# How does VQE performs the measurement on a given Hamiltonian?

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!

• 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. – KAJ226 Oct 15 '20 at 17:00
• I see you found your answer, but why would matter though which one it measure first? – A. Jahin Oct 15 '20 at 18:15

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.
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 $$Z\otimes Z$$