Given a self-adjoint operator $A$ on a separable Hilbert space, let $A''$ denote the commutative von Neumann algebra generated by $A$. According to [1] and [2], every commutative von Neumann algebra on a separable Hilbert space can be expressed this way. Of course, we can have $A'' = B''$ even if $A\neq B$.
The operator norm of $A-B$ is a natural way to quantify the difference between two operators $A$ and $B$ (whether or not they are self-adjoint). Is there a natural way to quantify the difference between two commutative von Neumann algebras $A''$ and $B''$? To be more specific, I'm looking for a map $$ \{A,B\}\mapsto d(A'',B'') $$ with these properties:
$d(A'',B'')$ is a non-negative real number that depends only on $A''$ and $B''$.
$d(A'',B'') = d(B'',A'')$.
$d(A'',B'')=0$ if and only if $A''=B''$.
$d(A'',B'')$ is a continuous function of $A$ and $B$. (I'm not sure I'm using the word "continuous" in a meaningful way here. Please let me know if it's ambiguous.)
Question: Does such a map exist?
One candidate that comes to mind is $d(A'',B'') = \text{inf}\|A-B\|$ where the infimum is over all possible self-adjoint generators $A$ and $B$ of $A''$ and $B''$, but I don't know if this is continuous.
References:
[1] EP10 on page 23 in Jones (2009), "Von Neumann Algebras," https://math.berkeley.edu/~vfr/VonNeumann2009.pdf
[2] Lemma 1 in Suzuki and Saitô (1963), "On the operators which generate continuous von Neumann algebras," https://projecteuclid.org/euclid.tmj/1178243811
Your candidate does not satisfy your third bullet point; actually, $d(A'',B'')=0$ for all $A,B$. Because if $A$ generates $A''$, so does $\varepsilon A$ for any $\varepsilon$. You could force the generators to have norm one, but that's still not enough: for instance consider $A=(1,0)$, $B=(1,2)$. Then $A''=\mathbb C\oplus 0$, while $B''=\mathbb C\oplus\mathbb C$. But $(1,\varepsilon)$ also generates $B''$ for any $\varepsilon>0$, so $d(\mathbb C\oplus 0,\mathbb C\oplus\mathbb C)=0$.
Looking in general, $d(\mathbb C\oplus 0,\mathbb C\oplus\mathbb C)=0$ for any $d$ satisfying your fourth point: because $$ d(\mathbb C\oplus 0,\mathbb C\oplus\mathbb C)=d((1,0)'',(1,1/n)'')\to d((1,0),(1,0))=0. $$
Without your fourth requirement your map trivially exists in a very useless way: define $$ d(A'',B'')=\begin{cases}0,&\ A''=B''\\ 1,&\ \text{ otherwise} \end{cases} $$