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I am very confused about how to design the VQLS circuit. For example, I know the A is: enter image description here

X is pauli matrix. And how to add these three X pauli gates into circuit? I know the overall circuit should look like this:

enter image description here

I don't know how to design it, so...: enter image description here I think this circuit is error, because I can't get the right result... Thanks for any help!

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3 Answers 3

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Note that, $A_l$ is followed by $A_{l^{\color{red} \prime}}^{\dagger}$ not $A_l^{\dagger}$. Otherwise, they will cancel each other and your circuit will be just the ansatz.

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  • $\begingroup$ I feel very shame...I misunderstanded these matrix...Al is orginal matrix, A†l is conjugate transpose matrix of Al. So what is the A†l′ ? According to the formual of VQLS, I've only seen these two matrices. Is there something wrong with my understanding? Thanks for your reply! $\endgroup$ Sep 1, 2022 at 4:18
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If you have a general linear system $Mx=b$, then a quantum computer can at best solve the corresponding $$A\left|x\right> = \left|b\right>,$$ where $\left|b\right> = \frac{b}{||b||}$, $A=\frac{M}{||M||}$ is the original matrix normalized so that $||A|| \leq 1$ and $\left|x\right> = \lambda x$.

For VQLS specifically you must further decompose the matrix as a linear sum of unitary matrices $A=\sum_{l} c_l A_l$. In your case, $$A = c_0 A_0 = 0.5 (X \otimes X \otimes X).$$ As $A_0A_0^{\dagger}=I$ the operations cancel out indeed and the measurement result will be $P(0)=1$.

The circuit you present in your question is only to compute $\left<\psi|\psi\right>$ though, which in your case is 1. As you need to evaluate a normalized cost $$C = \frac{\left< \psi | H | \psi \right>}{\left<\psi|\psi\right>}$$ you have to evaluate the Hamiltonian, global or local, $\left<\psi| H |\psi \right>$, in which case the $A_l$'s never cancel out.

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$A_l$ = $A_{l^\prime}^\dagger$ in your question and you can try it.

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  • $\begingroup$ Hi and welcome to Quantum Computing SE. Once you have enough reputation you can post comments. Please use the comment for the short answers like this one. $\endgroup$ Nov 18, 2022 at 6:40

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