I guess that the characters on B(H) are as the form $\phi_x$ where $\phi_x(T)=\langle Tx,x\rangle$. I couldn't prove that the function is homomorphism!
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The maps $\phi_x$ are pure states, but they are not multiplicative. In fact, there are no multiplicative functionals on $B(H)$. To see this, consider a nonzero multiplicative map $\phi:B(H)\to\mathbb C$.
Note that such a map satisfies $\phi(AB)=\phi(A)\phi(B)=\phi(B)\phi(A)=\phi(BA)$. Given any projection $P$, we have $\phi(P)=\phi(P^2)=\phi(P)^2$, so $\phi(P)$ is either $0$ or $1$. Also $\phi(I)=1$, or otherwise $\phi=0$.
Now consider a projection $P$ with infinite-dimensional rank and infinite-dimensional kernel. Then $P$ and $I-P$ are equivalent, i.e. there exists a partial isometry $V$ with $V^*V=P$, $VV^*=I-P$. Note that $1=\phi(I)=\phi(P+I-P)=\phi(P)+\phi(I-P)$. So we have $\phi(P)=1$ and $\phi(I-P)=0$, or the opposite. But then $$ 1=\phi(P)=\phi(V^*V)=\phi(VV^*)=\phi(I-P)=0, $$ a contradiction (and we get the same contradiction if $\phi(P)=0$, $\phi(I-P)=1$).
Martin Argerami
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Am I right that this still works for non-separable Hilbert spaces? – Norbert Nov 07 '13 at 18:42
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Yes, halving works in any properly infinite von Neumann algebra. – Martin Argerami Nov 07 '13 at 19:57