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Can someone explain the explicit calculations for:

$$(I \otimes ( |00\rangle \langle 00| + |11 \rangle \langle 11| ) ) \times ( (|00 \rangle \langle 00| + |11 \rangle \langle 11|) \otimes I) = |000 \rangle \langle 000| + |111 \rangle \langle 111|$$

At first I thought that I should do: $$( I \otimes |0\rangle \langle 0| \otimes |0\rangle \langle 0| + I \otimes |1\rangle \langle 1| \otimes |1\rangle \langle 1|) \times (|0\rangle \langle 0| \otimes |0\rangle \langle 0| \otimes I + |1\rangle \langle 1| \otimes |1\rangle \langle 1| \otimes I) $$ and then group them like $$I(|0\rangle \langle 0|) \otimes |0\rangle \langle 0||0\rangle \langle 0| \otimes (|0\rangle \langle 0|)I + I(|1\rangle \langle 1|) \otimes |1\rangle \langle 1||1\rangle \langle 1| \otimes (|1\rangle \langle 1|)I$$ But this doesn't seem to be the correct way to do this....It works for this particular calculation, but not for others.

Can someone explain how I should be calculating this?

(I can do this using matrices but don't want to have to use matrices)

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  • $\begingroup$ If $\times$ is just multiplication, then is the first line even correct? $\endgroup$
    – Condo
    Jan 12 at 16:48
  • $\begingroup$ The first line as in the question? Or my first line of attempt, because the first line of attempt I have assumed is incorrect, I just wanted to show what I'd tried $\endgroup$
    – am567
    Jan 12 at 17:36
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    $\begingroup$ Yes, I don't think the question is correct. The rank is multiplicative across tensor products. On the LHS you have a tensor product of rank 2 projections (since $(I\otimes P)(Q\otimes I)=(Q\otimes P)$), so it has rank 4, while the projection on the RHS clearly has rank 2. $\endgroup$
    – Condo
    Jan 12 at 18:47
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    $\begingroup$ $(I \otimes P)(Q\otimes I) = Q \otimes P$ is only when the sizes of the left $I$ and $Q$ are the same, this is not true here. $\endgroup$ Jan 12 at 21:37
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    $\begingroup$ @VladimirLysikov right, so the $I$ is $2\times 2$, thanks! $\endgroup$
    – Condo
    Jan 12 at 23:07

1 Answer 1

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The tensor product grouping is correct, but you have performed a version of the freshman's fallacy by assuming $$(A+B)(C+D)=AC+BD,$$ which is not true.

As such, you are missing the two cross terms here, which happen to both vanish. But in general the cross terms do not vanish, and that is why it works for this particular calculation and not others.

Specifically, here we have $A=I\otimes|0\rangle\langle 0|\otimes |0\rangle\langle 0|$, $B=I\otimes|1\rangle\langle 1|\otimes |1\rangle\langle 1|$, $C=|0\rangle\langle 0|\otimes |0\rangle\langle 0|\otimes I$, and $D=|1\rangle\langle 1|\otimes |1\rangle\langle 1|\otimes I$. You have correctly computed AC and BD. The remaining terms are (note that the order of multiplication matters; we don't have terms like $DA$) $$BC=I|0\rangle\langle 0|\otimes|1\rangle\langle 1||0\rangle\langle 0|\otimes |1\rangle\langle 1|I=|0\rangle\langle 0|\otimes|1\rangle\mathbf{0}\langle 0|\otimes |1\rangle\langle 1|=0$$ and $$AD=I|1\rangle\langle 1|\otimes|0\rangle\langle 0||1\rangle\langle 1|\otimes |0\rangle\langle 0|I=|1\rangle\langle 1|\otimes|0\rangle\mathbf{0}\langle 1|\otimes |0\rangle\langle 0|=0,$$ where I put the 0 in bold in the middle just to emphasize where it comes from.

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