Let $H$ be separable Hilbert space. Let $A$ be a maximal abelian von Neumann subalgebra of $B(H)$, and $B$ an abelian von Neumann algebra with $A\cap B={\mathbb C}I$, where $I$ is the indentity element of $B(H)$. Does there exist another maximal abelian von Neumann subalgebra of $B(H)$, say $C$, such that $C\supseteq B$ and $A\cap C={\mathbb C}I$?
This is only a partial answer, but it didn't fit in the comment box.
In finite dimension, say $\dim(H)=n$, a maximal abelian von Neumann algebra $A \subseteq B(H) \cong M_n(\mathbb{C})$ just comes down to (the set of matrices that are diagonal in) a choice of basis for $H$. Similarly, $B$ consists of diagonal matrices in some (second) basis, possibly with repeated eigenvalues. So maximality forces $C$ to consist of all diagonal matrices in some (third) basis that spans the eigenspaces of the second one. The question is whether this third basis can be chosen while respecting $A \cap C=\mathbb{C}I$. If each eigenspace of $B$ has dimension an integer power of a prime number, then mutually unbiased bases are known to exist, and the answer is affirmative.

$\begingroup$ Thanks! But I focus on infinite dimensional Hilbert space $H$. $\endgroup$ Jun 27 '12 at 6:45