To find the reduced density matrix, $\rho_A$, of a composite quantum system with two subsystems A and B, I've seen that the procedure is to take the partial trace of the full density matrix, $\rho_{AB}$, w.r.t the states of subsystem B;
$$ \rho_A = tr_B(\rho_{AB}) = \sum_i ⟨i_B|\rho_{AB}|i_B⟩ $$
However, on page 106 of 'Quantum Computing and Quantum Information' by Nielsen and Chuang, it is stated that
$$ tr(|b_1⟩⟨b_2|) = ⟨b_2|b_1⟩ $$
should there be a factor of 2 in this equation? This is repeated later as well, so I assume I am wrong.
Best Answer
The second equation seems to be about the trace over the space of the $\vert b_j \rangle$, so that it evaluates to \begin{align} tr(|b_1⟩⟨b_2|) &= \sum_i \langle i \vert ( \vert b_1 \rangle \langle b_2 \vert ) \vert i \rangle \\ &= \sum_i \langle i \vert b_1 \rangle \langle b_2 \vert i \rangle \\ &= \sum_i \langle b_2 \vert i \rangle \langle i \vert b_1 \rangle \\ &= \langle b_2 \vert \left( \sum_i \vert i \rangle \langle i \vert \right) \vert b_1 \rangle \\ &= \langle b_2 \vert b_1 \rangle, \end{align}
where $\{\vert i \rangle \}$ is a basis for the Hilbert space in which the $\vert b_j\rangle $ are in.
So no extra factor of 2.