I'm trying to understand the following paper, https://journals.aps.org/prresearch/abstract/10.1103/PhysRevResearch.3.013167, but I'm new to quantum computing. In it they use this expression: $$\langle y^{(n)}|\otimes𝟙 \,|\Psi_b\rangle.$$ What is this operator and what effect does it have on the state $\Psi_b$? I think it is some sort of projection of the state $y^{(n)}$ into $\Psi_b$ based on the context although that would be written as $|y^{(n)}\rangle\langle y^{(n)}|\Psi_b\rangle$. Thanks for the help.

  • $\begingroup$ are you referring to the equation at the end of page 3, first column? For doublestruck 1 see math.meta.stackexchange.com/a/33303/173147. Easiest way is to paste the unicode symbol: 𝟙 $\endgroup$
    – glS
    Dec 18, 2021 at 17:53
  • $\begingroup$ Yes that is the equation I was referring to. $\endgroup$
    – G Aker
    Dec 19, 2021 at 13:34

1 Answer 1


It likely simply refers to a partial projection. It will probably be clearer using some parentheses: $$\langle y^{(n)}|\otimes𝟙 \,|\Psi_b\rangle \equiv (\langle y^{(n)}|\otimes𝟙 )|\Psi_b\rangle$$ where 𝟙 is the identity matrix.

For example, you have $$(\langle 0|\otimes 𝟙)(|00\rangle+|01\rangle+|10\rangle) = |0\rangle+|1\rangle.$$


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