Expected value in a VQE qiskit

Dec 15 2020

I'm learning VQE(variational quantum eigensolver) of qiskit, but I have a question about how it measures the expected value of the energy ($\left \langle H \right \rangle$). I saw in other question and they comment that qiskit use $\left \langle H \right \rangle = \langle \psi | H |\psi \rangle = \sum_{i} \lambda_{i} P_{i} $ where $P_{i}=|\langle \phi_{i}|\psi \rangle|^2$. But my question is if qiskit needs the eigenvector of the operator, why does it use a VQE? Is qiskit has the diagonal representation already or how does qiskit do to measure the energy in a simulator and real device?

Respuestas

2 YitianWang Dec 15 2020 at 08:12

The output of variational quantum eigensolver(VQE) is a number(the ground state energy of molecules), see qiskit document.

Esta salida tiene la forma de un número, por lo que puede deducirse de unas pocas mediciones, o solo nos estamos enfocando en parte de la información que contiene el estado cuántico. Esta característica nos ayuda a reducir significativamente el requerimiento de recursos si necesitamos conocer el vector de estado detallado del estado, la técnica que necesitamos es la tomografía de estado cuántico (una técnica costosa). Por ejemplo, para un$n$-qubit estado, matemáticamente puede ser descrito por un $2^n$ vector dimensional normalizado uno, y lo que hace la tomografía de estado es extraer todos estos, digamos números.

So if you are using a classical computer to simulate a quantum computer, then VQE is not useful because you do have all the matrices and vectors. But if you are using real quantum devices then to avoid the resource-consuming quantum tomography then you have to choose your target wisely.

I am new to the field of quantum tomography, so if you have problems or my statement is confusing, give me some time, please.