The point is the following. If one takes a reduced tensor product of a pair of $W^*$-algebras $A_1,A_2$ with states $\phi_1$, $\phi_2$, the resulting tensor product does not depend on the choice of $\phi_i$ as long as they give rise to equivalent GNS representations of $A_i$ (e.g. the states are faithful). Note, however, that it is not as straightforward as one might think to write an isomorphism between $(A_1,\phi_1)\otimes (A_2,\phi_2)$ and $(A_1,\phi'_1)\otimes (A_2,\phi'_2)$, since in general there is no unitary from $L^2(A_i,\phi_i)$ to $L^2(A_i,\phi'_i)$ which intertwines the actions of $A_i$. For $A_i$ finite-dimensional, there is an invertible linear map (similarity) $S_i : L^2(A_i,\phi_i)\to L^2(A_i,\phi'_i)$ with this property, but it is not isometric. There are analogs of this more general situations, too.

For infinite tensor products, the situation is even more delicate. Indeed, the naive way to prove isomorphism between $\bigotimes_i (A_i,\phi_i)$ and $\bigotimes_i (A_i,\phi'_i)$ even for finite-dimensional $A_i$ runs into the problem that similarities $S_i$ do not give a bounded map between the infinite tensor products of Hilbert spaces: $\bigotimes S_i$ may not exist.

It turns out, for example, that for $A_i\cong M_{2\times 2}$, the algebras of $2\times 2$ matrices and $\phi_i(x) = Tr(d x)$, with $d$ a positive operator with eigenvalues $\lambda_1,\lambda_2$ satisfying $\lambda_1+\lambda_2=1$, $\lambda_1/\lambda_2 = \lambda\in (0,1]$, the infinite von Neumann tensor product $\bigotimes (A_i,\phi_i)$ depends on $\lambda$. The resulting von Neumann algebra is called an ITPFI factor, and is the unique type $III_\lambda$ (if $\lambda\neq 1$) or type $II_1$ (if $\lambda=1$) hyperfinite factor.