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?


If I am not clear, please help me. Thanks in advance.



Computing multiplicity function for self adjoint operator with nonatomic spectral measure
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 $\sigma(T)$.


Does that help?


