# Explicit calculation for multiplying two projection operators

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)

• If $\times$ is just multiplication, then is the first line even correct? Jan 12 at 16:48
• 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 Jan 12 at 17:36
• 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. Jan 12 at 18:47
• $(I \otimes P)(Q\otimes I) = Q \otimes P$ is only when the sizes of the left $I$ and $Q$ are the same, this is not true here. Jan 12 at 21:37
• @VladimirLysikov right, so the $I$ is $2\times 2$, thanks! Jan 12 at 23:07

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.
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.