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.
