The unitary implementation of $*$-isomorphism of $B(H)$ Is it possible to construct $*$-isomorphism of (factor von Neumann) algebra $B(H)$ which is not unitary implementable?
 A: There's a classic trick to produce the unitary. I was told once who this is due to, but I've unfortunately forgotten. Recall that, if $\xi,\eta \in H$, then $\eta \otimes \overline \xi$ denotes the rank-1 operator which sends $\zeta \mapsto \langle \xi,\zeta \rangle \eta$. Here, the inner-product is conjugate linear in the 1st slot. It's easy to check that $T \cdot ( \eta \otimes \overline \xi) = (T \eta) \otimes \overline \xi$ and $(\eta \otimes \overline \xi ) \cdot T^* = \eta \otimes \overline{T \xi}$ whenever $T \in B(H)$.
Let $\theta$ be a $*$-automorphism of $B(H)$.

"Special" case: $\theta$ fixes a rank-1 projection.
Let $\xi \otimes \overline \xi$ be a rank-1 projection fixed by $\theta$. Here, $\xi$ is a unit vector in $H$. We shall see that $\theta$ can be implemented by a unitary which fixes $\xi$. Imagine for a moment that we already have such an implementing unitary $U$. Then, for any $\eta \in H$, we have
$$ (U    \eta ) \otimes \overline \xi  = (U \eta ) \otimes \overline{U \xi} = U ( \eta \otimes \overline{\xi}) U^* = \theta(\eta \otimes \overline{\xi})$$
whence
$$U(\eta) = \theta(\eta \otimes \overline \xi) \xi.$$
Since formula above only depends on $\theta$, we can use it do define $U$ and then check $U$ is indeed a unitary with the desired properties.

I left out the details, but they aren't too bad.

General case:
Fix some rank-1 projection $e$. If $\theta$ does not already fix $e$, then choose a unitary $W$ such that $W \theta(e) W^* = e$. Then, $W \theta( \cdot ) W^*$ does fix $e$ and this new automorphism is unitarily implemented if and only if $\theta$ is.

A: Let $\theta:B(H)\to B(H)$ be a $*$-automorphism. Fix an orthonormal basis $\{\xi_j\}$ of $H$, and write $E_{jj}$ for the corresponding rank-one projections, i.e. $E_{jj}\xi=\langle\xi,\xi_j\rangle\,\xi_j$. We can expand $\{E_{jj}\}_j$ to a system of matrix units $\{E_{kj}\}_{k,j}$, where $E_{kj}\xi=\langle\xi,\xi_j\rangle\,\xi_k$. 
Next notice that $\theta(E_{11})$ is also a rank-one projection (because $\theta$ being an automorphism forces it to be minimal). Let $\eta_1$ be a unit vector in the range of $\theta(E_{11})$. 
Define $U:H\to H$, linear,  by $U\xi_j=\theta (E_{j1})\eta_1$. Then $$\langle U\xi_j,U\xi_k\rangle=\langle \theta (E_{1k}E_{j1})\eta_1,\eta_1\rangle=\delta_{kj}=\langle\xi_j,\xi_k\rangle. $$So $U $ is  a unitary and $\{\eta_j\}$, where $\eta_j=U\xi_j $, is an orthonormal basis (the observation that $\theta(E_{11})$ is minimal is necessary to guarantee that $U$ is onto). 
Note that $\theta(E_{kj})$ is a rank-one operator sending $\eta_j\to\eta_k$ (because $\theta(E_{kj})\theta(E_{j1})=\theta(E_{kj}E_{j1})=\theta(E_{k1})$); this means that $\theta(E_{kj})\xi=\langle\xi,\eta_j\rangle\,\eta_k$. Then
$$
UE_{kj}U^*\eta_t=UE_{kj}\xi_t=\delta_{j,t}U\xi_k=\delta_{j,t}\eta_k=\langle\eta_t,\eta_j\rangle\,\eta_k=\theta(E_{kj})\eta_t.
$$
As finite linear combinations of the $E_{kj}$ are weakly dense and $\theta$ is normal (which implies weak-continuous on bounded sets), we get $\theta=U\cdot U^*$.
