Let $x$ be a trace class operator on a Hilbert space $H$. Then $x$ induces unique normal functional on $B(H)$, which we denote it by $f\_x$. 


Let us consider the polar decomposition $x=u|x|$ and $f\_x=v|f\_x|$. 



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> Question: Is $|f\_x|$ the corresponded functional of $|x|$ and $u=v$? 
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Polar decomposition
It depends on the conventions that you pick, but if you pick them correctly, you'll get what you want.


First, for a normal linear functional $f$ on $B(H)$, there is a unique partial isometry $v\in B(H)$ and positive normal linear functional $|f|$ such that $f=v|f|$ and $v^\*v$ is the support projection of $|f|$, where we use the convention that $v|f|(a) = |f|(av)$ for all $a\in B(H)$. See Theorem 4.3.2 of <http://www.math.vanderbilt.edu/~peters10/teaching/spring2013/vonNeumannAlgebras.pdf> for more details on this.


Suppose $x\in B(H)$ is trace class, and consider the induced normal linear functional $f = \operatorname{Tr}(\,\cdot\,x)$ on $B(H)$. Now let's look at the polar decomposition of $x = u|x|$, which satisfies $\ker(u)=\ker(|x|)=\ker(x)$, so $u^\*u$ is the support projection of $|x|$, and $u^\*x=|x|$. Then $|x|$ induces a positive normal linear functional on $B(H)$, which we'll call $g$. One sees that for all $a\in B(H)$,
\begin{align\*}
ug(a) &= g(au) = \operatorname{Tr}(au|x|)=\operatorname{Tr}(ax) = f(a)
\\
u^\*f(a) &= f(au^\*) = \operatorname{Tr}(au^\*x)=\operatorname{Tr}(a|x|) = g(a).
\end{align\*}
Thus $ug = f$ and $u^\*f = g$, so we see 
$$
g(a)=[u^\*f](a)=[u^\*(ug)](a)=[ug](au^\*)=g(au^\*u),
$$ 
and $u^\*u$ dominates the support projection of $g$.
We easily verify that for all $b\in B(H)$,
$$
g(b(1-u^\*u))=\operatorname{Tr}(b(1-u^\*u)|x|)=\operatorname{Tr}(b(|x|-u^\*u|x|))=\operatorname{Tr}(b0)=0,
$$
so $u^\*u$ is exactly the support projection of $g$.
This means that if $f=v|f|$ is the polar decomposition of $f$, then by the uniqueness of the polar decomposition, we must have that $|f|=g=\operatorname{Tr}(\,\cdot\,|x|)$ and $v=u$.


