Normal states are dense in $B(H)$

Question

Can someone explain/sketch the proof of the fact stated in the title : The set of normal states (in some $B(H)$) is weakly dense in $B(H)$ ?

Or, if possible, some reference. Thank you very much.

Answer 1

The normal functionals of $B(H)$ can be identified with the elements of the predual of $B(H)$, which are the trace-class operators $T(H)$, via the duality $$ T(H)\ni X\longmapsto \text{Tr}(X\ \cdot). $$ So the question is why $T(H)$ is weakly dense in $B(H)$.

There are several ways of proving this. The easiest is to notice that $T(H)$ contains all rank-one operators, which implies that $T(H)'=\mathbb C I$. Then $$ \overline{T(H)}^{w}=T(H)''=(\mathbb C I)'=B(H), $$ where the first equality is von Neumann's Double Commutant Theorem.

Note that the assertion as stated in the title cannot be true: weakly limits of positives are positive, so only positive operators can be limits of normal states.