_n, H_n)$$.
If $$T$$ is any bounded self-adjoint operator, one form of the spectral theorem says that there is a measurable Hilbert bundle over $$X = \sigma(T)$$, such that $$T$$ is unitarily equivalent to multiplication by $$x$$ acting on the Hilbert space associated to this bundle. The set $$X_n$$ is the multiplicity $$n$$ portion of $$\sigma(T)$$.
• Sir, but getting the decomposition of the spectrum i.e, getting $X_{n}$ s, without using multiplicity function is not understood by me. Can you please elaborate? – user136400 Feb 28 '19 at 9:17