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I got confused when reading an article about linear combination of unitary method. It shows its process as the following:

I can't figure out how the effect of $B^{\dagger}$ is calculated with the limited information of $B$.

Here is the article mentioned.

Thanks to you all!

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The trick is that you don't need to calculate the inverse of $B$. What you really want to evaluate is $$ (\langle 0|\otimes I)(B^\dagger \otimes I)(\text{select}(V))(B\otimes I). $$ So, the point is that you only need $\langle 0|B^\dagger$ which is the Hermitian conjugate of $B|0\rangle$, which you know.

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  • $\begingroup$ Thanks for the clear answer! I am getting it like this: since directly deducing the actual form of $W|0\rangle|\psi\rangle$ is hard, we split it into two parts: the part whose first register is $|0\rangle$ and the remaining part. For the part whose first register is $|0\rangle$, we reduce $|0\rangle$ to 1 by multiplying $\langle 0|\otimes I$ in the left, and can see that it can be deduced to $\frac 1s|0\rangle U|\psi\rangle$ in the end. Therefore the remaining part would be some $\sqrt {1-\frac 1{s^2}}|\Phi\rangle$ with $|\Phi\rangle$ orthogonal to $|0\rangle$. Am I right? $\endgroup$
    – Lucida
    Jul 17 '20 at 14:42
  • $\begingroup$ Exactly! ________ $\endgroup$
    – DaftWullie
    Jul 17 '20 at 14:47
  • $\begingroup$ Thanks a lot for helping me out! $\endgroup$
    – Lucida
    Jul 17 '20 at 16:16

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