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I have read some papers talking about Pauli operator grouping for simultaneous measurement in VQE. I was wondering can this "simultaneous measurement" approach be used in other variational quantum algorithms? Like variational quantum classifier?

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I don't see why not since in VQE the only thing that being done on the quantum computer is the evaluation of $ \langle H \rangle $, where $H$ is some Hermitian matrix. For quantum chemistry purpose, $H$ usually represents the electronic Hamiltonian of the molecular system.

Because $\langle H \rangle$ cannot be measured directly on a quantum computer, but instead it must be decompose into linear combination of Pauli strings $P_i \in \{I,X,Y,Z\}^{\otimes N}$. For example, $\langle H \rangle = \langle II \rangle + \langle ZZ \rangle + \langle ZI \rangle + \langle IZ \rangle + \langle XX \rangle $.

The number of Pauli strings scales as $O(N^4)$ (this can be reduced through various techniques) so to save the number of evaluations, we grouped the Pauli strings, $P_i$, that commutes with each other. For example, $IXZ$ and $ZXZ$ commute, since $[IXZ, ZXZ] = IXZ\cdot ZXZ - ZXZ \cdot IXZ = 0$. So you can find a whole family of these commutative terms. Then within this family, you can just measure in the $Z$ basis for 1st qubit, $X$ basis for 2nd qubit, and $Z$ basis for 3rd qubit.

Thus, this grouping Pauli opertator for simultaneous measurement is not specific to VQE or quantum chemistry application portion of VQE. As long as you are evaluating $\langle M \rangle$ for some matrix $M$ through evaluating the expectation of the linear combination of the Pauli strings then this technique can be applied.

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  • $\begingroup$ The number of Pauli strings scales generally as $O(4^N)$, with $N$ the number of qubits. You'll typically see quartic dependence $O(N^4)$ in chemistry applications, but this is only after approximating the Hamiltonian to include only one- and two-body terms. $\endgroup$
    – jecado
    Sep 23 '21 at 16:56
  • $\begingroup$ I was meant in term of translating a second quantize Hamiltonian to qubit representation here, of course. Each fermionic operators contribute $O(N)$. The tensor product of Pauli matrices formed a real vector space of Hermitian matrix and so yes for a general Hermitian matrix you would need all $4^N$ terms. $\endgroup$
    – KAJ226
    Sep 23 '21 at 18:54
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    $\begingroup$ To be clear, the general Hermitian matrix can be written in second-quantized form, with creation and annihilation operators, even in the context of quantum chemistry problems. Polynomial dependence is only obtained by constraining the number of terms in the Hamiltonian, either due to a simple model or an approximate one! $\endgroup$
    – jecado
    Sep 24 '21 at 15:51
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    $\begingroup$ ahh, I see what you mean. :) $\endgroup$
    – KAJ226
    Sep 24 '21 at 17:17
  • $\begingroup$ Thanks for pointing that out! $\endgroup$
    – KAJ226
    Sep 24 '21 at 17:34

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