What is the range of $A = tr(q[r, h]) tr(p[r,h])$ where $[m,n]=mn-nm$ is the commutator of $m$ and $n$,
knowing that
all matrices are hermitian $d \times d$.
$ q,p,r $ are trace one, PSD and idempotent, hence the range of $tr(qr)$ and $tr(pr)$ is in $[0,1]$;
h is involutionary, hence its trace is in $[-d,d]$.
I have trouble determining bounds given that $h$ is not PSD and I am unsure whether $qr$ or $pr$ are PSD again.
Presumably $d\ge2$. Let $k=rh-hr$. For any unit vectors $u\in\operatorname{range}(r)$ and $v\in\ker(r)=\operatorname{range}(r)^\perp$, we have $$ \begin{aligned} ku&=(rh-hr)u=rhu-hu=(r-1)hu,\\ kv&=(rh-hr)v=rhv.\\ \end{aligned} $$ Therefore $\|ku\|_2,\|kv\|_2\le1$ and $ku\perp kv$. It follows that $\|ik\|_2\le1$. Moreover, since $ik$ is Hermitian, $\operatorname{tr}\big(q(ik)\big)$ and $\operatorname{tr}\big(p(ik)\big)$ are real numbers whose absolute values are at most $1$. Hence $$ \operatorname{tr}(q[r,h])\operatorname{tr}(p[r,h]) =-\operatorname{tr}\big(q(ik)\big)\operatorname{tr}\big(p(ik)\big)\in[-1,1].\tag{$\ast$} $$ E.g. when $$ q=\pmatrix{\frac{1}{2}&-\frac{i}{2}\\ \frac{i}{2}&\frac{1}{2}\\ &&0_{(d-2)\times(d-2)}}, \quad r=\pmatrix{1&0\\ 0&0\\ &&0_{(d-2)\times(d-2)}}, \quad h=\pmatrix{0&1\\ 1&0\\ &&I_{d-2}}, $$ the lower bound $-1$ in $(\ast)$ is attained by $p=q$ and the upper bound $1$ is attained by $p=q^\top$.
