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I'm trying to understand, given a specific Hamiltonian, for example $H = Z\otimes Z+X\otimes Z$, does the VQE algorithm calculates the expectation value of $Z\otimes Z$ first or does it calculates the expectation value of $X\otimes Z$ first? couldn't understand it from the source code.

Thank you very much!

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    $\begingroup$ If you use IBM hardware, you can look at the order of the circuit being executed on the hardware and see which one is being executed first. $\endgroup$
    – KAJ226
    Commented Oct 15, 2020 at 17:00
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    $\begingroup$ I see you found your answer, but why would matter though which one it measure first? $\endgroup$
    – A. Jahin
    Commented Oct 15, 2020 at 18:15

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For a VQE, the circuit applies $Z \otimes Z$ and $X \otimes Z$ on different iterations, and their expectation values are added classically. So, it doesn't really matter which is run first on a quantum computer - they're entirely different circuits.

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Thank you for your answers. After a check it appears that the VQE algorithm processes first the rightmost expression and only then the leftmost. i.e. it first processes $X\otimes Z$ and then $Z\otimes Z$

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