# Partial Trace of Density Operator

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.

• Could you elaborate on why you think there should be a factor $2$ in the first place? It seems to me that your question is not specific to the partial trace, but you're rather asking why something like $\mathrm{Tr} |\psi\rangle \langle \psi| = \langle \psi|\psi\rangle$, no? Commented Dec 24, 2021 at 13:14
• @Jakob I was thinking that tr(|b1⟩⟨b2|)=∑i(⟨bi|b1⟩⟨b2|bi⟩)=⟨b1|b1⟩⟨b2|b1⟩+⟨b2|b1⟩⟨b2|b2⟩=2⟨b2|b1⟩. I understand Lucas' solution but if you could explain why what I did is wrong that would be helpful. Commented Dec 24, 2021 at 22:26
• Does this help? Commented Dec 24, 2021 at 23:01

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.
• Thanks, this clears up my question! Do you know why we can write the $tr_B(|a_1⟩⟨a_2| ⊗ |b_1⟩⟨b_2|)$ as $tr(|b_1⟩⟨b_2|)$ ? Commented Dec 24, 2021 at 12:21
• @Angus You probably should check the definition of $\mathrm {Tr}_B$ (i.e. the partial trace) again. Commented Dec 24, 2021 at 13:12
• @Lucas Baldo Could you explain what exactly you mean with total trace? I think here the $\mathrm{Tr}$ denotes only the trace in the subsystem $B$, since $|b\rangle \in \mathscr B$ (i.e. the Hilbert space of the subsystem $B$). Commented Dec 24, 2021 at 13:14
• @Jakob yes, you're probably right. I meant the space that the $\vert b_i \rangle$ belong to. I couldn't tell this was a subsystem of a bipartite system without looking at the reference. Commented Dec 24, 2021 at 13:32
• @Jakob Ahh yes, I misunderstood the 'total trace' as a trace over the whole system rather than the subsystem to which $|bi⟩$ belong, so was slightly confused with what Lucas had done, but this makes sense now. Commented Dec 24, 2021 at 14:45