Following this circuit: enter image description here

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

First, the system is:



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

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!


1 Answer 1


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.

  • $\begingroup$ Oh, thanks. But how about the state $\mid\psi^{'}\rangle$, why $\sqrt{p_0}$ is still here? $\endgroup$
    – Monad
    Commented Jun 11, 2021 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.