Let $\rho$ and $\sigma$ be two faithful, normal states on a von Neumann algebra $M$. Let us define two metrices on the unit ball $M_1$ of $M$ as follows: \begin{align*} d_1(x,y):= \rho((x-y)^*(x-y))^{\frac{1}{2}} \\ d_2(x,y):= \sigma((x-y)^*(x-y))^{\frac{1}{2}} \end{align*} How do we show that the metrices defined above are equivalent ?
N.B. We know that both the metrices induces the strong operator topology on the unit ball. Hence they are topologically equivalent.
They aren't. Let $M=B(H)$. Define trace-class positive injective operators $R,S$ by fixing an orthonormal basis $\{e_n\}$ and putting $$ Re_n=\tfrac1{n^2}\,e_n,\qquad Se_n=\tfrac1{n^4}\,e_n. $$ Then define $$ \rho(x)=\operatorname{Tr}(Rx),\qquad \sigma(x)=\operatorname{Tr}(Sx). $$ As $R,S$ are positive an injective, $\rho$ and $\sigma$ are faithful positive linear functionals (that we could normalize to normal states if we needed to). Consider the sequence $\{x_n\}$ given by $$ x_n=E_n, $$ where $E_n$ is the orthogonal projection onto $\mathbb C e_n$. Then $$ d_1(x_n,0)=\operatorname{Tr}(RE_n)^{1/2}=\frac1{n}, $$ while $$ d_2(x_n,0)=\operatorname{Tr}(SE_n)^{1/2}=\Big(\frac1{n^4}\Big)^{1/2}=\frac1{n^2}. $$ So the two metrics are not equivalent, since $$\frac{d_1(x_n,0)}{d_2(x_n,0)}=n.$$