He is not claiming that the commutator is equivalent to the divergence, what he is saying is that [xl',Hl] = 0 is equivalent to div H = 0, which is different. If you click on the annotation on the left you will see that [xl',Hl] = 0 means that Hl is not a function of xl and so ∂_lH_l will be zero and consequently div H = Sum ∂_lH_l is also zero.
I know, I forgot to check the formula before clicking the submit button. I meant it as you explain. The problem is that this commutator equation implies nothing about partial derivatives of H with respect to spatial coordinates that EM fields are function of, because Dyson's x_l are not spatial coordinates (which are always real numbers), but they are non-relativistic Heisenberg operators of coordinates of considered particle (infinite matrices).
The difference is as important as the difference between coordinate of a space point and coordinate of a particle at some definite time. This confusion was probably caused by the unfortunate choice of notation, where particle coordinates were denoted as x_l, notation better spent on the general spatial coordinates. If we use, say, r_l to denote Heisenberg operators for the particle coordinates, it is clear that (not being a function of operators r_l) implies nothing about (not being a function of position x_l).
Let me simplify this argument. What Dyson is saying seems no different from saying that if you have a field v_l(x) giving velocity of water at general point x, this field is obviously not a function of coordinates r_l of a test particle and it follows that sum of derivatives ∂_lv_l is zero. Behold, we arrive at the conclusion that the flow must be incompressible, no need to assume anything from experience. All water must be incompressible! The absurdity of this conclusion is pretty obvious, and the reason it was obtained is clear: v_l is independent of r_l, but it depends on x_l, so nothing about partial derivatives ∂/∂x_l can be easily inferred.
