Consider the following fragment from Takesaki's book "Theory of operator algebra I":
Why is $f_x$ well-defined? It seems to depend on the choice of decomposition $\omega = \sum_{n=1}^\infty \alpha_n \omega_{\xi_n, \eta_n}$ (with $\{\alpha_n\}\in \ell^1$ and $\{\xi_n\}$ and $\{\eta_n\}$ orthonormal sequences). I think such a decomposition need not be unique in general.
Edit: $\mathscr{L}(\mathfrak{H})$ denotes bounded operators on the Hilbert space $\mathfrak{H}$ and $\mathscr{L}\mathscr{C}(\mathfrak{H})$ denotes compact operators on $\mathfrak{H}$.
Best Answer
There is no uniqueness in the sense you say, but that's not what's needed. If $$ \omega = \sum_{n=1}^\infty \alpha_n \omega_{\xi_n, \eta_n},\qquad \omega' = \sum_{n=1}^\infty \alpha_n' \omega_{\xi_n', \eta_n'} $$ and $\omega=\omega'$, this means that for any $x\in\mathscr C(\mathfrak H)$ you have $$ \sum_{n=1}^\infty \alpha_n \langle x\xi_n, \eta_n\rangle=\sum_{n=1}^\infty \alpha_n' \langle x\xi_n', \eta_n'\rangle, $$ which, as compacts are sot-dense in $\mathscr L(\mathfrak H)$, is exactly what you need.