Does $\mathrm{tr}(A \otimes B) = \mathrm{tr} (A) \otimes \mathrm{tr}(B)$ hold for partial trace?

Question

I was reading this question from this site answered by DaftWullie. I would like to request you to read the question there. The answer says

However, in this particular case, the calculation is much simply. Let $|\phi\rangle$ be the Bell pair such that $$ |\psi\rangle=|\phi_{12}\rangle|\phi_{34}\rangle. $$ Because there's a separable partition between (1,2) and (3,4), this is not changed by the partial trace. Thus $$ \text{Tr}_B|\psi\rangle\langle\psi|=\left(\text{Tr}_2|\phi\rangle\langle\phi|\right)\otimes \left(\text{Tr}_3|\phi\rangle\langle\phi|\right). $$

I don't understand where this expression comes from. There shouldn't be any $\otimes$ in there, right? Can anybody derive it step by step?

Because trace is applied over $|\psi\rangle\langle\psi|$ or $|\phi_{12}\rangle|\phi_{34}\rangle\langle\phi_{34}|\langle\phi_{12}|$ or $(|\phi_{12}\rangle \otimes |\phi_{34}\rangle)( \langle\phi_{34}|\otimes \langle\phi_{12}|)$. This is matrix multiplication of $(|\phi_{12}\rangle \otimes |\phi_{34}\rangle)$ and $( \langle\phi_{34}|\otimes \langle\phi_{12}|)$.

Answer 1

DaftWullie's answer is correct. The key identity they are using is

$$ \mathrm{tr}_B(\rho_A\otimes\sigma_{BC}) = \rho_A \otimes (\mathrm{tr}_B\sigma_{BC})\tag1 $$

which says that we can pull out tensor factors that do not act on the system being traced over. Using $(1)$ and the symbols defined in the linked question and answer, we have

$$ \begin{align} \mathrm{tr}_B |\psi\rangle\langle\psi| &= \mathrm{tr}_2 \mathrm{tr}_3 |\psi\rangle\langle\psi| \\ &= \mathrm{tr}_2 \mathrm{tr}_3\left(|\phi_{12}\rangle\langle\phi_{12}| \otimes |\phi_{34}\rangle\langle\phi_{34}|\right) \\ &= \mathrm{tr}_2 \left(|\phi_{12}\rangle\langle\phi_{12}| \otimes \mathrm{tr}_3|\phi_{34}\rangle\langle\phi_{34}|\right) \\ &= (\mathrm{tr}_2 |\phi_{12}\rangle\langle\phi_{12}|) \otimes (\mathrm{tr}_3|\phi_{34}\rangle\langle\phi_{34}|) \end{align} $$

where in the first equality we use $\mathrm{tr}_B = \mathrm{tr}_2 \circ \mathrm{tr}_3$ since Bob has qubits $2$ and $3$, in the second equality we use the definition $|\psi\rangle = |\phi_{12}\rangle|\phi_{34}\rangle$, in the third we use $(1)$ with $A=12$, $B=3$ and $C=4$ and in the fourth we use $(1)$ once again with $A=4$, $B=2$ and $C=1$.

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Remark on operator domains

Note that $\mathrm{tr}_2 |\phi_{12}\rangle\langle\phi_{12}|$ is an operator acting on the Hilbert space $\mathcal{H}_1$ of subsystem $1$ and $\mathrm{tr}_3|\phi_{34}\rangle\langle\phi_{34}|$ is an operator acting on the Hilbert space $\mathcal{H}_4$ of subsystem $4$. Moreover, $\mathrm{tr}_B |\psi\rangle\langle\psi|$ is supposed to be an operator acting on the Hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_4$ of the two qubits owned by Alice. This confirms that there should be $\otimes$ between the factors in the formula from DaftWullie's answer.

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Proof of $(1)$

Recall that $\mathrm{tr}_B\rho = \sum_i\langle i_B|\rho|i_B\rangle$ where $|i_B\rangle$ is an orthonormal basis of the Hilbert space of the subsystem $B$. Calculate

$$ \begin{align} \mathrm{tr}_B(\rho_A\otimes\sigma_{BC}) &= \sum_i \langle i_B|\rho_A\otimes\sigma_{BC}|i_B\rangle \\ &= \sum_i \rho_A\otimes \langle i_B|\sigma_{BC}|i_B\rangle \\ &= \rho_A\otimes \sum_i \langle i_B|\sigma_{BC}|i_B\rangle \\ &= \rho_A \otimes (\mathrm{tr}_B\sigma_{BC}). \end{align} $$