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Following this circuit: enter image description here

With $\mathcal{G}, A$ being unitary matrices and $|\psi\rangle$ the initial state.

First, the system is:

$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\;\otimes|\psi\rangle$

Next:

$\frac{1}{\sqrt{2}}(|0\rangle\otimes\mathcal{G}|\psi\rangle+|1\rangle\otimes A|\psi\rangle)$

This paper shows that:

A measurement of the ancilla selects one of the two branches and results in either the state $\mid\psi^{'}_0\rangle=\frac{1}{2 \sqrt{p_0}}(\mathcal{G}+A)\mid\psi\rangle$ with probability $$p_0=\frac{1}{4}\langle\psi\mid(\mathcal{G}+A)^\dagger(\mathcal{G}+A)\mid\psi\rangle$$

This sentence makes me confused about how to get the amplitude when it is a matrix? And how about the state $\mid\psi^{'}_0\rangle$, I think it should be $\frac{1}{2}(\mathcal{G}+A)$

Thanks for reading!

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The statement made in the research paper is right.

The initialization of the $\psi$ state is more than one qubit. For n qubits, it's state is in $2^n$ dimensions. Coming back to your question, compute your state with the operator $G+A$, i.e. $\frac{1}{2}(G+A) |\psi\rangle=|\phi\rangle$, for some state $|\phi\rangle$.

Now the probability of $|\phi\rangle$ is $\langle\phi|\phi\rangle=p_0=\frac{1}{4}\langle\psi|(G+A)^{\dagger}(G+A)|\psi\rangle$.

The state $|\psi^{'}_0\rangle $ is normalized, that's why it has co-efficient $\frac{1}{2\sqrt{p_0}}$.

$\langle\psi^{'}_0|\psi^{'}_0\rangle$= $\frac{1}{4p_0}$$\langle\psi|(G+A)^{\dagger}(G+A)|\psi\rangle$=1.

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  • $\begingroup$ Oh, thanks. But how about the state $\mid\psi^{'}\rangle$, why $\sqrt{p_0}$ is still here? $\endgroup$ – Tuấn Hải Vũ Jun 11 at 2:14

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