On the uniqueness of the Lebesgue decomposition of normal states on $B(H)$

TL;DR

This paper characterizes when the Lebesgue decomposition of normal states on the algebra of all bounded operators on a Hilbert space is unique, showing it depends on the finite rank of the representing operator.

Contribution

It provides a necessary and sufficient condition for the uniqueness of Lebesgue decomposition of normal states on B(H), linking it to the finite rank property of the representing operator.

Findings

Decomposition is unique if the representing trace class operator has finite rank.

Non-uniqueness is not solely due to non-commutativity.

Characterization applies specifically to normal states on B(H).

Abstract

The non-commutative theory of the Lebesgue-type decomposition of positive functionals is originated with S. P. Gudder. Although H. Kosaki's counterexample shows that the decomposition is not unique in general, the complete characterization of uniqueness is still not known. Using the famous operator-decomposition of T. Ando, we give a necessary and sufficient condition for uniqueness in the particular case when the underlying algebra is $B(H)$, the $C^*$-algebra of all continuous linear operators on a Hilbert space $H$. Namely, given a normal state $f$, the $f$-Lebesgue decomposition of any other normal state is unique if and only if the representing trace class operator of $f$ has finite rank. Some recent results tell that the decomposition is unique over a large class of commutative algebras. Our characterization demonstrates that the lack of commutativity is not the real cause of non-uniqueness.