Quantum Computing Asked by fagd on January 1, 2021
In this paper (nature version), the authors state
We group the Pauli operators into tensor product basis sets that require the same post-rotations.
As a result, they have the table S2 in the suppl. I don’t understand this statement, could anyone please explain it, say for H2?
Let me show an example for grouping for this Hamiltonian:
$$H = 5 cdot XI + 3 cdot XZ - 2 cdot YI + 1.5 cdot IY$$
The expectation value:
$$langle H rangle = 5 cdot langle XI rangle + 3 cdot langle XZ rangle - 2 cdot langle YI rangle + 1.5 cdot langle IY rangle$$
Here I will group them in this way: the first group $XI$ and $XZ$, the second group $YI$ and $IY$. Note that (it is important) the members of the same group should commute with each other. Also, I should mention that this is not the only way of grouping.
For the first circuit:
begin{align} &langle X I rangle = p(text{00 or 01 measurements}) - p(text{10 or 11 measurement}) &langle X Z rangle = p(text{00 or 11 measurements}) - p(text{10 or 01 measurement}) end{align}
For the second circuit:
begin{align} &langle Y I rangle = p(text{00 or 01 measurements}) - p(text{10 or 11 measurement}) &langle I Y rangle = p(text{00 or 10 measurements}) - p(text{01 or 11 measurement}) end{align}
where $p$ denotes a probability of a measurement outcome described in parenthesis. The main idea here is: For a given Pauli term $P$ the expectation value is equal to:
$$langle P rangle = p_+ - p_-$$
where $p_+$ ($p_-$) is the probability of having an eigenstate that has eigenvalue $+1$ ($-1$). More details about this can be found in this answer about expectation value estimation. About why $HS^{dagger}$ is applied in the second circuit can be understood from this answer.
Correct answer by Davit Khachatryan on January 1, 2021
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