looking for help with a trace/norm inequality

Question

I'm trying to understand a derivation that seems to claim that $\left\vert\text{Tr}\left[\rho U^\dagger\left[U,O\right]\right]\right\vert\leq\|\left[U,O\right]\|$, where $\rho$ is Hermitian and has norm no greater than one (it's a density operator), $U$ is unitary, and $O$ is positive (I don't think it should matter, but this is used in the equation two before Eq. (1) in the paper linked to below). I know that for normal operators, the norm is just the spectral radius, but since $\rho$, $U$, and $O$ don't necessarily commute (and so products aren't necessarily normal), this doesn't seem very useful. In addition, the statement $\left\vert\text{Tr}\left[A\right]\right\vert\leq\|A\|$ is certainly $\textbf{not}$ true in general, for instance when $A$ is the identity on any space of dimension greater than one. Can somebody point me in the right direction to understand this inequality? I think I must be overlooking something very straightforward.

http://arxiv.org/pdf/quant-ph/0603121v1.pdf

Answer 1

$$\mathrm{Tr}[\rho U^{\ast }[U,O]]=<\rho ,U^{\ast }[U,O]> $$ defines a bounded linear functional on the trace class on some separable Hilbert space. Since the dual of the trace class (norm $||..||_{1}$) is the set of bounded operators (norm $||..||$), we have the inequality $$ |<\rho ,U^{\ast }[U,O]>|\leqslant ||\rho ||_{1}||U^{\ast }[U,O]||=\mathrm{tr}% \rho ||U^{\ast }[U,O]||=||U^{\ast }[U,O]||\leqslant ||U^{\ast }||||[U,O]||=||[U,O]|| $$

Answer 2

There's something off here, because the left side is positively homogeneous in $\rho$ but the right side doesn't depend on $\rho$. Are you missing an assumption on the trace norm of $\rho$?

Answer 3

Observe that $Tr(\rho\ \cdot\ )$ defines a state, say $\omega_\rho$, hence you have the estimate $$|\omega_\rho(U^*[U,O])|\leq\Vert U^*[U,O]\Vert\leq\Vert[U,O]\Vert.$$