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I want to know if a partial trace keeps the cyclic property of the trace.

The partial trace is defined as

$$ tr_B: \mathcal{B}_1(\mathcal{H}_A\otimes \mathcal{H}_B) \longrightarrow \mathcal{B}_1(\mathcal{H}_A) \\[3mm] \text{such that} \\[3mm] tr_B [A \otimes B] = tr[B] \ A $$

When is this partial trace cyclic?

And can I cycle anything in the example $\ tr_B [(A_1 A_2 \otimes B_1 B_2) \ \rho_{AB}] \ $ if $\rho_{AB}$ is not pure (aka cannot be written as a tensor product $\rho_{AB} = \rho_{A} \otimes \rho_{B}$)?


Also, as far as I understand it, the full trace is cyclic in each subsystem (I think). Therefore:

$$ tr[A B C \otimes D E F] = tr[C A B \otimes D E F] = tr[A B C \otimes E F D]$$

Is this right?
My reasoning is that we can represent each operator $O_i=A,B,C$ as $(O_i \otimes \mathbb{I})$ and cycle these freely, since we only need to keep the order with respect to each subsystem.

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  • $\begingroup$ If I made any wrong claim, please correct me! $\endgroup$ – João Bravo Jul 1 at 17:11
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Let's for simplicity restrict to the finite dimensional case. In the partial trace you can cyclically permute factors in the part you are taking the trace over, but not the other part, as follows directly from the definition you gave:

$$\operatorname{tr}_{B}[A_1 \otimes B_1]=\operatorname{tr}[B_1] A_1.$$

We see immediately that

$$\operatorname{tr}_{B}[A_1 \otimes B_1B_2] = \operatorname{tr}[B_1B_2] A_1 = \operatorname{tr}[B_2B_1] A_1 = \operatorname{tr}_{B}[A_1 \otimes B_2B_1]$$

but as soon as $A_1A_2 \ne A_2A_1$ and $\operatorname{tr}[B_1] \ne 0$, we have

$$\operatorname{tr}_{B}[A_1A_2 \otimes B_1] = \operatorname{tr}[B_1] A_1A_2 \ne \operatorname{tr}[B_1] A_2A_1 = \operatorname{tr}_{B}[A_2A_1 \otimes B_1].$$

For the full trace of a tensor product of operators we have

$$\operatorname{tr}[A \otimes B] = \operatorname{tr}[A]\operatorname{tr}[B]$$

(see Kronecker product: abstract properties under spectrum), so in this case we can indeed perform all cyclic permutations of factors in each subsystem, as you stated.

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  • $\begingroup$ Ah of course! Very clear answer, thank you :) $\endgroup$ – João Bravo Jul 2 at 11:45
  • $\begingroup$ One question though. In this trace $$tr_B [(A_1 \otimes B_1 B_2) \ \rho_{AB}]$$ if $\rho_{AB}$ is not pure (aka cannot be written as a tensor product $\rho_{AB} =\rho_{A} \otimes \rho_{B}$), can we cycle anything? $\endgroup$ – João Bravo Jul 2 at 11:48
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    $\begingroup$ The only thing that you can do (if I didn't make a mistake) is $$\operatorname{tr}_B[(A_1\otimes B_1B_2)\rho_{AB}] = \operatorname{tr}_B[(A_1\otimes B_2)\rho_{AB}(I\otimes B_1)].$$ This can be seen by writing $\rho_{AB}$ as a sum of pure tensors and using the linearity of the partial trace. $\endgroup$ – doetoe Jul 2 at 14:42
  • $\begingroup$ Yeah makes sense. Thank you $\endgroup$ – João Bravo Jul 2 at 17:02

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