2
votes
1answer
83 views

Eigenvalues and Compact Resolvent

For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
2
votes
2answers
236 views

A question on unbounded operators

Assume that $H$ is a separable Hilbert space. Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?: Every densely defined operator $A:D(A)\to ...
6
votes
2answers
347 views

C*-algebraic representation of observables vs self-adjoint operators one

I am trying to reconcile the "physicist" definition of an observable: self-adjoint operator on a Hilbert space, and the operational one as given by Strocchi in "An introduction to the mathematical ...
4
votes
5answers
1k views

Measurable functions and unbounded operators in von Neumann algebras

How do you define unbounded measurable functions for a general von Neumann algebra? For the commutative algebra $L^\infty(X,\mu)$, we can consider the space of all measurable functions that are ...