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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.

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    $\begingroup$ 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? $\endgroup$ Commented Dec 24, 2021 at 13:14
  • $\begingroup$ @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. $\endgroup$
    – Angus
    Commented Dec 24, 2021 at 22:26
  • $\begingroup$ Does this help? $\endgroup$ Commented Dec 24, 2021 at 23:01

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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.

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  • $\begingroup$ 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 $tr(|b_1⟩⟨b_2|)$ ? $\endgroup$
    – Angus
    Commented Dec 24, 2021 at 12:21
  • $\begingroup$ @Angus You probably should check the definition of $\mathrm {Tr}_B$ (i.e. the partial trace) again. $\endgroup$ Commented Dec 24, 2021 at 13:12
  • $\begingroup$ @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 $B$). $\endgroup$ Commented Dec 24, 2021 at 13:14
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    $\begingroup$ @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. $\endgroup$ Commented Dec 24, 2021 at 13:32
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    $\begingroup$ @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. $\endgroup$
    – Angus
    Commented Dec 24, 2021 at 14:45

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