How is partial trace simplified?

Question

I'm currently reading papers about the effect of repeated measurements, such as "Purification through Zeno-Like Measurements", arXiv:quant-ph/0301026 (DOI: 10.1103/PhysRevLett.90.060401).

It says

After N such measurements have been done, the survival probability of finding system A still in its initial state is represented by $$P^{(\tau)}(N)=\mathrm{Tr}\left[\left(Oe^{-iH\tau}O\right)^{N}\rho_{0}\left(Oe^{iH\tau}O\right)^{N}\right]=\mathrm{Tr}_{B}\left[\left(V_{\phi}(\tau)\right)^{N}\rho_{B}\left(V_{\phi}^{\dagger}(\tau)\right)^{N}\right] $$ Notice that the quantity $V_{\phi}(\tau)=\left\langle\phi\right|e^{-iH\tau}\left|\phi\right\rangle$ is an operator acting on the Hilbert space of system B.

where $O=\left|\phi\right\rangle\left\langle\phi\right|\otimes I_B $ is the projection operator onto an eigenstate of subsystem A, and $\rho_{0}=\left|\phi\right\rangle \left\langle \phi\right|\otimes\rho_{B} $ is the initial density matrix.

My question is: how the trace is reduced to the partial trace with respect to system B?

Answer 1

In this case, it appears as if the $\text{Tr}_B$ notation is to emphasize that the argument is an operator on system B only. One can think of the "full" trace as being the partial trace over all relevant systems, so that $\text{Tr} = \text{Tr}_{AB} = \text{Tr}_A \text{Tr}_B$. The argument you quote thus is an explicit application of the $\text{Tr}_A$ part of the "full" trace, leaving a trace over only system B.